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ST334 ACTUARIAL METHODS
version 2016/01

These notes are for ST334 Actuarial Methods
...

Actuarial CT1 is called ‘Financial Mathematics’ by the Institute of Actuaries
...

Contents
1 Simple and Compound Interest
...
1
...
1
...

factor 10
...
5 Exercises 15
...
6 Time dependent interest rates 16
...
1

Discount rate and discount
1
...

1
...
21
2
...
2
...
2
...
2
...
2
...
2
...

2
...
2
...

3 Perpetuities, Annuities and Loan Schedules
...
1

Perpetuities 43
...
2

Annuities 44
...
3

Exercises 51
...
4

Loan schedules 56
...
5

Exercises 60
...
67

Markets, interest rates and ﬁnancial instruments 67
...
2 Fixed interest government borrowings 68
...
3 Other ﬁxed interest borrowings 69
...
4 Investments with uncertain returns 71
...
5 Derivatives 72
...
6 Exercises 77
...
1

5 Bonds, Equities and Inﬂation
...
5
...

of interest 89
...
5 Exercises 91
...
1

5
...

5
...
99
6
...
6
...

6
...
6
...
6
...

6
...

7 Arbitrage
...
1

Arbitrage 129
...
2

Forward contracts 131
...
3

Exercises 136
...
139
Exercises 1
...
Exercises 1
...
Exercises 1
...
Exercises 2
...
Exercises 2
...

Exercises 2
...
Exercises 3
...
Exercises 3
...
Exercises 4
...
Exercises 5
...

Exercises 5
...
Exercises 6
...
Exercises 6
...
Exercises 6
...
Exercises 7
...

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ST334 Actuarial Methods ⃝R
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Page i

Page ii

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

Prologue
Order of Topics
...
This has not been done in order to avoid starting with a long list
of formulæ on nominal and effective interest rates and annuities
...

Exercises
An enormous number of exercises is included
...
Only you can decide how many you need to do in
order to achieve the required proﬁciency to pass the tests and the examination
...

Brealey, R & S
...
(2003, ﬁfth edition) How to Read the Financial Pages Random House
McCutcheon, J
...
& W
...
Scott (1986) An Introduction to the Mathematics of Finance Butterworth
Ross, S
...
(2002, second edition) An Elementary Introduction to Mathematical Finance CUP (Chapter 4)
Steiner, R
...
(2001, fourth edition) The Financial Times Guide to Using the Financial Pages Prentice Hall

Most of the exercises have been taken from past examination papers for the Faculty and Institute of Actuaries
...
actuaries
...
uk/students/pages/past-exam-papers and they are
copyright The Institute and Faculty of Actuaries
...
j
...
ac
...

Notation
symbol
a ∞ or a ∞ ,i
¨
a ∞ or a ∞ ,i
¨
(m)
a ∞ and a(m)
¨∞
a n or a n ,i
a n or a n ,i
¨
¨
a(m) or a(m)
n
n ,i
a(m) or a(m)
¨n
¨ n ,i
a∞
an
¨
k| a n and k| a n
(m)
¨ (m)
k| a n and k| a n
k| a n
ACT/365
A(p)

page
43
43
52
44
44
45
45
49
49
47
47
49
1
4

meaning
present value of perpetuity
value at time 1 of a perpetuity
value at time 0 and at time t = 1/m of a perpetuity payable m times per year
present value of an annuity with n payments
value at time 1 of an annuity with n payments
value at time 0 of an annuity with m payments per year for n years
value at time 1/m of an annuity with m payments per year for n years
value at time 0 of a continuously payable perpetuity with ρ(t) = 1
value at time 0 of a continuously payable annuity with ρ(t) = 1 for t ∈ (0, n)
annuity payable annually for n years and delayed by k years
annuity payable m times per year for n years and delayed by k years
value at time 0 of a continuously payable annuity with ρ(t) = 1 for t ∈
(k, k + n)
actual number of days divided by 360
accumulated value at time p of an investment of 1 at time 0

C
CD
c(i)
CP

79
67
111
67

redemption value of a bond
certiﬁcate of deposit
convexity
commercial paper

d=1−ν
d(m)
dp
(Da) n
DCF
d(i)
dM (i)
DPP
δ = ln(1 + i)

10
12
12
49
22
110
109
33
6

discount rate
nominal discount rate payable m times per year
nominal discount rate payable per time period, p
present value of a decreasing annuity for n years
discounted cash ﬂow
effective duration or volatility or modiﬁed duration
duration or Macaulay duration or discounted mean term
discounted payback period
nominal rate of interest compounded continuously or
force of interest corresponding to the effective annual interest, i
force of interest function

δ : (0, ∞) → [0, ∞)

16

f
FRA
fT,k

79
74
101

FT,k = ln(1 + fT,k )
ft,T,k

101
102

Ft,T,k = ln(1 + ft,T,k )
Ft,T = limk→0 Ft,T,k

103
103

g=
i
i

f r/
C

81
1
3

face or par value of a bond
forward rate agreement
forward interest rate p
...
for money at time T in the future for length of
time k
forward force of interest
forward interest rate p
...
at time t for investment starting at time T for length
of time k
forward force of interest
instantaneous forward rate at time t for investment starting at time T
annual coupon divided by redemption value of a bond
simple interest
compound interest

c
ST334 Actuarial Methods ⃝R
...
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Jan 28, 2016(12:36)

Page iii

iep
i(m) = i1/m
iM and iR
ip
(Ia) n and (I a) n
¨
(I a) n
¯
(Ia) n
(Is) n and (I s) n
¨
(I s) n
¯
(Is) n
IRR

4
4
90
4
48
50
50
48
50
50
29

the effective interest rate over the period p
nominal rate of interest compounded m times per year
money and real rate of interest
nominal rate of interest per period p
value at time 0 and at time 1 of an increasing annuity for n years
increasing continuously payable annuity with a step function ρ
increasing continuously payable annuity with ρ(t) = t for t ∈ (0, n)
value at time n and at time n + 1 of an increasing annuity for n years
increasing continuously payable annuity with a step function ρ
increasing continuously payable annuity with ρ(t) = t for t ∈ (0, n)
internal rate of return

LIBID
LIBOR
LIMEAN
LIRR

68
68
68
38

London Inter-Bank Bid Rate
London Inter-Bank Offered Rate
average of LIBOR and LIBID
linked internal rate of return

MWRR

38

money weighted rate of return

NPV
ν = 1/(1 + i)
ν : [0, ∞) → [0, ∞)

22
10
17

net present value
discount factor
present value function

P (n, i)

79

current price of abond with n years to maturity and yield i

r
ρ : (0, ∞) → R

79
25

coupon rate per year of a bond
continuous payment stream

s n or s n ,i
¨
s n or s n ,i
¨
(m)
s n or s(m)
n ,i
¨ n ,i
s(m) or s(m)
¨n
sn

45
45
45
45
49

value at time n of an annuity with n payments
value at time n + 1 of an annuity with n payments
value at time n of an annuity with m payments per year for n years
value at time n + 1/m of an annuity with m payments per year for n years
value at time n of a continuously payable annuity with ρ(t) = t for t ∈ (0, n)

TWRR

38

time weighted rate of return

yt
Yt = ln(1 + yt )

99
99

t-year spot rate of interest
t-year spot force of interest

CHAPTER 1

Simple and Compound Interest
1 Simple interest
1
...
The existence of a market where money can be borrowed and lent allows people to transfer
consumption between today and tomorrow
...

Interest on short-term ﬁnancial instruments is usually simple rather than compound
...
2 The simple interest problem over one time unit
...
The repayment is the sum of the principal and interest and is sometimes called
the future value
...
Then the gain
(or interest) on the investment c0 is c1 − c0
...
The last formula
encapsulates the maxim “a pound today is worth more than a pound tomorrow”
...
2a
...
Then the repayment in

c1 = (1 + 0
...
08

Of course, the transaction can also be viewed from the perspective of the person who receives the money
...
The interest that the borrower must pay on the loan
is c1 − c0
...
3 The simple interest problem over several time units
...
The return to the investor will be
cn = c0 + nic0 = c0 (1 + ni)
It follows that present value (at time t0 ) of the amount cn at time t0 + n is
1
c0 =
cn
1 + ni

1
...
Interest calculations are usually based on the exact number of days as a
proportion of a year
...
This
convention is denoted ACT/365 which is short for actual number of days divided by 365
...
J
...
J
...
4a
...
a
...
a
...
a
...
(a) 1000 × 0
...
44
...
1 = 100
...
1 × 366/365 = 100
...

Under ACT/365, a deposit at 10% for one year which is a leap year will actually pay slightly more than 10%—it
will pay:
366
10% ×
= 10
...

1
...
In general, an amount of money has different values on different dates
...
The time value is represented in the following
diagram:
n years
n years

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...
This ﬁxed time
point is sometimes called the focal point or valuation date
...
a
...
But the value x at time t0 − 2 has value x(1 + i) = c0 (1 + i)/(1 + 2i) one year
later at time t0 − 1
...
In fact, the amount c0 at time t0 had value c0 /(1 + i) at time t0 − 1
...

Example 1
...
A person takes out a loan of £5000 at 9% p
...
simple interest
...

What is the amount outstanding one year after the loan was taken out (time t0 + 1)?
Solution
...
Hence we must take t0 + 1 as the focal point
...
09] − 1000 1 + 0
...
09
− 500 1 + 0
...
625
12
12
2
Here is another solution: £1,000 is borrowed for 3 months, £750 for 5 months, £500 for 6 months and £2,750 for
12 months
...
09
0
...
09
× 3 + 750 ×
× 5 + 500 ×
× 6 + [2750 × 0
...
625
1000 ×
12
12
12
The total capital outstanding after 12 months is £2,750 and hence the total amount due is £2,750 + £320
...
625
...
63
...
6
Summary
...
1 Explanation of compound interest
...

Compound interest means that interest is given on any interest

Example 2
...
Suppose an investor invests £720 for two years at a compound interest rate of 10% p
...
Find the
amount returned
...
The amount in pounds is c2 = 720(1 + i)2 = 720 × 1
...
20
...
a
...
This formula assumes the interest rate is constant and
hence does not depend on the time point or the amount invested
...
2 Simple versus compound interest
...
Exercise 25 shows that for the investor, simple is preferable
to compound for less than 1 year whilst compound is preferable to simple for more that 1 year
...

2
...
Suppose t1 ≤ t2 then an investment of c/(1 + i)t2 −t1 at time t1 will produce a return of c
at time t2
...

In particular, the expression “the present value (time 0) of the amount c at time t” means exactly the same as
“the discounted value at time 0 of the amount c at time t
...

(1 + i)t
1+i
Example 2
...
Inﬂation adjusted rate of return
...
Suppose further that an
investment produces a return of i1 per annum
...
04 (or 4% per annum) and the inﬂation rate is i0 = 0
...

Solution
...
After 1 year, the accumulated value will be c0 (1 + i1 )
...

For our particular case ia = 0
...
02 = 0
...
96%
...

2
...
Loan companies often quote a rate
such as 15% per annum compounded monthly
...
15 12
£100 1 +
= £100 × 1
...
08
...
08%
...
a
...
Then the
corresponding effective annual rate of interest is deﬁned to be the rate which would produce the same amount
of interest per year; i
...

(
)
0
...
01)12 − 1 = 0
...
68%
...
J
...
4a
...
What is the
amount returned?
Solution
...
05)4 = £1215
...

2
...
In general, i(m) and i1/m denote the nominal
rate of interest per unit time which is compounded m times per unit time and which corresponds to the effective
interest rate of i per unit time
...
5a)
i= 1+
m
[
]
The quantity i(m) = m (1 + i)1/m − 1 is also described as a nominal rate of interest per unit time paid mthly,
or convertible mthly, or with mthly rests
...
Writing equation (2
...
5b)
(1 + i)p = 1 + pip where p =
m
The quantity iep = pip = i(m) /m is called the effective interest rate over the period p
...

Clearly, equivalent effective annual interest rates should be used for comparing nominal interest rates with
different compounding intervals
...
5a
...
a
...
Find
(a) the effective annual rate of interest;
(b) the effective rate of interest over 3 months
...
In this case p = 1/4 and ip = 0
...
The answer to part (b) is iep = 0
...
025
...
025)4 − 1 = 1
...
38%
...
6 Terminology
...

In general, for a situation of less than one year, the equivalent simple interest rate per annum is usually quoted
...
13 where c0 is
his initial investment
...
In this case, the quoted rate is the one year
rate which is compounded
...
Thus the
phrase “a 3 year investment at 10%” used above is just shorthand for “a 3 year investment at 10% p
...
”
In general, to specify an interest rate, we need to specify the basic time unit and the conversion period
...
When the conversion period does not equal the basic time period,
the interest rate is called nominal
...
a
...
There is one interest payment after 2 years and this payment has size equal to 12% of the
capital1
...
a
...

In general, for both p ∈ (0, 1) and p > 1, an effective interest rate iep over the period p is also described as a
nominal interest rate of ip = iep /p per annum convertible 1/p times a year
...
a
...

Simple and Compound Interest

Jan 28, 2016(12:36)

Section 2 Page 5

Calculations should usually be performed on effective interest rates—the nominal rate is just a matter of presentation
...

For many problems, you should follow the following steps (or possibly a subset of the following steps):
• From ip , the nominal rate for period p, calculate iep , the effective rate for period p, by using nominal
rate = frequency × effective rate (where frequency is 1/p)
...

• Convert to the nominal rate for period q by using nominal rate = frequency × effective rate where
frequency is 1/q
...

2
...

Example 2
...
What is the simple interest rate per annum which is equivalent to a nominal rate of 9% compounded
six-monthly, if the money is invested for 3 years?
Solution
...
045)6 = 1
...
If the annual rate of simple interest is r then we need
1 + 3r = (1
...
100753 giving a simple interest rate of 10
...

Example 2
...
Suppose the effective annual interest rate is 4 1/2%
...
(a) Now (1 + iep )4 = 1 + i = 1
...
011065 for one quarter
...
04426 or nominal 4
...
a
...
(b) (1 + iep )12 = 1
...
04410
...
410% p
...
payable monthly
...
045 implies iep = 0
...
Hence
answer (c) is ip = iep /2 = 0
...
6% p
...
payable every 2 years
...
7c
...
Between
1 April 2001 and 1 October 2002, the account paid 5% per annum effective
...
Interest is added to the account after close of business on 31 March each year
...

Solution
...

• Calculate the amount in the account at every time point where there is a transaction
...
0154
...
0154 − 1,000
...
0154 − 1000) × 1
...
051/2 × 1
...
041/2 = 4110
...
7a)
• Convert every amount into the value at 1 October 2003 and then add these amounts together
...
0154 × 1
...
04 on 1 October 2003
...
053/2 × 1
...
And so on
...
In
fact, the second approach is just the same as expanding out equation (2
...

2

The term factor is used because the accumulated value is c0 A(p) if the initial capital is c0
...
We then haveA(0, t2 ) = A(0, t1 )A(t1 , t2 ) and so
A(0, t2 ) A(t2 )
A(t1 , t2 ) =
=
A(0, t1 ) A(t1 )
Hence there are no real advantages in using this alternative notation A(t1 , t2 ) and we stick with A(t)
...
J
...
7d
...
a
...

(a) What is the effective interest rate over 2 years
...
(a) The effective interest rate over 2 years is 5%
...
Then (1 + i)2 = 1
...
0247
...
47%
...
7e
...

Solution
...
0975/12) = 1008
...
09875/4) = 1024
...
0975
− 1 = 0
...
20%
...
09875
1+
− 1 = 0
...
25%
...

If both rates corresponded to the same effective annual rate of interest, then the 3 month rate could be obtained from
the 1 month rate by doing the calculation:
[(
]
)3
0
...
09829433
12
which is slightly less than the rate of 9 7/8% in the table above
...
7f
...

Solution
...
Then the effective rate of interest for time period n is:
(1 + in) − (1 + i(n − 1))
i
=
1 + i(n − 1)
1 + i(n − 1)
which decreases with n
...

Example 2
...
Show that a constant rate of compound interest implies a constant rate of effective interest
...
Let i denote the rate of compound interest per unit time
...
8 Continuous compounding: the force of interest
...
5a) we have
[
]
i(m) = m (1 + i)1/m − 1

(2
...
Using this result on equation (2
...
Denote this limit by δ
...

It follows that if an amount c0 is invested at a rate δ per annum compounded continuously, then after 1
year it will have grown to c1 = c0 eδ and after p years it will have grown to c0 epδ , where p is not necessarily
an integer
...
8a
...

Find the amount returned after 91 days
...

(
)
91
Solution
...
06 ×
= £10,149
...

365
(
)91
0
...
70
...
06 after 1 year and hence grows to c0 e0
...
This has value
£10,150
...

The answer to part (c) could be derived from ﬁrst principles as follows: suppose each day is split into n pieces and
the interest is compounded after each piece
...
06
ct = c0 1 +
365n
(
)n
(
)91
x
Using
lim 1 +
= ex gives
lim ct = c0 e0
...
06×91/365 as above
...
8b
...

(
)1/91
91
0
...
(a) 1 + 0
...
000163 (b)
= 0
...
06/365 − 1 = 0
...
8c
...
Find the equivalent effective annual interest rates
...
06
Solution
...
06 ×
− 1 = 0
...
0618 (c) e0
...
0618
365
365
Example 2
...
Continuation of the previous example
...

[(
]
)91
365
0
...
06×91/365
Solution
...
06 (b)
1+
− 1 = 0
...
060451
91
365
91

The general formulæ are left as exercises
...
8e
...
4% per annum
...

Solution
...

Hence, assuming an investment of £c0 , the instrument pays the amount
(
)
0
...
The effective annual rate is
)365/91
(
91
− 1 = 0
...
064 ×
365
which is 6
...
The nominal continuously compounded rate is
(
)
365
91
δ = ln(1 + i1 ) =
× ln 1 + 0
...
0635 which is 6
...

91
365
Alternatively, equating the accumulated value after 91 days gives
91
e91δ/365 = 1 + 0
...
J
...
9
Summary
...
Now
nominal rate = frequency × effective rate
...

Hence i(m) denotes the nominal rate of interest compounded m times per unit time
...
Suppose the force of interest (or the nominal rate of interest compounded
continuously) is δ per annum
...

Also, if i denotes the equivalent effective annual rate of interest, then:
1 + i = eδ
and
δ = ln(1 + i)

3 Exercises

(exs1-1
...
(a) Suppose a loan of £5000 is to be repaid by the amount £6000 at the end of two months
...
a
...
The simple yield on a deposit is 9
...
What is the equivalent simple yield on the ACT/365
basis?
3
...
a
...
How much interest is earned (a) during
the ﬁrst year; (b) during the second year?
4
...
What is the
equivalent nominal annual interest rate convertible monthly?
5
...
a
...
One way to repay the loan is £10 at the end of
each month for 11 months with a ﬁnal payment of £1010 at the end of the twelfth month
...
Suppose the simple interest rate quoted for a 13 week (91 day) investment under ACT/365 is 10% p
...
What is the
effective annual rate?
7
...
How much did A owe one year ago?
8
...
What is the effective annual interest rate?
9
...
Find (a) the equivalent nominal
interest rate per annum payable every 6 months; (b) the equivalent effective annual interest rate; (c) the equivalent
effective interest rate payable quarterly
...
(a) Find the effective annual interest rate equivalent to a nominal rate of 6% payable every 6 months
...
a
...
a
...
Suppose the nominal interest rate p
...
payable quarterly is 8%
...
a
...
Consider the effective interest rate of 1% per month
...

(b) Find the equivalent nominal interest rate p
...
(i) payable monthly (ii) payable every 3 months; (iii) payable
every 6 months; (iv) payable every year; (v) payable every 2 years
...
Suppose an investment of £1,500 returns a total (principal plus interest) of £1,540 after 91 days under ACT/365
...
Suppose someone has 3 outstanding loans: loan 1 is cleared by a repayment of £300 in 2 years, loan 2 by a repayment
of £150 in 4 years and loan 3 by a repayment of £500 in 5 years
...
a
...

(a) What single payment in 6 months’ time will pay off all 3 loans?
(b) At what time could a repayment of £950 clear all 3 debts (to the nearest month)?
15
...
When an interest payment is received, it is
immediately invested to the end of the 3 year period
...
a
...
5% p
...
at the end of the second year
...
(a) A loan has a ﬁxed nominal rate of 6% p
...
payable every 3 months
...
Now suppose the loan is repaid by 10 equal payments at the end of each year; what
is the value of these annual payments?
(b) Now suppose the loan has a ﬁxed nominal rate of i% p
...
payable every 3 months and the loan is paid off by 20
payments of £x every 6 months for 10 years
...

17
...
)
18
...
5% p
...
under ACT/360
...
(a) Suppose a deposit of £500 leads to a repayment of £600 after 4 years
...

(i) If the loan is extended for one further year, calculate the new repayment
...
(a) Suppose an investment for k days has a continuously compounded interest rate of r
...

(b) Suppose an investment for k days is quoted as having a simple interest rate of i
...
Suppose the force of interest is δ = 0
...

(a) Find the equivalent nominal rate of interest per annum compounded
(i) every 7 days (assume 365 days in the year);
(ii) monthly
...
What is the amount returned?
22
...
By using the inequality 1 < t1/n < x1/n for t ∈ (1, x) and n > 0, show that ln x < n(x1/n − 1) <
x1/n ln x
...

[
]
23
...
Show that f ↓ as m ↑
...
Suppose i denotes the effective annual interest rate and ip denotes the nominal interest rate per period p
...
Show that
A(p) − 1
δ = lim ip = lim
p↓0
p↓0
p
25
...
Hence A1 (t) = 1 + ti and A2 (t) = (1 + i)t
...

For the investor, simple is preferable to compound for less than 1 year; compound is preferable to simple for more
that 1 year
...
On some ﬁnancial instruments, compound interest is used but with linear interpolation between integral time units
...
After 1 year and 57 days how much does he
owe? (Use ACT/365
...

Page 10 Section 4

Jan 28, 2016(12:36)

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ST334 Actuarial Methods ⃝R
...
Reed

27
...
After 30 days the investor sold
the bill to a second investor for a price of £93
...
The second investor held the bill to maturity
when it was redeemed at par
...

(Institute/Faculty of Actuaries Examinations, September 2004) [3]
28
...
The annual effective rate of
interest over the whole of the next 10 years will be 7%, 8% or 10% with probabilities 0
...
5 and 0
...

(a) Calculate the single premium
...

(Institute/Faculty of Actuaries Examinations, September 2000) [4]
29
...
5% per month
...

(Institute/Faculty of Actuaries Examinations, September 2002) [2+2=4]
30
...
50
...

Determine which investor receives the higher rate of return
...
An investor purchases a share for 769p at the beginning of the year
...
Capital gains tax of 30% is paid on the difference between
the sale and the purchase price
...

(Institute/Faculty of Actuaries Examinations, September 2007) [4]
32
...

(b) Explain why the amount takes longer to accumulate in (i)(a)
...
Calculate the time in days for £3,000 to accumulate to £3,800 at:
(a) a simple rate of interest of 4% per annum;
(b) a compound rate of interest of 4% per annum effective
...
An investor pays £120 per annum into a savings account for 12 years
...
In the second four years, the payments are made quarterly in advance
...

The investor achieves a yield of 6% per annum convertible half-yearly on the investment
...

(Institute/Faculty of Actuaries Examinations, April 2015) [7]

4 Discount rate and discount factor
4
...
Suppose the quantity c0 accumulates to c1 in
one time unit
...
It is also called
the discount on the repayment c1
...
The last formula can be remembered in the form:
interest rate × present value = discount rate × future value
...

1+i
1+i
Hence c0 is the present value (at time t0 ) of the amount c1 at time t0 + 1, and
c0 =

present value = discount factor × future value

Simple and Compound Interest

Jan 28, 2016(12:36)

Section 4 Page 11

It is easy to confuse the terms ‘discount rate’ (denoted d) and ‘discount factor’ (denoted ν), but the context
usually provides some assistance
...
Note that a discount rate
(like an interest rate) is usually expressed as a percentage
...

Example 4
...
Suppose the amount of £100 is invested for one year at 8% per annum
...
08) × 100 = 108
...
08
The discount factor is 1/1
...
9259
...
074 or 7
...

4
...
If the amount cn is due after n time units and a simple discount
rate of d per time unit applies, then the amount that needs to be invested now is c0 = (1 − nd)cn
...
2a
...
a
...
Find the amount of the discount
...
The number of days is 61
...
05 = 835
...

The interest of £835
...
62
...
62
100,000 × 0
...
05
=
=
= 0
...
62
100,000 − 835
...
05
Alternatively, using (1 + ni)(1 − nd) = 1 gives i = d/(1 − nd) = 0
...

Note that the simple interest rate is not equal to the simple discount rate d = 0
...

4
...
The coupon is the interest payment made by the issuer of a security
...
The face value is generally repaid (or redeemed) at maturity, but sometimes it
is repaid in stages
...

The yield is the interest rate which can be earned on an investment
...

A discount instrument is a ﬁnancial instrument which does not carry a coupon
...
Treasury bills are sold in
this way
...

Example 4
...
A Treasury bill for £10 million is issued for 90 days
...

Solution
...
29
c0 =
1 + ni 1 + (90/365) × 0
...
The term market price denotes the present value
...
J
...
4 The key formulæ for simple interest and simple discount are cn = (1 + ni)c0 and c0 = (1 − nd)cn
...

Some books state the results in words as follows:
mathematical formula
formula in words
cn
maturity proceeds
c0 =
market price =
1 + ni
1 + (yield × length of investment)
d
discount rate
i=
yield =
1 − nd
1 − (discount rate × length of investment)
i
yield
d=
discount rate =
1 + ni
1 + (yield × length of investment)
c0 = cn (1 − nd)
market price = face value × (1 − discount rate × length of investment)
cn − c0 = cn dn
amount of discount = face value × discount rate × length of investment
Example 4
...
A Treasury bill for £10 million is issued for 90 days
...
The key formulæ are cn = (1 + ni)c0 and c0 = (1 − nd)cn
...
Here is one way: from (1 − nd)(1 + ni) = 1 we get
i
0
...
09759 or 9
...
a
...
1 × 90/365 3740
Hence the amount of the discount is
90
365
90,000,000
cn − c0 = ndcn =
×
× 10,000,000 =
= 240,641
...
5 Effective and nominal discount rates
...

Suppose d(m) denotes the nominal discount rate per annum payable m times per year
...
Hence it is equivalent to the effective discount d per annum
where
(
)m
d(m)
1−d= 1−
m
Example 4
...
The effective discount rate is 1% per month
...

Solution
...

If p = 1/m, the quantity d(m) is also denoted dp and is called the nominal discount rate payable per time
period p
...
As for interest rates,
we have the relations d = d(1) = de1 and so 1 − d = (1 − dep )1/p
...

The key results for nominal discount rates are:

nominal = effective × frequency
(
)
(
)(
)
(m) m
d
i(m)
d(m)
1/p
1−d= 1−
= (1 − dep )
1+
1−
= (1 + iep )(1 − dep ) = 1
m
m
m
For many problems, the procedure is as follows (or possibly a subset of the following steps):
• From the nominal rate for period p, calculate dep , the effective rate for period p by using nominal rate
= effective rate × frequency, where frequency is 1/p
...

• Convert to the nominal rate for period q by using nominal rate = effective rate × frequency where
frequency is 1/q
...
5b
...
What is the effective interest rate per annum?
(b) The nominal discount rate is 6% per annum payable quarterly
...
(a) Nominal 6% p
...
payable quarterly is equivalent to an effective rate of 1 1/2% for a quarter, which is
equivalent to 1
...
06136, or 6
...

(b) Nominal discount rate 6% p
...
payable quarterly
is equivalent to an effective discount rate of 1 1/2% for a quarter
...
015)4 = 0
...
Hence we have an effective discount rate of 5
...
a
...
5c
...

m→∞

Solution
...
For the second relation use 1 + i(m) /m = (1 + i)1/m
...
8
...

Example 4
...
Suppose the effective interest rate is 1% per month
...

(b) Find the nominal discount rate (i) payable every 3 months; (ii) payable annually; (iii)payable every 2 years
...
(a) The effective interest rates are (i) 1
...
0112 − 1 p
...
(iii) 1
...

Hence the values of dep are: (i) 1 − (1/1
...
01)12 ; (iii) 1 − (1/1
...

(b) The values of d(m) = m × dep are: (i) 4(1 − (1/1
...
01)12 ; (iii) 0
...
01)24 )
...
6 Relation between interest payable mthly in advance and in arrear
...
6a
...

• Suppose further that the borrower repays the interest by equal instalments of x at the end of each mth subinterval
(i
...
at times 1/m, 2/m, 3/m,
...
Find x
...
Let ν = 1/(1 + i) denote the value at time 0 of the amount 1 at time 1
...
, 1 should be equivalent to a single payment of i at
(
)
time 1
...

x=

Hence i(m) is the total interest paid for the loan schedule above and i(m) /m is the size of each of the m payments
...

• Now suppose that the borrower repays the interest by equal instalments of y at the start of each mth subinterval
(i
...
at times 0, 1/m, 2/m, 3/m,
...
Find y
...
The cash ﬂow is now:
Time
Cash Flow

0
−1 + y

1
m

2
m

y

y

In this case,
1=

m−1
∑
i=0

and so

yν i/m +

1
1−ν
=y
+ν
1+i
1 − ν 1/m

···
···

m−1
m

y

1
1

Page 14 Section 4

Jan 28, 2016(12:36)
y = 1 − ν 1/m = 1 − (1 − d)1/m =

d(m)
= dep
m

c
ST334 Actuarial Methods ⃝R
...
Reed
where p = 1/m
...
, (m − 1)/m should be equivalent to a single payment of
0,
)
i at time 1
...

Hence d(m) is the total interest paid for the loan schedule above and d(m) /m is the size of each of the m
payments
...

If the annual effective interest rate is i and d is given by (1 − d)(1 + i) = 1, then the preceding example shows
that the 4 series of payments in the following table (4
...

Time

0
d
d(m)
m

1
m

2
m

d(m)
m
i(m)
m

d(m)
m
(m)
i
m

···

···
···

m−2
m

m−1
m

d(m)
m
i(m)
m

d(m)
m
(m)
i
m

1

i(m)
m
i

(4
...
7 Continuous compounding: the force of interest
...
8) that 1 + i = eδ for continuous
compounding where δ denotes the force of interest per annum and i denotes the equivalent effective annual
rate of interest
...
Hence the discount factor is e−δn/365
...

Example 4
...
Suppose an investment of 91 days has an interest rate of 6% per annum continuously compounded
...
The corresponding simple interest rate per annum is:
)
365 ( 0
...
06045 which is 6
...

91
Now c1 = c0 e0
...
Hence the discount factor is
e−0
...
98516

4
...

• Discount rate (d) and discount factor (ν ):
1
ν=
d=1−ν
1+i
• Simple discount over several time units
...
They are sold for
less than the face value—i
...
they are sold at a discount
...

• Compound interest
...

m
m
m
)(
)
(
d(m)
i(m)
1−
= (1 + iep )(1 − dep ) = 1
1+
m
m

Simple and Compound Interest

Jan 28, 2016(12:36)

Exercises 5 Page 15

5 Exercises

(exs1-2
...
If the simple interest rate on a 91 day investment is 8%, what is the discount factor?
2
...
What is the 3 year
discount factor?
3
...
a
...
Suppose £720 is invested for 2 years at a compound interest rate of i = 1/6 per annum
...
Suppose the size of an initial investment at time 0 is c0 = 275 and the discount rate is 33 1/3% per annum
...

6
...
Show that the discount rate is also increased
and the discount factor is reduced
...
Consider the simple interest problem
...
Prove that the
discount rate is not increased by as much as k times
...
Assume i > 0
...
(a) Suppose the nominal discount rate is 2 1/2% p
...
payable every 2 years
...
(i) What is the equivalent nominal annual discount rate
payable quarterly? (ii) What is the equivalent nominal annual discount rate payable monthly? (iii) What is the
equivalent nominal annual discount rate payable every 3 years?
10
...
Show that
(1 + ieq )1/q (1 − dep )1/p = 1
11
...

(a) Find the effective interest rate for a period of (i) k months; (ii) p years
...

12
...
What is the equivalent nominal discount rate
payable every 6 months?
13
...

(a) How many years does it take for the accumulated value to double at a rate i per annum compounded annually?
(b) How many years does it take for the accumulated value to double at a nominal rate i per annum compounded
continuously
...

(d) Suppose the answer to part (a) is denoted by na and the answer to part (b) is denoted by nb
...

14
...
e
...
Suppose the corresponding compound interest
rate is i1 per annum
...
(ii) Show that d2 < 2d1 where d2 is the simple accumulated
discount rate and d1 is the annual discount rate
...
Let i1 denote the corresponding annual compound interest rate
...
Show that it > 2is and dt < 2ds
...
(a) Suppose the discount rate is 1% for one month
...

(b) Suppose a bank has an effective discount rate of 2% for a duration of 3 months
...
Let δ = ln(1 + i)
...

(b) Show that i > i(2) > i(3) > · · · and d < d(2) < d(3) < · · ·
...

Page 16 Section 6

Jan 28, 2016(12:36)

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...
Reed

17
...
Calculate the effective
rate of return per annum received by an investor who purchases the Bill at issue and holds it to maturity
...
An investment is discounted for 28 days at a simple rate of discount of 4
...
Calculate the annual
effective rate of interest
...
A government issues a 90-day Treasury Bill at a simple rate of discount of 5% per annum
...
(Use
(Institute/Faculty of Actuaries Examinations, September 2000) [3]
ACT/365
...
The rate of discount per annum convertible quarterly is 8%
...

(b) The equivalent rate of discount per annum convertible monthly
...
The rate of interest is 4
...

(i) Calculate
(a) the annual effective rate of discount;
(b) the nominal rate of discount per annum convertible monthly;
(c) the nominal rate of interest per annum convertible quarterly;
(d) the effective rate of interest over a ﬁve year period
...

(Institute/Faculty of Actuaries Examinations, September 2013) [5+2=7]
22
...
One is a 91-day deposit which pays a rate of interest of 4% per annum
effective
...

Calculate the annual simple rate of discount from the treasury bill if both investments are to provide the same effective
rate of return
...
A 182-treasury bill, redeemable at $100, was purchased for $96
...
The rate of return received by the initial purchaser was 4% per annum
effective
...

(ii) Calculate the annual simple rate of return achieved by the second investor
...

(Institute/Faculty of Actuaries Examinations, April 2015) [2+2+2=6]

6 Time dependent interest rates
6
...
Suppose an amount c0 is invested at
a rate δ per annum compounded continuously
...

Now suppose that the interest rate varies with time, t
...
For t ≥ 0, let
A(t) denote the accumulated value at time t of an investment of 1 at time 0
...

Assuming A(t) is differentiable implies
lim ip (t) = lim
p↓0

p↓0

A(t + p) − A(t)
pA(t)

exists for all t > 0
...
Hence
A′ (t)
for all t > 0
...

Equation (6
...
1a)

Simple and Compound Interest

Jan 28, 2016(12:36)

(∫

t2

A(t2 ) = A(t1 ) exp

)
δ(s) ds

Section 6 Page 17

for t2 ≥ t1

(6
...

(6
...
1a
...
Show that
A(t) = A(t1 ) · · · A(tn )
(∫
)
t
Solution
...
This result is true for ordinary constant compound interest
...
2 Present and discounted value
...
Then
Deﬁnition 6
...
The present value function corresponding to the force of interest function, δ, is the

function ν : [0, ∞) → [0, ∞) deﬁned by

( ∫ t
)
ν(t) = exp −
δ(s) ds

(6
...

Using A(t) = 1/ν(t) gives equation (6
...
1b)
...

Example 6
...
Express ip (t), the nominal rate of interest per period p, in terms of
(b) ν(t), the present value function
...

Solution
...
1c) gives
(∫
)
t+p
exp t δ(s) ds − 1
ip (t) =
p
(b) Using A(t + p) = A(t)[1 + pip (t)] and A(t) = 1/ν(t) gives
[
]
ν(t)
1
−1
ip (t) =
p ν(t + p)

6
...
For further results on the present value of a mixture of discrete and continuous payment
streams, see section 2 of the next chapter
...
4
Summary
...

• The accumulated value at time t of the amount 1 at time 0 is
(∫ t
)
A(t) = exp
δ(s) ds
for t > 0
...
J
...
tex)

1
...
05 per annum
...

(b) Now suppose £1,000 is invested for 6 months at an effective rate of interest of 5% p
...
Find the accumulated
value
...
Suppose the force of interest function, δ, is given by
δ(t) = 0
...
01t for t ≥ 0
...

3
...
Denote the present
value of the amount 1 at time t ≥ 0 by ν(t)
...

0

(a) Given ν, how can δ be determined?

(b) If ν(t) = e−bt , ﬁnd δ(t)
...
Let δ(t) denote the average of δ(t) over the interval (0, t)
...

t 0
We also set δ(0) = δ(0)
...
Find A(t), ν(t) and δ(t)
...
Suppose the function δ : (0, ∞) → R is nondecreasing
...

6
...
Denote the present
value of the amount 1 at time t > 0 by ν(t)
...

0

Prove that the function δ : (0, ∞) → R, is non-decreasing if and only if
(
)α
ν(αt) ≥ ν(t)
for all 0 ≤ α ≤ 1 and for all t > 0
...
Suppose δ(t) = 0
...
7t where time t is measured in years
...

8
...
Show that ip (t), the nominal interest rate per period p
for an investment starting at time t is given by:
[
(∫ t+p
)
]
1
ip (t) =
exp
δ(u) du − 1
p
t
(b) Suppose that δ(t) = 0
...
7t
...

9
...
The force of interest δ(t) at time t is a + bt2 where a and b are constants
...

(Institute/Faculty of Actuaries Examinations, September 2005) [5]
Determine a and b
...
(a) Calculate the present value of £100 in ten years’ time at the following rates of interest/discount
...
91 and is redeemed at $100
...

(Institute/Faculty of Actuaries Examinations, September 2005) [4+3=7]

Simple and Compound Interest

Jan 28, 2016(12:36)

Exercises 7 Page 19

12
...
04 + 0
...
07
if 8 ≤ t
...

(b) Calculate the present value at time t = 0 of £100 due at time t = 10
...
The force of interest δ(t) is a function of time and at any time t, measured in years, is given by the formula:
{
0
...
09 − 0
...
01t − 0
...

(b) Determine the constant effective rate of interest per annum, to the nearest 1% which would lead to the same
result as in (a) being obtained
...
The nominal rate of discount per annum convertible quarterly is 8%
...

(ii) Calculate the equivalent effective rate of interest per annum
...

(Institute/Faculty of Actuaries Examinations, September 2012) [1+1+2=4]
15
...
05 + 0
...
15
for t > 5
(i) Calculate the present value of £1,000 due at the end of 12 years
...

(Institute/Faculty of Actuaries Examinations, September 2008) [5+2=7]

Page 20 Exercises 7

Jan 28, 2016(12:36)

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ST334 Actuarial Methods ⃝R
...
Reed

CHAPTER 2

Cash Flows, Equations of Value
and Project Appraisal
1 Cash Flows
1
...
When assessing the value of an investment, it often helps to list the times and values of
any cash ﬂows
...

Here are some examples
...
This is a security which provides a speciﬁed cash amount at some speciﬁed future time
...
Then the
cash-ﬂow sequence can be represented by the following table:
Time
Cash ﬂow

n
S

0
−c0

Fixed-interest security
...
, n and a lump sum S (called the redemption value) at time n
...

Index-linked security
...
This means that the sizes of the future cash
ﬂows are unknown at time 0, although they may be known in real terms
...
An investor pays a premium of c0 at time 0 and receives payments of c at times 1, 2,
...

Time
Cash ﬂow

0
−c0

1
c

···
···

n−1
c

n
c

Equity
...
The size of the dividend is
determined each year by the company
...

Term Assurance
...
If death occurs before time n, then the beneﬁciary receives
an amount c
...
There are many variations—for example, the
policy may be purchased by a sequence of annual premiums
...
The borrower initially receives a loan and then repays it by a series of payments which
consist of partial repayment of the loan capital and interest
...
The insurance company receives a premium (positive cash ﬂow) at the
start of the year
...
The negative
cash ﬂows may occur after the end of the period of cover—especially for personal injury claims
...
J
...
J
...
1 Net present value
...
The cash amounts must be discounted to allow for the fact that cash received tomorrow is worth
less than cash received today—the reason for the term discounted cash ﬂow
...
1a
...
Find the
net present value at time 0
...
The cash-ﬂow sequence can be represented by the following table:
Time
Cash ﬂow, a

0
−c0

1
c1 = c0 (1 + i)

The net present value at time 0 is

NPV(a) = −c0 +

1
c1 = 0
1+i

In general, suppose c = (c0 , c1 ,
...
, n
...
Then the net present value is
n
∑ cj
c2
cn
c1
+ ··· +
=
(2
...
1a) is sometimes called the discounted cash ﬂow or (DCF ) formula
...
Also, it decreases faster when later payments such as cn are larger
...

Example 2
...
Consider the two cash ﬂow sequences a = (12, 12, 12, 20, 24) and b = (20, 18, 14, 12, 12) at
times t = 0,
...
Find the net present values of the two cash-ﬂows assuming an interest rate of (a) 3% and (b) 10%
...

Solution
...
59 where ν = 1/1
...
Similarly
NPV(b) = 72
...

(b) For 10%, we have NPV(a) = 64
...
15 and so b is preferable to a
...

Example 2
...
Consider the following two investment projects:
• Project A
...

• Project B
...
a
...

Solution
...
Then
−95 + 6ν + 6ν 2 + 6ν 3 + 6ν 4 + 6ν 5 + 6ν 6 + 6ν 7 + 106ν 8 = 0
which leads to 101ν + 100ν 8 − 106ν 9 = 95 or 95(1 + r)9 − 101(1 + r)8 − 100r + 6 = 0
...
06832, or an effective annual rate of interest of 1 + i = (1 + r)2 = 1
...
141308 which is 14
...

Using this yield on the cash ﬂows in project A we get the NPV:
300,000 325,000 345,000 360,000
+
+
+
= 19,374
...

However, the above analysis has not allowed for the differential risks of the two projects
...
1d
...
The annuity pays out the amount £a at the end of
each of the years 1, 2,
...
Suppose the investment gives a yield of 7%
...

Solution
...
07
...
65 using r = 0
...

Future cash ﬂows and interest rates are usually uncertain
...

2
...
Consider the cash-ﬂow sequences a = (a0 ,
...
, bn )
...

(a) If aj ≥ bj for every j = 0,
...

Proof: This is easy
...
, n

k=0

then NPV(a) ≥ NPV(b) for all interest rates i ∈ [0, ∞)
...
) The proof is by induction on n
...

We now assume the result is true for n and we shall prove this implies the result is true for n + 1
...
, an+1 ) and b = (b1 ,
...
Then
n+1
∑ ak − bk
NPV(a) − NPV(b) =
(1 + i)k
k=0

Consider separately the two cases: an+1 ≥ bn+1 and an+1 < bn+1
...
Second, suppose an+1 < bn+1 ; then
an+1 − bn+1
an+1 − bn+1
≥
n+1
(1 + i)
(1 + i)n
and so
n−1
∑ ak − bk an − bn an+1 − bn+1
NPV(a) − NPV(b) =
+
+
(1 + i)k (1 + i)n
(1 + i)n+1
≥

k=0
n−1
∑
k=0

ak − bk an − bn + an+1 − bn+1 ∑ a′ − b′
k
k
+
=
(1 + i)k
(1 + i)n
(1 + i)k
n

k=0

where a′ = ak and b′ = bk for k = 0, 1,
...
Now
n
n
k
k
j
∑
k=0

a′ ≥
k

j
∑

b′ for all j = 0,
...

(1 + i)k
k=0

Hence NPV(a) − NPV(b) ≥ 0
...

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2
...

Example 2
...
Replacing a piece of capital equipment
...
For j = 1, 2,
...
Typically the wj decrease with j due to depreciation, and the cj increase
with j
...
Also, its operating
cost for year j of its operation is Cj
...

By considering the sequence of cash ﬂows over the ﬁrst 5 years, decide whether the machine be replaced immediately,
or at the end of year 1, or at the end of year 2, etc
...
Suppose the machine is replaced at time 0 (the beginning of year 1)
...
The best choice is the one with the smallest value for NPV
...
3b
...

Suppose someone plans to save the same amount £a at the beginning of every month for the next 240 months
...
Assume the
interest rate is i per annum compounded monthly
...

Solution
...
Let α = 1 + i/12
...
At time 0, the value of the savings is:
a
a
α240 − 1
a + + · · · + 239 = a 239
α
α
α (α − 1)
At time 0, the value of the withdrawals is:
c
c
c
α360 − 1
+ 241 + · · · + 599 = c 599
α240 α
α
α (α − 1)
Setting these quantities equal gives the same result as before
...
3c
...
Suppose the capital borrowed is c0 and the interest rate is i per annum
compounded monthly
...

(a) Express x in terms of n, c0 and i
...
(a) Let α = 1 + i/12
...
, n − 1
...

This type of calculation is considered in depth in section 4 of chapter 3 on page 56
...
4 The NPV of a discrete cash ﬂow
...
Hence ν(0) = 1
...
, and cn at time tn
...

...
5 The NPV of a continuous cash ﬂow
...
This means that if m(t) denotes the total payment received in [0, t], then
m′ (t) = ρ(t)
It follows that m can be obtained from ρ by using the relation:
∫ t
m(t) =
ρ(u) du
0

Now the cash received in (t, t + h] is m(t + h) − m(t) = hρ(u) for some u ∈ [t, t + h]
...
This motivates the following deﬁnition:

The net present value (time 0) of the continuous rate of payment ρ : (0, t) → R is
∫ t
NPV(ρ) =
ρ(u)ν(u) du

(2
...

2
...
Now suppose we have the discrete
payments of paragraph 2
...
To get the net present value, we
just need add the net present values for the discrete and continuous payments:
∫ t
n
∑
NPV(a, ρ) =
cj ν(tj ) +
ρ(u)ν(u) du
j=1

0

If there are inﬁnitely many discrete payments and ρ : (0, ∞) → R, then
∫ ∞
∞
∑
NPV(a, ρ) =
cj ν(tj ) +
ρ(u)ν(u) du
j=1

0

provided the sum and integral converge—and they will converge for any sensible practical example
...
6a)

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If there are both incoming and outgoing payments, then the net present value is just the difference between the
net present values of the positive and negative cash ﬂows
...
6a) becomes
∫ ∞
∞
∑ cj
ρ(u)
NPV(a, ρ) =
du
tj +
(1 + i)
(1 + i)u
0
j=1

For A(tL ), the accumulated value at time tL , just use:
A(tL ) =

NPV(a, ρ)
ν(tL )

and in general we have
(value at time t1 of cash ﬂow) × ν(t1 ) = (value at time t2 of cash ﬂow) × ν(t2 ) = NPV
For example, if we have a constant interest rate i and the project ends at time tL , where tL ∈ [tn , tn+1 ) and
ρ(u) = 0 for u > tL , then A(tL ), the accumulated value of the project at time tL is:
∫ tL
n
∑
tL
tL −tj
(1 + i) NPV(a, ρ) =
cj (1 + i)
+
ρ(u)(1 + i)tL −u du
(2
...
7 The NPV in terms of the force of interest, δ
...
Hence equation (2
...
In general, for the cash ﬂow (a, ρ), the accumulated value at time tL
is
(∫
) ∫
( ∫ tL
)
∞
tL
∞
NPV(a, ρ) ∑
=
cj exp
A(tL ) =
δ(u) du +
ρ(u) exp
δ(s) ds du
ν(tL )
tj
0
u
j=1

( ∫t
)
2
...
Differentiating the relation ν(t) = exp − 0 δ(u) du shows that ν ′ (t) =
−δ(t)ν(t) for all t ≥ 0
...

(2
...
5a) shows that equation (2
...
Hence equation (2
...
The term interest stream means a
continuous payment stream
...
8a) implies
∫ ∞
δ(u)ν(u) du = 1
0

This says that the net present value of the interest stream δ : (0, ∞) → R is 1
...
9
Summary
...
9a)

0

where ν(t) denotes the present value (time 0) of the amount 1 at time t for t ≥ 0
...
9a) becomes
∫ t
ρ(u)
NPV(ρ) =
du
u
0 (1 + i)
• The accumulated value at time t of the continuous rate of payment ρ : (0, t) → R is
(∫ t
)
∫ t
∫ t
ν(u)
A(t) =
ρ(u)
du =
ρ(u) exp
δ(s) ds du
ν(t)
0
0
u
For a constant interest rate i, equation (2
...
9b)

(2
...
9d)

0

3 Exercises

(exs2-1
...
Consider the cash-ﬂows a = (100, 2500) and b = (2500, 10) at times 0 and 1
...
Show that NPV(a) > NPV(b)
...
Show that NPV(a) < NPV(b)
...
Show that the cash ﬂow a can be transformed by borrowing and investing at
rate 1% into a cash-ﬂow c = (c0 , c1 ) with c0 ≥ b0 and c1 ≥ b1
...
Show that the cash ﬂow b can be transformed by borrowing and investing at
rate 1% into a cash-ﬂow c = (c0 , c1 ) with c0 ≥ a0 and c1 ≥ a1
...
, an ) and b = (b0 ,
...
Show that the cash-ﬂow sequence a can be transformed by investing and borrowing at rate i
into a cash-ﬂow sequence c = (c0 ,
...
, cn ≥ bn
...
Suppose a fund has value A(0) at time 0
...

(a) Find A(t), the value of the fund at time t
...
Express A(t) in
terms of i and ρ
...
An individual makes an investment of £4m per annum in the ﬁrst year, £6m per annum in the second year and £8m
per annum in the third year
...
Calculate the accumulated
value of the investments at the end of the third year at a rate of interest of 4% per annum effective
...
The force of interest δ(t) at time t is given by
{
0
...
05 + 0
...

Calculate the accumulated value at time t = 12 of a continuous payment stream of £100 per annum payable from
(Institute/Faculty of Actuaries Examinations, April 2000) [5]
time t = 0 to time t = 6
...
A particular ﬂoating note pays interest linked to an index of local ﬂoating interest rates
...
The nominal value of the note
is 100 and the note paid interest of 6
...
6% of the nominal value on
31 December 2000
...

(Institute/Faculty of Actuaries Examinations, April, 2002) [5]
6
...
05
if 0 < t < 8
δ(t) =
0
...
0004t2 if 8 ≤ t ≤ 15
Calculate the accumulated value at time t = 15 of a continuous payment stream of £50 per annum payable from time
t = 0 to t = 8
...
J
...
The force of interest δ(t) is a function of time and at any time, measured in years, is given by the formula
{
0
...
01t if 0 ≤ t ≤ 4;
δ(t) = 0
...
01t if 4 < t ≤ 8;
0
...

Calculate the present value at time t = 0 of a payment stream, paid continuously from time t = 9 to t = 12, under
(Institute/Faculty of Actuaries Examinations, April 2007) [6]
which the rate of payment at time t is 50e0
...

8
...

Starting 2 years after the initial outlay, it is estimated that income will be received continuously for 4 years at a rate
of £5,000 per annum, increasing to £9,000 per annum for the next 4 years, then increasing to £13,000 per annum for
the following 4 years and so on, increasing by £4,000 per annum every 4 years until the payment stream stops after
income has been received for 20 years (i
...
22 years after the initial outlay)
...

Calculate the net present value of the project at a rate of interest of 9% per annum effective
...
The force of interest, δ(t), is a function of time and at any time t (measured in years) is given by
{
0
...
005t for t ≤ 8
δ(t) =
0
...

(b) Calculate the present value at time t = 0 of a continuous payment stream at the rate of £200e0
...

(Institute/Faculty of Actuaries Examinations, April 2005) [3+5=8]
10
...
The payments under the grant are as follows:
Year 1
£5,000 per annum paid continuously
Year 2
£5,000 per annum paid monthly in advance
Year 3
£5,000 per annum paid half-yearly in advance
Calculate the total present value of these payments at the beginning of the ﬁrst year using a rate of interest of 8% per
annum convertible quarterly
...
The force of interest δ(t) is a function of time and at any time t, measured in years, is given by the formula:
{
0
...
008t
if 5 < t ≤ 10;
0
...
0003t2 if 10 < t
...

(ii) Calculate the effective annual rate of interest over the 12 years
...
05t per
unit time between time t = 2 and time t = 5
...
The force of interest δ(t) at time t is at + bt2 where a and b are constants
...

(i) Calculate the values of a and b
...

(iii) At the force of interest calculated in (ii), calculate the present value of a continuous payment stream of 20e0
...
(Institute/Faculty of Actuaries Examinations, September 2006) [5+2+4=11]
13
...
04
for 0 < t ≤ 5;
0
...

(i) Calculate the present value of a unit sum of money due at time t = 10
...

(iii) (a) In terms of t, determine an expression for ν(t), the present value of a unit sum of money due during the
period 0 < t ≤ 5
...
04t
...
The force of interest δ(t) is a function of time and at any time, measured in years, is given by the formula
{
0
...
01t 0 ≤ t ≤ 8
δ(t) =
0
...

(ii) (a) Calculate the present value of £500 due at the end of 15 years
...

(iii) A continuous payment stream is received at rate 10e−0
...
Calculate
the present value of the payment stream
...
The force of interest, δ(t), is a function of time and at any time t (measured in years) is given by
{
0
...
01t for 0 ≤ t ≤ 10
δ(t) =
0
...

(ii) (a) Calculate the present value of £1,000 due at the end of 15 years
...

(iii) A continuous payment stream is received at the rate of 20e−0
...

Calculate the present value of the payment stream
...
The force of interest per unit time at time t, δ(t), is given by:
{
0
...
005t for t < 6;
δ(t) =
0
...

(i) Calculate the total accumulation at time 10 of an investment of £100 made at time 0 and a further investment of
£50 made a time 7
...
05t per unit time received
(Institute/Faculty of Actuaries Examinations, April 2013) [4+5=9]
between time 12 and time 15
...
The force of interest, δ(t), is a function of time and at any time t, measured in years, is given by the formula:
{
0
...
10 − 0
...
01t − 0
...

(ii) Calculate the constant rate of interest per annum convertible monthly which leads to the same result as in (i)
being obtained
...
02t
...
The force of interest, δ(t), at time t is given by:
{
0
...
003t2 for 0 < t ≤ 5;
δ(t) = 0
...
03t
for 5 < t ≤ 8;
0
...

(i) Calculate the present value (at time t = 0) of an investment of £1,000 due at time t = 10
...

(iii) Calculate the present value (at time t = 0) of a continuous payment stream payable at the rate of 100e0
...

19
...
03 + 0
...
06
for 9 < t
...

(ii) (a) Calculate the present value of £5,000 due at the end of 15 years
...

A continuous payment stream is received at rate 100e−0
...

(iii) Calculate the present value of the payment stream
...
J
...
1 The equation of value and the yield or IRR
...
Then the rate of return is 0
...
Similarly, if the principal c0 grows to c1 after one year, then the rate of
return r is equal to
c1
r=
−1
c0
Alternatively, it is the solution of
c1
c0 =
1+r
If the prevailing rate of interest is lower than the internal rate of return of a project, then that suggests the
project is proﬁtable
...
The equation can also be written as an equation for i:
n
∑ cj
NPV(a) =
=0
(1 + i)tj
j=1

and is then also called the yield equation for i or the equation of value for i
...
Then the equation of value for δ is:
∫ ∞
n
∑
−δtj
NPV(a, ρ) =
cj e
+
ρ(t)e−δt dt = 0
0

j=1

and the yield equation or equation of value for i is
∫ ∞
n
∑ cj
ρ(t)
+
dt = 0
NPV(a, ρ) =
tj
(1 + i)
(1 + i)t
0
j=1

Deﬁne f : [0, ∞) → R by
f (i) =

n
∑
j=1

cj
+
(1 + i)tj

∫

∞
0

ρ(t)
dt
(1 + i)t

In general, the equation f (i) = 0, where f (i) = NPV(a, ρ) is considered as a function of i, is called the equation
of value for a project
...

4
...
Consider the following setup: we have the sequence of cash ﬂows: a =
(−c0 , b1 ,
...
This corresponds to an initial investment of c0 which leads to repayments b1 ,
...
The internal rate of return, r∗ , of the cash-ﬂow sequence a = (−c0 , b1 ,
...

Suppose c0 > 0, bj ≥ 0 for j = 1,
...
Let
n
∑ bj
g(r) =
for r ∈ (−1, ∞)
(1 + r)j
j=1

Then
lim g(r) = ∞,

r→−1

lim g(r) = 0,

r→∞

and g(r) ↓↓ as r ↑↑ for r ∈ (−1, ∞)
...
Using g(0) =
r∗ > 0 if

n
∑

bj > c0 ,

r∗ < 0 if

j=1

n
∑

∑n

and r∗ = 0 if

bj < c0

j=1 bj
n
∑

gives the result that

bj = c0

j=1

j=1

The last result is just common sense: the rate of return is positive iff the total amount returned exceeds the
initial investment
...
Now suppose the
interest rate is i
...
In words, if the interest rate is
greater than the internal rate of return, then the net present value is negative
...
, −cm , b0 , b1 ,
...
< t′ where all cj > 0 and all bj > 0
...

n
1
Example 4
...
An investment of £100 returns £70 at the end of year 1 and £70 at the end of year 2
...
The sequence of cash-ﬂows is a = (−100, 70, 70)
...
Hence 70x2 + 70x − 100 = 0
...
796 and r∗ = (1/x) − 1 =
0
...
The internal rate of return is 25
...

Three payments leads to a cubic, and so on
...

4
...
Starting values for a numerical search are often required
...
Then the equation
c1
cn
c2
+ ··· +
c0 =
+
2
1 + r (1 + r)
(1 + r)n
gives the following approximation r0 for r∗ :
n
n
∑
∑
c0 =
cj − r0
jcj
j=1

∑n
or

r0 =

j=1

j=1
∑n

cj − c0

j=1 jcj

• (b) Convert to c0 (1 + r)n = c1 (1 + r)n−1 + c2 (1 + r)n−2 + · · · + cn
...

• (c) Assume all future cash-ﬂows equal the average cash-ﬂow
...

...

...
This leads to
[
]
[
]
x
1
1
1
c0 =
1−
= xa n ,r
where a n ,r =
1−
r
(1 + r)n
r
(1 + r)n
can be found in actuarial tables
...

• (d) Suppose the cash ﬂow has the following form:
Time
Cash ﬂow

0
−c0

1
x

2
x

...

n−1
x

n
cn

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There is an interest payment of x for each of the n years plus a ﬁnal payment of cn − x
...
So the
approximate rate of return is
x + (cn − x − c0 )/n
i=
c0
Assuming there is a positive capital gain, we will have x/c0 < i < [x + (cn − x − c0 )/n]/c0
...

Example 4
...
(a) Consider the following cash-ﬂow sequence
Time
Cash ﬂow

0
−50

1
4

2
4

3
4

4
4

5
4

6
4

7
4

8
4

9
4

10
59

Use method (d) described above to ﬁnd an initial approximation to the internal rate of return, r∗
...
The return per year is 4 + (55 − 50)/10 = 4
...
Hence r∗ ≈ 4
...
09
...
0866868
...
3b
...
Use the
ﬁrst 3 methods described above to ﬁnd initial approximations to the internal rate of return, r∗
...
1c) on page 22
...
(a) The ﬁrst method gives r0 =
cj / jcj = 18/365 = 0
...

The second method gives r0 =
18/293 = 0
...
For the third method, we have x = 105/4 = 26
...
So we want r with a 4 ,r = 87/26
...
312
...
387 ∑ a 4 ,0
...
312; so take r = 0
...
The actual value of r∗ is 0
...

and
From actuarial tables, a 4 ,0
...
052
...
0895
...
5
...
5 = 190/37 = 5
...
From
actuarial tables, a 8 ,0
...
146 and a 8 ,0
...
0455, so take r = 0
...

The actual value is r = 0
...
The third method gives a poor approximation because the mean of 18
...

4
...
If the interest rate is i, then NPV(i) is the present value of the cash-ﬂows of the
project
...

Suppose r∗ , the yield or IRR exists, and NPV(i) > 0 for i < r∗ and NPV(i) < 0 for i > r∗
...

4
...
Consider the cash-ﬂow sequence a = (c0 , c1 ,
...
Let x∗ =
1/(1 + r∗ )
...
Any root for x which satisﬁes x > 0 leads to a solution for r with r > −1
...
Hence the internal rate of return cannot be deﬁned
...
It then follows that the property that
NPV is positive for interest rates i on one side of r∗ and negative for interest rates on the other side of r∗ may
not hold
...

Example 4
...
Consider the following sequence of cash-ﬂows:
Time
Cash ﬂow

0
−32

1
128

2
−166

3
70

Find the internal rate of return, r∗
...
The equation c0 + c1 x + c2 x2 + · · · + cn xn = 0 gives −32 + 128x − 166x2 + 70x3 = 0
...
Hence we have 0 = (x − 1)(70x2 − 96x + 32) = (x − 1)(7x − 4)(10x − 8)
...
Hence r∗ = 0, 0
...
25
...
Also NPV(r) < 0 for r ∈ (0, 0
...
25, 0
...

Example 4
...
Find the IRR for the following cash-ﬂow
...
We have 2 − 4x + 3x2 = 0 and this has no real roots
...

• The IRR measure does not use the value of the currently available interest rates
...
In particular, if an investment accumulates funds
for a ﬁnal payment, then the IRR calculation assumes that the sinking fund earns interest at the internal rate of
return
...

• NPV and IRR may lead to different conclusions—an example is given in the next section
...

5 Comparing Two Projects
5
...

Example 5
...
Consider the following two cash-ﬂow sequences
...
This suggests that project A
is preferable to project B
...
04
...

Solution
...
This gives (20 + x + x2 )(−4 + 5x) = 0
...

For project B, solve −80 + 10x + 10x2 + 90x3 = 0
...
Hence x∗ = 8/9 and
∗
rB = 1/x∗ − 1 = 1/8
...
04 = 1/25 then NPV(a) = 17
...
87 and project B is preferable to project A!!
(c) This is shown in ﬁgure (5
...
The value of i where the two graphs cross is called the cross-over rate
...
0

0
...
10

0
...
20
interest rate

0
...
30

Figure 5
...
Plot of NPVi (a) and NPVi (b) as function of i for i ∈ (0, 0
...
The sum of the returns is 102 for project A and 110 for project B
...
Hence, if interest rates are high, then project A is preferable
because these early returns can be invested
...
1a) are possible—it is possible to have more than one cross-over point
...
J
...
2 Different interest rates for lending and borrowing
...
The size of the differential depends on
his credit worthiness, size of the amounts, time of loan, etc
...

5
...
For many projects, there is an initial cash outﬂow followed by a sequence
of cash inﬂows
...
Let t1 denote the smallest value of t with
A(t) ≥ 0
...
Equivalently, the DPP is the smallest time t such that the
NPV of payments up to time t is positive
...

If iB = 0 then the quantity corresponding to t1 is called the payback period
...
It is similar to the DPP—but it ignores
discounting
...
It is therefore an inferior measure and should
rarely be used
...

• If A(k) < 0 (or equivalently NPViB (a, ρ) < 0) then the project is not proﬁtable and the DPP does not exist
...
3a
...

(a) Find an expression for the discounted payback period, t1
...

Assume an annual interest rate of i1
...
Now A(0) = −C and A(1) = C(1 + i1 ) + x
...
, n}
...

(b) If t1 ≤ n, then the accumulated proﬁt at time n is
[
]
A(t1 )(1 + i1 )n−t1 + x (1 + i1 )n−t1 −1 + · · · + (1 + i1 ) + 1

The DPP and payback period are measures of the time it takes for a project to become proﬁtable
...
The DPP is especially useful if capital is scarce
...
However, the other project which has a longer DPP may have a higher NPV! This
can occur if the other project has a late large cash inﬂow
...
It may be sensible to use a higher interest rate for discounting projects which are “longer” in order
to allow for the uncertainty in forecasting future interest rates
...

Because some of the quantities used in the calculation of the NPV are estimates, it is sensible to vary these
quantities one at a time to see how the value of the NPV varies with the assumptions
...
4
Summary
...

• Internal Rate of Return (IRR): disadvantages
...
Linear interpolation
...
Payback period
...
tex)

1
...
cashflow which represents the vector of cash
ﬂows, and r which represents a vector of interest rates
...

Thus the call
irr( c(-87,25,-40,60,60), seq(0
...
3,by=0
...
010
0
...
434862816
14
...
many lines omitted
[46,]
[47,]
[48,]

0
...
056
0
...
288410285
0
...
275585819

...
299 -43
...
300 -43
...
056
...
Compare the following two investment projects:
• Project A: Investment costing £50,000 with returns £8,000 each half-year for the next 4 years
...

3
...
, −cm , b0 , b1 ,
...
< t′ where all cj > 0 and all bj > 0
...

n
4
...
Both involve outlays of £1 million
...
7 million
...
321 million after 9 years, £0
...
245 million after
11 years
...

(ii) By general reasoning or by illustrative calculation, show that at a positive rate of interest i∗ where i∗ < i′ ,
project B will have a higher net present value than project A
...
A small technology company set up a new venture on 1 January 2001
...
5 million required on 1 August 2001
...
e
...
3
million per annum and that the rate will increase by £0
...
It
is assumed that the net income will be received continuously throughout the project
...

Calculate the net present value of the venture on 1 January 2001 at a rate of interest of 6% per annum, convertible
(Institute/Faculty of Actuaries Examinations, April 2001) [8]
half-yearly
...
A computer manufacturer is to develop a new chip to be produced from 1 January 2008 until 31 December 2020
...
The cost of development comprises £9 million payable on 1 January 2006
and £12 million payable continuously during 2007
...

(i) Calculate the discounted payback period at an effective rate of interest of 9% per annum
...

(Institute/Faculty of Actuaries Examinations, April 2005) [6+2=8]

Page 36 Exercises 6

Jan 28, 2016(12:36)

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...
Reed

7
...
The company will invest £10
million per annum for the ﬁrst 2 years of the project, the investment being made continuously during this period
...
The amount of payment at the end of year 3 will be £8
million, reducing by £0
...
When the payments have reduced to zero, the company’s involvement in the
project will end
...

(ii) Explain whether the internal rate of return achieved on this project will be greater or less than 10% per annum
(Institute/Faculty of Actuaries Examinations, September 2002) [7+2=9]
effective
...
A motor manufacturer is to develop a new car model to be produced from 1 January 2002 for 6 years until 31 December 2007
...

The production cost of each car is assumed to be incurred at the beginning of the calendar year of production and
will be £9,000 during 2002
...
Both the production costs and the sale prices are assumed to increase by 5% on each 1 January, the ﬁrst
increase occurring on 1 January 2003
...

The sale price of each car produced in 2002 is £12,100
...

(ii) Without doing any further calculations, explain whether the discounted payback period would be greater than,
equal to, or less than the period calculated in (i) if the effective rate of interest were substantially less than 9%
per annum
...
An investment project gives rise to the following cash ﬂows
...
From the beginning of the ﬁrst year until the end of the the twenty-ﬁfth
year, net revenue will be received continuously
...
The rate of payment is assumed to grow continuously at a rate of 6% per annum effective
...

(ii) Calculate the discounted payback period of the project at an effective rate of interest of 7% per annum
...

(Institute/Faculty of Actuaries Examinations, September 2000) [6+5+6=17]
10
...
Explain why the discounted payback period is a poorer decision
criterion than net present value assuming the investor is not short of capital
...
Project A involves the investment of £1 million at the outset
...
5 million after ten years
...
Income will be received from this project continuously
...
08 million, in the second year £0
...
10 million, with
the rate increasing by £0
...

(ii) Calculate the net present value of both investment projects at a rate of interest of 4% per annum effective
...

(iv) In the light of your answer to (i) above, explain which project is the more desirable to an investor with unlimited
capital, and why
...
A car manufacturer is to develop a new model to be produced from 1 January 2016 for six years until 31 December 2021
...

It is assumed that 6,000 cars will be produced each year from 2016 onwards and that all will be sold
...
All production costs are assumed to be incurred at the beginning of each calendar year
...
All revenue from sales is assumed to be received at the end of each calendar year
...

(ii) Without doing any further calculations, explain whether the discounted payback period would be greater than,
equal to, or less than the period calculated in part (i) if the effective rate of interest were substantially less than
9% per annum
...

Jan 28, 2016(12:36)

Section 7 Page 37

(i) In respect of an investment project, deﬁne
(a) the discounted payback period
(b) the payback period
(ii) Discuss why both the discounted payback period and the payback period are inferior measures compared with
the net present value for determining whether to proceed with an investment project
...
The project will be regarded as
viable if it provides a positive net present value at a rate of interest of 10% per annum effective
...

Cost of making bid
0
...

Revenue
Sale of television rights
To be received continuously at a rate of 0
...

Other revenue from sale of merchan- Assumed to be received in the middle of each year from 2004 to 2015
dise, marketing rights, tickets, etc
inclusive
...
1 per
annum and increase each year by 0
...
The same
revenue is expected in 2012 as in 2011
...
2 per annum until 2015, after which year no further revenue
will be received from this source
...

13
...
The company has to make an initial investment of
3 payments, each of £105,000
...

After 15 years, it is assumed that a major refurbishment of the infrastructure will be required, costing £200,000
...
Thereafter, the income is
expected to increase by 3% per annum (compound) at the start of each year
...
The cash ﬂow within each
year is assumed to be received at a constant rate
...
a
...

(ii) Show that there is no discounted payback period within the ﬁrst 15 years, assuming an effective rate of interest
of 8% p
...

(iii) Calculate the discounted payback period for the project, assuming an effective rate of interest of 8% p
...

(Institute/Faculty of Actuaries Examinations, May 1999
...
) [8+8+6=22]

There are further exercises on project appraisal in exercises 3 of chapter 3 on page 54
...
1 Background
...

Example 7
...
(a) An investor buys 1 share for £50 at time 0
...
He then buys a second share (for £53)
...
Find the internal rate of return
...
Hence his total
holding is then 11 shares
...

Solution
...

Part (a):

Time, t

0

1

2

Fund value at time t − 0
Dividend receipts at time t

0
0

53
2

Investment at time t

50

Net cash ﬂow (out) at time t

50

Part (b):

0

1

2

108
4

0
0

53
2

594
22

53

0

50

530

0

51

−4

50

528

−22

Page 38 Section 7

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Reed

For part (a), the equation of value at t = 0 is
51
4
108
−
=
1 + i (1 + i)2 (1 + i)2
This quadratic equation in i has solution i = 0
...
12%
...
02%
...
This
corresponds to 10%
...
This corresponds to
5
...
The internal rate of return is 7
...

It is inﬂuenced more by the performance in year 2 when 2 shares are held than the performance in year 1 when
only 1 share is held
...
02% because 11 shares are held at time 2
...
Also, the equation for the
MWRR may not have a unique or indeed any solution
...
1 and in year 2 the rate of return is i2 = 0
...
The TWRR, iT is deﬁned
by
(1 + iT )2 = (1 + i1 )(1 + i2 )
In our particular case, this becomes
√
55 112
− 1 = 0
...
81% for both parts (a) and (b)
...
2 General results
...
The MWRR is more appropriate for measuring the performance of
an individual investor who does determine the sizes and timings of investments
...
3 The linked internal rate of return (LIRR)
...

The linked internal rate of return (LIRR) is a combination of the MWRR and TWRR approaches: the MWRR
(equivalently, the internal rate of return) is calculated at frequent intervals and then the TWRR formula is used
to combine these quantities
...
Then the LIRR is i where
(1 + i)tn = (1 + i1 )t1 (1 + i2 )t2 −t1 (1 + i3 )t3 −t2 · · · (1 + in )tn −tn−1
The LIRR is useful if the fund is not valued every time there is a cash ﬂow
...

The values of LIRR and TWRR will be the same if the lengths of the intervals (tr−1 , tr ) are the same in both
calculations
...

Cash Flows and Project Appraisal

Jan 28, 2016(12:36)

Exercises 8 Page 39

Example 7
...
An investment fund is valued once a year at the beginning of April
...

(University of Warwick ST334 Examinations, 2012)
Solution
...
Let i1 denote the annual effective rate of interest earned by the fund in the interval
from April 1, 1996 to April 1, 1997
...
This is the quadratic 50x2 + 20x − 71 = 0
which has solution x = 1
...
Hence 1 + i1 = x2 = 1
...
Also 1 + i2 = 97/71
...
01666889 × 97/71 and hence 1 + iL = 1
...
Hence
the LIRR is 17
...

7
...

• Money Weighted Rate of Return (MWRR): same as internal rate of return
...

ft −0 ft
ft −0 ft2 −0
··· n
(1 + i)t = 1
f0
ft1
ftn−1 ftn

(7
...

8 Exercises
1
...

Calendar Year
1997
1998
£ millions
£ millions
Value of fund at 30 June
—
460
Net cash ﬂow received on 1 July
—
50
Value of fund at 31 December
400
550

(exs2-3
...

(Institute/Faculty of Actuaries Examinations, April 2001) [3]
2
...

2000
Calendar Year
£ million
Value of fund on 1 January before cash ﬂow
100
Net cash ﬂow received on 1 January
20
80
Value of fund on 31 December

2001
£ million
80
30
200

2002
£ million
200
10
200

Calculate the effective time weighted rate of return over the three year period
...
You are given the following information in respect of a pension fund:
Calendar
Value of fund at
Value of fund at
Net cash ﬂow received on
Year
1 January
30 June
1 July
1997
£180,000
£212,000
£25,000
1998
£261,000
£230,000
£18,000
1999
£273,000
£295,000
£16,000
2000
£309,000
Calculate the annual effective time weighted rate of return earned on the fund over the period from 1 January 1997
to 1 January 2000
...
J
...
A fund has a value of £120,000 on 1 January 2001
...
Immediately before receipt of the ﬁrst cash ﬂow,
the fund had a value of £137,000 and immediately before the second cash ﬂow the fund had a value of £173,000
...

(i) Calculate the annual effective time weighted rate of return earned on the fund for the period 1 January 2001 to
31 December 2002
...

(Institute/Faculty of Actuaries Examinations, April 2004) [3+3=6]
5
...
2 million on 31 December 2001 and £4
...

It had received a net cash ﬂow of £1
...

The money weighted rate of return and the time weighted rate of return for the period from 31 December 2001 to
31 December 2004 are equal (to two decimal places)
...

(Institute/Faculty of Actuaries Examinations, April 2005 ) [7]
6
...
A net cash ﬂow of £5,000 was received on 1 July 2004 and a further
cash ﬂow of £8,000 was received on 1 July 2005
...
The
value of the fund on 1 July 2006 was £38,000
...

(ii) Calculate the annual effective time weighted rate of return earned on the fund over the period 1 July 2003 to
1 July 2006
...

(Institute/Faculty of Actuaries Examinations, April 2007) [3+3+2=8]
7
...
It received net income of £40m on 1 January 2004 and £100m on 1 July 2004
...

(i) Calculate, for the period 1 January 2003 to 31 December 2004, to three decimal places:
(a) the time weighted rate of return
...

(ii) Explain why the linked internal rate of return is higher than the time weighted rate of return
...
A pension fund had assets totalling £40 million on 1 January 2000
...
The value of the fund totalled:
£43 million on 31 December 2000; £49 million on 30 June 2001 and £53 million on 31 December 2001
...

(ii) State both in general, and in this particular case, when the linked internal rate of return will be identical to the
time weighted rate of return
...
A pension fund makes the following investments(£m):
1 January 2004
1 July 2004
1 January 2005
12
...
6
7
...
0

1 January 2004 to
1 July 2004 to
1 January 2005 to
1 January 2006 to
30 June 2004
31 December 2004
31 December 2005
31 December 2006
5%
6%
6
...

(i) Calculate the linked internal rate of return per annum over the 3 years from 1 January 2004 to 31 December 2006,
using semi-annual sub-intervals
...

(iii) Calculate the money weighted rate of return per annum over the 3 years from 1 January 2004 to 31 December 2006
...

(Institute/Faculty of Actuaries Examinations, September 2007) [3+3+4+2=12]

Cash Flows and Project Appraisal

Jan 28, 2016(12:36)

Exercises 8 Page 41

10
...
3 million
...
9 million
...
9 million
...
8 million
...

(ii) Calculate the annual effective money-weighted rate of return (MWRR) for 2012 to the nearest 1%
...

(Institute/Faculty of Actuaries Examinations, April 2013) [2+3+2=7]
11
...
3 million
...
9 million and, on 1 May 2011, there was a net cash inﬂow to the fund of £1
...

On 31 December 2011, the value of the assets was £4
...

(i) Calculate the annual effective time-weighted rate of return (TWRR) for 2011
...
1%, the annual effective money-weighted rate of return (MWRR) for 2011
...

(Institute/Faculty of Actuaries Examinations, April 2012) [2+4+2=8]
12
...
Suppose the price in pence
of the units in two accumulation funds, fund A and fund B, at various dates was as follows:
Date
Fund A: unit price
Fund B: unit price

1/4/2001
80
70

1/4/2002
90
67

1/4/2003
98
66

1/4/2004
180
100

1/4/2005
190
110

1/4/2006
160
130

Suppose the rate of inﬂation3 is given by the index in the following table:
Date
Inﬂation index

1/4/2001
100

1/4/2002
102

1/4/2003
105

1/4/2004
117

1/4/2005
120

1/4/2006
122

An individual invests £1000 in fund A on 1/4/2001 and a further £1000 in fund A on 1/4/2002
...

(a) Calculate the real time weighted rate of return of the total investment for the period 1/4/2001 to 1/4/2006
...
Use this equation to determine whether the real money weighted rate of return is greater or less than
the real time weighted rate of return
...
An investment fund is valued at £120 million on 1 January 2010 and at £140 million on 1 January 2011
...
On 1 July 2012, the value of the fund is
£600 million
...

(ii) Explain why the money weighted rate of return would be higher than the time weighted rate of return
...

3

(i) State the strengths and weaknesses of using the money weighted rate of return as opposed to the time weighted
rate of return as a measure of an investment manager’s skill
...
He invested a further £12,000 in this
account on 1 August 2006
...

Assuming that the investor made no further deposits or withdrawals in relation to this account, calculate the
annual effective time weighted rate of return for the period 1 January 2006 to 31 December 2007
...

Page 42 Exercises 8

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CHAPTER 3

Perpetuities, Annuities
and Loan Schedules
1 Perpetuities
1
...
A perpetuity gives a ﬁxed amount of income every year forever
...

Suppose the rate of interest per unit time is constant and equal to i
...
The value at time 0 of the following cash-ﬂow sequence:
Time
Cash ﬂow, c

0
0
↑
a∞

1
1
↑
a∞
¨

2
1

3
1

...

is denoted a ∞ or a ∞ ,i and is given by
ν
1
=
1−ν i
or a ∞ ,i and is given by
¨

a ∞ = ν + ν2 + ν3 + · · · =
The value of this perpetuity at time 1 is denoted a ∞
¨

a ∞ = (1 + i)a ∞ = 1 + ν + ν 2 + ν 3 + · · · =
¨

(1
...
1b)

where d = 1 − ν is the discount rate
...
1a
...
How
much must a benefactor give a University in order to establish a permanent post?
Solution
...
04

1
...
Suppose the rate of interest per unit time is constant and equal to i
...
Hence ν = 1 − d where d is the discount
rate
...
This can be interpreted as
follows:
• The principal £(1 − d) is returned at time 1 together with the interest £i(1 − d) = £d
...

OR
• £1 is borrowed at time 0
...

The interest £d paid at time 0 is the present value at time 0 of the quantity i at time 1
...

Thus a ∞ can be regarded as the value of a perpetuity when payments are made in arrears and a ∞ can be
¨
regarded as the value of a perpetuity when payments are made in advance
...

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...
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Jan 28, 2016(12:36)

Section 1

Page 43

Page 44 Section 2

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...
Reed

Jan 28, 2016(12:36)

1
...
Consider the following increasing perpetuity:
Time
Cash ﬂow, c

0
0
↑
(Ia) ∞

1
1
↑
(I a) ∞
¨

2
2

3
3

...

The value at time 0 is denoted (Ia) ∞ or (Ia) ∞ ,i and is given by
(Ia) ∞ = ν + 2ν 2 + 3ν 3 + · · · =

ν
1
=
2
(1 − ν)
id

(1
...
Clearly, (Ia) ∞ = ν (I a) ∞
...
3b)

2 Annuities
2
...
Suppose n ≥ 1 and the payment 1 is received at
time points 1, 2,
...

Time
Cash Flow, c

0
0
↑
an

1
1
↑
an
¨

2
1

···
···

n−1
1

n
1

(2
...
1a) at time 0 is denoted a n or a n ,i and is called the present
value of the immediate annuity-certain a
...

If i = 0 then clearly a n = n
...
The deﬁnition is extended to all n ≥ 0 as follows:
a n = ν + ν2 + · · · + νn =

Deﬁnition 2
...
Suppose n ≥ 0 and i ≥ 0
...
Then

a n ,i


n
= 1 − νn

i

if i = 0
(2
...
1a) at the time of the ﬁrst payment is denoted a n or a n ,i and is
¨
¨
called the present value of the annuity-due a
...

If i = 0 then clearly a n = n
...
1b
...
Then a n ,i = (1 + i) a n ,i
¨

Clearly if i > 0 then
a ∞ = lim a n =
n→∞

5

1
i

and

a ∞ = lim a n =
¨
¨
n→∞

1
d

The word certain is used to distinguish it from annuities which cease to pay out when the holder dies
...
1c)

Annuities and Loan Schedules

Jan 28, 2016(12:36)

Section 2 Page 45

Example 2
...
Consider a loan of £1,000 at 12% per annum compounded monthly
...
How much are the monthly payments?
Solution
...
01 per month
...
01
...
01 = 44
...
25
a 60 ,0
...
25
1−ν
i
1 − ν 60

The process of gradually repaying a debt is called amortisation
...
2 Annuities with constant payments: accumulated values
...
1a) at time n, the time of the last payment, is denoted s n or s n ,i
and is called the accumulated value of the immediate annuity-certain a
...
1a) at time n + 1, one time unit after the time of the last payment,
is denoted s n or s n ,i and is called the accumulated value of the immediate annuity-due a
...
Otherwise,
¨
s n = (1 + i)n−1 + (1 + i)n−2 + · · · + 1
(1 + i)n − 1
= (1 + i)n a n =
i

(2
...
2b)

n+1
0
↑
sn
¨

It follows that s n is the accumulated value at time n of the annuity if payments are made in arrears, whilst
s n is the accumulated value at time n if the n payments are made in advance at times 0, 1,
...
The
¨
deﬁnitions are extended to all n ≥ 0 as follows:
Deﬁnition 2
...
Suppose n ≥ 0 and i ≥ 0
...
3 Annuities payable mthly
...
So the cash ﬂow is as follows:
Time
Cash Flow, c

0
0
↑

a(m)
n

1/m
1/m
↑
(m)
an
¨

2/m
1/m

···
···

(nm − 1)/m
1/m

nm/m
1/m
↑
(m)
sn

(nm + 1)/m
0
↑
(m)
sn
¨

(2
...
3a)
...
We suppose m and n are positive integers and let ν = 1/(1 + i)
...

Page 46 Section 2

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...
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We can derive this result another way
...

Time
0
1/m
2/m
···
(m − 2)/m
(m − 1)/m
1
d
(m) /m
d
d(m) /m
d(m) /m
···
d(m) /m
d(m) /m
i(m) /m

i(m) /m

···

i(m) /m

i(m) /m

i(m) /m
i

Hence the sequence of m equal payments of i(m) /m at times 1/m, 2/m,
...
Hence the sequence of nm equal payments of 1/m at times 1/m, 2/m,
...
, n
...
, n gives
i
a(m) = (m) a n ,i
(2
...
3b) is a very important result—it enables fast calculation of a(m) for standard values of i, m and n
n ,i
by using tabulated values of i/i(m) and a n ,i
...
3a
...

i
i
Solution
...
The quantity (4) is given in the tables for 7%—it is in the constants on the
9
i
i
left hand side of the page and is given as 1
...
The quantity a 9 is given in the main table as 6
...
Hence
the present value of our annuity is 1
...
515 232
...
We can work out
¨n
¨ n ,i
the formula for this quantity from ﬁrst principles as follows:
a(m) =
¨n

nm−1
1 ∑ j/m 1 1 − ν n
1 − νn
ν
=
= (m)
m
m 1 − ν 1/m
d

(2
...
, n − 1
...
, n − 1
...
3b
...
Let ν = 1/(1 + i)
...
, 21/4
...
, 32
...
For example, if n = 2
...
75 years; i
...
a(2) + 4 ν 2
...
But this is not the value of the formula in deﬁnition 2
...

2
...
3a)
...

Annuities and Loan Schedules

Jan 28, 2016(12:36)

Section 2 Page 47

Deﬁnition 2
...
Suppose n ≥ 0, m ≥ 0 and i ≥ 0
...
3d)

Note that a(1) = a n ,i , a(1) = a n ,i , s(1) = s n ,i and s(1) = s n ,i
¨ n ,i ¨
¨ n ,i ¨
n ,i
n ,i
Generalisation
...
, j for every j = 1, 2,
...
Then the new net present value is
n
i ∑ j
ν xj
i(m)
j=1

2
...
Consider the following cash ﬂow (where k and n are positive integers):
Time
Cash Flow, c

0
0

1
0

···
···

k
0

···
···

k+1
1

k+n−1
1

k+n
1

This can be considered as an immediate annuity-certain delayed for k time units
...
This
value is denoted k| a n and is
¨
k+n−1
∑
NPV(c; t = 1) = k| a n =
¨
νj
j=k

Clearly we have
¨
k| a n

= νka n
¨

Deferred annuities payable mthly
...
This leads
¨n
n
to the following deﬁnitions:
Deﬁnition 2
...
Suppose k ≥ 0, n ≥ 0, m ≥ 0 and i ≥ 0 and ν = 1/(i + 1)
...

(m)
k| a n

= a(m) − a(m)
n+k
k

and

¨ (m)
k| a n

= a(m) − a(m)
¨ n+k
¨k

Page 48 Section 2

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...
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Jan 28, 2016(12:36)

2
...
Consider the following increasing annuity:
Time
Cash Flow, c

0
0
↑
(Ia) n

1
1
↑
(I a) n
¨

···
···

2
2

n
n
↑
(Is) n

n+1
0
↑
(I s) n
¨

The present value (time 0) of this annuity is denoted (Ia) n and the present value at time 1 is denoted (I a) n
...

¨
¨
Clearly:
(Ia) n = NPV(c) = ν + 2ν 2 + · · · + nν n
]
[
1 1 − νn
n
− nν
=
i 1−ν

(2
...
5b)

and
(I a) n = NPV(c; t = 1) = 1 + 2ν + 3ν 2 + · · · + nν n−1
¨
= (1 + i)(Ia) n
a n − nν n a n − nν n+1
¨
=
=
1−ν
ν(1 − ν)

(2
...
5a) by using the results (Ia) ∞ = 1/i(1 − ν) and a ∞ = 1/i on pages 43 and 44 above
...
6 Decreasing annuities
...
, 1
...

2
...
Suppose the force of interest is constant and
equal to δ
...

Present Value
...
and let a n ,i or a n denote the value at time 0 of an annuity payable continuously between times 0 and n when the rate of payment ρ(t) is constant and equal to 1
...
Otherwise, by using 1 + i = eδ and ν = 1/(1 + i), we get
∫ n
∫ n
n
νt
a n ,i =
ρ(t)ν(t) dt =
ν t dt =
ln ν t=0
0
0
n
1−ν
i
=
= a n ,i
if δ ̸= 0
(2
...
7a
...
Let i = eδ − 1
...
7b)

Accumulated value
...
Then
∫ n
ρ(t)eδ(n−t) dt = eδn a n = (1 + i)n a n
sn =
(2
...
7b
...
Then s n ,i = (1 + i)n a n ,i

Clearly


 a n ,i = n
s n ,i =

 (1 +

i)n

−1

δ

if i = 0
i
= sn
δ

if i ̸= 0

Deferred continuously payable annuities
...
Then
∫ k+n
∫ k+n
ρ(t)ν(t) dt =
e−δt dt
k| a n =
k
k
∫ k+n
∫ k
=
e−δt dt −
e−δt dt
0

= a k+n − a k
= νka n

0

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2
...
There are two common payment rates, ρ:
• the step function ρ(t) = j for t ∈ (j − 1, j] and j = 1, 2,
...

For the step function, it can be shown (exercise 25) that:
∫ n
n
∑∫
(I a) n =
¯
ρ(t)ν(t) dt =
0

j

jν t dt
j−1

j=1

an −
¨
where δ = − ln ν
...

¯
¯
nν n

=

(2
...
8b)

For this case, the notation for the accumulated value is (Is) n = (1 + i)n (Ia) n
...
9
Summary
...

Net present value at time 0:
a n = ν + ν2 + · · · + νn
1 − νn
if i ̸= 0
...

Net present value at time 0:
a n = (1 + i)a n
¨
Accumulated value at time n + 1:

¨
s n = (1 + i)n a n
¨

• Annuities with constant payments payable m times per year for n years with nm payments at times
1/m, 2/m,
...
(Set m = 1 to get results for a n , a n , s n , and s n
...

Net present value at time 0:

a ∞ = lim a n =

Net present value at time 1:

1
i

a ∞ = (1 + i)a ∞
¨

n→∞

• Increasing Annuities
...

1
and
(I a) ∞ = (1 + i)(Ia) ∞
¨
i(1 − ν)
• Deferred annuities, k| a n , and decreasing annuities, (Da) n
...
continued

Annuities and Loan Schedules

Jan 28, 2016(12:36)

Exercises 3 Page 51

Summary (continued)
...

an
s n = (1 + i)n a n

Net present value at time 0:
Accumulated value at time n:

• Increasing continuously payable annuities:
(I a) n for the step function ρ(t) = j for t ∈ (j − 1, j] and j = 1, 2,
...

3 Exercises
1
...
tex)

(b) s n+1 = s n + 1
¨
n

for n ≥ 2

(c) a n = a n−1 + 1
¨
n

(d) ia n + ν = 1
(e) da n + ν = 1
¨
(f) is n + 1 = (1 + i)
(g) d¨ n + 1 = (1 + i)n
s
Give interpretations of these results: for example, (a) is the result that s n is the value of the same series of payments
¨
as s n but 1 time unit later
...
The force of interest, δ, is 6%
...

(Institute/Faculty of Actuaries Examinations, September 1999) [3]
3
...

(Institute/Faculty of Actuaries Examinations, April 1998) [3]
4
...
Find the value of s(12)
...
What is the value of s(4) at an effective rate of interest is 7 1/2% per annum
...
An annuity due starts at £200 per year and thereafter decreases by £10 per year
...
Express the present value, x, of the annuity in terms of a 8 and (I a) 8
...

(Institute/Faculty of Actuaries Examinations, April 1999, adapted)
7
...
It is expected to yield an income of £20m per annum
annually in arrears
...

(Institute/Faculty of Actuaries Examinations, September 1999) [3]
8
...
The ﬁrst payment will be £10
...
The last payment will be made at the beginning of the tenth year
...
An annuity certain with payments of £150 at the end of each quarter is to be replaced by an annuity with the same
term and present value, but with payments at the beginning of each month instead
...

(Institute/Faculty of Actuaries Examinations, April 2006) [3]
10
...
Show
that the present value of the annuity can be expressed as
2(Ia) n − a n − n2 ν n+1
1−ν
(Institute/Faculty of Actuaries Examinations, April 2000
...
) [4]
11
...
The investment earns interest at 8% per annum convertible half yearly
...

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12
...

Calculate the accumulated fund at the end of the term if the interest rate is 6% per annum convertible monthly for
the ﬁrst 15 years and 6% per annum convertible half-yearly for the ﬁnal 10 years
...

(i) Calculate s(12) at an effective rate of interest of 13% per annum
...
5

(Institute/Faculty of Actuaries Examinations, April 2000) [5]

(ii) Explain what your answer to (i) represents
...
Recall that a n = a(1) , a n = a(1) , s n = s(1) and s n = s(1)
...
(a) Suppose m and n are positive integers
...
The interest rate is 1%
per month effective
...

16
...
02t for 0 ≤ t ≤ 5
...

17
...
So we have the cash
ﬂow:
Time
0
1
2
3

...

(a) Show that NPV(c) = 1/(i − g) for g < i
...
Find the NPV
...
Prove the following result:
lim a(m) = lim a(m) = a n
¨n
n

m→∞

m→∞

19
...

¨
¨n
n

(All equal 1 − ν n
...
Show that

(
a(m) =
¨ n ,i

i
i(m)

+

i
m

)
a n ,i

21
...
Prove that
1
i(m)
a(m) = a nm ,j
where j =
n ,i
m
m
22
...
Let a(m) denote the present value (at time 0) of a perpetuity where constant payments of
¨∞
1/m are made m times per year in advance
...

...

Time
Cash ﬂow

0
0

1/m
1/m

and

a(m) =
∞

...

2/m
1/m

(a) Show that
a(m) =
¨∞

1
d(m)

(b) Hence show that
1
d(m)

=

1
1
+ (m)
m i

1
i(m)

Annuities and Loan Schedules

Jan 28, 2016(12:36)

Exercises 3 Page 53

23
...

...
Suppose further that nm is an integer
...
Suppose the interest rate is i, the number of payments per year is m and the number of increases per year is q where
q divides m
...
)
Time
Cash Flow

0
1/(mq)

1/m
1/(mq)

···
···

1/q − 1/m
1/(mq)

···
···

1/q
2/(mq)

2/q − 1/m
2/(mq)

···
···

Denote the present value (time 0) by (I (q) a)(m)
...
Prove the following results about increasing continuously payable annuities:
a n − nν n
¨
a n − nν n
(I a) n =
¯
and
(Ia) n =
δ
δ
26
...
Prove the following results:
(a)

¨ (m)
k| a n

= ν k−1/m a(m)
n

(b)

(m)
k| a n

= a(m) − a(m)
n+k
k

(c)

¨ (m)
k| a n

= a(m) − a(m)
¨ n+k
¨k

27
...
Consider the following cash ﬂow:
Time
Cash ﬂow, c

0
0

1
n

2
n−1

···
···

n
1

Let (Da) n = NPV(c)
...

28
...
Payments then continue at £60 per annum until the 20th payment under the annuity
has been made
...
An investor is considering the purchase of an interest in a warehouse for £250,000
...
Maintenance and management
costs are also ﬁxed for the 5 year period at a rate of £7,000 per annum payable by the investor at the beginning of
each year
...

(i) Calculate the annual effective rate of return
...

(Institute/Faculty of Actuaries Examinations, September 1999) [7]
30
...
The ﬁrst option
involves 24 payments of £20 paid at the beginning of each month starting immediately
...
50 paid at the end of each month starting immediately
...

Determine which, if any, of the payment options the customer will accept
...
J
...
A ﬁnancial regulator has brought in a new set of regulations and wishes to assess the cost of them
...

The costs are estimated as follows
...

• The cost to ﬁnancial advisers who will have to set up new computer systems and spend more time ﬁlling in
paperwork is expected to be incurred at a rate of £60m in the ﬁrst year, £19m in the second year, £18m in the
third year, reducing by £1m every year until the last year, when the cost incurred will be at the rate of £1m
...

The beneﬁts are estimated as follows
...

• The beneﬁts to companies who will spend less time dealing with complaints from customers is estimated to be
received at a rate of £12m per annum for twenty years
...
Assume that all costs and beneﬁts occur continuously throughout the year
...
A bank offers two repayment alternatives for a loan that is to be repaid over 10 years
...

Determine which terms would provide the best deal for the borrower at a rate of interest of 4% per annum effective
...
The force of interest, δ(t), is a function of time and at any time t (measured in years) is given by
{
0
...
12 − 0
...
05
for t > 9
...

(ii) Calculate the present value at t = 0 of a payment stream, paid continuously from t = 10 to t = 12, under which
the rate of payment at time t is 100e0
...

(iii) Calculate the present value of an annuity of £1,000 paid at the end of each year for the ﬁrst 3 years
...

34
...
Show that
the value at time 0 of such an annuity payable until time n, which is represented by (Ia) n is given by
a n − nν n
(Ia) n =
δ
∫n
where δ is the force of interest per unit of time and a n = 0 exp(−δt) dt
...
The consortium
forecasts that:
(1) the cost of reopening the mine will be £700,000, and this will be incurred continuously throughout the ﬁrst
twelve months
(2) after the ﬁrst twelve months the revenue from sales of coal, less costs of sale and extraction, will grow
continuously from zero to £2,500,000 at a constant rate of £250,000 per annum
(3) when the revenue from sales of coal, less costs of sale and extraction, reaches £2,500,000 it will then decline
continuously at a constant rate of £125,000 per annum until it reaches £250,000
(4) when the revenue declines to £250,000 production will stop and the mine will have zero value
Additional costs are expected to be constant throughout at £200,000 p
...
excluding the ﬁrst year
...

What price should the consortium pay to earn an internal rate of return (IRR) of 20% p
...
effective?
(Institute/Faculty of Actuaries Examinations, September 1998) [14]

Annuities and Loan Schedules

Jan 28, 2016(12:36)

Exercises 3 Page 55

35
...
It is anticipated that the ofﬁces will take six months to build
...
It is expected that rental income from the ofﬁces
will be £1 million per month, which will be received at the start of each month beginning with the seventh month
...
The block of ofﬁces is expected to be sold 25 years
after the start of the project for £60 million
...

(ii) Without doing any further calculations, explain whether your answer to (i) would change if the effective rate of
interest were less than 10% per annum
...
An investor borrows £120,000 at an effective interest rate of 7% per annum
...
Once the loan is paid off, the investor
can earn interest at an effective rate of 5% per annum on money invested from the annuity payments
...

(ii) Determine the proﬁt the investor will have made at the end of the term of the annuity
...
A piece of land is available for sale for £5,000,000
...
The total cost of
development would be £7,000,000 which would be incurred continuously over the ﬁrst two years after purchase of
the land
...

The developer has three possible strategies
...

The developer also believes that she can obtain a rental income from the housing between the time that the development is completed and the time of the sale,
...
Thereafter, the rental income is expected to increase by £50,000 per annum at
the beginning of each year that the income is paid
...

(ii) Determine which strategy would be optimal if the discounted payback period were to be used as the decision
criterion
...
5% per annum
...

(iv) Suggest reasons why the developer may not achieve an internal rate of return of 17
...

(Institute/Faculty of Actuaries Examinations, April 2006) [9+2+6+2=19]
38
...

(ii) An insurance company is considering setting up a branch in a country in which it has previously not operated
...
It therefore has two
decision criteria
...

The following cash ﬂows are generated in the development and operation of the branch
...
Between the present time and the opening of the branch in three years’ time, the insurance
company will spend £1
...
This outlay is
assumed to be a constant continuous payment stream
...
3m per annum
paid quarterly in advance for twelve years starting in three years’ time
...
05m in the second year, rising by 5% per annum each year thereafter
...

Cash Inﬂows
...
9m per annum in the next three years and to £2
...
This net income is assumed to be received continuously throughout each year
...

Determine which, if any, of the decision criteria the project fulﬁls
...
An investor is considering investing in a capital project
...
The project is expected to provide a continuous income at a rate of £80,000 in the

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ﬁrst year, £83,200 in the second year and so on, with income increasing each year by 4% per annum compound
...

It is assumed that, at the end of 15 years, a further investment of £300,000 will be required and that the project can
be sold to another investor for £700,000 at the end of 25 years
...

(ii) Without doing any further calculations, explain how the net present value would alter if the interest rate had
been greater than 11% per annum effective
...
A pension fund is considering investing in a major infrastructure project
...
No other costs will be incurred by the pension
fund
...

• In the second year, 50,000 vehicles a day will use the road, each paying a toll of £1
...

• In the third year, both the number of vehicles using the road and the level of tolls will rise by 1% from their level
in the second year
...

• At then end of the 20th year, it is assumed that the road has no value as it will have to be completely rebuilt
...
Calculate the net present value of the investment in the road at a rate of interest of 8% per annum effective
...
An investor is considering two projects, Project A and Project B
...
Rent is received quarterly in arrear for 25 years, at an initial rate of £100,000 per annum
...
Maintenance and other expenses are incurred quarterly in arrear, at a rate of £12,000 per annum
...

Project B involves the purchase of an ofﬁce building for £1,000,000
...
It is assumed that the rent will increase to £90,000 per annum after 20 years
...
After 25 years the property reverts to its original owners for no payment
...

(ii) Calculate the annual effective internal rate of return for Project B
...

(iii) Discuss the extent to which the answers to parts (i) and (ii) above will inﬂuence the investor’s decision over
(Institute/Faculty of Actuaries Examinations, April 2012) [5+4+3=12]
which project to choose
...
A property development company has just purchased a retail outlet for $4,000,000
...

An agreement has been made with a prospective tenant who will occupy the outlet beginning one year after the
purchase date
...
The initial rent will be $360,000 per annum
...
The rental income is
received quarterly in advance
...

(Institute/Faculty of Actuaries Examinations, April 2015) [9]

4 Loan Schedules
4
...

Example 4
...
Suppose A borrows £1,000 for 3 years at an effective interest rate of 7% per annum
...
Find x
...
The cash ﬂow is as follows:
Time
Cash ﬂow

0
−1,000

1
x

2
x

Hence 1000 = x(ν + ν 2 + ν 3 ) = xa 3
...
05
...
95
l2 = 356
...
05 il0 = 70
x1 − il0 = 311
...
05 il1 = 48
...
83
x3 = 381
...
93 x3 − il2 = 356
...
95
l2 = 356
...
2 General notation
...
, n, let
lk = loan outstanding at time k immediately after repayment at time k
xk = repayment at time k
fk = capital repaid at time k
bk = interest paid at time k
Hence
1
l0 = x1 ν + x2 ν 2 + · · · + xn ν n
where ν =
1+i
The loan schedule can be represented in a table:
year

loan outstanding
before repayment

repayment

interest due

capital repaid

loan outstanding
after repayment

1
2

...

...

...

x1
x2

...

...

...

x1 − il0
x2 − il1

...

...

...

k

...

...

...

xk

...

...

...

xk − ilk−1

...

...

...

n

ln−1

xn

iln−1

xn − iln−1

ln = ln−1 − (xn − iln−1 ) = 0

4
...
We can use either a prospective loan calculation or a retrospective loan calculation
...

At time t = 1,
the interest due is b1 = il0
the capital repaid is f1 = x1 − il0
leaving the outstanding loan of l1 = l0 − (x1 − il0 ) = l0 (1 + i) − x1
And in general, for k = 1, 2,
...
, n
lk = l0 (1 + i)k − x1 (1 + i)k−1 − x2 (1 + i)k−2 − · · · − xk
In words, equation (4
...
3a)

Page 58 Section 4

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...
Reed

Prospective loan calculation
...
, n − 1:
the capital repaid is fk = lk−1 − lk
hence the payment at time k must be xk = ilk−1 + (lk−1 − lk ) = (1 + i)lk−1 − lk
and hence lk−1 = νxk + νlk
These recurrence relations can be solved for lk as follows:
ln−1 = νxn
ln−2 = νxn−1 + νln−1 = νxn−1 + ν 2 xn
and by induction for k = 0, 1
...
3b)

In words, equation (4
...
3a
...
Suppose the effective annual interest
rate is 12%
...

Solution by the prospective method
...
400(ν +
ν 2 + ν 3 + ν 4 ) = 400a 4 ,0
...
94
Solution by the retrospective method
...
This is
(1
...
12 = (1
...
12 − 400s 6 ,0
...
94

Example 4
...
A loan of £10,000 is to be repaid by 10 equal annual repayments of £1,400 plus a further ﬁnal
repayment one year after the last regular repayment
...
Find the size of the
last repayment
...
Let ν = 1/1
...
The equation of value gives 10000 = 1400a 10 ,0
...
Hence x = 7269
...

Example 4
...
A loan of £9,000 is to be repaid by eight equal annual repayments of £1,400 plus a further ﬁnal
repayment one year after the last regular repayment
...
Find the outstanding
loan after the eighth payment
...
Using the retrospective method shows that the outstanding loan is the value at time 8 of the original loan
minus the accumulated value at time 8 of the repayments
...
128 − 1400s 8 ,0
...
10

4
...
Suppose a loan is repaid by n equal instalments of size x paid in arrears
...

Time
Cash ﬂow, c

0
−l0

1
x

2
x

...

n−1
x

n
x

Hence l0 = xa n
...

After the kth payment, we have lk = xa n−k by using the prospective method
...

So we have the general result that the kth repayment consists of an interest payment of x(1 − ν n−k+1 ) and a
capital repayment of xν n−k+1 for k = 1, 2,
...

Annuities and Loan Schedules

Jan 28, 2016(12:36)

Section 4 Page 59

4
...
Now suppose the loan is repaid by nm instalments of x1/m at time 1/m,
x2/m at time 2/m,
...

Using the retrospective method, the loan outstanding after the kth repayment has been made is equal to
(value at time k of original loan) − (accumulated value at time k of repayments)
Hence for k = 1, 2,
...
, nm − 1 we have
lk = ν 1/m x(k+1)/m + ν 2/m x(k+2)/m + · · · + ν (n−k)/m xn
At time k/m the loan outstanding is lk
...
5a
...
, nm/m =
n
...
Find an expression for the capital repayment in the
k th instalment
...
Let ν = 1/(1 + i)
...
, nm
...
5b
...
, nm/ = n or it is repaid in advance by nm equal instalments of size y at times 0, 1/ , 2/ ,
...

m
m
m
m
m
m
Suppose the interest rate charged is i% per annum effective
...

Solution
...
Then use
[
]
[
]
l0 = x ν 1/m + ν 2/m + · · · + ν (nm)/m = y 1 + ν 1/m + ν 2/m + · · · + ν (nm−1)/m

Example 4
...
(a) The interest rate on a £5,000 loan is a nominal 12% per annum convertible 6-monthly
...
Find the amount of each
repayment
...
The new interest rate is then ﬁxed at a
nominal 11% per annum convertible 6 monthly
...

Solution
...
The monthly rate of interest is j = 1
...
009759
...
0610
...
49
...
Deﬁne j1 by (1 + j1 )6 = 1
...
Hence the new
regular payment is y where ya 24 ,j1 = l12 − 1,000 = xa 48 ,j − 1,000
...
64
...
6 Flat rate and APR Banks and other credit institutions often quote a ﬂat rate of interest
...
Let xR denote the arithmetic sum of the
total repayments
...
The ﬂat rate is always less than the true
effective rate of interest charged on the loan
...
1%; this is called the annual percentage rate or APR
...
J
...
7
Summary
...

Retrospective loan calculation
lk = (value at time k of original loan) − (accumulated value at time k of repayments)
Prospective loan calculation:
lk = value at time k of all future repayments
...

5 Exercises

(exs3-2
...
The annual rate of payment on a three year consumer credit contract is £2,000
...

What is the ﬂat rate of interest per annum?
(Institute/Faculty of Actuaries Examinations, September 1999) [2]
2
...
84 p
...
payable annually in arrear for
two years
...

(Institute/Faculty of Actuaries Examinations, May 1999
...
) [3]
3
...
Find the interest element in the 5th payment
...
A customer borrows £4,000 under a consumer credit loan
...

Instalments are paid monthly in arrears for 5 years
...

(Institute/Faculty of Actuaries Examinations, September 1998) [4]
5
...
The loan is to be repaid over a 4 year term by level monthly instalments
of £130, payable in arrears
...
An individual purchases a car for £15,000
...
The annual rate of interest is 12
...

Calculate the ﬂat rate of interest of the loan
...
A 20 year loan of £100,000 is granted, to be repaid in monthly instalments paid in arrears on the basis of a rate of
interest of 6% per annum convertible monthly
...

(i) Calculate the initial monthly instalment
...
5% per annum convertible
monthly
...

(Institute/Faculty of Actuaries Examinations, September 1999) [5]
8
...
a
...
The cost of the
annuity is calculated using an effective rate of interest of 10% p
...

(i) Calculate the interest component of the ﬁrst instalment of the sixth year of the annuity
...

(Institute/Faculty of Actuaries Examinations, September 1997) [6]
9
...
The annual amount of the
annuity is calculated at an effective rate of interest of 10% per annum
...

(ii) Calculate the capital and interest instalments in the 25th payment
...
A bank makes a loan of £100,000 to an individual
...
The instalments are such that the borrower pays interest at an effective rate of 5% per
annum on the loan
...

(ii) (a) Calculate the capital outstanding after 12 years, immediately after payment of the instalment then due
...

(Institute/Faculty of Actuaries Examinations, September 2003) [3+4=7]

Annuities and Loan Schedules

Jan 28, 2016(12:36)

Exercises 5 Page 61

11
...
Suppose borrower A borrows £5,000 at an effective interest rate of 10% per
annum
...

(a) What is the annual interest payment?
(b) Suppose A pays a regular annual amount into a sinking fund which accumulates at an effective rate of 8%
per annum
...

(c) Compare the total annual payment of (a) and (b) with the regular payment that would be needed if the loan was
repaid by the normal method of amortisation
...
A company has borrowed £50,000 from a bank
...
The annual repayments are calculated at an effective rate of interest
of 10% per annum
...

(ii) At the beginning of the third year, immediately after the second payment has been made, the company asks for
the loan to be rescheduled over a further 5 years from that date
...

(a) Calculate the amount of the new quarterly payment
...
A company has borrowed £800,000 from a bank
...
The annual repayments are calculated at an effective rate of
interest of 8% per annum
...

(ii) At the beginning of the 8th year, immediately after the 7th payment has been made, the company asks for the
term of the loan to be extended by 2 years
...

(a) Calculate the amount of the new quarterly payment
...

(Institute/Faculty of Actuaries Examinations, April 2007) [3+6=9]
14
...
The annuity starts at a rate of £100 p
...
and increases by £10 per
annum
...
Repayments are calculated using a rate of interest of 8% p
...
effective
...

(ii) Construct a loan schedule showing the capital and interest elements in and the amount of loan outstanding after
the 6th and 7th payments
...

(Institute/Faculty of Actuaries Examinations, September 1998) [10]
15
...
The loan is due to be repaid on
1 January 2010 but the borrower can repay the loan early if he wishes
...
The loan instalments only cover the interest on the loan
...
The individual was told that premiums in
the investment policy were expected to earn a rate of return of 7% per annum effective
...
The borrower agrees to increase his monthly payments into the
investment policy to £5,000 per annum for the ﬁnal 10 years
...

(b) Calculate the amount by which the investment policy would have fallen short of repaying the loan had extra
premiums not been paid for the ﬁnal 10 years
...

(d) Suggest another course of action the borrower could have taken which would have been of higher value to him,
explaining why this higher value arises
...

(Institute/Faculty of Actuaries Examinations, September 2006) [11]

Page 62 Exercises 5
16
...
J
...

(b) Calculate the capital outstanding in the above annuity at the end of the 15th year after the payment then due
has been received
...

(Institute/Faculty of Actuaries Examinations, September 2000) [12]
(Ia) n =

17
...
The ﬁrst instalment is 50, the second 48 and so on
with the payments reducing by 2 per annum until the end of the 15th year after which there are no further payments
...

(i) Calculate the amount of the loan
...

(iii) Calculate the amount of the capital repaid in the instalment at the end of the 14th year
...
A loan of £80,000 is repayable over 25 years by level monthly repayments in arrears of capital and interest
...

Calculate
(i) (a) The capital repaid in the ﬁrst monthly instalment
...

(c) The interest included in the ﬁnal monthly payment
...

19
...
The mortgage is repayable
by monthly instalments paid in advance
...
After 10 years, the borrower has the option to repay any outstanding loan and take out a new loan, equal
to the amount of the outstanding balance on the original loan, if interest rates on 15 year loans are less than 6% per
annum effective at that time
...

(i) Calculate the amount of the monthly instalments for the original loan
...

(iii) The borrower takes advantage of the option to repay the loan after 10 years
...
The ﬁrst loan is repaid and a new loan is
taken out, repayable over a 15 year period, for the same sum as the capital outstanding after 10 years on the
original loan
...

(b) Calculate the accumulated value of the reduction in instalments at a rate of interest of 2% per annum
effective, if the borrower exercises the option
...

(i) A loan is repayable over 20 years by level instalments of £1,000 per annum made annually in arrear
...

Show that the amount of the original loan is £12,033
...
(Minor discrepancies due to rounding will not be
penalised
...
e
...

Loan outstanding at the
Instalment paid at the end of the year
beginning of the year
Interest
Capital
Year x
£8,790
...
52
£560
...

(iii) At the beginning of year 11, it is agreed that the increase in the rate of interest will not take place, so that the
rate remains at 5% per annum effective for the remainder of the term
...

(a) Calculate by how many years, if any, the repayment schedule is shortened
...

(c) Calculate the reduction in the total interest paid during the existence of the loan as a result of the interest
(Institute/Faculty of Actuaries Examinations, April 2005) [2+4+7=13]
rate not increasing
...
A loan is repayable by a decreasing annuity payable annually in arrears for 20 years
...
The repayments were calculated using
a rate of interest of 9% per annum effective
...

(ii) Construct the schedule of repayments for years eight (after the seventh payment) and nine, showing the outstanding capital at the beginning of the year, and the interest element and the capital repayment in each year
...

Calculate the amount of the tenth payment if subsequent payments continue to reduce by £200 each year, and
the loan is to be repaid by the original date, i
...
20 years from commencement
...
A loan was taken out on 1 September 1998 and was repayable by the following increasing annuity
...
Thereafter, payments were made on 1 November,
1 March and 1 July until 1 March 2004 inclusive
...
The effective
rate of interest throughout the period was 6% per annum
...

(ii) Calculate the amount of capital repaid on 1 July 1999
...

(Institute/Faculty of Actuaries Examinations, April 2004) [5+3+7=15]
23
...
The loan is repayable by an annuity payable monthly in arrears for
15 years
...

(i) Calculate the initial amount of monthly repayment
...

(iii) Calculate the amount of loan which will remain outstanding after the monthly repayment due on 1 April 2002
has been made
...
Calculate the amount of the revised
annual payment on this basis if the interest rate remains unchanged
...

(i) Prove

a n − nν n
¨
i
A loan is repayable by an increasing annuity payable annually in arrears for 15 years
...
The repayments were calculated using
a rate of interest of 8% per annum effective
...

(iii) Construct the capital/interest schedule for years 9 (after the eighth payment) and 10, showing the outstanding
capital at the beginning of the year, the interest element and the capital repayment
...

Calculate the amount of the eleventh payment if subsequent payments continue to increase by £200 each year,
and the loan is to be repaid by the original date, i
...
15 years from commencement
...
An actuarial student has taken out two loans
...
715% per annum
...

The student has a monthly disposable income of £600 to pay the loan interest after all other living expenses have
been paid
...
After 2 years, the student is approached by a representative of Freeloans who offers the student a 10 year
loan on the capital outstanding which is repayable by equal monthly instalments of capital and interest in arrear
...

(i) Calculate the ﬁnal disposable income (surplus or deﬁcit) each month after the loan payments have been made
...
J
...

(iii) Estimate the capital repaid in the ﬁrst month of the third year assuming that the student has taken out the new
loan
...

(Institute/Faculty of Actuaries Examinations, April 2006) [5+5+5+2=17]
26
...

Product A
...
25 annually in arrear
...

Product B
...
In addition, the
customer also has to pay an arrangement fee of £6,000 at the beginning of the mortgage and an exit fee of £5,000 at
the end of the twenty-ﬁve year term of the mortgage
...

(Institute/Faculty of Actuaries Examinations, April 2008) [8]
27
...
The
loans are typically of short duration and to high risk consumers
...
Campaigners on behalf of the consumers and campaigners on behalf of the banks
granting the loans are disputing one particular type of loan
...
Repayments are made
at an annual rate of £2,400 payable monthly in advance for 2 years
...
The consumers’ association asserts that, on this particular type of loan, consumers
who make all their repayments pay interest at an annual effective rate of over 200%
...
The banks state that, on the same loans, 40% of the consumers default on all their remaining
payments after exactly 12 payments have been made
...
The banks
also argue that it costs 30% of each monthly repayment to collect the payment
...
Furthermore, with inﬂation of 2
...
463% per annum effective
...

(b) State why the ﬂat rate of interest is not a good measure of the cost of borrowing to the consumer
...

(Institute/Faculty of Actuaries Examinations, September 2007) [4+10=14]
28
...
The initial instalment is £5,000 with each instalment
decreasing by £200
...

(i) Calculate the amount of the original loan
...

After the 12th instalment is paid, the borrower and lender agree to a restructuring of the debt
...
Instead, future instalments will remain at the level of the 12th
instalment and the remaining term of the debt will be shortened
...

(iii) (a) Calculate the remaining term of the revised loan
...

(c) Calculate the total interest paid during the term of the loan
...
A company has borrowed £500,000 from a bank
...
The annual instalments are calculated at an effective rate of
interest of 9% per annum
...

At the beginning of the eighth year, immediately after the seventh instalment has been made, the company asks for
the loan to be rescheduled over a further four years from that date
...

(ii) (a) Calculate the amount of the new quarterly instalment
(b) Calculate the interest content of the second quarterly instalment of the rescheduled repayments
...
A bank makes a loan to be repaid by instalments paid annually in arrear
...
The rate of interest charged is 4% per annum effective
...

(ii) Calculate the capital and interest components of the ﬁrst payment
...
The bank agrees to
reduce the capital by 50% of the loan outstanding after the eighth repayment
...
The bank changes the rate
of interest to 8% per annum effective
...

(Institute/Faculty of Actuaries Examinations, September 2013) [3+2+5=10]
31
...
The ﬁrst repayment will be £200 and the repayments will increase
by £100 per annum
...
The repayments are calculated using
a rate of interest of 6% per annum effective
...

(ii) (a) Calculate the interest component of the seventh repayment
...

(iii) Immediately after the seventh repayment, the borrower asks to have the original term of the loan extended to
15 years and wishes to repay the outstanding loan using level annual repayments
...
Calculate the revised annual repayment
...
An individual takes out a 25-year bank loan of £300,000 to purchase a house
...
He
then pays only the interest for the remaining 10 years, quarterly in arrear, and repays the other half of the capital as
a lump sum at the end of the term
...
a
...
Calculate the level contribution he must make monthly in advance to the savings
account in order to repay half the capital after 15 years
...
However, over the ﬁrst
15 years, the effective rate of return earned on the savings account was 10% per annum
...
After the ﬁrst 15 years, the effective rate
of interest changed to 7% per annum
...

(Institute/Faculty of Actuaries Examinations, September 2008) [5+4+5=14]
33
...
Under this repayment schedule, the instalment at the end of January 2016 will be X, the
instalment at the end of February 2016 will be 2X, and so on, until the ﬁnal instalment at the end of December 2020
will be 60X
...

(i) Prove that
a n − nν n
¨
(Ia) n =
i
(ii) Show that X = £26
...

The student is concerned that she will not be able to afford the later repayments and so she suggests a revised
repayment schedule
...
She would now repay the
loan by 60 level monthly instalments of 36X = £958
...

(iii) Calculate the APR on the revised loan schedule and hence determine whether you believe the bank should
accept the student’s suggestion
...

(Institute/Faculty of Actuaries Examinations, April 2015) [3+4+5+2=14]

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CHAPTER 4

Basic Financial Instruments
1 Markets, interest rates and ﬁnancial instruments
1
...
The term primary market refers to the original issue of a security, whilst the term secondary
market refers to the market where the security is bought and sold after it is issued
...

The term money market refers to the market in short-term ﬁnancial instruments which are based on an interest
rate
...

The term capital market refers to all long-term ﬁnancial instruments
...
Bond calculations are considered in the next chapter
...

This is not an exhaustive list of markets—for example, there are various commodities markets, the equity
market, and so on
...

1
...
A security is negotiable if it can be bought and sold in a secondary
market; otherwise it is non-negotiable
...

If a security is a bearer security then any payment is made to whoever is holding the security at the time
...

1
...
The term money market refers to all short-term ﬁnancial instruments
...

Some of the main instruments in the money market are1
• Treasury bill or T-bill
...
A negotiable bearer security
...

Bills issued by the government and denominated in euros are called Euro bills
...
Borrowing by a bank
...

• Certiﬁcate of deposit or CD
...
Negotiable and usually a bearer security
...

• Commercial paper or CP
...
A negotiable bearer security
...

• Bill of exchange
...
Used as an IOU between companies in order to ﬁnance trade
...

• Repurchase agreement or repo
...
Non-negotiable
...

• Futures contract
...

In the UK, the term gilt or gilt-edged security denotes a UK Government liability in sterling, issued by HM
Treasury and listed on the London Stock Exchange
...
dmo
...
uk) is
the government agency responsible for managing the government debt
...
These can be downloaded from their web site—just choose
the Publications link
...

The money market fulﬁls several functions:
• One bank may have a temporary shortage and another may have a temporary surplus of money
...

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Section 1

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• Every bank will want to hold a proportion of its assets in a form in which it can quickly get at the money
...

• The money market forms a bridge between the government and the private sector
...
These instruments include bills
of exchange, gilt repos and Euro bills
...
Its decisions are widely reported in the press
...
4 Interest rates in the money market
...
Deposit rates and borrowing rates for individuals are calculated from these
bank base rates
...
The London Inter-Bank Bid Rate or LIBID is the
rate at which a bank bids for money from another bank
...

Clearly, LIBOR > LIBID
...
For example, it is linked to the
capital market by the ability to create long-term instruments from a series of short-term instruments
...
1 Bonds
...
In most developed
economies, government bonds form the largest and most liquid section of the bond market
...
In the UK, bonds guaranteed by the government are called
gilt-edged securities or gilts
...
)
In general, a borrower who issues a bond agrees to pay interest at a speciﬁed rate until a speciﬁed date, called
the maturity date or redemption date, and at that time pays a ﬁxed sum called the redemption value
...

The interest rate on a bond is called the coupon rate
...
The face and redemption values are often the
same
...

The values of P and i vary throughout the lifetime of the bond
...

Also, the price of a bond (and hence its yield) depends on the prevailing interest rates in the market
...

When a bond is sold between two coupon dates, the seller will be entitled to some accrued interest
...
8 of chapter 5 on page 83
...

Basic Financial Instruments

Jan 28, 2016(12:36)

Section 3 Page 69

2
...
Some bonds have variable redemption dates—the actual redemption date is chosen by the
borrower (or occasionally the lender) at any interest date within a certain period or at any interest date after a
certain date
...

Some banks allow the interest and redemption proceeds of a bond to be bought and sold separately: this creates
zero coupon bonds and bonds with zero redemption value
...
Hence, governments sometimes
issue bonds such that the coupon rate and redemption value are linked to an inﬂation index
...
(So this means there is no inﬂation protection during the lag period
...
3 Government bills
...
They are
also called Treasury bills or T-bills and are negotiable bearer securities
...
They are often used as
a benchmark risk-free short-term investment
...

2
...
There are no US or UK government zero-coupon bonds
...
Thus a bank may decompose a 3-year bond which has a semi-annual coupon into 7 components: the
6 coupon payments and the ﬁnal repayment of the principal
...

3 Other ﬁxed interest borrowings
3
...
Debentures (or corporate bonds) form part of the loan capital
or long term borrowing of a company
...

Debentures are clearly not as secure as government bonds and so are less marketable and investors will expect
a higher yield
...

3
...
A samurai bond is a yen bond issued in Japan by a foreign (i
...
nonJapanese) organisation; a yankee bond is a dollar bond issued in the USA by a foreign (i
...
non-US) organisation and a bulldog bond is a bond issued in sterling in the UK by a foreign (i
...
non-UK) organisation
...

Deposits of dollars in a European bank are called eurodollars
...
The preﬁx euro just indicates that the currency is being deposited outside of its country
of origin and the general term is eurocurrency
...

Eurobonds are bonds issued by large companies, syndicates of banks or governments
...
Thus a eurobond allows a company to borrow US dollars in the
Swiss ﬁnancial market, etc
...
Eurobonds provide medium or long-term borrowing free
from the controls of any government
...
Yields are typically less than the yields from conventional bonds from the same issuer with the same
terms
...
The terminology is confusing: a “euro bond”, as opposed to a eurobond, is a bond denominated
in euros which could be issued domestically in the euro zone or internationally
...
J
...
3 Certiﬁcates of deposit
...
They usually last from 28 days to 6 months and interest is paid on maturity
...

There is an active secondary market and marketability depends on the status and credit rating of the issuing
bank
...
Because of
this ﬂexibility, the yield on a CD will typically be less than the yield on a deposit with the same bank and for
the same term
...
4 CD calculations: CD with a single coupon
...

Let
f = face value of the CD
ic = the coupon rate
d = number of days interest the CD earns
dT = 360 for ACT/360 or 365 for ACT/365
The maturity proceeds are:

)
(
d
mp = f × 1 + ic
dT

Clearly, ic is also the yield on the CD when it is issued
...
Let ia and ib denote the corresponding yields at the times da and
db
...

...

...

...

...

...

...

...

...

...
Substituting into equation (3
...
4a)

(3
...
4a
...

(a) After 30 days, its yield has fallen to 5
...
What is its price?
(b) After a further 30 days its yield has risen back to 6%
...
(Assume ACT/365
...
06
= 507,397
...
Then
(
)
mp
365
mp
−1
= 0
...
22
pa
60
1 + 0
...
4b) gives
(
)
365 + 0
...
05473 or 5
...

365 + 0
...
5 Repurchase agreements or repos
...

They are non-negotiable and the difference between the purchase and repurchase prices implies the yield
...
The seller sells a gilt but agrees to buy it back at a speciﬁed later
date at an agreed price
...
If the
speciﬁed later date is the next day then it is called an overnight repo ; otherwise it is called a term repo
...

In summary, a repo is a purchase or sale of a security at the present time together with an opposite transaction
later
...
Repos are used extensively by
ﬁnancial institutions to ﬁnance positions and they provide a link between the bond market and the money
market
...
1 Ordinary shares or equities
...
This payout is
called the dividend
...
Shareholders are the owners of a company and, in
principle,2 can vote at the company’s Annual General Meeting
...
The
expected return is higher than for most other asset classes—but the risks of loss are also higher3
...

If a share is bought ex-dividend or xd then the seller, and not the buyer, receives the next dividend payment
...
The usual procedure with ﬁxed
interest stocks is different—the buyer of a ﬁxed interest stock usually compensates the seller for accrued
interest
...
2 Preference shares
...
Like equities, they offer a stream of dividend income
...
They rank above ordinary shareholders for payment
of dividends and in winding-up but do not normally have voting rights unless the dividend is in arrears
...
in the title and the company cannot afford to pay the
dividend, then the dividends are carried forward and paid when the company is able to do so
...
stands for cumulative
...
occurs in the name of a preference share, then the shares are redeemable and will be
repaid at some speciﬁed future date
...

4
...
Companies sometimes raise money by issuing convertible loan stock
...

Convertible loan stock can be converted into ordinary shares at a ﬁxed price at a ﬁxed date in the future (or at
one of a series of dates at the option of the holder)
...

Convertible loan stock behaves like a cross between ﬁxed interest stock and ordinary shares
...

2

Because private investors usually hold their shares in a nominee account at a stockbroker, voting is often impossible
...
Hence, loan stock has lower risk than preference shares; preference shares have lower risk than ordinary
shares
...
J
...
It combines the lower
risk of a debt security with the potential large gains of an equity
...
Investors accept the lower rate because of the possibility of capital gains from
the rise in share prices
...
)
Also, if the company had issued ordinary shares, then its earnings and dividends would have been immediately
diluted over a larger share capital base
...
4 Property
...
The
returns from investing in property arise from the rental income and any capital appreciation achieved on its
sale
...
However, capital values can ﬂuctuate substantially in the short term
...
Typically rents increase every 3 to 5 years
...

The running yield of an investment is the annual income from the investment divided by its current market
price; it ignores any capital growth
...
So ranking investments highest to lowest by running yield normally gives
the order: bonds, then property then equities
...
The purpose of the derivatives market is to redistribute risk
...

There are two main types of people using derivatives: those who want to hedge or guard against a risk and the
traders or speculators who are prepared to accept a high risk in return for the possibility of a high gain
...
Market risk is the risk that market conditions change
adversely—it is the risk that market conditions will change in such a way that the present value of the outgoings speciﬁed under an agreement increases
...

The main exchange-traded derivatives are futures and options
...

5
...

A long position in a security such as a bond (or equivalently to be long in a security) means the investor owns
the security and will proﬁt if the price of the security rises
...
Short selling or shorting or going short in a security means selling a security that has been
borrowed from a third party with the intention of repurchasing it in the market at a later date and returning it
to the third party
...

Similarly the holder of a long position in a derivative will proﬁt if the price of the derivative rises whilst the
holder of a short position in a derivative makes a proﬁt if the price of the derivative falls
...
2 Forward contract
...
The contract is usually tailor-made between two ﬁnancial institutions or between
a ﬁnancial institution and a client
...
Such contracts are not
normally traded on an exchange
...
Forward contracts are subject to credit risk—the risk of default
...
3 Futures contract
...
Futures contracts can be traded on an exchange—this implies that futures
contracts have standard sizes, delivery dates and, in the case of commodities, quality of the underlying asset
...
A futures contract based on a ﬁnancial asset is called a ﬁnancial future
...
Then
B is said to hold a short forward position and A is said to hold a long forward position
...
Thus the long side proﬁts if the price rises and the short side proﬁts if the price falls
...

Each party to a futures contract must deposit a sum of money called the margin with the clearing house
...
When the contract is ﬁrst struck, the
initial margin is deposited with the clearing house
...

Note that futures contracts are guaranteed by the exchange on which they are traded
...
Futures markets enable a trader to speculate on price movements
for only a small initial cash payment and they also enable a trader to take a short position if he believes a price
is going to fall—he would sell a future contract
...

• Forward contracts are tailor-made contracts between the two parties
...
)
• Settlement of forward contracts occurs at maturity
...

• For most futures contracts, delivery of the underlying asset is not normally made
...
With forward contracts, delivery is usually carried out
...
Farmers in the Midwest shipped their grain to Chicago
for sale and distribution
...
It made sense for farmers to agree to a forward
contract in which they agreed to supply an agreed amount of grain at an agreed price at some time in the future
...
This led to the establishment of an exchange
which laid down rules for contracts which could be traded and also checked the ﬁnancial status of both parties to the
contract—this led to futures contracts
...
I believe that Chicago is still the second largest futures exchange in the world
...
C
...
Brett (page 290 onwards)
...
4 Some special cases of futures contracts
...
Government bond futures contracts are one of the most popular futures contracts
...
Some contracts are usually speciﬁed in terms of a notional bond and then there needs to be an agreed
list of bonds which are eligible for fulﬁlling the contract
...
The price paid by the long side may have to be adjusted to allow for the fact

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that the coupon on the bond which is actually delivered may not be equal to the coupon on the notional bond
speciﬁed in the contract
...

Suppose a fund manager has some UK gilts and expects a rise in interest rates
...
If the rise in
interest rate occurs, then the price of the gilts and the futures price of the bonds will fall
...

Example 5
...
On January 10, 2001, the settlement price of the “US Treasury Bond $100,000; pts
...
06
...
01%
...
32nds of 100%” implies that 123
...

The listed price of 123
...
50
...
06 then the long side gains $1,000 and the short side loses
the same amount
...
Note that the value of the bond has only increased by 1/(123 6/32) × 100% = 0
...

A basis point is always 0
...
The value of a tick
varies from instrument to instrument: it can be a basis point or 1/2 basis point or 1/4 basis point, etc
...
The contract provides for a transfer of assets underlying a stock index at a speciﬁed
price on a speciﬁed date
...

For example, there are futures based on the FTSE 100 index and these provide a way of betting on the movement of the market as a whole
...

A bull will buy a FTSE futures contract: each rise of 1 point will give him a gain of £25
...

• The currency future
...

5
...

• Forward rate agreement or FRA
...
It is an interest rate
hedge which allows the borrower to ﬁx the cost of money he intends to borrow in a few months’ time
...
An FRA is a forward contract and is an OTC equivalent of an
interest rate future—these are explained below
...
The interest on the loan
is set when the contract is arranged and the amount of the loan is notional—this means that no loan is ever
extended4
...

Example 5
...
Consider a 3 × 9 FRA for £1,000,000 with an FRA rate of 3
...
Suppose the reference rate is
LIBOR and the 6-month LIBOR on the effective date is 3
...
Assume ACT/360 and the loan is for a period of
180 days
...

Solution
...
037 − 0
...
75
1 + 0
...
037 − 0
...
This quantity is divided by 1 + 0
...

£1,000,000

Because the loan is notional, an FRA is an off-balance sheet instrument
...
For such a contract, the principal of the loan is paid
...
An interest rate future is similar to an FRA but instead is a futures contract—hence it
is traded on an exchange and there is the process of “marking to market” instead of a single cash payment
...
Typically, the price
is stated as 100 − x where x is the interest rate
...
25 then the contract is priced at 93
...

On expiry, the purchaser will have made a proﬁt or a loss depending on the difference between the ﬁnal
settlement price and the original dealing price
...

A borrower who wishes to hedge against a rise in interest rates should sell an interest rate future
...

Note that FRAs and futures move in opposite directions
...
Thus a trader who sells an FRA will also
sell an interest rate future in order to cover his position
...
5b
...
02 for March, 2001
...
02 = 2
...

The phrase “points of 100%” explains the pricing unit
...
01%
...
01% over 3 months which gives 1,000,000 × 0
...
25 = 25 DM (we use 0
...
This smallest possible price movement is called a tick
...
23 or moves up 21 ticks, then this is equivalent to 21 × 25 = 525 DM
...
5c
...
29
...
22
...

Solution
...
0007 × 0
...

The price of an interest rate future is determined by the market
...
22)% = 3
...
He just makes a proﬁt
or loss depending on interest rate at the time of settlement
...

5
...
This is an agreement to exchange a speciﬁed series of interest payments and a speciﬁed
capital sum in one currency for a speciﬁed series of interest payments and a capital sum in another currency
...
Conversely, another
company may require French francs but can borrow US dollars more cheaply
...
In practice, both will
use a bank which arranges many swaps of this type
...
7 Interest rate derivatives
...
In return, the other party pays a series of variable payments—these are the interest on the same (notional)
deposit at a ﬂoating rate
...
This is the
proﬁt margin for the issuer
...
Market makers will often hedge market risk by making an
offsetting agreement
...

Example 5
...
Company A can borrow at a ﬂoating rate of LIBOR+0
...
0% p
...
Company B
is not so creditworthy and can borrow at a ﬂoating rate of LIBOR + 0
...
5% p
...
Company A
prefers to borrow at a ﬂoating rate and company B prefers to borrow at a ﬁxed rate
...
3% p
...

The cost to company A is 8
...
3% + LIBOR = LIBOR − 0
...

The cost to company B is LIBOR + 0
...
3% − LIBOR = 9
...

Hence both sides gain by 0
...
a
...
J
...
For example, a company may wish to swap the interest
on a ﬂoating-rate loan it has arranged for a ﬁxed rate of interest
...

A company may wish to arrange a cap—this means that the bank will pay any interest above a speciﬁed agreed
rate
...
Arrangements that include a cap
and a ﬂoor are called a collar or cylinder
...
Of course, the bank also needs
to hedge its risk by trying to match up the different swaps it issues, etc
...
8 Options
...

A put option gives the right (but not the obligation) to sell a speciﬁed asset for a speciﬁed price (the strike price
or exercise price) at a speciﬁed time in the future
...
The holder of a put option has just purchased the right to sell the asset at a certain price at
some time in the future
...

An American option means that the option can be exercised on any date before its expiry; a European option
means that the option can only be exercised on the expiry date
...

Financial options are traded in standard units called contracts; for example, a contract for 100 shares
...

Thus the owner of a traded option may sell the option to someone else, exercise the option to buy (or sell) the
underlying asset at the ﬁxed price, or abandon the option with no delivery of the underlying asset
...
A put option is said to be
in-the-money if the price of the underlying asset is below the exercise price and is said to be out-of-the-money
if the price of the underlying asset is above the exercise price
...
8a) shows the proﬁt/loss at expiry for the holder of a call option
...
Theoretically, the
price of the asset could rise to any amount, and so the potential proﬁt is unbounded above
...
The proﬁt/loss diagram for the writer
of a call option is the mirror image in the y-axis of this graph—this means that his potential loss is unlimited if
he does not own the asset
...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...
8a
...
8a) is the proﬁt/loss at expiry for the holder of a put option
...
The holder of the put option will then exercise his right to sell the asset

Basic Financial Instruments

Jan 28, 2016(12:36)

Exercises 6 Page 77

at the exercise price
...
In that situation, the holder of the put option will not
exercise his right to sell the asset and so he will just lose the premium he has paid
...

5
...
Trading in futures can be more dangerous than trading in the underlying asset
because of gearing or leverage
...
Thus it is
possible to lose large amounts (and gain large amounts) with a relatively small initial outlay
...

6 Exercises

(exs4-1
...
A CD (Certiﬁcate of Deposit) is issued for £1,000,000 on 12 March for 90 days with a 5% coupon
...

(a) What are the proceeds on maturity?
(b) On 4 April, what should the secondary market price be in order that the yield is then 4
...
By 4 May,
the yield has dropped to 4%
...
)
2
...

(b) Explain why convertibles have “option-like” characteristics
...
Describe how cash ﬂows are exchanged in an “interest rate swap”
...
State the main differences between a preference share and an ordinary share
...
Describe the features and risk characteristics of a “Government Bill”
...

(i) Deﬁne the characteristics of a government index-linked bond
...

(Institute/Faculty of Actuaries Examinations, September 2003) [2+2=4]

7
...
Under the terms of the swap, the company makes ﬁxed annual
payments equal to 6% of the principal of the swap
...
5% per annum
...

(b) Explain which of the risks described in (a) are faced by the company
...
State the characteristics of an equity investment
...
Describe the characteristics of the following investments:
(a) Eurobonds
(b) Certiﬁcates of Deposit
(Institute/Faculty of Actuaries Examinations, April 2008) [4]
10
...
Describe the characteristics of commercial property (i
...
commercial real estate) as an investment
...
Explain why the running yield from property investments tends to be greater than that from equity investments
...
J
...
1 Notation
...

The interest rate on a bond is called the coupon rate
...
The face and redemption values are often the
same
...

The values of P and i vary throughout the lifetime of the bond
...

Also, the price of a bond and hence its yield depend on the prevailing interest rates in the market and the risk
of default
...
2 Finite redemption date
...
2a
...
The coupon rate is
10% p
...
convertible semi-annually
...
a
...
a
...
For this example, f = C = 500,000; r = 0
...
Each coupon payment is £25,000
...
5
1
...
5
2
...
5
3
...
5
4
...
10
...
1 1
...
5 1
...
12
...
13 1
...
5 1
...
1
...
14 − 1
2
8
8
8
P = 25(α + α + · · · + α ) + 500α = 25α
+ 500α =
25 1/2
+ 500 = 503
...
14
1
...
1 and ν = 1/(1 + i), then
i
500
= 50 (2) a 4 ,i + 500ν 4 = 503
...

Note that commonly used values of i/i(m) , ν n and a n are provided in the tables
...
154 − 1
1
25
+ 500 = 433
...
154
1
...
792
...

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

Section 1

Page 79

Page 80 Section 1

Jan 28, 2016(12:36)

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ST334 Actuarial Methods ⃝R
...
Reed

Here is the general result: suppose the coupon rate is nominal r% p
...
payable m times per year
...
2a)
n ,i
n
(1 + i)
m
(1 + i′ )mn
i(m)
where (1 + i′ )m = 1 + i and i′ =
is the effective interest rate per period
...

If P = f = C, then using the relation a(m) = (1 − ν n )/i(m) and equation (1
...
This
n
means that if the face value, redemption value and price are all the same, then the nominal yield convertible
mthly equals the coupon rate
...
2a) becomes:
P = f r(1 − t1 )a(m) +
n ,i

C
(1 + i)n

(1
...
2b
...

The bond is redeemable at 105%
...
05
...

(b) Assume the income tax rate is 35%
...
a
...
For this example, we have f = 100, C = 105, r = 0
...
07 and n = 15
...
Hence
105
i
105
= 6 (2) a 15 ,i +
≈ 93
...
Hence the bond should be issued at a discount
...
65 × a(2) ,i +
15

105
105
i
= 3
...
19
15
(1 + i)
i
1
...
3 Gross yield, redemption yield and ﬂat yield
...
This is the
same as the annual coupon (f r) divided by the current price, P
...

This is the same as (1 − t1 )f r/P , the annual coupon after tax divided by the current price, P
...
3a
...
Then the
gross interest yield (also known as the ﬂat yield or running yield) is 10/103 = 0
...
71%
...
If there are redemption proceeds, then we should also take these into account when calculating the
yield
...
1 and ignores
taxation
...

If the gross redemption yield on a bond is an effective 3% every six months, then the gross redemption yield is
6
...
032 − 1 = 0
...
Alternatively, it is described as a gross redemption yield of
6% convertible 6-monthly
...

2

In Europe, the gross redemption yield is the same as the internal rate of return as deﬁned in paragraph 1
...
However, in
the USA and the UK, the gross redemption yield of a bond with 6-monthly coupons is deﬁned to be twice the effective
yield for 6 months—what we have deﬁned as the gross redemption yield convertible 6-monthly
...
09% per annum effective” or “a
gross redemption yield of 6% per annum convertible 6-monthly”
...
bankofengland
...
uk/education/Pages/ccbs/handbooks

Bonds, Equities and Inﬂation

Jan 28, 2016(12:36)

Section 1 Page 81

Example 1
...
Consider a US treasury bond which has 14 years to redemption and a coupon of 12% p
...
payable
6-monthly
...

Solution
...

Time
Cash ﬂow

0
−P

1/2
6

1
6

3/2
6

2
6

5/2
6

···
···

27/2
6

14
106

So let ν = 1/1
...
04
...
326
1−α

In practice, the most common problem is ﬁnding the gross redemption yield when the current market price is
given
...
Let i′ denote the effective yield per 6 months and let α = 1/(1 + i′ )
...
3a)
2
2
1−α
2
i′
Finding i′ may require successive approximation on a calculator
...
3a) implies
(
)
fr
α2n
f r 100 − P
2i′
100 − P 1
+
(100 − P ) = P and so 2i′ =
+
≈ fr +
(1
...
This approximation for i′ is useful as a starting value when
ﬁnding an approximate solution
...
This approximation will be greater than
the exact value
...
4 Perpetuity
...
Assume the
coupon is r% payable half-yearly
...
Then the price P to achieve an effective yield of i per annum is:
f r(1 − t1 )
P = f r(1 − t1 )a(2) =
∞
i(2)

1
...
Suppose we have a bond with face value f , redemption
value C, coupon rate r% payable m times per year for n years
...
2b):
g=

P (n, i) = (1 − t1 )f ra(m) + Cν n
n ,i

(1
...
5b)
(1
...
It is levied only
when the security is sold
...
Let the new price of the bond be P2 (n, i)
...
5c) shows that:
• If i(m) ≤ (1 − t1 )g, then P ≥ C
...

• If i(m) > (1 − t1 )g, then P (n, i) < C
...
Hence
[
]
P2 (n, i) = (1 − t1 )f ra(m) + Cν n − t2 C − P2 (n, i) ν n
n ,i
g(1−t1 )
[C
i(m)

− Cν n ] + Cν n ; this is called Makeham’s formula
...
5b) is also equal to

4

Aide m´ moire
...

Page 82 Section 1

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...
Reed

Jan 28, 2016(12:36)

Hence
P2 (n, i) =
and so

(1 − t1 )f ra(m) + (1 − t2 )Cν n
n ,i
1 − t2 ν n

 (1 − t )f ra(m) + Cν n

1

n ,i

P2 (n, i) =

if i(m) ≤ (1 − t1 )g

 (1 − t1 )f ra(m) + (1 − t2 )Cν n
n ,i


1 − t2 ν n

if i(m) > (1 − t1 )g

Example 1
...
A bond with face value £1,000 is redeemable at par after 10 years and has a coupon rate of 6%
payable annually
...
If the current price of the bond
is £800, what is the net yield?
Solution
...

...
5% of the purchase price
...
Hence
the net yield is greater than 4
...
The gain on redemption is £140; if this amount was paid in equal instalments
over the 10 years, then the payment each year would be £50, which is 6
...
3b)
...
5 < i < 6
...
Now use tables:
if i = 0
...
339; if i = 0
...
660; and if i = 0
...
854
...
055 and i = 0
...
055)/(0
...
055) = (f (x) − f (0
...
06) − f (0
...
055 + 0
...
660)/(789
...
66) = 0
...
8%
...
6 Effect of term to redemption on the yield
...
Equation (1
...

Suppose bond A is redeemed after n1 years and bond B is redeemed after n2 years where n1 < n2
...

• If P < C then i1 > i2 and so bond A with the shorter term has the higher yield
...

• If P = C then i1 = i2
...
It is clearly better to receive this amount sooner rather than later
...
7 Optional redemption date
...
The redemption
date may be at the borrower’s option 5
...
The last possible redemption date is called the ﬁnal redemption date; if
there is no such date then the security is said to be undated
...

An investor who purchases a bond with a variable redemption rate at the option of the borrower (or issuer)
cannot know the yield that will be obtained
...
The quantities f , r, t1 , C and m are ﬁxed known
numbers
...

We think of P as a function such that if we are told n and i, then we can ﬁnd P (n, i) by evaluating equation (1
...

5

It is also possible for a security to be redeemable at the lender’s option, but this is much less common
...

Then for a ﬁxed price, a longer term implies a higher yield—see paragraph 1
...
So suppose n1 < n∗
and P (n1 , i) = P (n∗ , i∗ ) for some i and i∗ ; then i∗ > i
...

If i(m) < (1 − t1 )g, the purchaser of the bond will receive no capital gain and should price the bond
on the assumption that redemption will take place at the earliest possible time n1
...

Second case: suppose i(m) > (1 − t1 )g, equivalently suppose P (n, i) < C
...
6 above
...

If i(m) > (1 − t1 )g, the purchaser of the bond will receive a capital gain and should price the bond on
the assumption that redemption will take place at the latest possible time
...

Third case: suppose i(m) = (1 − t1 )g, equivalently suppose P (n, i) = C for all n
...

• The minimum yield i if the price is P
...
By the above remarks, we have the following:
If P < C then it is better for the investor that the bond is redeemed sooner rather than later
...

If P > C then it is better for the investor that the bond is redeemed later rather than sooner
...

If P = C then the yield will be i where i(m) = (1 − t1 )g whatever the redemption date
...
8 Clean and dirty prices
...
Then B will be the registered owner of the bond at the next coupon date and so receive the coupon
...
He therefore sells the bond for the dirty price—this
is the NPV of future cash ﬂows from the bond6
...
So we have
dirty price = NPV of future cash ﬂows
clean price = dirty price − accrued interest
The price quoted in the market is the clean price but the bond is sold for the dirty price
...
For U
...
treasury bonds, the convention
is ACT/ACT
...
This implies that the accrued interest on the 31st of the month is
the same at the accrued interest on the 30th of the month
...
In the UK and Japan, the convention is ACT/365
...
8a
...
Suppose the
coupon rate is 10% per annum convertible semi-annually
...
Assume ACT/365
...

Page 84 Section 1

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ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

Solution
...
Hence accrued interest is 10 × 68/365 =
1
...

Example 1
...
A bond has a 9% coupon paid annually on August 11
...
The current
market yield for the bond is 8%
...

Assume 30/360
...
(a) Using 30/360 gives 63 days between 8/06/00 and 11/08/00
...
08
...
48
1−ν
The number of days from 11/08/99 to 8/06/00 is 297
...
425
...
48 − 7
...
06
...

For part (b), we have 10 days instead of 63 and 350 days instead of 297
...
75, accrued
interest of 8
...
00
...
UK gilts normally go ex dividend 7 working days before a coupon payment
...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

coupon
date

coupon
date

time

Figure 1
...
Plot of dirty price against time
(PICT X)
E

1
...
Bond calculations
...
a
...
Then the price P for an effective
yield of i per annum is:
C
P = f r(1 − t1 )a(m) +
n ,i
(1 + i)n
annual coupon f r
• Gross running yield =
=
;
current price
P
annual coupon after tax (1 − t1 )f r
• Net running yield =
=
current price
P
• Gross redemption yield (or yield to maturity) and net redemption yield
...
Suppose g = annual coupon/redemption value, i is the yield and
m is the number of times per year the coupon is paid
...

• Optional redemption date: at option of issuer
...

If i(m) > (1 − t1 )g there is a capital gain; purchaser should assume late redemption
...

Bonds, Equities and Inﬂation

2 Exercises

Jan 28, 2016(12:36)

Exercises 2 Page 85
(exs5-1
...
Consider a bond with face value f , redemption value C, current price P which has a coupon rate of r% payable
m times per year for n years
...
Find an equation for the yield to maturity
...
Suppose bond A is redeemed after n1 years and bond B is redeemed after n2 years where n1 < n2
...

Let i1 and i2 denote the yields of bonds 1 and 2 respectively
...
(b) If P < C then i1 > i2
...

3
...
The
coupon rate is r% nominal p
...
payable m times per year and the income tax rate is t1
...
A bond with an annual coupon of 8% per annum has just been issued with a gross redemption yield of 6% per annum
effective
...
The gross redemption yield on bonds of all terms to maturity is 6% per
annum effective
...

(a) Is the bond likely to be redeemed earlier or later than was assumed at issue?
(b) Is the gross redemption yield likely to be higher or lower than was assumed at issue?
(Institute/Faculty of Actuaries Examinations, September 1998 (adapted)) [3]
5
...
There
are 8 days until the next coupon payment and the bond is ex-dividend
...

Calculate the purchase price to provide a yield to maturity of 6% per annum
...
A ﬁxed interest stock bears a coupon of 7% per annum payable half yearly on 1 April and 1 October
...

On 1 July 1991 an investor purchased £10,000 nominal of the stock at a price to give a net yield of 6% per annum
effective after allowing for tax at 25% on the coupon payments
...

(i) Calculate the price at which the stock was bought by the original investor
...

(Institute/Faculty of Actuaries Examinations, April 2000) [4]
7
...
Coupons at an annual rate of 4%
are paid annually in arrears
...

Calculate the gross redemption yield from the bond
...
A new issue of a ﬁxed interest security has a term to redemption of 20 years and is redeemable at 110%
...

An investor who is liable to tax on all income at a rate of 23% and on all capital gains at a rate of 34% bought all
the stock at the date of issue at a price which gave the investor a yield to maturity of 8% per annum effective
...

(i) Describe the risk characteristics of a government-issued, conventional, ﬁxed-interest bond
...

Annual coupons are paid in arrears of 8% of the nominal value of the bond
...
The capital is repaid at par
...
After
the end of the 5th year, coupons are only paid on that part of the capital that has not been repaid
...

Calculate the purchase price of the bond per £100 nominal, at issue, to provide a purchaser with an effective net
rate of return of 6% per annum
...

(Institute/Faculty of Actuaries Examinations, September 2003) [2+5=7]

Page 86 Exercises 2

Jan 28, 2016(12:36)

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ST334 Actuarial Methods ⃝R
...
Reed

10
...
Capital is to be redeemed at 103 on a single coupon date between 15 and 20 years after the date of issue,
inclusive
...

An investor who is liable to income tax at 20% and capital gains tax of 25% wishes to purchase the entire loan at
the date of issue
...

(Institute/Faculty of Actuaries Examinations, April 2005) [9]
11
...
The loan is to be redeemed with a capital payment of 110 per £100 nominal on a coupon date between 10
and 15 years after the date of issue, inclusive, the date of redemption being at the option of the borrower
...

(i) Determine whether the investor would make a capital gain if the investment is held until redemption
...

(iii) Calculate the maximum price which the investor should pay
...
An investor purchased a bond with exactly 15 years to redemption
...
It pays coupons of 4% per annum, half yearly in arrear
...

(i) Calculate the price paid for the bond
...
The bond is
purchased by the second investor to provide a net return of 6% per annum effective
...

(b) Calculate, to one decimal place, the annual effective rate of return earned by the ﬁrst investor during the
period for which the bond was held
...
Ten million pounds nominal of loan stock was issued on 1 March 1990
...
The stock pays interest at 10% per annum, payable half-yearly in
arrears on 1 September and 1 March
...
What price did the investor pay
per £100 nominal on the date of issue?
(ii) On 1 March 1998 the ﬁrst investor, having received the interest due on that date, sold the stock to a second
investor who was subject to 40% tax on both income and capital gains
...
What price did the second
investor pay per £100 nominal?
(iii) Assuming that the ﬁrst investor sold the whole of the stock of his loan to the second investor, what was the
effective net yield, after tax, earned on the ﬁrst investment?
(Institute/Faculty of Actuaries Examinations, April 1998) [13]
14
...
The bond, redeemable at par, has a gross
redemption yield of 6%
...
The investor does not pay tax
...

(ii) After exactly 10 years, immediately after payment of the coupon then due, this investor sells the bond to another
investor
...
The bond is purchased by the second
investor to provide a net rate of return of 6
...

(a) Calculate the price paid by the second investor
...

15
...
The security will
be redeemed at par on any 1 January between 1 January 2006 and 1 January 2011 inclusive, at the option of the
borrower
...
The investor pays tax at 40% on
interest income and 30% on capital gains
...

(i) Calculate the price per £100 nominal at which the investor bought the security
...

(iii) Calculate the net yield per annum convertible half-yearly, which the investor actually received over the two
years the investor held the security
...
A loan of nominal amount £10,000,000 is to be issued bearing a coupon of 8% per annum payable quarterly in
arrears
...
An institution not subject to
either income or capital gains tax bought the whole issue at a price to yield 9% per annum effective
...

(ii) Exactly 5 years later, immediately after the coupon payment, the institution sold the entire issue of stock to an
investor who pays both income and capital gains tax at a rate of 20%
...

(iii) Calculate the net running yield per annum earned by this investor
...

(Institute/Faculty of Actuaries Examinations, September 1999) [14]
17
...
The loan is to be
redeemed at £110 per £100 nominal in 13 years’ time
...
Calculate the issue price per £100 nominal of
the stock
...
The
potential buyer is prepared to buy the stock provided she will obtain a net redemption yield of at least 8% per
annum effective
...

(b) Calculate the net effective annual redemption yield (to the nearest 1% per annum effective) that will be
obtained by the original investor if the loan is sold to the buyer at the price determined in (ii)(a)
...
A loan of nominal amount of £100,000 is to be issued bearing interest payable quarterly in arrear at a rate of 8%
...
a
...

(i) An investor who is liable to income tax at 40% and tax on capital gains at 30% wishes to purchase the entire
loan at the date of issue
...
a?
Exactly 10 months after issue the loan is sold to an investor who pays income tax at 20% and capital gains at
30%
...
a
...

(ii) Explain which price this second investor should pay to achieve a yield of at least 6% p
...

(Institute/Faculty of Actuaries Examinations, April 1997) [17]
19
...
The security will be
redeemed at par on any 1 January from 1 January 2017 to 1 January 2022 inclusive, at the option of the borrower
...
The investor pays tax at 30% on interest income and 25% on capital gains
...

(i) Calculate the price per £100 nominal at which the investor bought the security
...

(iii) Show that the effective net yield that the investor obtained on the investment was between 8% and 9% per
annum
...
A loan of nominal amount £100,000 is to be issued bearing coupons payable quarterly in arrear at a rate of 7%
per annum
...
The date of redemption is at the option of the borrower
...

Calculate the price which the investor should pay to ensure a net effective yield of at least 5% per annum
...
A ﬁxed-interest bond pays annual coupons of 5% per annum in arrear on 1 March each year and is redeemed at par
on 1 March 2025
...
158% per annum effective
...

On 1 March 2012, immediately after the payment of the coupon then due, the gross redemption yield on the bond
was 5% per annum
...

Page 88 Section 3

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

A tax-free investor purchased the bond on 1 March 2007, immediately after payment of the coupon then due, and
sold the bond on 1 March 2012 immediately after payment of the coupon then due
...

(iv) Explain, without doing any further calculations, how your answer to part (iii) would change if the bond were
due to be redeemed on 1 March 2035 (rather than 1 March 2025)
...

(Institute/Faculty of Actuaries Examinations, April 2012) [3+1+2+3=9]
22
...

(i) Calculate the price per £100 nominal to provide a gross redemption yield of 3% per annum convertible half
yearly
...

23
...

Both bonds provide a capital repayment of e100 together with a ﬁnal coupon payment of e6 in exactly one year
...
He believes
that there are four possible outcomes for the bond from country B, shown in the table below
...
1
Capital payment received, but no coupon payment received
0
...
3
Both coupon and capital payments received in full
0
...

(i) Calculate the price of the bond issued by country B to give the same expected return as that for the bond issued
by country A
...

(iii) Explain why the investor might require a higher expected return from the bond issued by country B than from
(Institute/Faculty of Actuaries Examinations, September 2013) [3+1+2=6]
the bond issued by country A
...
1 Calculating the yield
...
The sizes of these
dividend payments are uncertain and depend on the performance of the company
...

Consider the following special case of annual dividends (the results for six-monthly dividend payments are
obtained in the same manner)
...
Suppose dividends are paid annually and dk is the estimated gross dividend in k years time
...

Cash ﬂow
−P
d1
d2
d3

...

Cash ﬂow
−P
d1
d1 (1 + g)
d1 (1 + g)2

...
1a
...
Subsequent
dividends will be paid annually
...
Assume also that dividends subsequent to the fourth grow at the rate of 5% per annum
...

Solution
...
1

5/2
6 × 1
...
13

9/2
6 × 1
...
05

...

Let ν = 1/(1 + i) = 1/1
...
Hence

NPV = 6ν 1/2 + 6(1
...
1)2 ν 5/2 + 6(1
...
1)3 (1
...
1)ν

3/2

2 5/2

+ 6(1
...
1) ν

+ 6(1
...
05
1 − 1
...
2 Other measures
...
Consequently,
other numerical measures for comparing equities are usually used—mainly the dividend yield and the P/E
ratio
...

The P/E ratio or price/earnings ratio is given by
current share price
P/E ratio =
net annual proﬁt per share
The net annual proﬁt per share is the last annual proﬁts of the company divided by the number of shares
...
Then the P/E ratio is 20; i
...
the
shares are on a multiple of 20, or the shares sell at 20 times earnings
...
If the share price rises, the yield falls and the P/E ratio rises
...
(Belief that a company is a growth company implies belief in higher dividends in the future and hence
a higher share price
...

4 Real and Money Rates of Interest
4
...
The real rate of interest is the rate of interest after allowing for inﬂation; the money rate of interest
is the rate of interest without allowing for inﬂation
...
1a
...
Inﬂation is 5% p
...
Find the money and real rates of
interest
...
The money rate of interest is 20% p
...
In real terms, £100 grows to £120/1
...
29 in 1 year
...
29% p
...

Suppose iM denote the money rate of interest, iR denotes the real rate of interest and q denotes the inﬂation
rate, then
iM − q
1 + iM = (1 + q)(1 + iR ) or, equivalently, iR =
(4
...

Suppose £1 grows to £A(t) over the interval (0, t)
...
The rate of interest needed for £1 to grow to £A∗ (t) is the real rate
of interest and is, in this case, less than the money rate of interest
...
The real rate of interest is
then higher than the money rate of interest
...
J
...
1b
...
The cash ﬂow should be adjusted as follows:
a0 = −100

Cash ﬂow

a1 = 8 ×

a2 = 8 ×

Q(0)
Q(1)

This leads to a real rate of interest iR given by
8 × 150
8 × 150 2 108 × 150 3
100 =
νR +
ν +
νR
156
166 R
175
and hence iR = 2
...

tn
Cash ﬂow
0
Ct1
Ct2

...

Q(tn )
The equation for the yield is given by
n
n
∑
∑ Ct
Q(0) tk
k
Ctk
ν = 0 which leads to
ν tk = 0
Q(tk )
Q(tk )
k=1

(4
...
Equation (4
...

4
...
Suppose we assume that the rate of inﬂation will be q%
per annum
...

tn
Cash ﬂow
0
Ct1
Ct2

...

(1 + q)tn
The real yield iR is given by
n
∑ Ct
1
k
ν tk = 0
where νR =
tk R
(1 + q)
1 + iR
k=1

The money yield iM is given by
n
∑

tk
Ctk νM = 0

where νM =

k=1

1
1 + iM

Note that νR = (1 + q)νM which is just another version of equation (4
...

4
...
Suppose we have the series of payments c = {ct1 ,
...
This means that the payment at time tk
will become
Q(tk )
cq,tk = ctk
Q(0)
Hence the NPV of the series of inﬂation adjusted payments cq = {cq,1 ,
...
If iR denotes the real rate of interest corresponding to the
money rate of interest iM , then 1 + iM = (1 + iR )Q(1)/Q(0) and in general (1 + iM )tk = (1 + iR )tk Q(tk )/Q(0)
...

This result generalises as follows: suppose we have both a discrete payment stream, c, and a continuous
payment stream, ρ
...

(4
...

If i0 is the yield of a project when inﬂation is ignored, then NPVi0 (c, ρ) = 0
...
3a) implies that
NPVi1 (cq , ρq ) = 0 where (1 + i1 ) = (1 + i0 )(1 + q)
...

Projects which appear unproﬁtable when interest rates are high, may become proﬁtable when only a small
allowance is made for inﬂation
...
4 Index linked bonds
...
However, sometimes
the calculation is more complicated than this
...
The real yield must be calculated using
the inﬂation index at the times the actual payments are made
...
4a
...
The coupon rate is r% per annum convertible semi-annually
...
If the current price of the bond is P , ﬁnd an equation for the money yield iM
...
The cash ﬂow is as follows:
Time
0
1/2
1
3/2
Cash ﬂow (unadjusted)
−P
f r/2
f r/2
f r/2
Inﬂation Index
Q(0)
Q(1/2)
Q(1)
Q(3/2)
The equation for the money yield, iM is
2n
∑ f r Q(k/2) k/2
Q(n) n
P =
νM + C
ν
2 Q(0)
Q(0) M

...

...
5
Summary
...

• Real rate of interest and money rate of interest: (1 + iM ) = (1 + q)(1 + iR )
...

5 Exercises

(exs5-2
...
On the assumption of constant future price inﬂation of 2
...
Explain
why the real yield to redemption will be lower if the actual constant rate of inﬂation is higher than the assumed 2
...

2
...
The country has experienced deﬂation (negative inﬂation) of 2% per annum
effective during the period
...

(Institute/Faculty of Actuaries Examinations, September 2005) [2]
3
...
A dividend
of £5 per share has just been paid
...
Find the value of the share to the nearest £1,
assuming an effective rate of interest of 8% p
...

(Institute/Faculty of Actuaries Examinations, May 1999) [3]
4
...
A dividend of £2 per
share has just been paid
...
Find the value of the share to the nearest £1 assuming
(Institute/Faculty of Actuaries Examinations, April 1999) [3]
an effective rate of interest of 7% per annum
...

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

(i) Deﬁne the characteristics of a government index-linked bond
...

(Institute/Faculty of Actuaries Examinations, September 2003) [2+2=4]

6
...
Exactly one year later, the investment is worth £11
...
An index of prices
has a value of 112 at the beginning of the investment and 120 at the end of the investment
...
Calculate:
(a) The money rate of return per annum before tax
...

(c) The real rate of return per annum after tax
...
An investor purchased a holding of ordinary shares two months before payment of the next dividend was due
...
The
investor anticipates that dividends will grow at a constant rate of 4% per annum in perpetuity
...

(Institute/Faculty of Actuaries Examinations, April 2003) [4]
8
...
Dividends from the share will be paid
annually
...
The second dividend is expected
to be 8% greater than the ﬁrst dividend and the third dividend is expected to be 7% greater than the second dividend
...

Calculate the present value of this dividend stream at a rate of interest of 7% per annum effective
...
A loan of £20,000 was issued, and was repaid at par after 3 years
...
The value of the retail price index at various times was as follows:
At the date the loan was made
245
...
2
Two years later
282
...
5
(Institute/Faculty of Actuaries Examinations, April 1997) [5]
Calculate the real rate of return earned on the loan
...
An equity pays annual dividends and the next dividend, payable in 10 months time, is expected to be 5p
...
Inﬂation is expected to be 2% per annum throughout
...
5% per annum, convertible half-yearly
...
An investment which pays annual dividends is bought immediately after a dividend payment has been made
...
If the dividend payment expected at the end of the ﬁrst year is d per unit invested and if it is assumed the
investment is held indeﬁnitely, show that the expected real rate of return per annum per unit invested is given by
i′ = (d + g − e)/(1 + e)
...
An ordinary share pays annual dividends
...
It is expected that subsequent dividends will grow at a rate of 5% per annum compound and that
inﬂation will be 3% per annum
...

Calculate the expected effective real rate of return per annum for an investor who purchases the shares
...
A share has a price of 750p and has just paid a dividend of 30p
...
Two analysts agree
that the next expected dividend will be 35p but differ in their estimates of long-term dividend growth
...

(i) Derive a formula for the value of a share in terms of the next expected dividend (d1 ), a constant rate of dividend
growth (g), and the rate of return from the share (i)
...

(Institute/Faculty of Actuaries Examinations, September 1999) [3+3=6]
14
...
The next dividend is expected to be 5p per share and is due in exactly
3 months’ time
...
5% per annum
...

Calculate the effective real rate of return per annum for an investor who purchases the share
...
An investor is considering investing in the shares of a particular company
...
The next
dividend is expected to be d1 , the purchase price of a single share is P , and the annual effective rate of return
expected from the investment is 100i%
...
Dividends are expected to be paid in perpetuity
...
50
...

(Institute/Faculty of Actuaries Examinations, April 2001) [7]
16
...

(ii) The prospective dividend yield from an ordinary share is deﬁned as the next expected dividend divided by the
current price
...
Inﬂation
is expected to be 2% per annum
...
Dividends will be paid annually and the next dividend is expected to be paid in 6 months’ time
...

(Institute/Faculty of Actuaries Examinations, September 2004) [4+4=8]

17
...

Investment A: £10,000 was placed in a special savings account with a 5-year term
...

Investment B: £10,000 was placed in a zero coupon bond with a 5-year term in which the redemption proceeds were
the amount invested multiplied by the ratio of the Retail Price Index for the month two months prior to that in which
redemption fell to the Retail Price Index for the month two months prior to the date of investment; this amount was
further increased by 2 3/4% compound for each complete year money was invested
...

The Retail Price Index at various times was as follows:
November 1992
237
...
0
January 1994
250
...
4
January 1996
266
...
4
November 1997
274
...
6
(i) What is the amount of the annual income yielded by the annuity?
(ii) Calculate the real rate of return per annum earned on investment A and on investment B over the period 15 January 1993 to 15 January 1998
...

(iii) Determine which of the three investments yielded the highest real rate of return per annum over the period
(Institute/Faculty of Actuaries Examinations, April 1998) [1+4+3=8]
15 January 1993 to 15 January 1998
...
An investor purchases a bond 3 months after issue
...
The investor pays tax of 25% on both income and capital gains (with
no relief for indexation)
...

(ii) The real rate of return expected by the investor from the bond is 3% per annum effective
...

(Institute/Faculty of Actuaries Examinations, April, 2002) [6+2=8]
19
...

A ﬁxed interest bond is issued on 1 January 2003 with term of 25 years and is redeemable at 110%
...

An investor, who is liable to tax on income at a rate of 25% and on capital gains at a rate of 30%, bought £10,000
nominal of the stock at issue for £9,900
...

(ii) Without doing any further calculation, explain how and why your answer to (i) would alter if tax were collected
on 1 June instead of 1 April each year
...
J
...
An ordinary share pays annual dividends
...
This dividend is
expected to be £1
...
Dividends are expected to grow at a rate of 5% per annum compound from this level
and are expected to continue in perpetuity
...
The price of the share is £21
...

Calculate the expected effective annual real rate of return for an investor who purchases the share
...
A government issued a number of index-linked bonds on 1 June 2000 which were redeemed on 1 June 2002
...
The actual coupon and redemption payments were indexed according to the increase in the retail price index
between 6 months before the bond issue date and 6 months before the coupon or redemption dates
...
The issue price
was £94 per £100 nominal
...

(ii) The investor is subject to income tax at a rate of 25% and capital gains tax at a rate of 35%
...

(a) Calculate the investor’s capital gains tax liability in respect of this investment
...

(Institute/Faculty of Actuaries Examinations, April 2004) [3+6=9]
22
...
The purchase price paid by the investor was £25,000
...
7
183
...
0
200
...
1%, the following effective rates of return per annum achieved by the investor from
her investment in the annuity:
(a) the real rate of return;
(b) the money rate of return
...

23
...
A dividend of 25p per share has just been paid
...
Thereafter, dividends are expected to grow at 6% per annum
compound in perpetuity
...

(ii) Calculate the present value of the dividend stream described above at a rate of interest of 9% per annum effective
from a holding of 100 ordinary shares
...
20 each
...
He then sells them for £9 immediately after the second dividend is paid
...
5%
during the second year assuming dividends grow as expected
...
On 9 October 1997 an investor, not liable to tax, had the choice of purchasing either:
(A) 7 1/2% Treasury Stock at a price of £107 per £100 nominal repayable at par in 2006
...
The RPI base ﬁgure for indexing
was 69
...
5
...
)
Assume both stocks are redeemable on 8 October 2006; and that both coupons are paid half-yearly in arrear on
8 April and 8 October
...
5% per annum from its latest
known value,
(i) Show that the real yield from stock (A) as at 9 October 1997 is 4% per annum effective
...

(iii) Calculate the real yield as at 9 October 1997 from stock (B)
...

Jan 28, 2016(12:36)

Exercises 5 Page 95

(i) Describe the characteristics of an index-linked government bond
...
Coupons are paid
half-yearly in arrears on 1 January and 1 July each year
...
Interest and capital
payments are indexed by reference to the value of an inﬂation index with a time lag of 8 months
...
0
112
...
2
113
...
An investor, paying tax at the rate of 20% on coupons only, purchased the stock on 1 July 2003,
just after a coupon payment has been made
...

(Institute/Faculty of Actuaries Examinations, September 2006) [3+10=13]
26
...
Coupons are payable
half-yearly in arrears, and the annual nominal coupon rate is 3%
...

A tax-exempt investor purchased the stock at £111 per £100 nominal on 16 September 1999, just after the coupon
payment had been made
...
5
March 1996
112
...
7
September 1999
127
...

(ii) Calculate the effective real annual yield to the investor on 16 September 1999
...

(iii) Without doing any further calculations, explain how your answer to (ii) would alter, if at all, if the inﬂation
index for July 1995 had been more than 110
...
(Institute/Faculty of Actuaries Examinations, April 2001) [17]
27
...
Coupons are payable
half-yearly in arrears, and the annual nominal coupon rate is 4%
...

The retail price index value in September 1996 was 200 and in March 1997 was 206
...
a
...
a
...

(i) (a) Derive the formula for the price of the bond at issue to a tax-exempt investor
...
53%
...
The retail price index
increases continuously at 5% per annum from March 1997
...
Tax is only due if the
real capital gain is positive
...

(Institute/Faculty of Actuaries Examinations, September 1997) [12+7=19]
28
...
60 each, and the company has just paid a dividend of 12p per share
...
The investor estimates the assumed inﬂation rate from equating the price of the share with the present value
of all estimated gross dividend payments using an effective interest rate of 6% per annum
...

(ii) Suppose that the actual inﬂation rate turns out to be 3% per annum effective over the following 12 years, but
that all the investor’s other assumptions are correct
...
You may assume that the investor pays no tax
...
J
...
An investor is interested in purchasing shares in a particular company
...
Future dividends are expected to grow at the rate of 5% per annum
compound
...

(ii) Without doing any further calculations, explain whether the maximum price paid will be higher, lower or the
same if:
(a) after consulting the managers of the company, the investor increases his estimate of the rate of growth of
future dividends to 6% per annum
...

(c) general economic uncertainty means that, whilst the investor still estimates future dividends will grow at
5% per annum, he is now much less sure about the accuracy of this assumption
...
An ordinary share pays dividends on each 31 December
...

The dividend growth is expected to be 3% in 2012, and a further 5% in 2013
...

(i) Calculate the present value of the dividend stream described above at a rate of interest of 8% per annum effective
for an investor holding 100 shares on 1 January 2012
...
20 each on 1 January 2012
...

(ii) Calculate the investor’s expected real rate of return
...
0
112
...
2
113
...
Mrs Jones invests a sum of money for her retirement which is expected to be in 20 years’ time
...
At retirement, the individual
requires sufﬁcient money to purchase an annuity certain of £10,000 per annum for 25 years
...

(i) Calculate the sum of money the individual needs to invest at the beginning of the 20 year period
...

(ii) Calculate the annual effective real return the individual would obtain from the zero coupon bond
...

(iii) Calculate the net annual effective real return to the investor over the 20 year period before the annuity commences
...

32
...

Over the 91 days, an index of consumer prices rises from 220 to 222
...

(Institute/Faculty of Actuaries Examinations, September 2008) [3]
33
...
The client pays tax at 40% on income
and 40% on capital gains
...
1 million per year in arrears for
10 years
...
At the end of the 10 years, the investment of £1 million is returned
...

At the end of 10 years, the accumulated value of the investment is returned to the investor after deduction of capital
gains tax
...
The
index of consumer prices is expected to increase by 4% per annum compound over the period
...

(ii) Explain why the expected rate of return is higher for Investment C than for Investment B and is higher for
Investment B than for Investment A
...
An ordinary share pays annual dividends
...
It is expected that subsequent dividends will grow at a rate of 6% per annum compound and that
inﬂation will be 4% per annum
...

Calculate the expected effective real rate of return per annum for an investor who purchases the share
...
J
...
1 Basic idea
...
The term structure of interest rates is concerned with the analysis of the dependence of interest rates on
the length of the investment
...
2 Spot rates and zero rates; zero coupon bonds
...
A zero coupon bond is
also called a pure discount bond
...

Let Pt denote the price at issue of a £1 zero coupon bond which matures after a time period of length
t years
...
This yield, yt , is sometimes called the t-year spot rate of interest
...
2a)
and hence
−1/t
yt = Pt
−1
Usually, yt ̸= yv when t ̸= v
...
An example
of a yield curve is given in ﬁgure 1
...

0
...
07

Figure 1
...
A Yield Curve
...

Yield curves usually rise gradually because long bonds
have higher yields than short bonds
...
06

0
...

Recall that δ, the nominal rate of interest compounded continuously or the force of interest, is deﬁned by
i = eδ − 1 or δ = log(1 + i) where i is the effective rate of interest per annum
...
2a) becomes
etYt Pt = 1

and hence

1
Yt = − log Pt
t

(1
...
Clearly, yt = eYt − 1
...

Every ﬁxed interest investment can be regarded as a combination of zero coupon bonds as the following example illustrates
...
J
...
J
...
2a
...
, n and has ﬁnal redemption value C at
time n
...
, Pn where P1 , P2 ,
...

(b) Express the current price of the bond in terms of x, C and y1 , y2 ,
...
, yn are the corresponding yields of the bonds
...
, yn } ≤ i ≤ max{y1 , y2 ,
...

Solution
...
, n and one zero coupon bond with length n and maturity value C
...
2b
...
, 10
...
, y10
...
The cash ﬂow is as follows:
Time
Cash ﬂow

0
0

1
8

2
8

3
8

4
8

5
8

6
8

7
8

8
8

9
8

10
108

The price is
10
∑
k=1

100
8
+
(1 + yk )k (1 + y10 )10

1
...
There are few zero coupon bonds and so the zero rate must also be calculated
from the prices of other ﬁnancial instruments
...
To ﬁnd y2 , suppose we observe the price P
of a two year bond with redemption value C, face value f and coupon rate r payable annually
...
Then the value of y3 can be determined from the price of a
three year bond, and so on
...

Suppose bond A is a ten year bond with price PA , coupon rA , face value and redemption value C
...

Without loss of generality, assume rA > rB
...
This portfolio will have a face value of
(rA − rB )C, a price of P = rA PB − rB PA and zero coupon
...

Interest Rate Problems

Jan 28, 2016(12:36)

Section 1 Page 101

1
...
The n-year par yield of a bond is the coupon rate that causes the current bond price
to be equal to its face value, assuming the bond to be redeemed at par
...
Assume the coupon is payable
annually for n years
...

...
This implies that
the n-year par yield, r, satisﬁes 1 = r

n
∑
k=1

1
1
+
k
(1 + yk )
(1 + yn )n

(1
...

The par yields give an alternative measure of the relationship between the yield and the term of an investment
...
Thus the
n-year coupon bias is r − yn where r is the n-year par yield
...
5 Forward interest rates
...
This is called a forward interest rate
...

period of investment

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...
Also £1 grows to £(1 + yT )T at time T
...
Hence
(1 + yT +k )T +k = (1 + yT )T (1 + fT,k )k
In particular
(1 + y2 )2 = (1 + y1 )(1 + f1,1 )
(1 + y3 )3 = (1 + y2 )2 (1 + f2,1 ) = (1 + y1 )(1 + f1,1 )(1 + f2,1 )
and so on
...
5a)
where f0,1 = y1 is the rate per annum for 1 year starting at time 0; f1,1 is the rate per annum for 1 year starting
at time 1, etc
...
5a) are called implied forward
rates as opposed to the market forward rates
...
5a
...
07, or 7%, and the two year zero rate is 0
...
Find the
implied one year forward rate for a period of length 1 year
...
Recall that the “two year zero rate” is the rate per annum for an investment of length two years
...
07 and y2 = 0
...
Using (1 + y2 )2 = (1 + y0 )(1 + f1,1 ) gives f1,1 = 1
...
07 − 1 = 0
...

1
...
For continuous compounding, we have
[
]k
PT
= 1 + fT,k = ekFT,k
PT +k
where the forward force of interest is FT,k = ln(1 + fT,k )
...
2b) on page 99
...

(1
...
J
...
6a
...
Find the implied force of interest for an investment starting in 6 months’ time and lasting
for 6 months
...
The general result is
e(t+k)F0,t+k = etF0,t × ekFt,k or (t + k)F0,t+k = tF0,t + kFt,k
In our case, t = k = 0
...
07 and F0,0
...
06
...
5,0
...
08
...

1
...

• Yields which depend on time and the length of the investment
...
So, for 0 ≤ t ≤ T , let P (t, T ) denote the value at time t of a zero coupon bond with face value 1 and
maturing at time T
...

This function P of two variables will be complicated: for example, the prices of 10 year bonds and 15 year
bonds will tend to move together in the short term
...
fs Corresponding to this function of two variables, we have the yield to maturity, y(t, T ) deﬁned
by
for all 0 ≤ t ≤ T
...
7a)
[1 + y(t, T )]T −t P (t, T ) = 1
Note that y(t, T ) denotes the yield to maturity from a zero coupon bond that starts at time t and ends at time T
...
Figure (1
...

This model reduces to the previous model if we assume the price just depends on the length of time between
issue and maturity and not on the date of issue
...

Then set Pt = P (0, t)
...
Set yt = y(0, t)
...

With continuous compounding, we have
ln P (t, T )
T −t
• Forward rates
...

e(T −t)Y (t,T ) P (t, T ) = 1

where

eY (t,T ) = 1 + y(t, T )

and so

Y (t, T ) = −

period of investment

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...

...
Hence the investment matures at time T + k
...

The discrete time forward rate can be related to the spot rate as follows: suppose we invest £1 at time t for
T − t years and agree to invest the accumulated amount at time T for a further k years
...
This amount will accumulate to £ [1 + y(t, T )]T −t 1 + ft,T,k at
time T + k
...
Hence
[
]k
(1
...
7a)
[
]k [1 + y(t, T + k)]T +k−t
P (t, T )
1 + ft,T,k =
=
T −t
P (t, T + k)
[1 + y(t, T )]

Interest Rate Problems

Jan 28, 2016(12:36)

Section 1 Page 103

More generally, repeated applications of equation (1
...
7c)

Note that ft,t,1 = y(t, t + 1) is the one-year spot interest rate—it is the one-year riskless rate prevailing at time t
for repayment at time t + 1
...
7c) are called implied forward rates as opposed to
the market forward rates
...
Suppose ft,T,k = f0,T,k for all t; i
...
the forward rate just depends on the start
and length of the investment and not when the agreement is made
...
The model then reduces to the simpler model at the start of this chapter
...
For continuous compounding, we have
[
]k
P (t, T )
= 1 + ft,T,k = ekFt,T,k
P (t, T + k)
where the forward force of interest is Ft,T,k = ln(1 + ft,T,k )
...
7a)
...

(1
...
The instantaneous forward rate at time t for an investment commencing at
time T is deﬁned by
Ft,T = lim Ft,T,k
k→0

Using equation (1
...
7e)

Conversely, using P (T, T ) = 1 and integrating equation (1
...

Note that any one of the functions P (t, T ), Y (t, T ) or Ft,T determines the other two
...
Hence, rt , which is equal to Ft,t , is the instantaneous
forward rate at time t for an investment commencing at time t
...

Page 104 Section 1

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

0
...
07

Figure 1
...
An Increasing Yield Curve
...
06

0
...
07

yield

0
...
05

Figure 1
...
A Decreasing Yield Curve
...
04

0
...
07

yield

0
...
05

Figure 1
...
A Humped Yield Curve
...
04

0
...
8 Explaining the term structure of interest rates
...

• Decreasing yield curve
...

• Increasing yield curve
...
This is the most common
shape
...
Long term bonds have lower yields than short term bonds, but the very short term
bonds (less than 1 year) have lower yields than 1 year bonds
...
This theory postulates that the relative attraction of long and short term bonds
depends on the expectations of future movements in interest rates
...
Similarly, if investors expect interest rates to rise, then long term bonds will become
less attractive and so lead to an increasing yield curve
...
This theory asserts that investors usually prefer short term investments over long
term ones
...
Hence purchasing a short
term bond lessens the risk
...

Interest Rate Problems

Jan 28, 2016(12:36)

Exercises 2 Page 105

(3) Market Segmentation
...

Pension funds are interested in long term bonds because their liabilities are long term
...

1
...
The simple model: interest rates constant in time
...
This is the effective interest rate p
...
for an investment
which lasts for t years
...
Hence
Pt (1 + yt )t = 1
• ft,k denotes the forward rate for an investment starting at time t for a period of k years
...

• Ft,k denotes the forward force of interest for an investment starting at time t for k years
...
By deﬁnition
d
1 dP
ln(Pt ) − ln(Pt+k )
Ft = lim Ft,k = lim ln(1 + ft,k ) = lim
= − ln Pt = −
k→0
k→0
k→0
k
dt
P dt
Conversely

)
( ∫ t
Ft du
Pt = exp −
0

The more sophisticated model: interest rates varying with time
...

• P (t, T ) denotes the price in £ at time t of a £1 zero coupon bond which matures at time T
...

• ft,T,k denotes the forward rate per annum at time t of an investment starting at time T and maturing
at time T + k
...

Par yield of a bond
...
tex)

1
...
a
...
a
...
a
...
The following n-year spot rates apply at time t = 0:
1 year spot rate of interest: 4 1/2% per annum effective
2 year spot rate of interest: 5% per annum effective
3 year spot rate of interest: 5 1/2% per annum effective
Calculate the 2 year forward rate of interest from time t = 1 expressed as an annual effective rate of interest
...
J
...
The 1-year forward rates for transactions beginning at times t = 0, 1, 2 are fi , where
f0 = 0
...
065
f2 = 0
...

(Institute/Faculty of Actuaries Examinations, September 1997) [3]
4
...
You are
given:
f0,1 = 6
...
5% per annum; f1,2 = 6
...

Determine the 3-year par yield
...

(i) The one-year forward rate applying at a particular point in time t is deﬁned as ft,1
...

(ii) Deﬁne the instantaneous forward rate Ft
...
In a particular bond market, the two-year par yield at time t = 0 is 4
...
40 per £100
nominal
...

(Institute/Faculty of Actuaries Examinations, April 2004) [6]
7
...
25 + 0
...

(a) Calculate the 2-year forward rate of interest from time t = 3 expressed as an annual effective rate of interest
...

(c) Without explicitly calculating the gross redemption yield of a 4-year bond paying annual coupons of 3
...

(Institute/Faculty of Actuaries Examinations, April 2006, adapted) [7]
8
...

(ii) Calculate the two-year forward rate of interest at time t = 3
...

(Institute/Faculty of Actuaries Examinations, April 1998) [2+2+4=8]
9
...
04 + 0
...

(b) Calculate the accumulated value of a unit sum of money accumulated from time t = 0 to time t = 9
...

(ii) Using your results from (i), or otherwise, calculate:
(a) The 8-year spot rate of interest from time t = 0 to time t = 8
...

(c) f8,1 , where f8,1 is the one-year forward rate of interest from time t = 8
...
Three bonds paying annual coupons in arrears of 7% and redeemable at 105 per £100 nominal reach their redemption
dates in exactly one, two and three years’ time, respectively
...

(i) Determine the gross redemption yield of the 3-year bond
...

(Institute/Faculty of Actuaries Examinations, September 2002) [3+5=8]
11
...

If at time t = 0 he invests £1,000 for 2 years, he will receive £1,118 at time t = 2
...
However, if at time t = 0 he
agrees to invest £1,000 at time t = 1 for one year, he will receive £1,058 at time t = 2
...

(b) The two year spot rate at time t = 0
...

(ii) Calculate the three year par yield at time t = 0 in this market
...
The n-year spot rate of interest, yn is given by:

n
for n = 1, 2 and 3
...

(ii) Assuming that coupon and capital payments may be discounted using the same discount factors, and that no
arbitrage applies, calculate:
(a) The price at time t = 0 per £100 nominal of a bond which pays annual coupons of 3% in arrears and is
redeemed at 110% after 3 years
...

yn = 0
...
The forward rate from time t − 1 to time t has the following values:
f0,1 = 4
...
5%, f2,3 = 4
...

(ii) Explain why the bond with a higher coupon would have a lower gross redemption yield, for the same term to
(Institute/Faculty of Actuaries Examinations, April, 2002) [7+2=9]
redemption
...
Bonds paying annual coupons of 6% annually in arrears and redeemable at par will be redeemed in exactly one year,
two years and three years respectively
...

(i) Determine the gross redemption yield of the 3 year bond
...

(iii) Calculate f0,1 , f1,2 and f2,3 where ft−1,t is the forward interest rate from time t − 1 to t
...
For a particular bond market, zero coupon bonds redeemable at par are priced as follows:
bonds redeemable in exactly one year are priced at 97;
bonds redeemable in exactly two years are priced at 93;
bonds redeemable in exactly three years are priced at 88;
and bonds redeemable in exactly four years are priced at 83
...

(ii) Explain why the four-year spot rate is greater than the rate of return from a bond redeemable at par in exactly
4 years’ time paying a coupon of 4% annually in arrears
...

(Institute/Faculty of Actuaries Examinations, September 2004) [8+2+2=12]
16
...

The gross redemption yield from a one year bond with a 6% annual coupon is 6% per annum effective; the gross
redemption yield from a 2 year bond with a 6% annual coupon is 6
...
6% per annum effective
...

(i) (a) Calculate i1 , i2 and i3 assuming no arbitrage
...

(ii) Explain why the forward rates increase more rapidly with term than the spot rates
...

(i) (a) Explain what is meant by the “expectations theory” explanation for the shape of the yield curve
...

(ii) Short-term, one-year annual effective interest rates are currently 10%; they are expected to be 9% in one year’s
time, 8% in two year’s time, 7% in three years’ time and to remain at that level thereafter indeﬁnitely
...

(b) Draw a rough plot of the yield curve for zero coupon bonds using the data from part (ii)(a)
...
)
(c) Explain why the gross redemption yield curve for coupon paying bonds will slope down with a less steep
gradient than the zero coupon yield curve
...
J
...
In a particular bond market, n-year spot rates per annum can be approximated by the function 0
...
04e−0
...

Calculate:
(i) The price per unit nominal of a zero coupon bond with term 9 years
...

(iii) The 3-year par yield
...
The annual effective forward rate applicable over the period t to t + r is deﬁned as ft,r where t and r are measured
in years
...
25%, f2,1 = 4
...
Calculate the following:
(i) f3,1
(ii) All possible zero coupon (spot) yields that the above information allows you to calculate
...

(iv) Explain why the gross redemption yield from the 4-year bond is lower than the 1-year forward rate up to
time 4, f3,1
...
The n-year spot rate of interest, in , is given by in = a − bn for n = 1, 2 and 3, and where a and b are constants
...
1% per annum effective and 6
...
The 4-year par yield is 7% per annum
...

(Institute/Faculty of Actuaries Examinations, April 2008) [4+5=9]
21
...
The price of each bond is £103 per £100
nominal
...

(ii) Calculate, to 3 decimal places, all possible spot rates implied by the information given, as annual effective rates
of interest
...

22
...
06 − 0
...
1n
...
Show all workings
...

23
...

(ii) (a) State the characteristics of a certiﬁcate of deposit
...
A one-month certiﬁcate of deposit provides a rate of return of 12% per annum convertible monthly
...

Calculate the forward rate of interest per annum convertible monthly in the second month, assuming no
arbitrage
...
Three bonds each paying annual coupons in arrear of 6% and redeemable at £103 per £100 nominal reach their
redemption dates in exactly one, two and three years’ time respectively
...

(i) Calculate the gross redemption yield of the 3-year bond
...

(Institute/Faculty of Actuaries Examinations, April 2013) [3+3=6]
25
...
05 + 0
...

(i) Calculate the accumulated value of a unit sum of money:
(a) accumulated from time t = 0 to time t = 7;
(b) accumulated from time t = 0 to time t = 6;
(c) accumulated from time t = 6 to time t = 7
...

(iii) Explain why your answer to part (ii)(c) is higher than your answer to part (ii)(a)
...
01t+0
...

Interest Rate Problems

Jan 28, 2016(12:36)

Section 3 Page 109

26
...

Over time t (measured in years), the spot rate of interest is equal to:
i = 0
...

An insurance company in this country has a group of annuity policies which involve making payments of £1m per annum for 4 years and £2m per annum in the ﬁfth year
...

(i) Calculate the value of the company’s liabilities
...

(iii) Calculate the forward rate of interest from time t = 3
...
5
...
1 Background
...
Assume that VA ≥ VL at time 0
...
If the rate of interest rises, then both VA and
VL will fall
...
Clearly the institution wants to
ensure that VA ≥ VL however the interest rate behaves
...
The concept of immunisation is a strategy for immunising a portfolio to changes in interest rates
...
2 Duration or Macaulay duration or discounted mean term
...

...
Suppose the force of interest is δ and so the annual
effective rate of interest is i = eδ − 1
...
2a)

k=1

Hence dM (i) is a weighted mean of the values tk where the weights are ctk /(1 + i)tk
...

Now 1 + i = eδ
...
2b)
dδ
di
dδ di
dδ
P di
Clearly dM (i) is a measure of the sensitivity of P to the interest rate i
...
2a
...

(a) Find the Macaulay duration, dM (i)
...

(c) Find the Macaulay duration of an n-year zero coupon bond with redemption value of £100
...
(a) The cash ﬂow is as follows:
Time
Cash ﬂow

0
−P

1
fr

2
fr

3
fr

...

n−1
fr

n
C + fr

Page 110 Section 3

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

Let ν = 1/(1 + i) and α = f r/C
...
2c)

k=1

αν(1 − ν n − n(1 − ν)ν n ) + (1 − ν)2 nν n
=
αν(1 − ν)(1 − ν n ) + (1 − ν)2 ν n
nαν + 1 − n(1 − ν)
1
− νn
=
1−ν
αν(1 − ν n ) + (1 − ν)ν n
1 n(i − α) − (1 + i)
=1+ +
i α[(1 + i)n − 1] + i
after some algebra
...
Hence
dP
f r(Ia) n + nCν n
= f r(1 + 2ν + · · · + nν n−1 ) + nCν n−1 =
dν
ν
and hence
−

(3
...
2e)

[
]
dP
dP dν
dP
=−
= ν2
= ν f r(Ia) n + nCν n
di
dν di
dν

and so on
...
Substituting these equalities in equation (3
...
Hence
]
[ n
n
∑
1 − νn
1 ∑
k
n
kν k + nν n =
kf rν + Cnν = i
dM (i) =
P
1−ν
k=1

k=1

Alternatively, just substitute α = i = (1 − ν)/ν in equation (3
...

(c) The only payment is £100 at time n
...
2b
...
Let i denote the effective annual yield and deﬁne i(2) in the usual way by (1 + i(2) /2)2 = 1 + i
...

(b) Show that
(
)
i(2) 1 dP
dM (i) = − 1 +
2 P di(2)
Solution
...
2a) gives
[
]
2n
1 f r ∑ k/2
Cn
dM (i) =
+
P 2
(1 + i)k/2 (1 + i)n
k=1

For part (b), note that
di
i(2)
=1+
di(2)
2
Using this result in equation (3
...
3 Effective duration or modiﬁed duration or volatility
...
Then:
n
∑ ct
k
P =
(1 + i)tk
k=1

The effective duration, d(i) or volatility or modiﬁed duration of the cash ﬂow is deﬁned to be

Interest Rate Problems

Jan 28, 2016(12:36)

Section 3 Page 111

n
1 dP
1 ∑ tk ctk
d(i) = −
=
P di
P
(1 + i)tk +1

(3
...

Thus the effective duration is also a measure of the rate of change of the net present value with i which does
not depend on the size of the net present value
...

Example 3
...
Suppose the modiﬁed duration of a bond is equal to 8
...
(A basis point is 0
...
) Find the approximate change in the price of the bond
...
The approximate change is −ϵd(i) = −0
...
008
...
8% approximately
...
4 Some results about duration
...
2a)
...

• The duration of a bond increases as the coupon rate decreases (other things being equal)—by equation (3
...

Duration is a measure of the riskiness of a bond; it is important for three reasons:
• it is a summary measure of the average payments which are due;
• it is a measure of the sensitivity of the price to interest rate changes;
• it is useful concept for immunising portfolios against interest rate changes
...
4a
...

...
Now P (i) = ν + ν 2 + ν 3 + · · · = ν/(1 − ν)
...
4b
...
P (i) = ν + ν 2 + · · · + ν n and
ν + 2ν 2 + · · · + nν n
1
nν n
1+i
n
dM (i) =
=
−
=
−
2 + · · · + νn
ν+ν
1−ν
1 − νn
i
(1 + i)n − 1

1
i

Page 112 Section 3

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

3
...
The convexity of the cash ﬂow above is deﬁned to be
c(i) =

n
1 d2 P
1 ∑ tk (tk + 1)ctk
=
P di2
P
(1 + i)tk +2
k=1

The duration (ﬁrst derivative) and convexity (second derivative) together give information about the local dependence of P on i; they show how P changes when there is a small change in i
...
This is illustrated in ﬁgure (3
...
If two bonds have the the same price, the same
yield and the same same duration but different convexities, then the investor should prefer the bond with the
higher convexity—hence they should not have the same price!
price, P
A
C

P0
D
B
dP
di

i0

at i = i0

yield, i

Figure 3
...
Suppose the price of a bond is P0 when the yield is i0
...

(wmf/convexity,0pt,0pt)

Finally, note that if the yield falls by the amount ∆, then the rise in the price is larger than the fall in the price
if the yield increases by the amount ∆
...
6 Immunisation
...
Let dA (i)
denote the effective duration and cA (i) denote the convexity
...
Let dL (i) denote the effective duration and cL (i) denote the
convexity
...
The fund is immunised against small changes of
size ϵ in the interest rate if VA (i0 + ϵ) ≥ VL (i0 + ϵ)
...
Taylor’s theorem gives
S(i0 + ϵ) = S(i0 ) + ϵS ′ (i0 ) +
Using S(i0 ) = 0 shows that

ϵ2 ′′
S (i0 ) + · · ·
2

Interest Rate Problems

Jan 28, 2016(12:36)

S(i0 + ϵ) = ϵS ′ (i0 ) +

Section 3 Page 113

ϵ2 ′′
S (i0 ) + · · ·
2

Dividing by VA (i0 ) = VL (i0 ) gives
S(i0 + ϵ)
S ′ (i0 ) ϵ2 S ′′ (i0 )
=ϵ
+
+ ···
VA (i0 )
VA (i0 ) 2 VA (i0 )
′
′
Now if we insist effective durations are equal at i0 , we must have VA (i0 ) = VL (i0 ) and so
S ′ (i0 )
=0
VA (i0 )

and hence

ϵ2
S(i0 + ϵ) ϵ2 S ′′ (i0 )
=
+ · · · = [cA (i) − cL (i)] > 0
VA (i0 )
2 VA (i0 )
2

provided that cA (i0 ) > cL (i0 )

We have shown that if
(1) VA (i0 ) = VL (i0 ), the net present values are the same
(2) dA (i0 ) = dL (i0 ), the effective durations are the same
(3) cA (i0 ) > cL (i0 ), the convexity of the assets is greater than the convexity of the liabilities
then we have Redington immunisation at interest rate i0
...
)
Redington immunisation is not usually a feasible method for organising assets
...

3
...
Recall equation (3
...

dδ
di

It also leads to
di dP
d2 P
d dP
dP
di d2 P
=
+ (1 + i)
= (1 + i)
+ (1 + i)
dδ 2
dδ di
dδ di
di
dδ di2
and hence
1 d2 P
+ dM (i) = (1 + i)2 c(i)
P dδ 2
Hence the three conditions for Redington immunisation are equivalent to:
(1) VA (i0 ) = VL (i0 ), the net present values are the same
(2) dM (A) (i0 ) = dM (L) (i0 ), the Macaulay durations are the same
(3) cδA (i0 ) > cδL (i0 ), the convexity of the assets is greater than the convexity of the liabilities where now the
1 2
convexity is cδ = P d P
...

k=1

Page 114 Exercises 4

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

3
...

• Macaulay duration or discounted mean term
...

n
1 ∑ tk ctk
1 dP
1 dP
=−
= −(1 + i)
dM (i) =
tk
P
(1 + i)
P dδ
P di
k=1

• Effective duration or modiﬁed duration or volatility
...

c(i) =

n
1 d2 P
1 ∑ tk (tk + 1)ctk
=
P di2
P
(1 + i)tk +2
k=1

• Immunisation
...

Alternative conditions are:
n
n
n
n
n
n
∑ lt
∑ tk at
∑ tk lt
∑ t2 at
∑ t2 lt
∑ at
k
k
k
k
k k
k k
=
=
>
(1 + i)tk
(1 + i)tk
(1 + i)tk
(1 + i)tk
(1 + i)tk
(1 + i)tk
k=1

k=1

k=1

k=1

k=1

4 Exercises

k=1

(exs6-2
...
A particular bond has just been issued and pays coupons of 10% per annum paid half yearly in arrears and is redeemed
at par after 10 years
...

(Institute/Faculty of Actuaries Examinations, September 1999) [3]
2
...
2a), the Macaulay duration can be regarded as the expectation of a random variable, X, with the
following distribution:
ct ν tk
X = tk
with probability ∑n k
for k = 1, 2,
...

tk
k=1 ctk ν
Show that the derivative of dM (i) with respect to i is equal to −νσ 2 where σ 2 = var[X]
...

3
...
The payment
stream will be at a rate of £10 million per annum throughout the period
...

(Institute/Faculty of Actuaries Examinations, September 2004) [4]
4
...
State the conditions that need
to be met if the investor is to be immunised from small, uniform changes in the rate of interest
...

(c) A perpetuity pays annual coupons in arrears
...
(Institute/Faculty of Actuaries Examinations, September 2004) [6]
5
...
They discover that the company
has liabilities of £15 million due in 13 years’ time and £10 million in 25 years’ time
...
425 million in 12 years’ time and the other paying £12
...

The current interest rate is 8% per annum effective
...

(Institute/Faculty of Actuaries Examinations, April 2003) [8]
2

See also page 81 of Bond Pricing and Portfolio Analysis by Olivier de La Grandville
...

Jan 28, 2016(12:36)

Exercises 4 Page 115

(i) State the features of a eurobond
...
Coupons are paid
annually in arrear
...
The rate of return from the
bond is 5% per annum effective
...

(b) Calculate the duration of the bond
...
An insurance company has liabilities of £10 million due in 10 years’ time and £20 million due in 15 years’ time,
and assets consisting of two zero-coupon bonds, one paying £7
...
834 million in 25 years’ time
...

(i) Show that Redington’s ﬁrst two conditions for immunisation against small changes in the rate of interest are
satisﬁed for this insurance company
...
5% per annum effective
...

(Institute/Faculty of Actuaries Examinations, April 2004) [5+2+2=9]
8
...
Its assets
consist of two zero coupon bonds, one paying £66,850 in 4 years’ time and the other paying £X in n years’ time
...

(i) Calculate the discounted mean term and convexity of the liabilities
...

9
...
The
manager of the fund has sold the assets previously held and is creating a new portfolio by investing in the zerocoupon bond market
...

(i) Explain whether it is possible for the manager to immunise the fund against small changes in the rate of interest
by purchasing a single zero-coupon bond
...
43 million in 15 years’ time and the other
paying £7
...
The current interest rate is 7% per annum effective
...

(Institute/Faculty of Actuaries Examinations, April 2005) [2+8=10]
10
...
The fund holds two investments, X and Y
...

Investment Y is a zero-coupon bond which pays a lump sum of £R at the end of n years (where n is not necessarily
an integer)
...

(i) Investigate whether values of £R and n can be found which ensure that the fund is immunised against small
changes in the interest rate
...
275 at 8%
...
Calculate the revised present value of
the assets and liabilities of the fund
...

(Institute/Faculty of Actuaries Examinations, April 2006) [8+4=12]
11
...
, followed by £0
...
The government
bond with the longest duration in which it can invest its funds pays a coupon of 10% per annum in arrears and is
redeemed at par in 15 years’ time
...

(i) (a) Calculate the duration of the insurance company’s liabilities at a rate of interest of 6% per annum
...

(ii) (a) Explain why the insurance company cannot immunise its liabilities by purchasing government bonds
...
Explain why a loss would be made
...
J
...

(i) An investment provides income of £1 million payable at the end of each year for the next 10 years
...
If the interest rate is 7% per annum effective, show that the “discounted mean term” (or
“Macaulay duration”) of the investment is 4
...

(ii) An investment company has liabilities of £7 million due in 5 years’ time and £8 million due in 8 years’ time
...
Investment A is the investment described in part (i) and Investment B
is a zero coupon bond which pays £X at the end of n years (where n is not necessarily an integer)
...
Investigate whether values of £X and n can be found which ensure
that the investment company is immunised against small changes in the interest rate
...
451 at 7%
...

(i) Prove that

a n − nν n
¨
i
A government bond pays a coupon half-yearly in arrears of £10 per annum
...
The gross redemption yield from the bond is 6% per annum convertible half-yearly
...

(iii) Explain why the duration of the bond would be longer if the coupon rate were £8 per annum instead of £10 per
annum
...
A ﬁxed interest security was issued on 1 January in a given year
...
The security is redeemable at 110% 20 years after issue
...
Income tax is paid on coupons at the end of the
calendar year in which the coupon is received
...

(i) Calculate the price paid by the investor to give a net rate of return of 6% per annum effective
...

(Institute/Faculty of Actuaries Examinations, September 2004) [6+8=14]
15
...
It wishes to invest in two ﬁxed interest
securities in order to immunise its liabilities
...
Security B has a coupon rate of 3% per annum and a term to redemption of 25 years
...

(i) Calculate the present value of the liabilities at a rate of interest of 7% per annum effective
...

(iii) Calculate the nominal amount of each security that should be purchased so that both the present value and
discounted mean term of assets and liabilities are equal
...

(Institute/Faculty of Actuaries Examinations, September 2006) [2+4+7+2=15]
16
...
4t) at the end of year t, for t equal to 5, 10, 15, 20 and 25
...

An amount equal to the total present value of the liabilities is immediately invested in 2 stocks:
Stock A pays coupons of 5% per annum annually in arrears and is redeemable in 26 years at par
...

The gross redemption yield on both stocks is the same as that used to value the liabilities
...

(ii) Calculate the discounted mean term of the liabilities
...

(Institute/Faculty of Actuaries Examinations, September 2002) [3+3+9=15]
17
...
Cyber plc is assumed to pay its ﬁrst annual dividend
in exactly 6 years
...
After the sixth year, annual dividends are
assumed to grow at 10% per annum compound for a further 6 years
...
Boring plc is expected to pay an annual dividend of 4p per share in exactly one
year
...
5% per annum compound in perpetuity
...

(i) (a) Calculate the value of Cyber plc
...

Interest Rate Problems

Jan 28, 2016(12:36)

Exercises 4 Page 117

There is a general rise in interest rates and the analyst decides it is appropriate to increase the valuation rate of interest
to 7% per annum effective
...

(iii) Explain your answer to part (ii) in terms of duration
...
Let {Ctk } denote a series of cash ﬂows at times tk for k = 1, 2,
...

(i) (a) Deﬁne the volatility of the cash ﬂow series, and derive a formula expressing the volatility in terms of tk ,
Ctk and ν
...

(ii) A loan stock issued on 1 March 1997 has coupons payable annually in arrear at 8% p
...
Capital is to be redeemed
at par 10 years from the date of issue
...
a
...
71
...
247 years
...
a
...
a
...
53, state with reasons whether the investor
will be immunised against small movements in interest rates on that date
...
a
...

(Institute/Faculty of Actuaries Examinations, April 1997) [4+10+3=17]
19
...

The investor currently holds an amount of cash equal to the present value of these two liabilities valued at an effective
rate of interest of 7% per annum
...
The market prices of both bonds are calculated at an effective rate of
interest of 7% per annum
...
The remainder of the cash is invested in bond Y
...

(ii) Determine the term needed for bond Y and the redemption amount payable at the maturity date
...

20
...
, n and P (i) denote the present value of
=
n
these payments at an effective interest rate i so that P (i) = k=1 Ctk /(1 + i)tk
...

(b) Deﬁne the volatility (or effective duration ) of the cash ﬂows in terms of P (i) and show that for this series
of cash ﬂows the discounted mean term = volatility × (1 + i)
...

The fund has earmarked cash assets exactly equal to the present value of the payments and the fund manager
wants to invest these in two zero coupon bonds: Bond A which is repayable at the end of 5 years and Bond B
which is repayable at the end of 20 years
...
To achieve this, how
much should the manager invest in each of the bonds, given that an effective rate of interest of 7% per annum is
used to value both assets and liabilities?
(iii) State without doing any further calculations, what further condition would be required to ensure that the fund is
immunised against possible losses due to small changes in the effective rate of interest
...
A pension fund expects to make payments of £100,000 per annum at the end of each of the next 5 years
...

The rate of interest is 5% per annum effective
...

(b) Show that the duration of the liabilities is 2
...

(ii) Calculate the nominal amounts of the 2 zero coupon bonds which must be purchased if the pension fund is to
equate the present value and duration of assets and liabilities
...
J
...

(b) Without calculating the convexity of the liabilities, comment on whether you think Redington’s immunisation has been achieved
...
A variable annuity, payable annually in arrears for n years is such that the payment at the end of year t is t2
...
The fund
values these payments using an effective interest rate of 7% per annum
...

The fund invests an amount equal to the present value of these liabilities in the following two assets:
(A) a zero coupon bond redeemable in 25 years, and
(B) a ﬁxed interest bond redeemable at par in 12 years’ time which pays a coupon of 8% per annum annually in
arrears
...

(b) Calculate the amount of cash that should be invested in each asset if the duration of the assets is to equal that of
the liabilities
...
An investor purchases a ﬁxed interest security
...
a
...
The investor is subject to tax on the coupon payments at a rate of 25%
...
a
...
76%
...
a
...

(ii) The investor has two liabilities
...
a
...
The amount of
each of the two liabilities is the same
...

(a) Show that the ﬁrst liability must be due in 9
...

(b) Calculate the convexity of the total of the two liabilities described above
...

(iii) One year after purchasing the ﬁxed interest security, the price at which the security can be sold is still at a level
that will yield a rate of return of 10% p
...
effective
...

Calculate the forward price based on a “risk-free” rate of return of 10% p
...
effective and no arbitrage
...
A pension fund has liabilities to pay pensions each year for the next 60 years
...
25m at the end of the third year and so on, increasing
by 5% each year
...
The bonds mature in 20 years’
time and pay an annual coupon of 4% in arrears
...

(ii) Calculate the nominal amount of the bond that the fund needs to hold so that the present value of the assets is
equal to the present value of the liabilities
...

(iv) Calculate the duration of the assets
...
5% per annum effective
...

(i) An investor is considering the purchase of an annuity, payable annually in arrear for 20 years
...
Using a rate of interest of 8% per annum effective, calculate the duration of the annuity when:
(a) the payments remain level over the term;
(b) the payments increase at a rate of 8% per annum compound
...

(Institute/Faculty of Actuaries Examinations, April 2008) [6+2=8]

26
...

The ﬁrst investment option involves setting up a branch in a foreign country
...
25 million, followed by investments of £0
...
2 million at the end of two years,
£0
...
The

Interest Rate Problems

Jan 28, 2016(12:36)

Exercises 4 Page 119

investment will provide annual payments of £0
...

There will be an additional incoming cash ﬂow of £5 million at the end of the 27th year
...
20 per share
...
05p per share in two years and so
on, increasing by 5% per annum compound
...
64 per share
...

(ii) Without doing any further calculations, determine which option has the higher discounted mean term at a rate
of interest of 7% per annum effective
...
A company has a liability of £400,000 due in 10 years’ time
...

This is also the interest rate on which current market prices are calculated and the interest rate earned on cash
...

(i) Calculate the nominal amounts of the zero-coupon bond and the ﬁxed-interest stock which should be purchased
to satisfy Redington’s ﬁrst two conditions for immunisation
...

(iii) Explain whether the company would be immunised against small changes in the rate of interest if the quantities
of stock in part (i) are purchased
...
A company has the following liabilities:
• annuity payments of £200,000 per annum to be paid annually in arrear for the next 20 years;
• a lump sum of £300,000 to be paid in 15 years
...

Security A has a coupon rate of 9% per annum and a term to redemption of 12 years
...

Both securities are redeemable at par and pay coupons annually in arrear
...

(i) Calculate the present value of the liabilities
...

(iii) Calculate the nominal amount of each security that should be purchased so that Redington’s ﬁrst two conditions
for immunisation against small changes in the rate of interest are satisﬁed for this company
...

29
...

The government of a particular country has just issued ﬁve bonds with terms to redemption of one, two, three, four
and ﬁve years respectively
...

(ii) Calculate the duration of the one year, three year and ﬁve year bonds at a gross redemption yield of 5% per annum effective
...

Four years after issue, immediately after the coupon payment then due the government is anticipating problems
servicing its remaining debt
...

Option 2: the redemption of the bond is deferred for 7 years from the original redemption date and the coupon rate
is reduced to 1% per annum for the remainder of the existing term and the whole of the extended term
...

(iv) Calculate the effective rate of return per annum from Option 1 and 2 over the total life of the bond and determine
which would provide the higher rate of return
...

(Institute/Faculty of Actuaries Examinations, September 2012) [7+6+2+6+2=23]

30
...

(i) Explain why comparing the two discounted payback periods or comparing the two payback periods are not
generally appropriate ways to choose between two investment projects
...
J
...
The incoming cash ﬂows from the two projects are
as follows:
Project A
...
5 million
...
55 million
...
Assume that all cash ﬂows are received in the middle of the year
...
Project B generates cash ﬂows of £0
...
Assume that all cash ﬂows are
received continuously throughout the year
...

(b) Calculate the discounted payback period from project B at a rate of interest of 4% per annum effective
...

(iv) Calculate the duration of the incoming cash ﬂows from projects A and B at a rate of interest of 4% per annum
effective
...

31
...
The
assets consist of two zero-coupon bonds, one paying £X in 5 years’ time and the other paying £Y in 20 years’ time
...
The insurance company wishes to ensure that it is immunised
against small changes in the rate of interest
...

(ii) Demonstrate that the third condition for Redington’s immunisation is also satisﬁed
...
A pension fund has liabilities to meet annuities payable in arrear for 40 years at a rate of £10 million per annum
...
The ﬁrst security pays annual coupons of 5% and is redeemed
at par in exactly 10 years’ time
...
The present value of the assets in the pension fund is equal to the present value of the liabilities of the
fund and exactly half the assets are invested in each security
...

(i) Calculate the present value of the liabilities of the fund
...

(iii) Calculate the duration of the liabilities of the pension fund
...

(v) Without further calculations, explain whether the pension fund will make a proﬁt or loss if interest rates fall
uniformly by 1
...

(Institute/Faculty of Actuaries Examinations, September 2013) [1+6+3+4+2=16]
33
...

(i) Calculate the price of the security at issue
...

(iii) Explain how your answer to part (ii) would differ if the annual coupons on the security were 3% instead of 9%
...

(b) Explain the usefulness of effective duration for an investor who expects to sell the security over the next
few months
...
1 One investment
...
Suppose further that the interest rates
are i1 for time period 1,
...
Let £Sn denote the accumulated amount at time n
...
, n
...
, in are independent—this is clearly an unreasonable assumption!!! Then
n
n
∏
∏
E[Sn ] =
E[1 + ij ] =
(1 + µj )
j=1
2
E[Sn ]

=

n
∏

j=1

E[1 + 2ij +

i2 ]
j

j=1

and an expression for var(Sn ) can be obtained by using

=

n
∏
j=1

2
(1 + 2µj + σj + µ2 )
j

Interest Rate Problems

Jan 28, 2016(12:36)

Section 5 Page 121

2
var(Sn ) = E[Sn ] − E[Sn ]2
2
2
If µj = µ and σj = σ 2 for j = 1,
...
2 Repeated investments
...

n−1
n
Cash ﬂow, c
1
1
1
1

...
Then
n n
∑∏
An =
(1 + ik ) = (1 + i1 ) · · · (1 + in ) + · · · + (1 + in−1 )(1 + in ) + (1 + in )
j=1 k=j

and hence
An = (1 + in )(1 + An−1 )

(5
...
, in−1
...
Clearly
µ1 = 1 + µ and E[A2 ] = 1 + 2µ + µ2 + σ 2
...
2a) gives:
1
µn = (1 + µ)(1 + µn−1 )
for n ≥ 2
...

Thus E[An ] is just the accumulated value s n ,µ = (1 + µ)s n ,µ calculated at the mean rate of interest
...
2a) gives
A2 = (1 + 2in + i2 )(1 + 2An−1 + A2 )
n
n
n−1
Using the fact that in is independent of An−1 and taking expectations gives
E[A2 ] = (1 + 2µ + µ2 + σ 2 )(1 + 2µn−1 + E[A2 ]) for n ≥ 2
...
Then use var[An ] = E[A2 ] − µ2
...
2a
...

(a) For a single premium of £1, ﬁnd the mean and standard deviation of the accumulation after a term of 5 years
...
Find the mean and standard deviation of the accumulation after a term
of 5 years
...
In this case S5 = (1 + i1 )(1 + i2 )(1 + i3 )(1 + i4 )(1 + i5 ) where i1 , i2 ,
...
i
...
U [0
...
06]
...
04 and the density of i is fi (x) = 25 for i ∈ [0
...
06]
...
06
0
...
04)3
4
2
σ = var[i] =
25(x − 0
...
02
0
...
045 = 1
...
481157
...
03033
...
04 = 1
...
04
...
04s 5 ,0
...
63298
...
08173
...

1
Then E[A2 ] = α(1 + 2µ1 + E[A2 ]) = α(1 + 2
...
04 + E[A2 ]) = 4
...

1
2
1
Then E[A2 ] = α(1 + 2µ2 + E[A2 ]) = α(1 + 2
...
04 + E[A2 ]) = 10
...

3
2
2
Then E[A2 ] = α(1 + 2µ3 + E[A2 ]) = α(1 + 2
...
04 + E[A2 ]) = 19
...

4
3
3
Then E[A2 ] = α(1 + 2µ4 + E[A2 ]) = α(1 + 2
...
04 + E[A2 ]) = 31
...

5
3
4
Using var[A5 ] = E[A2 ] − E[A5 ]2 gives var[A5 ]1/2 = 0
...

5

5
...
Calculating the distribution function of a product of independent random
variables is usually not possible—however, there is one case where it is possible and that is the case when the
factors have independent lognormal distributions
...
The random variable Y : (Ω, F , P) → ( (0, ∞), B(0, ∞) ) has the lognormal distribution with
parameters (µ, σ 2 ) iff ln(Y ) has the normal distribution N (µ, σ 2 )
...
J
...

2
Suppose Yi ∼ lognormal(µi , σi ) for i = 1,
...
, Yn are independent
...
, Yn are i
...
d
...

• Mean and variance of a lognormal
...
Hence
1

E[Y ] = E[eZ ] = eµ+ 2 σ

2

and E[Y 2 ] = E[e2Z ] = e2µ+2σ

2

and so
2

2

var[Y ] = e2µ+σ (eσ − 1)
Example 5
...
Suppose £1 is invested at time 0 for n time units
...
, n and the i1 , i2 ,
...

(a) Let £Sn denote the accumulated amount at time n
...

(b) Let £Vn denote the present value of £1 due at the end of n years
...

∑n
Solution
...

∑n
∑n 2
2
But ln(1 + ij ) ∼ N (µj , σj ) and the 1 + ij are independent
...

Hence
n
n
∑
∑
2
Sn ∼ lognormal(
µj ,
σj )
j=1

j=1

∑n

(b) Now Vn (1 + i1 )(1 + i2 ) · · · (1 + in ) = 1 and hence ln Vn = − j=1 (1 + ij )
...
3b
...
Express µ and σ 2 in terms of α and β
...
We know that α = eµ+ 2 σ and β = e2µ+σ (eσ − 1)
...
tex)

1
...
06 and standard deviation 0
...
Find Pr[it ≤ 0
...

(Institute/Faculty of Actuaries Examinations, September 1999) [3]
2
...
The yield in any year is independent
of the yields in all previous years
...

(Institute/Faculty of Actuaries Examinations, September 1997, adapted) [3]
3
...

The investment manager wishes to invest the coupon payments on deposit until the bond is redeemed
...

Deriving the necessary formulæ, calculate the mean value of the total accumulated investment on 31 December 2006
if the annual effective rate of interest has an expected value of 5 1/2% in 2004, 6% in 2005 and 4 1/2% in 2006
...
The yields on an insurance company’s funds in different years are independent and identically distributed
...
075 and σ 2 = 0
...

Find the probability that a single investment of £2,000 will accumulate over 10 years to more than £4,500
...
An investment bank models the expected performance of its assets over a 5 day period
...
1% and standard deviation 0
...
Also 1+i is lognormally distributed
...
05
...
Let it denote the rate of interest earned in the year t − 1 to t
...
625,
4% with probability 0
...
125
...
Let S3 denote the accumulated value of 1 unit for 3 years
...

(Institute/Faculty of Actuaries Examinations, April 1998) [5]
7
...
Over that period, the return
on the bank’s portfolio, i, has a mean value of 0
...
3%
...

Calculate the value of j such that the probability that i is greater than or equal to j is 0
...

(Institute/Faculty of Actuaries Examinations, September 2003) [6]
8
...
In any year, the yield is
independent of the value in any other year
...

(i) Derive formulæ for the mean and variance of Sn if jt = j and st = s for all years t
...
06
...
06 and s = 0
...

(Institute/Faculty of Actuaries Examinations, September 2003) [5+3=8]
9
...
The annual effective returns are independent and (1 + it ) is lognormally distributed,
where it is the return in the tth year
...

(b) Calculate the probability that the accumulation of the investment will be less than 90% of the expected value
...
The rate of interest earned in the year from time t − 1 to t is denoted by it
...

The expected value of the rate of interest is 5%, and the standard deviation is 11%
...

(ii) Calculate the probability that the rate of interest in the year from time t − 1 to t lies between 4% and 7%
...
In any year, the rate of interest on funds invested with a given insurance company is independent of the rates of
interest in all previous years
...
The
mean and standard deviation of it are 0
...
20 respectively
...

(ii) (a) Determine the distribution of S15 , where S15 denotes the accumulation of one unit of money over 15 years
...
5
...
The annual rates of interest from a particular investment, in which part of an insurance company’s funds is invested,
are independent and identically distributed
...

it has mean value 0
...
02, the parameter µ = 0
...
0003493
...
It currently has assets of £950,000
...
Find, to two decimal places, the probability that the insurance company will
be unable to meet its liabilities if:
(a) All assets are invested in the investment with the guaranteed return
...

(ii) Determine the variance of return from the portfolios in (i)(a) and (i)(b) above
...
J
...
An insurance company has just written contracts that require it to make payments to policyholders of £1,000,000 in
5 years’ time
...
The insurance company is to invest
half the premium income in ﬁxed interest securities that provide a return of 3% per annum effective
...
The return from these assets in year t,
it , has a mean value of 3
...
The quantities
(1 + it ) are independently and lognormally distributed
...

(ii) A director of the company suggests that investing all the premiums in the assets with an uncertain return would
be preferable because the expected accumulation of the premiums would be greater than the payments due to
the policyholders
...

(Institute/Faculty of Actuaries Examinations, September 2005) [9+2=11]
14
...

The rate of interest is ﬁxed randomly at the beginning of each year and remains unchanged until the beginning of the
next year
...

During the ﬁrst year the rate of interest per annum effective will be one of 3%, 4% or 6% with equal probability
...
7 or 4% with
probability 0
...

(i) Assuming that interest is always reinvested in the account, calculate the expected accumulated amount in the
bank account at the end of 2 years
...

(Institute/Faculty of Actuaries Examinations, September 2002) [4+7=11]
15
...
In any year, the yield on the investment will be 4% with probability 0
...
2 and 8% with probability 0
...

(i) Calculate the mean accumulation at the end of 10 years
...

(iii) Without carrying out any further calculations, explain how your answers to parts (i) and (ii) would change (if at
all) if:
(a) the yields had been 5%, 6% and 7% instead of 4%, 6% and 8% per annum, respectively; or
(b) the investment had been made for 12 years instead of 10 years
...
The annual yields from a particular fund are independent and identically distributed
...
07 and σ 2 = 0
...

(i) Find the mean accumulation in 10 years’ time of an investment in the fund of £20,000 at the end of each of the
next 10 years, together with £150,000 invested immediately
...
99
...
In any year the yield on funds invested with a given insurance company has mean value j and standard deviation s,
and it is independent of the yields in all previous years
...

(ii) Let it be the rate of interest earned in the tth year
...
08 and σ = 0
...

Calculate the probability that a single investment of £1,000 will accumulate over 16 years to more than £4,250
...
In any year the yield on a fund of speculative investments has a mean value j and standard deviation s
...

(i) Derive formulæ for the mean and variance of the accumulated value after n years, Sn , of a unit investment at
time 0
...
Then if the mean value and standard deviation of i are given by j = 6% and s = 1%:
(a) Find the parameters µ and σ 2 of the lognormal distribution for (1 + i)
...
69869 and 0
...

(c) Calculate the expected value of S12 and calculate the probability that the accumulation at the end of 12 years
will be at least double the original investment
...
A company is adopting a particular investment strategy such that the expected annual effective rate of return from
investments is 7% and the standard deviation of annual returns is 9%
...
The company has received a premium of £1,000 and
will pay the policyholder £1,400 after 10 years
...

(ii) Calculate the probability that the accumulation of the investment will be less than 50% of its expected value in
10 years’ time
...
Calculate the probability that it will
have insufﬁcient funds to meet its liability
...
An investment fund provides annual rates of return, which are independent and identically distributed, with annual
accumulation factors following the lognormal distribution with mean 1
...
02
...

Calculate the amount of cash she should invest now in order that she has a 99% chance of having sufﬁcient cash
available in 5 years’ time from the investment to meet the payment
...

(Institute/Faculty of Actuaries Examinations, April 2004) [9+3=12]
21
...

Let Sn be the accumulated amount after n years of a single investment of 1 at time t = 0
...

(b) Show that var[Sn ] = (1 + 2j + j 2 + s2 )n − (1 + j)2n
...
No other
values are possible
...

(b) The accumulated value at time t = 25 years of £1 million invested with the bank at time t = 0 has expected
value £5
...
5 million
...

(Institute/Faculty of Actuaries Examinations, April 2005) [5+8=13]

22
...
The annual effective rate is assumed to be 2%, 4% and 7% with
probabilities 0
...
55 and 0
...

(a) Calculate the expected accumulated value of an annuity of £800 per annum payable annually in advance over
the next 10 years
...

(Institute/Faculty of Actuaries Examinations, April 2006) [4]
23
...
07 and variance 0
...
The value taken by the rate of interest in any one year is independent of its value in any other year
...

(ii) The variance of the accumulation at the end of 10 years, if one unit is invested at the beginning of 10 years
...

(Institute/Faculty of Actuaries Examinations, September 2006) [3+5+1=9]
24
...
Interest is always reinvested in the
account
...
The rate of interest applicable in any one year is independent of the rate applicable in any other year
...

During the second year, the annual effective rate of interest will be either 7% with probability 0
...
25
...
7 or 4% with probability 0
...

(i) Derive the expected accumulated amount in the bank account at the end of 3 years
...

(iii) Calculate the probability that the accumulated amount in the bank account is more than £97,000 at the end of
3 years
...
J
...
The expected effective annual rate of return from a bank’s investment portfolio is 6% and the standard deviation of
annual effective returns is 8%
...
Deriving any necessary formulæ:
(i) calculate the expected value of an investment of £2 million after 10 years;
(ii) calculate the probability that the accumulation of the investment will be less than 80% of the expected value
...
An insurance company holds a large amount of capital and wishes to distribute some of it to its policyholders by way
of two possible options:
Option A
...
5% and the standard deviation of annual returns 7%
...
The policy
holder will receive the accumulated investment at the end of ten years
...
£100 will be invested for each policyholder for ﬁve years at a rate of return of 6% per annum effective
...

This spot rate will be 1% per annum effective with probability 0
...
3,
6% per annum effective with probability 0
...
3
...

Deriving any necessary formulæ:
(i) Calculate the expected value and the standard deviation of the sum the policyholders will receive at the end of
the ten years for each of options A and B
...

(iii) Comment on the relative risk of the two options from the policyholders’ perspective
...
A pension fund holds an asset with current value £1 million
...
The annual investment return in the next year will be 7% with probability 0
...
5
...
3, 0
...
3, respectively
...

(ii) Calculate the standard deviation of the accumulated value of the asset after 10 years, showing all steps in your
calculation
...
3, 0
...
3 respectively
...
The annual yields from a fund are independent and identically distributed
...
05 and σ 2 = 0
...

(i) Calculate the expected accumulation in 20 years’ time of an annual investment in the fund of £5,000 at the
beginning of each of the next 20 years
...

(Institute/Faculty of Actuaries Examinations, April 2012) [5+5=10]
29
...
The interest rate he will
earn on the invested funds in the ﬁrst 10 years will be either 4% per annum with probability of 0
...
7
...
3 or 6% per annum with probability 0
...
However, the interest rate in the second
10 year period will be independent of that in the ﬁrst 10 year period
...

(ii) Calculate the expected value of the investment in excess of £200,000 if the amount calculated in part (i) is
invested
...

(Institute/Faculty of Actuaries Examinations, September 2012) [2+3+2=7]
30
...
The yield on the investment in any year will be
5% with probability 0
...
6 and 9% with probability 0
...

(i) Calculate the mean accumulation at the end of 15 years
...

(iii) Without carrying out any further calculations, explain how your answers to parts (i) and (ii) would change (if at
all) if:
(a) the yields had been 6%, 7% and 8% instead of 5%, 7% and 9% per annum, respectively
...

(Institute/Faculty of Actuaries Examinations, April 2013) [2+5+4=11]

Interest Rate Problems

Jan 28, 2016(12:36)

Exercises 6 Page 127

31
...
The total premiums paid by policy holders at the outset of the contracts amounted to £7
...

The insurance company is to invest the premiums in assets that have an uncertain return
...
5% per annum effective and a standard deviation of 4% per annum effective
...

(i) Calculate the mean and standard deviation of the accumulation of the premiums over the ﬁve-year period
...
(Note
...
)
A director of the insurance company is concerned about the possibility of a considerable loss from the investment
strategy suggested in part (i)
...

(ii) Explain the arguments for and against the director’s suggestion
...
In any year, the yield on investments with an insurance company has mean j and standard deviation s and is independent of the yield in all previous years
...

Each year the value of (1 + it ), where it is the rate of interest earned in the tth year, is lognormally distributed
...
04 and standard deviation of 0
...

(ii) (a) Calculate the parameters µ and σ 2 for the lognormal distribution of (1 + it )
...

(iii) Explain whether your answer to part (ii)(b) looks reasonable
...
See Mathematical Writing by Donald E
...
Roberts
...
J
...
1 Arbitrage
...
& v
...
Commerce
...
L19
...
i
...
Engage in arbitrage
...

In ﬁnance, an arbitrage opportunity means the possibility of a risk-free trading proﬁt
...

An arbitrageur is a trader who exploits arbitrage possibilities
...
The mechanics are as follows: suppose client A instructs a broker to short sell S
...

Eventually A pays the broker to buy S in the market who returns it to the account he borrowed it from
...
)
Example 1
...
Suppose an investor has a portfolio which includes security A and security B
...
He assesses the prices at time 1 will be S1 = 7 and S1 = 14 if the market goes
A
B
up and S1 = 5 and S1 = 10 if the market goes down
...

Solution
...
Hence he makes a gain of 1 at time 0
...

Clearly, investors will sell A and buy B
...

Example 1
...
Suppose an investor has a portfolio which includes security A and security B
...
He assesses the prices at time 1 will be S1 = 7 and S1 = 7 if the market goes up
A
B
and S1 = 5 and S1 = 4 if the market goes down
...

Solution
...
Hence he neither gains nor loses at time 0
...

Clearly, investors will buy A and sell B
...

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...
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Jan 28, 2016(12:36)

Section 1

Page 129

Page 130 Section 1

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

1
...
Suppose investment A costs the amount sA and investment B costs the
amount sB at time t0
...

Here are three situations where an arbitrage exists:
• If NPVA (t0 ) = NPVB (t0 ) and sA ̸= sB then there is an arbitrage
...

• If sA > sB and NPVA (t0 ) ≤ NPVB (t0 ) or if sA < sB and NPVA (t0 ) ≥ NPVB (t0 ) then there is an
arbitrage
...
1a) is an example of the ﬁrst formulation—apply it to two A and one B
...
1b) is an
example of the second formulation
...

Selling one A and buying one B at time 0 produces a cash surplus at time 0 and increases the value of the
portfolio at time t = 1
...
3 Option pricing
...
3a
...
Suppose further that at time t = 1 its price will rise
to 150 or fall to 50 with unknown probability
...

Suppose C denotes the price at time t = 0 of a call option for one share for the exercise price of 125 at the exercise
time of t = 1
...

Solution
...
This will cost 100x + Cy at time t = 0
...

Hence if y = −4x, the value of the portfolio will be 50x whether the share price rises or falls
...
Consider the following 2 possible decisions:
• Investment decision A: buy 1 share, invest the rest;
• Investment decision B: buy 4 options, invest the rest
...
For decision A, the value of the portfolio at time t = 1 is
{
{
150
100
(z − 100)(1 + r) +
= (z − 100)(1 + r) + 50 +
50
0
For decision B, the value of his portfolio at time t = 1 is
{
4 × 25 = 100
(z − 4C)(1 + r) +
0
For no arbitrage, we can equate these
...

4
At time t = 0, sell 4 options for 4C and borrow 50/(1+r)
...

Use 100 to buy one share and the rest is the arbitrage—the rest is the amount z − 100
...
If the share then has price 150, repay the 50 and use the remaining 100 for paying
for the 4 options; if the share has price 50, then the options are worthless and use the 50 to repay the loan
...

4
At time t = 0, short sell one share for 100; purchase 4 options for 4C and put the rest of the money, which is
100 − 4C > 50/(1 + r), in the bank
...
In
both cases, the obligation of repurchasing the share which was sold short is covered
...

t = 1: probability p1
150
100

t = 1: probability 1 − p1
50
0

Arbitrage

Jan 28, 2016(12:36)

Section 2 Page 131

2 Forward Contracts
2
...
The spot price for a commodity such as copper is the current price for immediate
delivery
...

Suppose a manufacturer requires a supply of copper
...
Similarly, the copper supplier does
not know how to plan his production if he does not know how much he will receive for the copper he produces
in 6 months’ time
...

2
...
A forward contract is a legally binding contract to buy/sell an agreed quantity of
an asset at an agreed price at an agreed time in the future
...
Thus forward contracts are over-the-counter
or OTC
...

Settlement of a forward contract occurs entirely at maturity and then the asset is normally delivered by the
seller to the buyer
...
3 Futures contracts
...
Futures contracts can be traded on an exchange—this
implies that futures contracts have standard sizes, delivery dates and, in the case of commodities, quality of
the underlying asset
...

Suppose the contract speciﬁes that A will buy the asset S from B at the price K at the future time T
...
Therefore, a futures
contract says that the short side must deliver to the long side the asset S for the exchange delivery settlement
price (EDSP)
...

Let ST denote the actual price of the asset at time T
...

If K < ST then the long side A makes the proﬁt ST − K
...
At the time the contract
is made, the long side is required to place a deposit called the initial margin in a margin account
...
If the amount falls below the maintenance
margin then the long side must top-up the account; the long side is also allowed to withdraw any excess over
the initial margin
...
Similarly, the short side
also has a margin account
...

Example 2
...
Suppose the date is May 1, 2000 and the current spot price of copper is £1,000 per tonne1
...
Suppose the current price of September copper is £1,100 per
tonne
...
This means that the current
price of a September copper futures contract is 25 × £1,100 = £27,500
...
This means that A agrees on
May 1 to pay £27,500 to B for 25 tonnes of copper on September 1 whilst B agrees to sell 25 tonnes of copper to A
on September 1 for £27,500
...
For current information, see the web site of the London Metal Exchange: www
...
com
...
J
...
4 Pricing a forward contract with no income
...
So A is the long side and B is the short side
...

If K > ST then the short side B makes the proﬁt K − ST
...

Assume that the force of interest available on a risk-free investment in (0, T ) is δ
...

Method I
...
At time 0:
• buy one unit of S at the current price S0
...
Value of portfolio at time T: one unit of S which has value ST
...
At time 0:
• enter forward contract to buy one S with forward price K at time T ;
• invest Ke−δT in a risk-free investment at force of interest δ
...
Value of portfolio at time T: one unit of S which has value ST
...
The size of the investment at time 0
is chosen to be Ke−δT in order to ensure that values of the two portfolios are equal at time T
...

We can simplify Method I as follows: equate the time 0 values
...
At time T , he then has the asset S which he values at K
...
The forward price will change over time
...
Then Kt = St eδ(T −t) where St is the price of the security at time t
...
Hence as t approaches the settlement time, T , the forward price tends to the
price of the underlying asset
...
4a
...
(a) What is the price of a futures contract on this stock with
a settlement day in 180 days’ time if the risk-free interest rate is 5% per annum
...
) (b) After 50 days, suppose the price of the stock has risen to 308p and the risk-free
interest rate has not changed
...

Solution
...
05
...
05180/365 = 307
...
(b) New price of futures contract is Kt = 308 × 1
...
40
...
40 − 307
...
09
...
5 Pricing a forward contract with a ﬁxed income
...

Example 2
...
First consider the following simpliﬁcation: suppose the interest rate is 0% and hence £1 tomorrow
is worth exactly £1 today
...
It pays a coupon of £1 at time 1
...
(a) Here is the justiﬁcation that K should be £9
...
He will receive £1 at time 1 and £K at time 2; hence he makes a risk-free proﬁt of £(K − 9) for
every 1 unit of S
...
Then an arbitrageur should sell S short now for £10 and enter into a long forward contract to buy S
at time 2 for £K
...
) and £K at time 2
...
Similarly,
anyone owning the asset S at time 0 should make use of the same arbitrage possibility
...

Suppose the security S is a bond which pays the coupon c at time t1 where t1 ∈ (0, T )
...
We
wish to decide what is the appropriate value for K under the assumption that the force of interest available on
a risk-free investment in (0, T ) is δ (this means that if we invest an amount b for a length of time h during the
interval (0, T ) then it grows to beδh )
...

Consider the following two investment portfolios:
Portfolio A
...

Value of portfolio at time 0: S0
...

Portfolio B
...

Value of portfolio at time 0: Ke−δT + ce−δt1
...
(The amount Ke−δT + ce−δt1 accumulates to K + ceδ(T −t1 ) and the amount K from the risk-free
investment pays for the forward contract
...
)
The no arbitrage assumption implies that
Ke−δT + ce−δt1 = S0
and hence K = S0 eδT − ceδ(T −t1 )
...

Equate the time 0 values
...
In return he gets the coupon c
at time t1 and has the asset S at time T which he values at K
...

Now suppose we have more than one coupon payment: suppose we have coupons of size ck at times tk for
k = 1, 2,
...
Then the argument of equating time 0 values gives
n
∑
−δT
Ke
= S0 −
ck e−δtk
k=1

Example 2
...
The price of a stock is currently 250p
...
What is the price of a futures contract on
this stock with a settlement date of 250 days’ time if the risk-free interest rate is 7% per annum
...
We have S0 = 250, T = 250/365 and eδ = 1 + i = 1
...
Equating time 0 values gives
n
∑
Ke−δT = S0 −
ck e−δtk = 250 − 10 × 1
...
07−215/365
k=1

leading to K = 236
...

Example 2
...
A bond has a current price of £600
...
The ﬁrst payment is due in exactly 6 months
...

By using the “no arbitrage” assumption, ﬁnd the price of a forward contract to purchase the bond in exactly one year
...
Let K denote the forward price
...
07 + 30e−0
...
07
...
We are given the force of interest rates
F0,0
...
06 and F0,1 = 0
...
Hence the forward force of interest F0
...
5 = 0
...
6a) on page 102)
...
The coupon received in 6 months’ time can be invested and hence will grow to 30e0
...
J
...
Hence 600e0
...
04 + 30 + K
...
Both give
K = 582
...

The price of K = 582
...
If K > 582
...
5 for 6 months
...

time
0
0
...

If K < 582
...

His cash ﬂow is
time
0
0
...

2
...
Now suppose the security S pays a known
dividend yield which is a nominal r% per annum compounded continuously—the value r is the average over
the life of the forward contract
...
Hence the actual size of the dividend payments over (0, T ) are unknown at
time 0 because the value of the stock price over the time interval (0, T ) is unknown at time 0
...

Portfolio A
...

Value of portfolio at time 0: Ke−δT
...
(The amount Ke−δT
accumulates to K which pays for the forward contract
...
At time 0:
• buy e−rT units of S at the current price S0 (and reinvest dividend income in further units of S)
...
Value of portfolio at time T: one unit of S
...
6a
...
It is expected to provide a dividend income of 2% of the asset
price which is received continuously during a six month period
...

Solution
...
07 and er = 1
...
The time 0 value of one unit of stock at time T is S0 e−rT
...
07−r)0
...
035 /1
...
83
...
7 Value of a forward contract at intermediate times
...
At time T ,
the value of the contract to the long side is ST − K0 , the value of the stock at time T minus the price of the
forward contract agreed at time 0
...
Then VT = ST − K0 , V0 = 0 and the value to the short side at any time t ∈ [0, T ]
is (−Vt )
...
At time t
• buy the forward contract agreed at time 0 for the price Vt
...

Value of portfolio at time t: Vt + K0 e−δ(T −t)
...
(Because the
amount K0 e−δ(T −t) grows to K0 which is the amount needed by the forward contract which returns one
unit of S
...
At time t:
• agree to a new forward contract with price Kt for one unit of S at time T ;

Arbitrage

Jan 28, 2016(12:36)

Section 2 Page 135

• invest Kt e−δ(T −t) in a risk-free investment at force of interest δ for the interval (t, T )
...
Value of portfolio at time T : one unit of S
...
Using Kt = St eδ(T −t) gives Vt = St − S0 eδt
...
Hence the value to the long side of such a forward
contract at time t is Kt − K0 which is in time T values
...

Using K0 = S0 eδT and Kt = St eδ(T −t) gives Vt = St − S0 eδt
...
7a
...
Suppose further that the price of a stock at time t0 was P0 and
was P1 at time t1
...
An investor entered into a long
forward contract for 1 unit of the stock at time t0 with the contract due to mature at time t4
...
a
...

Solution
...
05
...

Let K1 denote the price of a contract agreed at time t1 for 1 unit of the stock to be delivered at time t4
...

Hence the value of the original contract to the long side at time t1 is the difference in prices of these two contracts in
time t1 values; and that is (K1 − K0 )ν t4 −t1 = P1 − P0 /ν t1
...
8 Speculation, hedging, gearing and leverage
...

Example 2
...
This example is a continuation of example (2
...

Suppose A buys one September futures contract for copper from B
...
This may be as low as 1% for a ﬁnancial futures contract
...

Hence A must deposit 10% × £27,500 = £2,750
...
Hence A makes a proﬁt of 25 × (£13,000 − £11,000) = £5,000 for an outlay of £2,750
...

Suppose A believes the price of copper will fall
...
If the price of copper is only £900 per tonne on September 1, then he makes a proﬁt of 25 × £200 =
£5,000
...
Then A loses
25 × £1,100 = £27,500
...

Gearing and leverage
...

Leverage is the term used in the USA whilst gearing is the term used in the UK
...
8b
...
3a) and (2
...

Suppose A believes the price will rise
...
If the price rises to £1,300 per tonne on
September 1, then he makes a proﬁt of 25 × £300 = £7,500 from an investment of £25,000; this proﬁt is 30% of his
initial investment
...

Suppose instead of buying the actual product today, he buys a futures contract for 25 tonnes for £27,500
...
Of course,
he is also exposed to a potential loss of £27,500 which is 1,000% of his initial investment!

Hedging is the avoidance of risk
...

Example 2
...
A UK exporter has won a contract to supply machinery to the US
...
In order to remove the risk that an adverse
movement in the exchange rate makes the deal unproﬁtable, the exporter purchases a currency futures contract
...
In this way, the exporter ensures that the risk that the export deal is made unproﬁtable by movements
in the exchange rate is removed
...
The
term static hedge means that the hedge portfolio does not change over the period of the contract
...
J
...
8d
...
Find a static hedge for the short side of this simple forward contract
...

Solution
...
If the short side
just purchases one unit of S at time 0 and the price falls, then more will have been paid for S then was necessary; if
the short side just purchases one unit of S at time T to fulﬁl the contract and the price ST at time T is greater then K
(the amount he receives at time T ), then a loss will be made—indeed, this potential loss is theoretically unlimited
...
The price of this hedge portfolio at time 0 is −Ke−δT + S0 = 0
...
Hence this hedge portfolio means the short side is certain not to make a loss or a proﬁt
whatever the price of the stock at time T
...
tex)

1
...
The risk-free rate of interest is 5% per annum convertible quarterly
...

2
...

Dividends of £1
...

Calculate the forward price, assuming a risk-free rate of interest of 6% per annum convertible half-yearly and no
arbitrage
...

(i) Explain what is meant by the “no arbitrage” assumption in ﬁnancial mathematics
...
Dividends of
£2 per share are expected after 3 and 6 months
...

(Institute/Faculty of Actuaries Examinations, April 2000) [6]

4
...
20
...

(Institute/Faculty of Actuaries Examinations, September 2000) [3]
5
...
Your answer should include reference to the terms “short forward
position” and “long forward position”
...
Dividends
are received continuously and the dividend yield is 3% per annum
...

Assuming a risk-free force of interest of 5% per annum and no arbitrage, calculate the forward price per share
of the contract
...
A one year forward contract is issued on 1 April 2003 on a share with a price of 600p at that date
...
The 6-month and 12-month spot risk-free interest
rates are 4% and 4
...

Calculate the forward price at issue, assuming no arbitrage
...

(i) Explain what is meant by a “forward contract”
...

(ii) An investor entered into a long forward contract for £100 nominal of a security seven years ago and the contract
is due to mature in 3 years’ time
...
The risk-free rate of interest can be assumed to be 4% per annum effective during the contract
...
You should assume no arbitrage
...

Jan 28, 2016(12:36)

Exercises 3 Page 137

(i) Explain what is meant by the “no arbitrage” assumption in ﬁnancial mathematics
...
The current market value of the share
is £4
...
20 per share has just been paid
...

Assuming a risk-free force of interest of 5% per annum and no arbitrage, calculate the forward price
...
The risk-free force of interest δ(t) at time t is given by:
{
0
...
08 + 0
...

(i) (a) Calculate the accumulation at time 15 of £100 invested at time t = 5
...

(c) Calculate the accumulation at time 15 of £100 invested at time t = 14
...

(ii) Calculate the present value at time t = 0 of a continuous payment stream payable at a rate of 100e0
...

(iii) A one year forward contract is issued at time t = 0 on a share with a price of 300p at that date
...
Calculate the forward price of the share, assuming no arbitrage
...
A bond is priced at £95 per £100 nominal, has a coupon of 5% per annum payable half-yearly, and has an outstanding
term of 5 years
...

The continuously compounded risk-free rates of interest for terms of 6 months and one year are 4
...
2% per annum, respectively
...

(Institute/Faculty of Actuaries Examinations, April 2005) [5]
11
...

(ii) Short-term, one-year annual effective interest rates are currently 8%; they are expected to be 7% in one year’s
time, 6% in two years’ time and 5% in three years’ time
...

(b) The price of a coupon paying bond is calculated by discounting individual payments from the bond at the
zero coupon bond yields in (a)
...

(c) A two-year forward contract has just been issued on a share with a price of 400p
...

Calculate the forward price using the above spot rates of interest assuming no arbitrage
...
A share currently trades at £10 and will pay a dividend of 50p in one month’s time
...
70
...

(Institute/Faculty of Actuaries Examinations, April 2006) [4]
13
...
Short sales of both
securities are possible
...
In the event that a particular stock market index goes
up over the next year, it will pay 25p and, in the event that the stock market index goes down, it will pay 15p
...
In the event that the stock market index goes up over the next year, it will
pay 20p and, in the event that the stock market index goes down it will pay 12p
...

(ii) Find the market price of B such that there are no arbitrage opportunities and assuming the price of A remains
(Institute/Faculty of Actuaries Examinations, September 2006) [2+2=4]
ﬁxed
...

14
...
The price of the security was £95 ﬁve years ago and is now £145
...

Assuming no arbitrage, calculate the value of the contract now if
(i) The security will pay dividends of £5 in two years’ time and £6 in four years’ time
...
(Institute/Faculty of Actuaries Examinations, April 2007) [3+3=6]

Page 138 Exercises 3

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

15
...
Dividends of 50p
per share are expected on 30 September 2007 and 31 March 2008
...

Calculate the forward price at issue, stating any assumptions
...
An 11-month forward contract is issued on 1 March 2008 on a stock with a price of £10 per share at that date
...

Calculate the forward price at issue, assuming a risk-free rate of interest of 5% per annum effective and no arbitrage
...

(i) A forward contract with a settlement date at time T is issued based on an underlying asset with a current market
price of B
...
Show that the theoretical forward price is given by K = Beδt , assuming no
arbitrage
...
Calculate the price of a forward contract to be settled in exactly 6 months, assuming a risk free rate of interest of
8% per annum convertible quarterly
...

(i) Explain what is meant by the “no arbitrage” assumption in ﬁnancial mathematics
...
The price of the security was £7
...
45
...
5% per annum effective throughout the nine-year period
...
20
annually in arrear until maturity of the contract
...

(Institute/Faculty of Actuaries Examinations, April 2012) [2+3+3=8]

19
...
Dividends
of £1 per share are expected on 1 December 2012, 1 March 2013 and 1 June 2013
...

(ii) Explain why it is not necessary to use the expected price of the share at the time the forward matures in the
calculation of the forward price
...

(i) Explain the main difference:
(a) between options and futures
...

(ii) A one-year forward contract is issued on 1 April 2013 on a share with a price at that date of £10
...
Dividends
of £1
...
On 1 April 2013, the 6-month riskfree spot rate of interest is 4
...

Calculate the forward price at issue, stating any further assumptions you make
...
A nine-month forward contract is issued on 1 March 2012 on a share with a price of £1
...
Dividends of
10p per share are expected on 1 September 2012
...
(Institute/Faculty of Actuaries Examinations, September 2013) [3]
22
...

A 9-month forward contract is issued on 1 April 2015 on a stock with a price of £6 per share on that date
...
5% per annum
...

(iii) The actual forward price per share of the contract is £6
...

Outline how an investor could make an arbitrage proﬁt
...
tex)

1
...
2 with n = 2/12
...
2 or 120% p
...

(b) Using c1 = c0 (1 + ni)
with c1 = 5100, c0 = 5000 and i = 0
...

2
...
095 × n/360
...
095 × n/360 = c0 × i × n/365
...
095/360 = 0
...
63%
...
1000 × 1
...
82503 and so the answer to (a) is £126
...
Also, 1000 × 1
...
7346 and so the answer to (b) is £142
...

[
]
4
...
1)1/12 − 1 × 12 = 0
...
57%
...
The amount is 1000 × 1
...
83 assuming the lender rounds in its favour, which in this case means up
...
Denote the effective rate by i
...
1 ×
= (1 + i)
gives i = 1 + 0
...
10382
365
365
or 10
...

7
...
065/2 = £1156
...
Let i denote the effective annual interest rate
...
Hence i = 79/1521 = 0
...
19%
...
(a) 4% p
...
payable every 6 months
...
02)2 − 1 = 0
...
04%
...
022 gives
iep = 1
...
00995 or 1
...

10
...
032 −1 = 0
...
a
...
09% p
...
(b) Equivalent to an effective rate of 1
...
029563 for 6 months
...
02963 = 0
...
91% p
...
payable every 6 months
...
Equivalent to an effective interest rate of 2% per quarter, which is equivalent to an effective interest rate of 1
...
04% per 6 months, which is equivalent to nominal 8
...
a
...

12
...
013 = 1
...
016 = 1
...
0112 = 1
...
0124 = 1
...
03%; (ii) 6
...
68%; (iv) 26
...

(b) (i) 12%; (ii) 4 × 3
...
12%; (iii) 12
...
68%; (v) 13
...

13
...
1069597 and this is 10
...

365
1,500
91
1,365
Using
( )365/91
1,540
77
91/365
(1 + i2 )
=
leads to i2 =
= 0
...
13%
...
(a) The effective interest rate is 0
...
Let i = 1
...
Hence:
150
500
300
+
+
= 805
...
13
...
0253 1
...
0259
(b) The effective interest rate per month is i1 = 1
...
We want the smallest t with
300
150
500
300
150
500
950
>
+
+
=
+
+
(1 + i1 )t
(1 + i1 )24 (1 + i1 )48 (1 + i1 )60 1
...
0258 1
...
This gives (1 + i1 )t < 1
...
2032
...

15
...
Hence we have:
Year
Assets at start of year
Interest at end of year
1
£50,000 at 5% p
...

£2,500
}
£50,000 at 5% p
...

£2,500
2
total = £2,650
£2,500 at 6% p
...

£150
}
£50,000 at 5% p
...

£2,500
£2,500 at 6% p
...

£150
3
total = £2,795
...
5% p
...

£145
...
75 = £57,945
...

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

Answers Page 139

Page 140 Answers 1
...
J
...
(a) The interest rate is an effective 1
...
Hence the discount factor for 6 months is ν1 = 1/1
...
0154
...
Then
20
2
20
800ν1 (1 − ν1 ) yν(1 − ν 10 ) yν1 (1 − ν1 )
ℓ=
=
=
2
1 − ν1
1−ν
1 − ν1
[
]
2
Hence y = 800(1 + ν1 )/ν1 = 800(1 + 1
...
18
...

17
...
073 = £1632
...
07 × 182/365) = £1,932
...
Denote the answer by i
...
45/360 = i/365
...
45625 or 4
...
a
...
(a) 600 = 500(1 + i)4
...
27
...
Hence 5,500(1 + i)2 = 6011
...
45
...
50
...
(a) The amount c0 grows to c0 erk/365 after k days
...

21
...
12
1+
m
with m = 365/7 gives i(m) = 365 (e0
...
1201 or 12
...

7
(ii) m = 12
...
12/12 − 1) = 0
...
06% per annum compounded monthly
...
12×7/365 = £1002
...
12/12 = £1010
...

22
...
Dividing by t and then integrating this inequality over t between 1 and x
gives:
∫ x
∫ x
∫ x
dt
dt
dt
< x1/n
<
1−1/n
t
t
1
1
1 t
1/n
1/n
1/n
Hence ln x < n(x
− 1) < x ln x
...
Hence result
...
If i = 0 then f (m) = 0 for all m; hence result
...

Hence the function f ′ is strictly increasing
...
Hence f ′ (m) < 0 for all m ∈ (0, ∞)
...

[
]
24
...
Using limp↓0 ip = limm↑∞ i(m) = limm↑∞ m (1 + i)1/m − 1 = ln(1+i) = δ
by using question 22
...
Deﬁne g : [0, ∞) → R by g(x) = (1 + x)t −]1 − tx
...

[
Then g(0) = 0 and g ′ (x) = t (1 + x)t−1 − 1 > 0 for all x > 0
...

Similarly for 0 < t < 1 we have g ′ (x) < 0 for x > 0; hence for 0 < t < 1 and i > 0 we have (1 + i)t < 1 + ti
...
(a) 1,000 × 1
...
4949121 or £1,116
...
(b) Interpolate between A(1) = 1000 × 1
...
12 = 1210
...
We want A(1) + [A(2) − A(1)] 57/365 = 1,117
...

(c) Because 1 + it > (1 + i)t for 0 < t < 1 by question 25
...
First investor: £91 grows to £93
...
Let i1 denote the effective interest rate per annum
...
90/91
...
4647
...
90 grows to £100 in 60 days
...
Then
(1 + i2 )60/365 = 100/93
...
Hence i2 = 0
...
Hence second investor achieved a higher rate of return
...
(a) Expected rate of interest is 0
...
07 + 0
...
08 + 0
...
1 = 0
...
Hence present value is 10,000/(1
...
26
...
26(0
...
0710 + 0
...
0810 + 0
...
110 ) − 10,000 = 42
...
Use 1 + i(4) /4 = 1 + i
...
00512
...
0053 − 1 = 0
...
03%
...
06)1/2 = 1
...
Hence i(4) = 4 1
...
5707 or
5
...

30
...
50 in 45 days
...
50/96)365/45
...
50 grows to £100 in 45 days
...
50)365/45
...
50 = 1
...
50/96 = 1
...

31
...
Hence CGT is 9
...
Hence 769 grows to (800 − 9
...
7 in half a year
...
7/769)2 giving i = 0
...
80%
...
5 Page 141

(
)
32
...
Then 4000 = 3600 1 + 0
...
Hence n = 365/(9 × 0
...
9
...
(ii) Now 1
...
Hence 4t = ln(10/9)/ ln(1
...
Or t = 0
...
1015)
years, or 91
...
1015) = 645
...
So 646 days are needed
...
00512t = 10/9
...
0005) years or (365/12) × ln(10/9)/ ln(1
...
54 days
...

(b) For time periods longer than 1 year, compound interest pays more because the investor receives the beneﬁt of
interest on interest
...
(a) We need d with 3,000 1 + 0
...
Hence d = 7,300/3 = 2,433
...

(b) We
need t with 3,000 × 1
...
Hence t = ln(19/15)/ ln(1
...
The number of days is 365t = 2199
...

34
...
032
...
032 ; let ν1 = ν 1/4 =
1/1
...
031/6
...
Hence value at time t = 12 is 120(1 − ν 4 ) × 1
...
Value of payments in next 4 years
48
16
at t = 12 is 30(1 − ν1 ) × 1
...
Value of payments in ﬁnal 4 years at t = 12 is 10(1 − ν2 ) × 1
...

Hence answer is 894
...
0424 + 542
...
77
...
tex)

1
...
08 × 91/365) = 0
...

2
...
08)3 = 1/1
...
794
3
...
Hence c0 (1 + i)2 = 3380 giving c0 = 3380 × 12/13 × 12/13 = 2880
...
(a) 720(1 + 1/6)2 = 980

(b) Using (1 + i)(1 − d) = 1 gives d = 1/7 or 14
...

5
...
5
6
...

7
...
Then d1 = i1 /(1 + i1 ) and so d1 /d = k(1 + i)/(1 + ki) < k
...
(a) As i − d = i2 /(1 + i), it follows that i − d > 0
...
(a) Using 1 − d(m) /m
= 1 − d with m = 1/2 and d(m) = 0
...
95 = 0
...
53% p
...

(
)4
(b)(i) Now 1 − d(4) /4 = 1 − d = 1/(1 + i) = 1/1
...
Hence d(4) = 0
...
38% p
...
payable
(
)12
quarterly
...
045 gives d(12) = 0
...
39% p
...
payable monthly
...
Using 1 − d(1/3) /(1/3)
= 1 − d = 1/(1 + i) =
(1/3)
3
(1/3)
1/1
...
045 and hence d
= 0
...
12% p
...
payable every 3 years
...
(1 − dep )1/p = 1 − d = 1/(1 + i) = 1/(1 + ieq )1/q
...
(a) (i) (1
...
1)p −1
...
a
...

Then use (1 − dep )1/p = 1 − d
...
Hence dep = 1 − (10/11)k/12
...
Use 1 − d(m) /m
= 1 − d
...
02)4 = 0
...
Hence 1 − d(2) /2 = 0
...
Hence d(2) =
2(1 − 0
...
0792 or nominal 7
...

13
...
Hence we want (1 + i)k = 2 or k = ln 2/ ln(1 + i)
...
Hence we want eik = 2 or k = ln 2/i
...
66, 14
...
24, and 7
...
Continuous compounding: 69
...
86, 9
...
93
...
Also i − i2 /2 < ln(1 + i) < i for i > 0
...
(a) Now c0 (1 + i2 ) = c0 (1 + i1 )2
...
(ii) Now c0 = (1 − d2 )c2 = (1 − d1 )2 c2
...

1
(b) Now ct = c0 (1 + it ) = c0 (1 + i1 )t
...
Hence 1 + it = (1 + is )t/s
...
Similarly, (1 − ds ) = (1 − d1 )s and (1 − dt ) = (1 − d1 )t imply
1 − dt = (1 − ds )t/s
...

15
...
01 for 1 month
...
Hence
d2 = 1 − (1 − d1 )2 = 1 − 0
...
0199
...
99k
3
(b) 0
...
96

Page 142 Answers 1
...
J
...
(a) Now 1 + i(m) /m
= 1 + i = eδ
...
Using 1 + i(m) /m
1 − d(m) /m
= 1
(
)m
gives 1 − d(m) /m
= e−δ and hence the result
...
To show d < d(2) < d(3) < · · ·, let g(m) = m(1−e−δ/m )
...

But ex > 1 + x for x > 0 and hence g ′ (m) > 0 for m > 0
...
(c))By exercise 22 on page 9,
(
(
)
we have limn→∞ n(x1/n − 1) = ln x
...
From 1 + i(m) /m 1 − d(m) /m = 1 we get
1/d(m) = 1/m + 1/i(m) and hence limm→∞ d(m) = limm→∞ i(m)
...
Using c0 = (1 − nd)cn gives c0 = 1 − 90 × 0
...
Also cn = (1 + i)90/365 c0 where i is the
(
)365/90
required answer
...
06231 or effective 6
...
a
...
Using c0 = (1 − nd)cn gives c0 = 1 − 0
...
26 cn = 36,374cn /36,500
...
Hence i = 36,500/36,374
−1 = 0
...
61% p
...

19
...
05/365)cn = 721cn /730
...
a
...
Hence i(2) = 2 (1 + i)1/2 − 1 =
[
]
2 (730/721)365/180 − 1 = 0
...
09% p
...
converted 6-monthly
...
Using c0 = (1 − nd)cn gives c0 = (1 − 0
...
984 c1
...
Hence i(2) = 2(1/0
...
08246564 or nominal
8
...
a
...

(b) Now c0 = (1 − d(12) /12)12 c1 where d(12) is the required answer
...
98 ) = 0
...
05% p
...
convertible monthly
...
(i)(a) Now c1 = 1
...
Hence d = 0
...
045 = 0
...
306%
...
Hence d(12) = 0
...
394%
...
045
...
04426 or 4
...
(d) Now c5 = 1
...
24618193765c0 and hence 24
...

(ii) The discount rate d(12) is applied each month to a value which has already been discounted
...

22
...
) Now if c0 is the price of the
treasury bill, then c0 ×1
...
Hence if d is the annual simple rate of discount, then c0 = (1−91d/365)100
...
0390296 or 3
...

91
1
...
(i) Suppose d days
...
50 grows to $98 in d days
...
5 × 1
...
5)/ ln(1
...
54 or 144 days
...
Hence 98(1 + 38i/365) = 100 giving i = 0
...
603%
...
Hence i = (100/98)365/38 − 1 = 0
...
416%
...
tex)

1
...
05/2) = 1025
...
050
...
6951 or £1,024
...

(∫
)
(∫
)
6
6
2
...
04 + 0
...
2 + 0
...
99
3
...

4
...

∫t
5
...

dt
t
0
∫t
As δ(s) ↑ as s ↑, it follows that 0 δ(s) ds ≤ tδ(t)
...

∫t
ˆ
6
...
Hence δ(t) = −ln ν(t)/t
...
Then for all t > 0 and all α ∈ [0, 1] we
t
(
)
ˆ
ˆ and so ln ν(αt) ≥ α ln ν(t)
...
The proof of the converse is similar
...
Now ν(t) = exp − 0 δ(s) ds = exp −0
...
7t − 1)/ ln 0
...
The present value of £100 in 3 1/2 years’ time is
100 × ν(3
...
696882 or £88
...

(∫
)
t
8
...

∫t
(b) Now 0 au du = (at − 1)/ ln a
...
06 × 0
...
7p ) × ln 10 ) − 1
...
3 Page 143

(
)
9
...
05n
...
3 days
...

365
(b) 1,550 = 1,500e0
...
Hence n = 7,300 × ln(31/30) = 239
...
It takes 240 days
...
Now 210 = 200 exp 0 (a + bt2 ) dt = 200 exp(5a+125b/3)
...

12
...

14
...

1000b/3)
...

Hence 750b/3 = ln(23 × 20/212 ) = ln(460/441); hence b = (3/750) × ln(460/441) = 0
...
0001687
...
008351979 or 0
...

(
)120
(
)120
(a)(i) Using c0 1 + 0
...
716 or £60
...
(ii) For this case, c0 = 1 − 0
...

Hence £60
...

(iii) Using c0 e0
...
5 = 60
...
35
...
Let i denote the
equivalent effective annual interest rate
...
91
...
04494 or 4
...

∫t
Now 0 δ(s) ds equals 0
...
005t2 if t < 8 and 0
...
07 = 0
...
07t if t > 8
...
04t+0
...
08+0
...

(b) The present value is 100/A(10) = 100e−0
...
84
...
So we want 500 exp 2 δ(s) ds + 800 exp 9 δ(s) ds
...
01s − 0
...
065 and 2 δ(s) ds = 0
...
Hence we want 500e0
...
065 =
1559
...
(b) We want i where 500(1 + i)8 + 800(1 + i) = 1559
...
Now £1 grows to £e0
...
412 in 8 years
which corresponds to 4
...
So guess 4%: then 500(1 + i)8 + 800(1 + i) = 1516
...

Guess 5%: then 500(1 + i)8 + 800(1 + i) = 1578
...
So answer is 5% to nearest 1%
...
71%
...
02)4 c1 = 0
...
Hence δ = −4 ln 0
...
0808108
...
984 c1
...
984 −1 = 0
...
(iii) Now c0 = (1−d(12) /12)12 c1
...
981/3 ) =
0
...

Now∫
{
{
t
2
−0
...
01t2
for 0 ≤ t ≤ 5;
δ(s) ds = 0
...
01t for 0 ≤ t ≤ 5; and hence ν(t) = e0
...
15t
0
...
25
for t > 5
e
for t > 5
0
(i) We want 1,000ν(12) = 1,000e−1
...
24797383 or £212
...

(ii) We want d with 212
...
12103392 or 12
...

Answers to Exercises: Chapter 2 Section 3 on page 27

(exs2-1
...
(a) NPV(a) = 2575
...
90
...
73 and NPV(b) = 2509
...

(c) At time 0, borrow 2400 and take c0 = b0 = 2500
...
01) + 2500 = 76)
...
This transforms b = (2500, 10) into
c = (100, 2400(1 + 0
...

(e) Consider the two cases a0 ≥ b0 and a0 < b0 and use induction on n
...
This converts a into the sequence (b0 , (1 + i)(a0 − b0 ) + a1 , a2 ,
...

Let a∗ = ((1 + i)(a0 − b0 ) + a1 , a2 ,
...
, bn )
...
Similarly for a0 < b0
...
The value at time 0 is:
[ ∫
∫ t
NPV = A(0) +
ρ(s)ν(s) ds where ν(s) = exp −
0

]

s

δ(u) du

0

Hence value at time t is: [∫
]
[∫ t
] ∫ t
[∫ t
]
t
NPV × exp
δ(u) du = A(0) exp
δ(u) du +
ρ(s) exp
δ(u) du ds
0
0
s
∫t 0
−
δ(s) ds
Alternatively: Now A′ (t) = δ(t)A(t)+ρ(t)
...
Integrating the equation ν(t)A (t) = ν(t)δ(t)A(t) + ν(t)ρ(t) gives
∫ t
ν(t)A(t) − ν(0)A(0) =
ν(s)ρ(s) ds
0

and hence

Page 144 Answers 2
...

6
...

8
...

∫t
0

δ(s) ds

∫

∫t

t

e

+

s

δ(h) dh

ρ(s) ds

0

(1 + i)t−s ρ(s) ds

3
...
9b) gives

4
...
J
...

]
[
]
Hence NPV = 4(ν − 1) + 6(ν 2 − ν) + 8(ν 3 ] ν 2 ) / ln ν = 8ν 3 − 2ν 2 − 2ν − 4 / ln ν
...
043 + 2 × 1
...
08 − 8 / ln 1
...
935
...
Hence NPV =
[∫
]
∫6
∫ 12
12
100 0 exp [−0
...
36 )/0
...
Then use A(12) = NPV × exp 0 δ(u) du
...
06 × 6 + 6 (0
...
0002s2 ) ds = 0
...
3 + 0
...
7608
...
7608) = 1078
...

[
]
Solving 990(1 + i) − 65(1 + i)0
...
5 = 1
...
5 − 1 = 0
...
17%
...
9a) gives NPV = 0 50ν(u) du
...
05t)
...
05u) du = 50(1 − e−0
...
05 = 1,000(1 − e−0
...
Value at time t = 15 is
∫ 15
∫ 15
A(15) = NPV/ν(15)
...
4+ 8 (0
...
0004u2 ) du = 0
...
0004(153 −83 )/3 = 0
...
1452/3
...
2295 or £953
...

∫ 12
Use equation (2
...
01t) for 9 < t < 12
...
01u)ν(u) du where
( ∫u
)
∫u
ν(u) = exp − 0 δ(s) ds
...
16+0
...
48−0
...
06(u−8) = 0
...

∫ 12
Hence ν(u) = e−0
...
Hence NPV = 50 9 e−0
...
45 − e−0
...
6 (e0
...
8165
...
09 and use units of £1,000
...
Hence NPV = −60−25ν 2/3 +50ν 22 +ν 2 (−5−4ν 4 −4ν 8 −4ν 12 −4ν 16 +21ν 20 )/ ln(ν) =
7
...
49
...
1c)
...
56 − 0
...
12 = 0
...
16 =
2
0

0
...
Hence A(10) = 500e0
...
014 or £841
...
(b) Use equation (2
...
1u for 10 < u ≤
∫u
18
...
4 + 0
...
06u − 0
...
06u + 0
...

∫ 18
∫ 18
∫ 18
Hence NPV = 10 ρ(u)ν(u) du = 200 10 exp(0
...
06u + 0
...
08 10 e−0
...
08 ×
[ 0
...
8
]
e
− e0
...
04 = 5,000 e − e0
...
33
...
Let i denote the effective interest rate per annum; hence 1 + i = 1
...
024
...

Then
[∫
]
1
ν
ν2
2
11
1/2
u
NPV = 5,000
ν du + (1 + x + x + · · · + x ) + (1 + ν )
12
2
0
]
[
]
[∫
u=1
1
eu ln ν
ν 1 − x12 ν 2
ν 1 − ν ν2
1/2
1/2
u ln ν
+ (1 + ν ) = 5,000
+ (1 + ν )
= 5,000
e
du +
+
12 1 − x
2
ln ν u=0 12 1 − x 2
0
[
]
ν − 1 ν 1 − ν ν2
1/2
= 5,000
+
+ (1 + ν ) = 13,347
...
(i) Use equation (6
...
2 + 0
...
0025t2 + 0
...
2 + 0
...
005 × 22 +
0
...
6828
...
6828 = 0
...
(ii) We want i with (1 + i)12 = e0
...
6828/12) − 1 = 0
...
85%
...
9a)
...
05u and
∫u
δ(s) ds = 0
...
Hence
0
∫ 5
5
e−0
...
18 − e−0
...
09u du = −
= 2
...
09 2
0
...
3 Page 145

(∫
)
(
)
(3)
5
12
...
1c)
...
Similarly
2
3
2
3
( )
( )
( )
( )
( )
( )
( )
3
and 50a + 1,000b = ln 23
...
Hence b = 500 × ln 184 =
3
10
3 ( ) 10 (
2
10
16
405
405
( )
)
−0
...
Also 50a = 8 ln 3 − ln 23 = ln 32805 and hence a = 0
...

2
10
2944
(ii) We want δ with 100e10δ = 230 or δ = (1/10) × ln(23/10) = 0
...

(iii) Use equation (2
...
05u for 0 < u < 10 and ν(u) = e−δu where δ is given in part (ii)
...
5−10δ 200 23 − 10e0
...
05−δ)u du = 20
=
= 170
...
05
23 ln(23/10) − 0
...

13
...
25
δ(s) ds = 0
...
01
−
= 0
...
2 − 15
...
06446282 or 0
...

[∫
]
[
]
10
(ii) Now 1 + i = ν(9)/ν(10) = A(9, 10) = exp 9 0
...
01 × 485/6 = 2
...
24416 or 124
...

( ∫
)
∫5
∫5
t
(iii) (a) ν(t) = exp − 0 δ(s) ds = exp (−0
...

∫

10

∫t
∫t
14
...
03 + 0
...
03t + 0
...
At t = 8 this has value 0
...
Hence for t > 8 we
∫t
have 0 δ(s) ds = 0
...
05 = 0
...
16
...
03t − 0
...
16 − 0
...
(ii) (a) We want 500ν(15) = 500 exp(−0
...
75) = 500 exp(−0
...
26
...
Now 201
...
Hence 500 1 − d(4) /4 = 201
...
0602096 or 6
...
(iii) We want 10 ρ(t)δ(t) dt = 10 10e−0
...
16−0
...
16 10 e−0
...
86 − e−1
...
07 = 14
...

∫t
∫t
15
...
04 + 0
...
04t + 0
...
At t = 10 this has value 0
...
Hence for t > 10
∫t
we have 0 δ(s) ds = 0
...
05 = 0
...
4
...
04t − 0
...
4 − 0
...
(ii)(a) We want 1000ν(15) = 1000 exp(−0
...
75) = 1000 exp(−1
...
636769 or £316
...
(b) 1000(1 − d)15 = 316
...
Hence d = 1 − exp(−1
...
07380 or 7
...

[
]
∫ 15
∫ 15
∫ 15
(iii) We want 10 ρ(t)δ(t) dt = 10 20e−0
...
4−0
...
4 10 e−0
...
4 e−0
...
9 /0
...
3 )/3 = 31
...

16
...
6 − 0
...
07(t − 6) = 0
...
07t
0

Hence ν(7) = exp(−0
...
79)
...
79) +
50 exp(0
...
0235456 or £282
...

∫ 15
∫ 15
(ii) NPV = 12 ρ(u)ν(u) du = 50 12 exp(−0
...
02t) dt = 2500(e−0
...
39 ) = 104
...
67
...
(i) Now the value at time t = 0 of £1,000 at time t = 10 is 1,000ν(10) and this has value 1,000ν(10)/ν(5) at time
( ∫
)
( ∫
)
t
10
t = 5
...
Hence we want 1,000 exp − 5 δ(s) ds
...
1 − 0
...
01s − 0
...
2 − 0
...
51/2 − 0
...
215
...
215) = 806
...
54
...
54(1 + i)5 = 1,000
...
54(1 + i(12) /12)60 = 1,000
...
043077 or 4
...

∫
∫4
∫4
(iii) Value at time t = 0 is ρ(s)δ(s) ds = 0 100e0
...
06s ds = 100 0 e−0
...
16 )
...
16 )/ν(12)
...
24 + 0
...
165 + 0
...
2 = 0
...
65)
...
16 ) × exp(0
...
061523 or £708
...

18
...
2 + 0
...
03 + 0
...
04
2
5
(4)
40

= 0
...
Hence NPV = 1,000ν(10) = 1,000 exp−0
...
311099 or £375
...

(ii) Now 1,000(1 − d /4) = 375
...
Hence d(4) = 0
...
68%
...
01u ν(u) du
...
02u + 0
...
01u
so ν(u) = e−0
...
78
...
78 10 e
du = 10,000e−0
...
1 − e−0
...
900257 or
£318
...

Page 146 Answers 2
...
J
...
(i) If 0 ≤ t ≤ 9 then 0 δ(s) ds = 0
...
005t2
...
27 + 0
...
675
...
675 + 0
...
135 + 0
...

2
Hence ν(t) = e−0
...
005t for 0 ≤ t ≤ 9 and ν(t) = e−0
...
06t for t > 9
...
135−0
...
035 = 1776
...
13
...
13e15δ = 5,000
...
13)/15 = 0
...
069
...
02s e−0
...
06s ds = 100 11 e−0
...
08s ds = 100e−0
...
88 −
e−1
...
08 = 124
...
055
...
tex)

1
...
It also returns the interest rate and NPV just before the
sign changes (if one exists)
...
cashflow, r){
n <- length(vec
...
cashflow)
signs <- sign(NPV)
signchanges <- (1:numr)[(signs[ - numr] * signs[-1]) <= 0]
if(length(signchanges) == 0)
noroot <- T
else {
noroot <- F
firstchange <- signchanges[1]
}
results <- cbind(interest = r, NPV = NPV)
if(noroot) {
list(results)
}
else list(table = results, interest = r[firstchange], NPV = NPV[firstchange])
}

2
...
07 + 8/1
...
078 = −2
...
This
suggests that project A is not proﬁtable
...
Tke IRR is the solution for r of the equation
n
m
∑ bj
∑ cj
=
′
(1 + i)tj
(1 + i)tj
j=0

j=0

Deﬁne the function g : (−1, ∞) → R by
n
∑
g(r) =

bj

m
∑

cj (1 + r)tm −tj
(1 + r)
j=0
j=0
Then g is continuous with limr→−1 g(r) = ∞, limr→∞ g(r) = −∞ and g(r) ↓↓ as r ↑↑ for r ∈ (−1, ∞)
...
Use units of £1,000
...
Then 1,700ν 8 = 1,000ν 8 +321ν 9 +
229ν 10 +245ν 11 or 700(1+i)3 −321(1+i)2 −229(1+i)−245 = 0
...

Now the return on project A is 7%
...
07) = −0
...
071) = 1
...
Interpolation gives
i′ = 0
...
001 × 0
...
0128 + 1
...
7001 or 7
...

(ii) The cash is received earlier with project A than with project B
...

In particular, if i = 0
...
068 = 66
...
068 +
321/1
...
0610 + 245/1
...
3471
...
6 Page 147

5
...
032 − 1 = 0
...
0609
...
5

31 Dec 2011
11
3

We also have the continuous cash ﬂow of 0
...
4 for t ∈ (2, 3),
...
2 for t ∈ (10, 11)
...
5ν 7/12 + 3ν 11 + 0
...
2
ν t dt
1

= −2 − 1
...
5ν

7/12

= −2 − 1
...
1k
t=k−2

k=3
12
∑

∫
[

ν k−1 − ν k−2
0
...
1 [
−3ν − ν 2 − ν 3 − · · · − ν 10 + 12ν 11
ln ν

[
]
0
...
883476 + 4
...
10875427438 or £3,108,754
...
5ν 7/12 + 3ν 11 +

6
...
09
...
09
1
√
Hence NPVO = 19
...
Now let x = v = 1/1
...
Incomings (in £ million):
Time
1/1/06
1/7/06
1/1/07
1/7/07
1/1/08
1/7/08
1/1/09
···
Time in years
0
1/2
1
1
...
5
3
···
Cash ﬂow (£ million)
0
0
0
0
0
2
...
5
···

1/1/21
15
2
...
5[x5 + x6 + · · · + xk+4 ] = 2
...
Setting NPVI = NPVO gives
[
]/
NPVO (1 − x)
k ∗ = ln 1 −
ln(x) = 12
...

2
...
5 years after 1/1/06; that means the payment on 1 July 2014
...
Hence, increasing the interest rate will reduce NPVI
more than NPVO
...

2
∫2
2
−1)
νt
7
...
1
...

ln
0

Now NPVI = 8ν 3 + 7
...
5ν 6 + · · · + 4ν 11 + 3
...
Hence −NPVO + NPVI = 14
...

(ii) Incomings are received later than outgoings
...
Hence
the IRR is more than 10%
...
7%
...
Let α = 1
...
09
...
5α2
−45α3
60
...
5
1/1/06
−45α4

1/1/03
−45α
31/12/06
60
...
5α
1/1/07
−45α5

31/12/07
60
...
Also NPVI = 60
...
5ν]/ − (αν)k ]/[1 − αν]
...
5ν − 45ν ) ln(αν) = 3
...
This gives a date of 31/12/05 which is 6 years after the start of the project
...
Hence decreasing i will increase NPVI more than NPVO
...

9
...
Let α = 1
...
07
...
We also have the continuous cash ﬂow of 25αt for t ∈ (0, 25)
...
61779346 or £51,617
...

O
[
]
t
(ii) We want t with NPVO = 0 25(αν)u du = 25[(αν)t −1]/ ln(αν)
...
6

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

[
]
ln 7
...
4204905995 or 22
...

(iii) We want to ﬁnd α1 with
(α1 ν)25 − 1
ln(α1 ν)
25 25
2
Hence ν α1 − 7
...
If α1 = 1
...
00541105; if α1 = 1
...
01938711
...
05 + 0
...
05)/(f (1
...
06)) = 1
...
2%
...
052) = −0
...
Linear interpolation gives α1 = 1
...
002f (1
...
05) − f (1
...
05134051
or 5
...
In fact it is 5
...
)

NPVO = NPVI

180(1 + ν + ν 2 ) = 25

and so

10
...
It does not show how proﬁtable the project is
...
5
= 1
...

1
...
04
...
08 + 0
...
08 + 0
...
08 + 0
...
0073097664
log ν

or £7,309
...

(iii) DPP of project A is 10 (no payments are received before 10)
...

(iv) Project A is preferable because it has the higher NPV (even though the DPP is larger)
...
Let α = 1
...
Then we have (in £million):
Date
1/1/14 1/7/14 1/1/15 1/1/16
Cash ﬂow
-19
-9
-5
-57

1/1/17
−57α
75
...
6α

1/1/19
−57α3
75
...
6α3

1/1/21
−57α5
75
...
6α5

Let ν = 1/1
...
Assume production runs for k years where 1 ≤ k ≤ 6 or 31/12/16 to 31/12/21
...
6
αj−1 ν j+2
j=1

j=1

k
∑
= −19 − 9ν 1/2 − 5ν + (75
...
6ν 3 − 57ν 2 )
1 − αν
giving
(19 + 9ν 1/2 + 5ν)(1 − αν)
= 0
...
6ν 3 − 57ν 2
which gives k = 4 years of production or 31/12/19
...
The DPP will be shorter
...
At the other extreme, if the interest rate was very high, it would never be possible to recover the costs of
borrowing the initial amounts which are invested in the project
...
(i) (a) The DPP is the smallest time such that the accumulated value of the project becomes positive
...
(b) The payback period is the time
when the net cash ﬂow (incomings less outgoings) is positive
...

(ii) The payback period ignores discounting—and hence ignores the effect of interest
...

Drawbacks of the DPP:
• It shows when the project becomes proﬁtable using a forecast interest rate, but it does not show how large (or
small) the proﬁt is
...

(iii) Take time 0 to be 1 January 2004
...
1
...
2
∫ 7
∫ 7
...
25

ν u du = 0
...
2 +
2
7
...
3

NPV of TV rights = 0
...
5
1
...
1
0
...
5
0
...
5
10
ν 0
...
5
=
10

NPV =

2007
3
...
4

Answers 2
...
5
0
...
5
0
...
25 − ν 2
= 3
...
25 − ν 7
= 0
...
5
0
...
5
0
...
5
0
...
5
0
...
5
0
...
5
0
...
5 1 + 2ν 12 − 3ν 9
2ν 12 + ν 8
=
−
= 3
...
613811 − 3
...
038032 leading to C = 1
...
165 million
...
Use units of £1,000
...
08
...
Let N1 denote the net present value of the continuous income for years 2, 3 and 4
...
, 30}, let N(k) denote the net present value of the continuous income for years 5,
...
Let
α = 1
...
Then
∫ j
k
k
∑
1 − (αν)k−4
29(ν − 1) ∑ j−5 j−1 ν − 1
N(k) =
α ν
=
29ν 4
29αj−5
ν t dt =
ln ν
ln ν
1 − αν
j−1
j=5

j=5

Let NPV(k) = −NPVO + N1 + N(k)
...
30097626
...
98 (ii) Now NPV(15) = −NPVO + NPV1 + NPV(15) = −129
...
34) from NPVO , this quantity is still negative
...

(iii) NPV(29) = −1
...

Hence the DPP occurs in the ﬁnal year
...
We use equation (7
...
5, etc
...
113 =
400 510 550 540 600 710 400 510 540 710
and hence X = 715
...

2
...
4a) gives
(1 + i)3 =

80
200 200 800
×
×
=
120 110 210 693

leading to i = 0
...
90%
...
The TWRR, i is given by
(1 + i)3 =
Hence i = 0
...
54%
...
35086
180 237 261 248 273 311 180 237 248 311

(exs2-3
...
8

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

4
...

Time, t
1 Jan 2001
1 N ov 2001
1 M ay 2002
31 Dec 2002
Fund value at time t − 0
137
173
205
Net cash ﬂow at time t
20
48
Fund value at time t + 0
120
157
221
Hence the TWRR, i, is given by
137 173 205 4,858,705
(1 + i)2 =
×
×
=
120 157 221 4,163,640
Hence i = 0
...
02%
...
The disadvantage of the TWRR is that it requires a lot more information—it
requires knowledge of all cash ﬂows and the the value of the fund at the times of all cash ﬂows
...
Use units of £1,000,000
...
Then
Time, t
31 Dec 2001
31 Dec 2003
31 Dec 2004
Fund value at time t − 0
x
4
...
44
Fund value at time t + 0
2
...
44
Hence the TWRR, i, is given by
x
4
...
2 x + 1
...
44)
The MWRR, i, is given by 4
...
2(1 + i)3 + 1
...
Using the approximation
(1 + i)3 ≈ 1 + 3i leads to i ≈ 0
...

Let f (i) = 55i3 + 165i2 + 201i − 14
...
06) = −1
...
07) = 0
...
Using linear interpolation
gives (0
...
01 = 0
...
897365 + 1
...
66
...
44(1+i)3 / 21 − 11(1 + i)3 = 2
...
5 million
...
Use units of £1,000
...
Using the
approximation (1 + i)3 ≈ 1 + 3i and (1 + i)2 ≈ 1 + 2i leads to i = 0
...

Let f (i) = 21i3 + 68i2 + 81i − 4
...
049) = 0
...
04) = −0
...
Using linear interpolation
gives: (0
...
009 = 0
...
1347386 + 0
...
04745 or 4
...

(ii) The TWRR, i, is given by
24 32 38 1,216
(1 + i)3 =
×
×
=
21 29 40 1,015
leading to i = 0
...
21%
...
The
MWRR is biased towards the return in the ﬁnal year because the size of the fund is much larger in the ﬁnal year
...
Use units of £1,000,000
...
01015254 or 1
...

(i) (b) Effectively we have the following information:
Time, t
1 Jan 2003
1 Jan 2004
Fund value at time t − 0
450
Net cash ﬂow at time t
40
Fund value at time t + 0
600
490

1 July 2004

1 Jan 2005
800

100
600

For year 1 we have 1 + i1 = 450/600 = 3/4
...
This quadratic in (1 + i2 )1/2 leads to (1 + i2 )1/2 = (−10 ± 100 + 320 × 49)/98 =
√
(−5 ± 3945/49) = 1
...
Hence 1 + i2 = 1
...

Then (1 + i)2 = (1 + i1 )(1 + i2 ) = 0
...
39188 and hence i = 0
...
17%
...
8 Page 151

(ii) The fund performs well in the period 1 July 2004 to 31 Dec 2004, but not so well in the period 1 Jan 2004 to
1 July 2004
...
The LIRR for the year is based on
a single return for the whole year—and there is more invested in the fund in the ﬁnal 6 months
...
Use units of £1,000,000
...
Hence the TWRR is given by
(1 + i)2 = (43/40) × (49/47) × (53/51)
...
07921 and the TWRR is 7
...

For (b), this is equivalent to only having the following information
...
40 grows to 43
...
075
...
47 at time 0 and 2 at time 0
...
Hence 47(1 + i2 ) + 2(1 + i2 )0
...
If
√
x = (1 + i2 )0
...
Hence x = (−1 + 2 623)/47 and 1 + i2 = x2 = 1
...

Hence the LIRR is given by (1+i)2 = 1
...
083368 and so it is 7
...
It is slightly less than the value of TWRR
...
In this case, the 3 intervals 1/1/2000
to 1/1/2001, 1/1/2001 to 1/7/2001 and 1/7/2001 to 1/1/2002 would have to be used for the calculation of the LIRR
...
In this case, the
growth in the ﬁrst 6 months is 49/47 and is 53/51 in the second 6 months—these are very similar quantities and so
there is little difference between the values of TWRR and LIRR
...
We have the following information
...
5 × 1
...
125 20
...
7225525 38
...
5
6
...
0
8
...
5
19
...
9085 37
...
05 × 1
...
065 × 1
...
Hence i = 0
...
8794%
...
125 20
...
7225525 38
...
05 × 1
...
065 × 1
...
5 19
...
9085 37
...
06879398 or 6
...

(iii) MWRR is solution of
12
...
6(1 + i)2
...
854229
If f (i) denotes the left hand side, then f (0
...
86789
...
055) = 38
...
Interpolating: (i −
0
...
854229 − 38
...
005/(38
...
454467) leading to i = 0
...
98%
...
98353%
...
The values of
TWRR and LIRR are equal because the value of the fund is known at the times of all cash ﬂows
...
We have the following information:
Time, t
1 Jan 2012
Fund value at time t − 0
Net cash ﬂow at time t
Fund value at time t + 0
1
...
9
−0
...
0

31 December 2012
0
...
9 0
...
3 1
...
16923 or 16
...

(ii) The MWRR is the solution of 1
...
9(1+i)1/4 = 0
...
So let f (i) = 13i+5−9(1+i)1/4
...
3) = 8
...
31/4 = −0
...
4) = 10
...
41/4 = 0
...
35) = 9
...
351/4 =
−0
...
By linear interpolation we have i − 0
...
15121 × (0
...
05151) and so i = 0
...

(iii) The fund does well before the large withdrawal
...

Page 152 Answers 3
...
J
...
(i) (2
...
3) × (4
...
4) = 1
...
36%
...
3(1 + i) + 1
...
2
...
Using the approximation (1 + i)2/3 ≈ 1 + 2i/3 gives i ≈ 4/33 = 0
...
Then
f (0
...
062803 and f (0
...
263347
...
12 + 0
...
062803/(0
...
062803) = 0
...
2%
...
Hence
the MWRR will be less than the TWRR, because the TWRR ignores the amount in the fund
...
(a) Ignoring inﬂation we have
Time, t
Fund value at time t − 0
Net cash ﬂow at time t
Fund value at time t + 0

1/4/2001
0
0
1000
1000

1/4/2002
1000 × 90
80
= 1125
1000 )
(
1000 1 + 90
80

1/4/2003)
(
1000 1 + 90 98
80 90
= 2313
...
82

Hence the real TWRR is given by
130
473 160
1048
1
98
1
66 + 360 98
)
× (
=
×
= 1
...
22
561
1
...
12294104322 or 12
...

(b) The equation for the real MWRR is
[
]
130 473 160 ν 5
ν5
1000ν
= 1000
+
= 4114
...
02
66 360 98 1
...
22
or 19961ν 5 − 5802
...
22 = 0
...
12294 gives LHS > 0
...
(It is 12
...
)
13
...

1 January 2010
1 January 2011
1 July 2012
Time, t
Fund value at time t − 0
120
140
600
Net cash ﬂow at time t
0
200
Fund value at time t + 0
120
340
(i) Let i denote the annual time weighted rate of return
...
3348967 or 33
...
(ii) The rate of growth in the ﬁrst year is 16
...

Hence the rate of growth is higher when the fund contains more money; hence the money weighted rate of return
will be higher than the time weighted rate of return
...
87%
...
(i) MWRR affected by amounts and timings of cash ﬂows which are not under the control of the fund manager
...
But, TWRR requires knowledge of all the
cash ﬂows and the fund values at all the cash ﬂow dates
...
Then (1 + i)2 = (45/41) × (72/19) leading to i = 0
...
75%
...
tex)

∑n−1
∑n
∑n
1
...

¨
¨
(c) If i = 0, a n − a n−1 = n − (n − 1) = 1
...

¨
(d) If i = 0, ia n + ν n = 0 + 1 = 1
...

(e) Use (d) and ia n = da n
...
If i ̸= 0, use equation 2
...
(g) Use (f) and is n = d¨ n
¨
s
Interpretations: (b) (Value of n + 1 payments at time n + 1) = (value of ﬁrst n payments at time n + 1) + 1(for payment
at time n + 1)
...

(d) Value at time 0 of loan of size 1, interest rate i with n annual interest payments
...
(f) and (g) correspond to (d) and (e) but valued at time n
...
Now δ = 0
...
0618
...
86
(Ia) 10 = 10
δ
δ
δ2
δ
or, from ﬁrst principles:
∫ 10
∫ 10 t
10
10
10ν 10
νt
tν t
ν
10ν 10 1 − ν 10
t
dt =
−
+
= 33
...
3 Page 153

√
3
...
Then 1 + i = 1
...
Let x = 1/(1 + i)1/4 = 1/ 1
...

(
)
1 − x16
40
1
1−
NPV = 40(x + x2 + · · · + x16 ) = 40x
=√
= 504
...
068
1
...
3bwith i(4) = 4 × ( 1
...
118252
...
1236/i(4) ×
4 ,0
...
1236 = 167
...
1236 = 504
...
s(12) = (1 + i)n+1/m a(m) = 1
...
0712+1/12 × i/i(12) × a 12 = 18
...
s(4) = (1 + i)20 a(4) = (1 + i)20
...
07520
...
027701 × 10
...
316
¨ 20
¨ 20
6
...

...

...

¨
¨
(ii) Using a n = (1 + i)a n , (I a) 8 = (1 + i)(Ia) 8 and (Ia) 8 = (a n −
¨
¨
[
]
nν n+1 )/(1 − ν) gives NPV(c) = 210(1 + i)a 8 − 10(1 + i)(Ia) 8 = 1
...
82
...
We want smallest integer n with 100 ≤ 20a n ,0
...
Using tables shows that n = 10 is the least such integer
...

Time
Cash ﬂow, c
10a 10
¨
2(Ia) 9
8a 10
¨
2(I a) 10
¨

NPV(c) =

∑9
t=0

8ν t + 2

0
10
10
0
8
2
∑9

t=0 (t

1
12
10
2
8
4

2
14
10
4
8
6

3
16
10
6
8
8

4
18
10
8
8
10

5
20
10
10
8
12

6
22
10
12
8
14

7
24
10
14
8
16

8
26
10
16
8
18

9
28
10
18
8
20

+ 1)ν t
...

9
...
Then NPV = 600a(4),i = 12xa(12)
...
17243
1/12 (12)
1/12 i(4)
δ/4 − 1)
(1 + i)
e
4(e
e −1
a n ,i (1 + i)

So the answer is £49
...

10
...

11
...
5
7
7
...
5
9
9
...
5 11 11
...
04
...
04
(
)
1
...
04 = 100
+ 1 s 12 ,0
...
835
=
(1 + i1 ) + 1 12 ,0
...
04
OR, let x = 1/1
...
Then NPV = 100[x2 + x4 + · · · + x12 + x12 a 12 ] = 100[x2 (1 + x2 + · · · + x10 ) + x12 a 12 ]
...
0412 + 2
...
0412
2
...
3

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

12
...
0056 and β = 1
...
Thus α is the 6-month growth factor for t ∈ [0, 15] and β is the 6-month growth factor
for t ∈ (15, 25]
...

The ﬁrst 30 payments are made at times 0, 0
...
, 14
...
Their accumulated value at time 15 is c0 = 400(α30 +
α29 + · · · + α); and their accumulated value at time 25 is β 20 c0
...
5, 16,
...
5
...
Hence
[
]
30
β 20 − 1
20
20 α − 1
A(25) = β c0 + c1 = 400 β α
+β
= 46,702
...
(i) Using ﬁrst principles is probably the easiest method
...
131/12
...
131/12

1 66
α α66 − 1 1
...
1311/2 − 1
(α + α65 + · · · + α) =
=
= 7
...
131/12 − 1
(12)
(12)
(12)
(12)
¨
¨
− 1) and s 5
...
135
...
5 where a 5
...
131/12 a 5
...
131/12 (1 − 1/1
...
5 )/i(12)
...
88286
...
13) to get d(12) = 0
...
5 /d(12) =
¨ 5
...
5
(12)
5
...
13 − 1)/d = 7
...

14
...
Then use the ﬁrst result and s(m) = (1 + i)n a(m) and s n = (1 + i)n a n to get the
n
n
third result
...

¨
∑n
(b) We want x with 10,000 = xa 120 = x(a 100 + ν 100 a 20 )
...
(a) Just use a n = j=1 ν j
...
47
...

16
...
02t dt) + exp( 3 0
...
02t dt) + exp( 1 0
...

∫5
Now k 0
...
01(25 − k 2 )
...
09 + e0
...
21 + e0
...
773
...
02s ds) = exp(−0
...
Hence a 5 = ν(1) + ν(2) + ν(3) + ν(4) + ν(5) =
e−0
...
04 + e−0
...
16 + e−0
...
Finally, s 5 = a 5 /ν(5) = a 5 e0
...
773
∑∞
17
...
06 = 166,667
18
...
See example 4
...

s
s
19
...
If i ̸= 0, then is n = (1 + i)n − 1 = d¨ n
...
If i ̸= 0, then ia n = 1 − ν n = da n
...

20
...
If i = 0 then LHS = n = RHS
...
Let ν = 1/(1 + i) denote the discount factor
...
Hence
1
1
ν 1/m
¨∞
a(m) = (ν 1/m + ν 2/m + · · ·) = ν 1/m a(m) = (m) = (m)
∞
m
d
i
(b) follows immediately from the two tables of cash ﬂows
...
(a) Algebraically:

NPV(c) =

k
∑
j=1

ν jr = ν r

a
1 − ν kr
i
=
a = kr
1 − νr
(1 + i)r − 1 kr
sr

Or: recall that s r denotes the accumulated value at time r of the sequence of r equal payments of 1 at times
1, 2,
...
Hence the single payment of 1 at time r is equivalent to the r payments of 1/s r at times 1, 2,
...

Hence our annuity is equivalent to kr equal payments of 1/s r at times 1, 2,
...

(b) Set r = 1/m and k = nm (and hence kr = n) in previous result
...
3 Page 155

24
...

]
1 [
(I (q) a)(m) =
¨ ∞
(1 + ν 1/m + · · · + ν 1/q−1/m ) + 2(ν 1/q + ν 1/q+1/m + · · · + ν 2/q−1/m ) + · · ·
mq
]
1 [
=
(1 + x + · · · + xm/q−1 )(1 + 2y + 3y 2 + · · ·)
mq
[
][
]
1 1 − xm/q
1
1 1 − ν 1/q
1
1
1
1
1
=
=
=
mq 1 − x (1 − y)2 mq 1 − ν 1/m (1 − ν 1/q )2
m 1 − ν 1/m
q 1 − ν 1/q
1 1
= a(m) a(q) = (m) (q)
¨∞ ¨∞
d d
25
...
The second is done in exercise 34
...
(a) Just use k| a(m) = ν k a(m) and ν 1/m a(m) = a(m)
...
If i ̸= 0, RHS = (1 − ν n+k )/i(m) − (1 − ν k )/i(m) =
ν k (1 − ν n )/i(m) = LHS
...

¨ n+k
n+k
27
...
This decomposition gives (II):
0
1
2
3
0
200
180
160
0
200
200
200
0
0
-20
-40
0
0
0
0
This decomposition gives (III):
0
1
2
3
0
200
180
160
0
60
60
60
0
140
120
100

0
0
0
0

1
n
n+1
−1

4
140
200
-60
0
4
140
60
80

The decomposition in (I) is incorrect:
0
1
2
3
4
0
200
180
160
140
0
200
180
160
140
0
0
0
0
0
60
60
60
60
60
−60
−60
−60
−60
−60

5
120
200
-80
0
5
120
60
60

6
100
200
-100
0
6
100
60
40

5
120
120
0
60
−60

···
···
···
···

2
n−1
n+1
−2

6
100
100
0
60
−60

7
80
60
20
7
80
80
0
60
−60

n
1
n+1
−n

7
80
200
-120
0
8
60
60
0

8
60
200
-140
0
9
60
60
0

8
60
60
0
60
0

9
60
40
−60
60
0

9
60
200
0
-140

...

...

...

...

...

...

...

...
(i) We need i with

[
]
7a 5
i
= a 5 64(1 + i)1/4 (4) − 7(1 + i)
ν
i
ν 1/4
1/4
(4)
Using tables for a 5 , (1 + i) and i/i gives RHS = 253
...
05 and RHS = 248
...
06
...
055 gives RHS = 251
...

Using interpolation between i = 0
...
06 gives (x−0
...
06−0
...
055))/(f (0
...
055)) and hence x = 0
...
005(250 − 251
...
371 − 251
...
057
...
057 or 5
...

(ii) Using the general result that (1 + iR )(1 + e) = 1 + iM where iR is the real rate, e is the inﬂation rate and iM is the
money rate, we get iR = (i − e)/(1 + e) = (0
...
02)/1
...
0363 of 3
...

250 = 64a(4) − 7a 5 ,i =
¨ 5 ,i
¨

64a(4)
5

−

i
30
...
Using a(m) = d(m) a n gives 240a(12) = 240 × 1
...
859410 = 458
...

¨ 2 ,i
n ,i
2 ,0
...
05
...
But 246a(12) = 246 × 1
...
859410 = 467
...
Hence i > 0
...

2 ,i
2 ,0
...

Page 156 Answers 3
...
J
...
Let α = 1
...
04
...

ln ν t=k−1
ln ν
δ
1+i
k−1
NPV of costs to companies plus NPV of costs to consumers is
[∫
]
∫ 2
∫ 20
1
]
60i [
t
t
19 t
60
ν dt +
αν dt + · · · +
α ν dt =
ν + αν 2 + · · · + α19 ν 20
δ
0
1
19
=

NPV of costs to ﬁnancial advisers is
∫ 1
∫ 2
∫ 3
∫
ν t dt + 19
ν t dt + 18
ν t dt + · · · +
60
0

1

2

0

1

NPV of beneﬁts to companies is

∫

ν t dt = 12
0

Hence

NPV =

ν t dt =

20

ν t dt =
19

20

12

20

19

NPV of beneﬁts to consumers is
∫ 1
∫ 2
∫
30
ν t dt + 33
ν t dt + · · · + 87
νt
ln ν

60iν 1 − (αν)20
δ
1 − αν

20

= 12
t=0

]
i[
60ν + 19ν 2 + 18ν 3 + · · · + ν 20
δ

]
i[
30ν + 33ν 2 + · · · + 87ν 20
δ

1 − ν 20 12i
=
a
δ
δ 20

[
]
i
1 − (αν)20
12a 20 − 40ν + (10ν + 14ν 2 + 18ν 3 + · · · + 86ν 20 ) − 60ν
δ
1 − αν

Let S = 10ν + 14ν 2 + 18ν 3 + · · · + 86ν 20
...
133939 − 76
60 1 − (1
...
04)20
= 1
...
590326 −
+
−
= −354
...
04
0
...
04 1 − 1
...
04
Net cost is £354
...

32
...
04 and x = ν 1/4
...
208909 or £9,975
...
For the second alternative we have NPV = 2,520(ν 2 + ν 4 + ν 6 + ν 8 + ν 10 ) =
2,520ν 2 (1 − ν 10 )/(1 − ν 2 ) = 10019
...
34
...

∫t
33
...
For 0 ≤ t ≤ 4 we have α(t) = 0
...
Note that α(4) = 0
...
For 4 < t ≤ 9 we have
α(t) = 0
...
12(t − 4) − 0
...
08 = −0
...
12t − 0
...
Note that α(9) = 0
...
For t > 9 we have
α(t) = 0
...
05(t − 9) = 0
...
05t
...
08t)
for 0 ≤ t ≤ 4;
ν(t) = exp −
δ(s) ds = exp(0
...
12t + 0
...
145 − 0
...

∫
∫ 12
We have ρ(t) = 100 exp(0
...
03u) exp(−0
...
05u) du = 100 exp(−0
...
02u) du = 5,000(exp(−0
...
385)) = 138
...
849
...
08 + e−0
...
24 ) = 2561
...
89
...
(i) Using ln ν = −δ and a n = 0 ν t dt = (1 − ν n )/δ gives
∫ n t
∫ n
∫ n
ν
nν n 1 − ν n
nν n
t
−
dt =
+
ρ(t)ν(t) dt =
tν dt =
(Ia) n =
ln ν
ln ν
(ln ν)2
0 ln ν
0
0
n
n
n
nν
1−ν
a n − nν
=−
+
=
δ
δ2
δ
(ii) Amounts are in thousands
...

Hence present value of revenue stream is
∫ 29
∫ 11
∫ 29
NPV(ρ) =
ρ(u)ν u du =
250(u − 1)ν u du +
ρ(u)ν u du
1

∫

1
10

∫

(2500 − 125u)ν u du = 250ν(Ia) 10 + ν 11 (2500a 18 − 125(Ia) 18 )

uν u du + ν 11

= 250ν
0

11
18

0

Appendix

Jan 28, 2016(12:36)

Answers 3
...
2a 18 ,0
...
02) = 5
...
094
...
470
Hence answer is 4761
...
470 = 3212
...

35
...

6
Hence
1 − x6
x7 1 − xn−6 −40 + 37x + 17 x7 + x6 − 5 xn+1
1 − xn−5
6
6
NPV = −40 − 3x
+ x6
−
=
1 − x[
1−x
6 ]1 − x
1−x
[6 (
)]
6
Hence NPV ≥ 0 when xn+1 ≤ 5 37x + 17 x7 + x6 − 40 , or when n + 1 ≥ ln 5 37x + 17 x7 + x6 − 40 / ln x =
6
6
112
...
Hence the DPP is 112 months, or 9 years and
4 months
...
If the interest rate decreases, then
the NPV of the costs increases and the NPV of the income also increases
...
Hence the
DPP will increase
...

36
...
42646
...
Checking k = 13
...
5 = (1 − ν 13
...
554839
...
5 years
...
5 years of payments is 14a(2) = 14 i(2) a 13
...
So proﬁt is 14 i(2) a 13
...
07 − 120 at 7%,
13
...
0713
...
0511
...
5 − 120 at time t = 25
...
5 years at time t = 13
...
05 = 7
...
The accumulated value
11
...
0511
...
05 = 213
...

11
...
3105
...
We use units of £10,000
...
15 and x = ν 1/4
...
679238
= 500 + 350
ln ν t=0
ln 1
...
15
2050
=
(50 × 1
...
15 + 60)
+
= 1122
...
155
1 − x 1
...
15
1650
NPV = (50)
+ 1650v 3 =
× 50 ×
+
= 1120
...
153
1 − x 1
...
15
1800
NPV = (50 + 55ν)
+ 1800v 4 =
(50 × 1
...
40298
4
4
1−x
4 × 1
...
154
(i) NPV criterion suggests selling after 5 years
...

(iii) NPV of beneﬁts if sold after 5 years (where x = 1/1
...
175):
1
0
...
1753 + 55 × 1
...
175 + 65)
+
4 × 1
...
1756
and this must equal

NPV(costs) = 500 + 350
and hence

1 − ν2
ln 1
...
3

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

(
)
1 − ν2
(50 × 1
...
1752 + 60 × 1
...
175
X = 1
...
175
4
1−x
= 2566
...
Rental income may not increase at
projected rate
...
Maintenance and repair costs will increase as sale is
delayed
...

38
...
The NPV is the net present value of a sequence
of cash ﬂows
...

(ii) Inﬂows:
∫ 6
∫ 9
∫ 15
ν6 − ν3
ν9 − ν6
ν 15 − ν 9
t
t
NPV(in) =
ν dt + 1
...
5
ν t dt + 8ν 15 =
+ 1
...
5
+ 8ν 15
ln ν
ln ν
ln ν
3
6
9
]
ν3 − 1 [ 3
=
ν + 1
...
5ν 9 (ν 3 + 1) + 8ν 15
ln ν
Outﬂows (where x = ν 1/4 and α = 1/05):
∫ 3
0
...
5
ν t dt +
[x + x13 + · · · + x59 ] + ν 3 [1 + αν + · · · + α11 ν 11 ]
4
0
ν3 − 1
1 − ν 12
1 − (αν)12
= 1
...
075ν 3
+ ν3
ln ν
1−x
1 − αν
Using 9% gives NPV(in) = 12
...
32338
...

Now use 7% per annum on the ﬁrst 12 years
...
9
ν dt + 2
...
9
+ 2
...
9ν 3 + 2
...
34598
ln ν
∫ 3
0
...
5
ν t dt +
[x + x13 + · · · + x47 ] + ν 3 [1 + αν + · · · + α8 ν 8 ]
4
0
ν3 − 1
1 − ν9
1 − (αν)9
= 1
...
075ν 3
+ ν3
= 12
...

39
...
04 and ν = 1/1
...
Use units of £1,000
...
241851
=
ln ν
ln ν
1 − αν
The net present value of the discrete payments is 500 + (100ν + 110ν 2 + 120ν 3 + 130ν 4 + 140ν 5 ) + 300ν 15 − 700ν 25 =
947
...
Or, 500 + 90a 5 ,0
...
11 + 300ν 15 − 700ν 25 where (Ia) 5 ,0
...
11 − 5ν 6 )/(1 − ν)
...
236472 or £21,236
...

(ii) The net present value of the outgoings is £998,531
...
51
...
However, over half of the outgoings is the payment of £500,000 at time t = 0 which is unaffected by a
change in ν
...
For example, if ν = 1/1
...
82
...
Let ν = 1/1
...
012
...
This amount is received continuously over the 365 days of the year
...
1 × 365)ν t dt = 550 × 365ν
ln ν
1
1
Proceeding in this way gives the present value of the income is
[
]
]
1 − α19 ν 19
ν−1[
ν−1
2
2 3
18 19
400 + 550ν
365
400 + 550ν + 550αν + 550α ν + · · · + 550α ν
= 365
ln ν
ln ν
1 − αν
which is 2275323
...
80
...
80
...
5 Page 159

41
...
Then NPV of outgoings is 1,309,500 + 12,000a(4) ,i
...
0525 ν 25
100,000a(4) 1 + 1
...
0510 ν 10 + 1
...
0520 ν 20 = 100,000a(4)
5 ,i
5 ,i 1 − 1
...
0525 ν 25
13,095 + 120a(4) ,i = 1,000a(4)
25
5 ,i 1 − 1
...
033144 × 9
...
At 9%, the right hand side is 1000 ×
1
...
889651 × (1 − 1
...
0925 )/(1 − 1
...
095 ) = 14313
...
The NPV of the incomings is 85,000a(4) ,i + 90,000a(4) /(1 + i)20
...
Now if there was no increase in
¨ 20
¨ 5 ,i
i
rent, i would satisfy 200/17 = a(4) ,i = (1 + i)1/4 i(4) a 25 ,i which suggests i ≈ 0
...
Now f (0
...
217460361
...
07) = −39
...
Using linear interpolation gives i = 0
...
01×39
...
217460361+
39
...
0747 and hence the answer is 7
...

(iii) The answers to parts (i) and (ii) suggest project A is preferable, but:
the outlay for project A is much larger than the outlay for project B—more money may have to be borrowed and
interest rates will change over the period of 25 years;
the assumption of no increase in maintenance costs over 25 years for project A is questionable;
the assumption of the rise in rents made for both projects A and B is questionable—-an attempt should be made to
assess how likely each assumption will be met
...
Let ν = 1/1
...
The cash ﬂow is as follows (units
of $10,000):
1/
4/
5/
6/
7/
8/
23/
24/
Time
0
···
2
4
4
4
4
4
4
4
Cash ﬂow
−400
−90
9
9
9
9
9α
···
9α4
680
Hence 400 + 90ν 1/2 = (1 + γ + γ 2 + γ 3 )(9ν + 9αν 2 + 9α2 ν 3 + 9α3 ν 4 + 9α4 ν 5 ) + 680ν 6 = (1 + γ + γ 2 + γ 3 )9ν(1 + αν +
α2 ν 2 + α3 ν 3 + α4 ν 4 ) + 680ν 6
...
55965787
1 − αν
9ν(1 − ν)
The left hand side is a 5 ,β where 1/(1 + β) = αν
...
55965787
...
045 < β < 0
...
Linear interpolation gives β = 0
...
068344 or 6
...

Answers to Exercises: Chapter 3 Section 5 on page 60

(exs3-2
...
Total repayment is 3 × £2,000 = £6,000
...
Hence the ﬂat rate is
750/(3 × 5250) = 0
...
8%
...
The cash ﬂow is
Time
Cash Flow

0
−1000

1
703
...
84

This leads to 1000 = 703
...
The usual formula for the solution of a quadratic leads to ν = 0
...
261693 or 26
...

3
...
Then 5,000 = xa 12 ,0
...
Using the prospective method, the capital outstanding after the 4th
payment is y = xa 8 ,0
...
Hence the interest element of the 5th payment is 0
...
08 × 5000a 8 ,0
...
08 =
305
...

4
...
Using a(m) =
n
5 ,0
...
067016 × 3
...
193 or £93
...

The ﬂat rate of interest is (60x − 4000)/20000 = 0
...
96%
...
3b, and tables

5
...
Let
ν = 1/(1 + i); hence ν 1/12 = 1/(1 + j)
...
4615 for j
...
005 = 42
...
01 = 37
...

Interpolating gives (x − 0
...
01 − 0
...
005))/(f (0
...
005))
...
005 +
0
...
4615 − 42
...
973959 − 42
...
00947
...
00947 and so i ≈ 0
...
Hence APR is 11
...
1% rounded down)
...
5

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

6
...
Then 15,000 = 12xa(12)
¨ 2 ,0
...
1236
...
2717 which would be rounded up to £697
...
Hence ﬂat rate is (24x −
15,000)/30,000 = 0
...
782%
...
We need x with 100,000 = 12xa(12) where j is the effective rate of interest per annum corresponding to a nominal
20 ,j
6% per annum convertible monthly
...
5% effective per month; so let ν = 1/1
...
Hence 100,000 =
∑240 k
240
)/(1 − ν)
...
43 or £716
...

k=1 xν = xa 240 ,0
...
Hence capital outstanding is y = x(ν + ν 2 + · · · + ν 168 ) = xν(1 − ν 168 )/(1 −
ν) = x(1 − ν 168 )/0
...
319786 or £81,298
...
Let ν1 = 1/(1 + 0
...
Then y = z(ν1 + ν1 + · · · + ν1 ) = zν1 (1 − ν1 )/(1 − ν1 )
...
694960
or £694
...

8
...
Also i = 0
...

The capital outstanding after 5 years of payments is (prospective loan calculation) x = 10,000a(12) = 10,000 ×
15 ,0
...
045045 × a 15 ,0
...
9588 or £79,487
...
1
− 1 = 633
...

(ii) NPV of annuity over the whole 20 years is 10,000a(12) = 10,000 × 1
...
1 = 88,970
...

20 ,0
...
34
...
Hence total interest payment over the ﬁrst 5 years is 50,000 − (88,971 − 79,487
...
34
...
Let x denote the quarterly payment
...
1 = 4x×1
...
1 = 4x×1
...
513564
...
48
...
1 = 4x i(4) a 14 ,0
...
Hence the
14
[ 1/4
]
i
interest element is 1
...
1 = xia 14 ,0
...
31
...
31 =
149
...

i
10
...
Hence x = 100,000/(12 × 1
...
093945) =
25 ,0
...
138 or £578
...

(ii) (a) Using the prospective method, it is 12xa(12)
= 12×578
...
022715×9
...
05

66649
...
93
...
93 × (1
...
54
...
14 − 271
...
60
...
(a) The annual interest payment is £5,000 × 0
...
(b) The annual sinking fund repayment, x is given by
xs 10 ,0
...
Hence x = 345
...
(c) The total of (a) and (b) is 845
...
The amortisation repayment, y is given
by ya 10 ,0
...
Hence y = 813
...
(i) 50,000 = xa 5 ,0
...
790787 and hence x = 13189
...
87
...
35
...
1 = 32,801
...
We now want 32,801
...
12 = 4y × 1
...
12
3
...
Hence y = 2179
...
10
...
Hence
[
]
interest paid is z (1 + i)1/4 − 1 = z(1 − ν 1/4 )/ν 1/4 = y(1 − ν 19/4 ) = 907
...
12
...
(i) We want x with 800,000 = xa 10 ,0
...
710081 = 119,223
...
60
...

(ii) Capital outstanding at start of year 8 is xa 3 ,0
...
60 × 2
...
7819
...
78 = 4ya(4)
= 4y i(4) a 5 ,0
...
043938 × 3
...
Hence y = 20,411
...
74
...
12
(b) Interest payment is 307,250
...
121/4 − 1] = 8,829
...
Hence capital repayment is 20,411
...
57 =
11,582
...

14
...

...

...
08 + 10(Ia) 20 ,0
...
08 + 10(a 20 ,0
...
08 = 90a 20 ,0
...
08a 20 ,0
...
08 = 225a 20 ,0
...
71
(ii) Using the prospective method, the capital outstanding after the 5th payment is equal to the value of all future
¨
repayments: hence l5 = 140a 15 ,0
...
08 = 140a 15 ,0
...
08 − 15ν 15 )/0
...
08 +
10(1
...
08 − 15ν 15 )/0
...
08 − 1875ν 15 = 1762
...
78 × 0
...
02 and capital 8
...

Hence the 7th repayment of 160 consists of interest (1762
...
98) × 0
...
30 and capital 19
...

(iii) The last payment is £290
...
Hence the 20th payment

Appendix

Jan 28, 2016(12:36)

Answers 3
...
Hence (1+i)l19 = 290
...
519
and 20th payment consists of an interest payment of 21
...
52
...
Cash ﬂow:
Time
Loan
Interest payments
Investment premiums

1/1/80
−100,000
0
1060/12

1/2/80

1/3/80

1/4/80

...
07 = 1060 i(12) a 30 ,0
...
071/12 a 30 ,0
...

¨
i
Accumulated value at time 1/1/10 is 1
...
071/12 s 30 ,0
...
031691 × 1
...
460786 = 103885
...
69
...
041/12 s 30 ,0
...
018204 × 1
...
084938 =
60730
...
43
...
57
...
041/12 s 10 ,0
...
86400
...
86-£39,269
...
29
...
041/12 s 20 ,0
...
0410 + 5000 (12) 1
...
04
i
i
[
]
i
1/12
10
= (12) 1
...
04 × 1
...
04
i
[
]
=1
...
041/12 1060 × 29
...
0410 + 5000 × 12
...
2949
as before
...
Hence the investor should have paid off the
loan rather than investing in the policy
...
The effective interest rate is i with 1+i = 1
...

Let ν1 = ν 1/12
...
Hence 12x = 12 ×
30 ,i
100,000 × i/(1 − ν 30 ) = 6000/(1 − ν 30 ) = 7194
...
00512 − 1)
...
00512 − 1)/(1 − 1/1
...
79
...
(i) See derivation of equation 2
...
(ii) The cash ﬂow is as follows:
Time
Cash ﬂow, c

0
0

1
10

2
12

3
14

...

20
48

(a) Then NPV(c) = 10ν + 12ν 2 + 14ν 3 + · · · + 48ν 20 = 402
...
or use NPV(c) = 8(ν + ν 2 + · · · + ν 20 ) + 2(ν + 2ν 2 +
· · · + 20ν 20 ) = 8a 20 + 2(Ia) 20
(b) The remaining cash ﬂows are
Time
Cash ﬂow, c

15
0

16
40

17
42

18
44

19
46

20
48

Thus we want 40ν + 42ν 2 + 44ν 3 + 46ν 4 + 48ν 5 = 200
...
It is also 38a 5 + 2(Ia) 5
...
03 × 200
...
029
...
029 = 33
...

17
...
06
...
Then (1 − ν)s = 25 − 11ν 15 − ν + ν 2 + · · · + ν 14
...
5033
i
i
3
3
Hence size of loan is 370
...

(ii) First payment: interest component is 370
...
06 = 22
...
23 = 27
...

Remaining loan outstanding is 342
...

Second payment: interest component is 342
...
06 = 20
...
564 = 27
...

(iii) Using the prospective method: loan outstanding is 24ν + 22ν 2 = 118,600/(53 × 53) = 42
...
Hence interest
payment of 14th instalment is 42
...
06 = 2
...
2214 × 0
...
468
...
5

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

18
...
08
Hence x = 602
...

80,000 = 12xa(12)
25 ,0
...
08
− 1) = 514
...
Hence the capital repaid is x − 514
...
01
...
Hence the total
6 ,0
...
08 /i12) ) = 8751
...
(c) The
6 ,0
...
08
capital outstanding, c is given by c × 1
...
732; hence c = 598
...
73 − 598
...
85
...

i
19
...
Then 100,000 = 12xa(12)
¨ 25 ,0
...
061/12 i(12) ×

a 25 ,0
...
Hence x = 100,000/(12 × 1
...
027211 × 12
...
5466 or £631
...

Time
0
1/12
2/12
···
23/12
24/12
···
299/12
Loan
100,000
Repayments
x
x
x
···
x
x
···
x

300/12 = 25
0

i
(ii) Use the prospective method
...
06 = 95779
...
61
...
61 ×
[
]
1
...
212 and the capital repayment is 631
...
21 = 165
...

(iii) (a) After 10 years at time 120/12, the loan outstanding is 12xa(12)
...
06
[
]
(12)
(12)
(12)
(12)
ment
...
02 = 12xa 15 ,0
...
06 /a 15 ,0
...
061/12 /1
...
06
15 ,0
...
474900 or £487
...

(b) The reduction in instalments is x− y = 631
...
47 = 144
...
, 299/12
...
The value at time t = 119/12 is
12ya(12)
...
021/12 × 12ya(12)
and the value at time t = 300/12 = 25 is
15 ,0
...
02

12xa(12)
23 ,0
...
02181/12 × 12ya(12)
...
02 = 1
...
849264
...
02
15 ,0
...
810 and the value at time t = 300/12 is 30222
...

]
[
20
...
05 + a 10 ,0
...
0510 = 12033
...
61
...
52/8790
...
05; hence x ≤ 10
...
07 /1
...
05
...
48 gives 8
...
023582α10−k + 1−α
=
0
...
023582α
+ 20(1 − α
) where α = 1/1
...
Hence 11
...
976418α
; hence 10 − k =
log(11
...
976418) log(1/1
...
Hence k = 7 = x − 1 and so x = 8
...
Value of annuity over 20 years is 1000a 20 ,0
...
21, over 19 years is
1000a 19 ,0
...
321 and over 18 years is 11,689
...
Hence we need 18 payments of £1,000 with a reduced
ﬁnal payment
...
Then NPV of the payments is 11,869
...
0519 and we want this to
equal 12,033
...
32
...
61
...
32 £12,033
...
The reduction is £2,000-£869
...
68
...
The cash ﬂow is
Time
Cash Flow

0
−c

1
6000

···
···

2
5800

19
2400

20
2200

So c = 6000ν + 5800ν 2 + · · · + 2400ν 19 + 2200ν 20 = 200x where x = 30ν + 29ν 2 + · · · + 12ν 19 + 11ν 20 and
(1 − ν)x = 30ν − ν 2 (1 − ν 19 )/(1 − ν) − 11ν 21 and so x = 42415
...

(ii) Cash ﬂow after payment 7 is:
Time
Cash Flow

7
0

8
4600

9
4400

···
···

19
2400

20
2200

Hence l7 = 4600ν + 4400ν 2 + · · · + 2400ν 12 + 2200ν 13 = 200νy where y = 23 + 22ν + · · · + 12ν 11 + 11ν 12 = 27225
...

year loan outstanding repayment interest due
capital repaid
loan outstanding
before repayment
after repayment
8
l7 = 27225
...
26 x8 − il7 = 2149
...
39
9
l8 = 25075
...
79 x9 − il8 = 2143
...
18
(iii)After the ninth payment, the capital outstanding is l9 = 22932
...
07
...
18 = x(ν + ν 2 + · · · + ν 11 ) − (200ν 2 + 400ν 3 + · · · + 2000ν 11 ) and so xν(1 − ν 11 )/(1 − ν) = 22932
...
09
...
5 Page 163

4
22
...
05, ν = 1/1
...
The cash ﬂow is as follows:

Time
Cash Flow

1/9/98
−c

1/7/99
1000

1/11/99
1000α

1/3/00
1000α2

1/7/00
1000α3

1/11/00
1000α4

···
···

1/11/03
1000α13

1/3/04
1000α14

Hence

[ 10
]
14
18
22
26
62
66
c = 1000 ν12 + αν12 + α2 ν12 + α3 ν12 + α4 ν12 + · · · + α13 ν12 + α14 ν12
[
]
1 − α15 ν 5
10
2
3
4
13
14
= 1000ν12 1 + αν3 + α2 ν3 + α3 ν3 + α4 ν3 + · · · + α13 ν3 + α14 ν3 = 1000ν 5/6
1 − αν 1/3
and this is 17691
...

(ii) Loan outstanding on 1/7/99 is 17,692 × 1
...
277 or £18,572
...
Hence interest repaid in ﬁrst
instalment is 18,572
...
28 and so the capital repaid is 1000 − 880
...
72
...
Capital outstanding at 1/3/01 (just after the 6th payment is made)
is
[
]
1 − α9 ν 3
4
4
32
1000α6 ν12 1 + αν12 + · · · + α8 ν12 = 1000α6 ν 1/3
= 13341
...
57
...
57 × 1
...
67 and the capital repaid is
1000α6 − 261
...
43
...
Let i denote effective interest rate per annum
...
024
...
021/3
...
0220
...
It
follows that x(1 + α + α2 ) = 10,000(1 − ν)/ [(1 − α)ν] − 10α − 20α2 and so x = 87
...
hence the initial
payment will be set at £87
...

(ii) Interest at 1 May 1994 is 10,000(1
...
Hence capital repayment is 87
...
021/3 − 1) = 21
...

(iii) Loan outstanding=(value at time k of original loan)-(accumulated value at time k of repayments)
...
0232
...
021/3
...
31
...
Then 6,708
...
Hence y == 2,036
...

[
]
[
]
n
ν
24
...

¨
(ii) Let ν = 1/1
...
Measuring in hundreds, x = 30ν + 32ν 2 + 34ν 3 + · · · + 58ν 15
...
Hence x = 35,255
...
Or use x = 28a 15 + 2(Ia) 15
...
Then y = 46ν + 48ν 2 + 50ν 3 + · · · + 58ν 7
...
5415
...
15
...
Interest = £26,754
...
08 = £2,140
...

Capital repayment is £2,459
...

Start of year 10: loan outstanding = £24,294
...
Repayment = £4,800
...
48 × 0
...
56
...
04
...
04
...
3804

11
1
x

12
2
x+2

Hence
214
...
3804 − 2ν 2 1 + 2ν + 3ν 2 + 4ν 3 ×

13
3
x+4

14
4
x+6

15
5
x+8

[
]
1 − ν5
+ 2ν 2 1 + 2ν + 3ν 2 + 4ν 3
1−ν

i
= 47
...
59
...
First we ﬁnd the monthly repayments
...
Then (60x − 10000)/50000 = 0
...
Hence x = 255
...

Hence the repayment is £255
...

Let y denote the monthly repayment for loan B
...
12
1
...
10

Page 164 Answers 3
...
J
...
4303 or £324
...
Hence answer to (i) is £600−£255
...
43 = £19
...

(ii) First we ﬁnd the effective interest rate per annum for loan A
...
96 × a(12)
...
255717 which leads to i = 0
...
Hence capital outstanding under loan A is 12 × 255
...
2
12 × 255
...
088651 × 2
...
679 or £7,043
...

Capital outstanding under loan B is 12 × 324
...
43 × 1
...
486852 = 10117
...
10
£10,117
...

So interest paid in 25th month is 7,043
...
21/12 − 1] + 10,117
...
101/12 − 1] = 188
...
52
...
96 + £324
...
52 = £391
...

(iii) Total capital outstanding is £7,043
...
83 = £17,161
...
New monthly payment is £(255
...
43)/2 = £290
...
Hence we need i with 17161
...
20a(12) ; hence we need i with a(12) = 4
...

10 ,i
10 ,i
Using tables gives a(12) = 1
...
192472 = 4
...
067016 × 5
...
355107
...
2
10 ,0
...
15 + 0
...
355107 − 4
...
355107 − 4
...
17699 or
an effective rate of 17
...

Hence interest paid in 25th month is 17161
...
1771/12 − 1] = 234
...
66
...
20 − 234
...
54 or £55
...

(iv) The new loan from Freeloans has a higher interest rate than loan B and a lower interest rate than loan A
...
Under the new arrangements, the
repayments have been halved
...

26
...
We have 100,000 = 7,095
...
0939361
...
05 or 5% p
...

Product B
...
(i)(a) The ﬂat rate of interest is
xR − l0 4800 − 2000
=
= 0
...
(b) The ﬂat rate of interest is not a good measure because it take no account of the
fact that the loan outstanding decreases each month
...

Time
0
1/12
2/12
3/12
···
20/12
21/12
22/12
23/12
Cash Flow
−2,000 + 200
200
200
200
···
200
200
200
200
Hence
2000 = 2400a(12) = 2400
¨ 2 ,i

i
d(12)

a 2 ,i

or 5 = 6

i
d(12)

a 2 ,i

Now if i = 2, then a 2 ,i = (1 − 1/9)/2 = 4/9 and (1 − d /12) = 1/3 and hence d(12) = 1
...
Hence
2
i
4
6 (12) a 2 ,i = 6
= 5
...
049823 9
Hence the claim by the consumers’ association is correct
...

Time
0
1/12
···
11/12
12/12
···
17/12
18/12
···
23/12
Receipts
−2,000 + 200
200
···
200
120
···
120
60
···
60
Costs
−60
−60
···
−60
−60
···
−60
−60
···
−60
c1
60
60
···
60
60
···
60
60
···
60
c2
60
60
···
60
60
···
60
c3
80
80
···
80
(12)

Hence
2000 = c3 + c2 = 960a(12) + 720a(12),i =
1 ,i
1
...
5 ,i

d(12)

Now 1
...
025 = 1
...
At 4%,
]
[
]
i [
1 − ν 1
...
51706 < 2000
960a 1 ,i + 720a 1
...
021537 960 × 0
...
04
Hence the banks’ case is also correct
...
Work in units of £100
...
04
...
5 Page 165

1 − ν 19
− 50ν
1−ν

and hence
L=

2ν(25 − 6ν 20 ) 2ν 2 (1 − ν 19 )
−
= 456
...
386946
1−ν

Giving the answer £45,638
...

(ii) The 12th instalment is £2,800
...
Hence
a 9 − 9ν 10
= 152
...
67, and the interest component of the 12th instalment is 15,258
...
04 = 610
...
The capital repayment is therefore £2,800 − £610
...
65
...
67 − £2,189
...
02
...
6902 = 28a k + xν k+1
Hence k = 5 and xν 6 = 130
...
651016 = 6
...
Hence x = 7
...

We have established that the remaining term of the revised loan is 6 years with 5 repayments of £2,800 and one ﬁnal
repayment of £764
...

(iii) The total of all repayments is 50 + 48 + 46 + · · · + 28 + 5 × 28 + 7
...
6415
...
6415 − 456
...
254554 or £15,925
...

29
...
Then 500,000 = xa 10 ,0
...
041327 or £77,910
...
The
total amount of interest is 10 × £77,910
...
40
...
This is xa 3 ,0
...
Let y denote
the amount of the new quarterly payment
...
12 = xa 3 ,0
...
04 × 2
...
043938 × 3
...
16
...
Hence ν = 1/1
...

Capital outstanding after ﬁrst quarterly instalment is y(ν + · · · + ν 15 ); capital outstanding after second quarterly
instalment is y(ν + · · · + ν 14 )
...
16(1 − 1/1
...
411819 or £5,383
...

30
...
04
...

...

...
04 − 20(Ia) 15 ,0
...
04 − 20

a 15 ,0
...
64507035
1−ν

or £3,052
...

(ii) Interest component is £3,052
...
04 = £122
...
So capital repayment is £400 − £122
...
89
...
So the capital outstanding is
a 7 ,0
...
04 − 20(Ia) 7 ,0
...
04 − 20
= 1099
...
18
...
59
...

Time
0
1
2

...
59
x
x+2

...

x−2
x−2
c2
0
2
4

...
59 = (x − 2)a 10 ,0
...
08 leading to x = 2 + (549
...
686907)/6
...
162495
or £74
...

31
...

9
10
Cash Flow, c
0
00
300
400

...

900
1,000

Page 166 Answers 4
...
J
...
360087 − 10ν 11
c = c1 + c2 = 100 a 10 ,0
...
06 = 100 7
...
249451415
1−ν
or £4,432
...
(ii) Immediately after the sixth repayment (at time t = 6 + 0), the loan outstanding is 800ν + 900ν 2 +
1000ν 3 + 1100ν 4 = 700a 4 ,0
...
06 = 3266
...
Hence the interest component of the seventh repayment
is £3266
...
06 = £196
...

(iii) The loan outstanding after the seventh repayment is £3266
...
64
...
Then 2662
...
08
...
3386576 or £463
...

32
...
amount owing is 300,000 × 1
...
Hence monthly interest payment is 300,000 × (1
...
45 leading to a total interest payment over the ﬁrst 15 years of 180 × 2,046
...
00
...
66 = 123,626
...
Total is £491,987
...

(ii) Let i denote the effective annual rate of interest
...
0452
...

Let z denote the monthly contribution to the savings account
...
We want this to have value £150,000 after
15 years
...
04530 z(1 − ν 15 )/(1 − x) = 150,000 and z = 150,000(1 − x)/(1
...
369219 or
£399
...

(iii) Value of savings after 15 years is 399
...
115 − 1)/(1 − 1/1
...
465880 or £160,395
...
So
outstanding loan is £139,604
...
Let y denote the new monthly repayment
...
54 = 12ya(12)
...
07
y = 139604
...
031691 × 7
...
498842 or £1,605
...

33
...
(ii) Clearly 30,000 = X(Ia) 60 ,0
...
Easier from ﬁrst principles: 30,000 = X(ν + 2ν 2 + · · · +
[
]
60ν 60 ) = Xν (1 − ν 60 )/(1 − ν)2 − 60ν 60 /(1 − ν)
...
6222 or £26
...

(iii) Let i denote the
effective rate of interest per annum and let ν = 1/(1 + j) = 1/(1 + i)1/12
...
32ν 36 a 60 ,j and
hence ν 36 a 60 ,j = 31
...
Using tables shows LHS = 31
...
01
...
011; then
LHS=29
...
Linear interpolation gives j = 0
...
0128 or 12
...

(iv) The total
repayments are 1,830X = 48,714
...
20 under the second schedule
...

Answers to Exercises: Chapter 4 Section 6 on page 77

(exs4-1
...
(a) £1,000,000 1 + 0
...
8;
(b) £1,012,328
...
045 × 67/365) = £1,004,035
...
8/(1 + 0
...
Hence we want
[
]
[
]
price on 4 May
365
1 + 0
...
050960034
price on 4 April
30
1 + 0
...
1%
...
(a) A futures contract is a legally binding contract to buy or sell an agreed quantity of an asset at an agreed price at
an agreed time in the future
...
(b) Convertibles can be converted into ordinary shares
at a ﬁxed price at a ﬁxed date in the future
...

3
...

Party A pays a series of ﬁxed payments to party B for a ﬁxed term
...

4
...

Preference shares rank above ordinary shares for payments of dividends and if the company is wound up
...

Preference shares usually do not have voting rights; ordinary shares do
...
Short term—less than one year
...

Used to fund short term spending of a government
...

Sold at a discount and redeemed at par
...

Often used as a benchmark risk-free short term investment
...
2 Page 167

6
...
The term “gilt-edged market” refers to the market in government bonds
...

(ii) Most bonds are linked to an inﬂation index but there is a
time lag
...
If inﬂation increases over the period of the bond, then the payments will not keep up with
inﬂation
...
(a) Market risk
...
If the current rate of 5
...
If the interest rate moves above 6%, the other party would have to pay out more
than it receives
...

(b) At the current time, the company does not face a credit risk, because it is paying 6% but only receiving 5
...

8
...
Entitle holders to share in proﬁts (the dividend) after interest on loans, etc have been paid
...
Higher expected returns than most other asset classes—but the risk of capital loss is also
higher
...
Possible capital gain from increase in price of share
...
Voting rights in proportion to number held
...
(a) Issued by large companies, banks and governments
...
Regular interest payments and redeemed
at par
...
Yields depend on
issuer, size of issue and time to redemption
...
Usually a bearer security
...

(b) Certiﬁcates issued by banks and building societies stating that money has been deposited
...
Interest paid on maturity
...
Active secondary market
...
(i) Purchase the security at time 0
...
Receive periodic coupon payments (usually every
6 months or every year) and a capital repayment on maturity
...
Often they are linked to the value of the index some time previously so
that there is no delay in making the payments
...
Receive periodic dividends, usually every 6 months or
every year
...
Dividends cease if company fails
and then share may have no value
...

11
...
4
...
The running yield equals income divided by price
...

Answers to Exercises: Chapter 5 Section 2 on page 84
1
...
tex)

n = nm/m
C + f r/m − t1 f r

Hence i is a solution of the equation
P =

nm
n
∑
f r ∑ k/m
ν
+ ν n C − t1 f r
νk
m
k=1

k=1

= f ra(m) + Cν n − t1 f ra n
n ,i
2
...
For bond A we have P = f (n1 , i1 ) and for bond B we have P = f (n2 , i2 )
...
Hence i1 = i2
...
Also f (n, i1 ) ↘ as n ↗
...
But
1
f (n, i) ↘ as i ↗
...

(c) If P > C then i(m) < (1 − t1 )g
...
Hence f (n2 , i1 ) > f (n1 , i1 ) = P = f (n2 , i2 )
...

Page 168 Answers 5
...
J
...
The cash ﬂow is:
Time
Cash ﬂow

0
−P

t0
(1 − t1 )f r/m

t0 + 1/m
(1 − t1 )f r/m

t0 + 2/m
(1 − t1 )f r/m

···
···

t0 + 3/m
(1 − t1 )f r/m

) (1 − t )f rν t0
(1 − t1 )f r ( t0
1
1
ν + ν t0 +1/m + ν t0 +2/m + · · · =
= (1 − t1 )f rν t0 a(m)
¨∞
m
m
1 − ν 1/m
Recall (1 − d(m) /m)m = 1 − d = ν and hence d(m) = m(1 − ν 1/m )
...

P =

4
...
A higher yield means that money is more expensive
...
The current
yield on bonds is 10% p
...
and the coupons on the original bond are only 8% p
...
Hence, the bond issuer can make
a proﬁt by purchasing a new bond to pay the coupons on the existing bond rather than redeeming the existing bond
...

(b) Use equation 1
...
We assume no tax
...
08 − 0
...
02a n ,i
...
Hence later redemption implies means a higher yield
...
The cash ﬂow is as follows (where t1 = 8/365):
Time
0
t1
t1 + 1/2
t1 + 1
Cash ﬂow, c
−P
0
4
4

t1 + 3/2
4

t1 + 2
4

···
···

t1 + 13/2
4

t1 + 7
104

Because the bond is ex-dividend, the payment at time t1 is zero
...

7
Using tables gives
[
] 8 × 1
...
582381 + 66
...
682
7
1
...
068/365
Alternatively, if x = 1/1
...
5 , we have
[
]
[
]
1
1
1 − x14
2
13
14
14
13
P =
4 x + x + · · · + x + x + 25x =
4x
+ 25x
= 111
...
068/365
1
...
Now each coupon after tax is 350 × 0
...
50
...

1/10/2009
1/4/2010
Payment no
...

37
38
Cash ﬂow
−P
262
...
50

...
50
10,262
...
0591 and (1 − t1 )g = 0
...
07 = 0
...
As i(2) > (1 − t1 )g, assume bond will be redeemed at the
latest possible date (and then yield will be at least 6% whatever the redemption date)
...
06
...
50
ν 0
...
5k + 10000ν 18
...
75 = 9385
...
051/2 − 1 = 0
...
75 × 0
...
0525
...
Hence the cash ﬂow for the second purchaser is as
follows:
Time
1/4/99
1/10/99
1/4/00

...

0
1
2

...
50
262
...

262
...
50
= 262
...
25

Let ν = 1/1
...
Hence
P = 262
...
50ν 0
...

Time (years)
Cash ﬂow

0
−96

1
4

2
4

1 − ν5
+ 10000ν 5 = 10136
...

...

25ν
]
Using method (d) in paragraph 4
...
044
...
04, RHS = 25
...
045, RHS = 23
...
Using interpolation gives (x − 0
...
045 − 0
...
04))/(f (0
...
04))
and hence x = 0
...
005(24 − 25
...
374 − 25
...
043
...
3%
...
2 Page 169

8
...
08 and so i(2) = 2[1
...
078461
...
77×95/110 = 0
...
Hence i(2) > (1−t1 )g
and so CGT is payable
...
77 × 9
...
08 + 110ν 20 − 0
...
77 × 9
...
08 + (72
...
34P )ν 20
20
and so
P (1 − 0
...
77 × 9
...
08 + 72
...
79
...
(i) Credit risk: most governments will not default
...
Income stream may be volatile relative to
inﬂation
...
06
...
7a 5 + 50ν 5 + 4 × 0
...
6 + 2
...
6855
...
7 × 4(a 5 + a 10 ) + 50ν 5 (1 + ν 5 ) = 97
...

10
...
041/4 − 1] = 0
...
Also (1 − t1 )g = 0
...
0388
...

Using units of £1,000 we have P = 5 × 0
...
25(103 − P )ν 20 = 4a(4) + 77
...
25P ν 20
...
25ν 20 )P = 4a(4) + 77
...
07201 or £102,072
...

20
11
...
25, i(2) = 2(1
...
04939, and (1 − t1 )g = 0
...
0477
...

[
]
Using the formula P (n, i) = C + (1 − t1 )g − i(m) Ca(m) shows there is a capital gain
...

(iii) Assuming the loan is redeemed at the latest possible date, P = 0
...
5[x+x2 +· · ·+x30 ]+[110−0
...
051/2
...
625x(1 − x30 )/(1 − x) + [77 + 0
...
Hence (1 − 0
...
625x(1 −
x30 )/(1 − x) + 77x30 giving P = 107
...
80
...
25a(2) + 77ν 15 )/(1 − 0
...
75383 or £107,753
...

15
12
...
Let ν = 1/1
...
05 + 100ν 15 = 4 ×
15
1
...
379658 + 100/1
...
1330
...
051/2 − 1] = 0
...
75 × 4/100 = 0
...
Hence i(2) > (1 − t1 )g
+ 100ν 7 − (100 − P ) × 0
...
Hence
and CGT is payable
...
06; then P = 4 × 0
...
06
(1 − 0
...
5203
...
06
(b) The cash ﬂow was as follows:
Time (years)
0
0
...
5
2
2
...
5 · · · 7
...
1330
2
2
2
2
2
2
2
···
2
2 + 77
...
1330 1
...
5 1
...
5 1
...
5 1
...
5 1
...
5203
Let i denote the required effective rate of return per annum and let x = 1/(1+i)1/2
...
133 = 3a 8 ,i +77
...

[
]
Using method (d) in paragraph 4
...
5203 − 90
...
133 = 0
...

Deﬁne f (i) = 1
...
5203x16 where x = 1/(1 + i)1/2
...
015) = 91
...
02) =
88
...
Using linear interpolation gives (x − 0
...
133 − 91
...
02 − 0
...
2486 − 91
...
015 + 0
...
2207/3
...
016966
...
7% to the nearest 0
...
(More
precisely, it is 1
...
)
13
...
75 × 10a(2) ,0
...
26
1
...
26 per £100 nominal
...
6 × 10/110 = 6/110 = 0
...
061/2 − 1] = 0
...
Hence
i(2) > (1 − t1 )g and so the investor is liable for CGT and the price is given by
110 − 0
...
4P
= 6a(2)
P2 = 0
...
06
2 ,0
...
062
1
...
544
...
5a(2) + P2 /(1 + i)8
...
3 gives i ≈ 7
...
081
...
08 gives RHS = 102
...
Trying i = 0
...
689
...
08 +
0
...
08))/(f (0
...
08)) = 0
...
005 × (102
...
588)/(99
...
588) = 0
...
1% approximately
...
2

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

14
...
06 + 100/1
...
53
...
Now (1−t1 )g = 0
...
056 and i = 0
...
Hence i > (1−t1 )g
and CGT is payable
...
7 × a 10 ,0
...
3(100 − P2 )/1
...
5a 10 ,0
...
3P2 ]/1
...
5a 10 ,0
...
33
1 − 0
...
53 = 5a 10 ,i + 74
...

[
]
Using method (d) in paragraph 4
...
33 − 88
...
53 = 0
...
So try 4
...
4267
...
7692
...
045)/0
...
53 −
87
...
4267 − 90
...
Hence i = 0
...
335%
...
(i) Now t1 = 0
...
08
...
051/2 − 1) = 0
...
Hence g(1 − t1 ) = 0
...
6 =
0
...
So assume redemption at the latest time of 2011
...
Let ν = 1/1
...
6
...

1/7/2010
1/1/2011
Cash ﬂow
−P
4β
4β

...
3(100 − P )
Hence, if x = ν 1/2 = 1/1
...
6 × 8a(2) ,0
...
3P
1
...
668
1−x
(ii) In this case, g = 0
...
071/2 − 1) = 0
...
And so g(1 − t1 ) = 0
...
So assume
redemption at the earliest time of 2006
...
07 and x = ν 1/2 = 1/1
...

Time
1/1/03
1/7/03
1/1/04
1/7/04
1/1/05
1/7/2005
1/1/2006
Cash ﬂow
−P2
4
4
4
4
4
104
P (1 − 0
...
4x

Hence P2 = 8a(2)
+ 100ν 3 = 4x(1 − x6 )/(1 − x) + 100x6 = 102
...

3 ,0
...
4
2
...
4
2
...
3(P2 − P ) = 2
...
69
Hence 98
...
8a(2) + 101
...
4(x + x2 + x3 + x4 ) + 101
...
Using method (d) in
2 ,i [
]
paragraph 4
...
8 + (101
...
67)/2 /98
...
064
...
06
...
67 − 4
...
69ν 2 = −0
...

2 ,i
Try i = 0
...
Then 98
...
8a(2) − 101
...
1353
...
06)/0
...
764)/(0
...
764) and so i = 0
...

Hence i(2) = 2 × (1
...
0632
...
We work in multiples of 100,000
...
09 +
= 8 (4) a 15 ,0
...
821947
15
15
1
...
0915
and hence £9,682,194
...

(ii) Now (1 − t1 )g = 0
...
4/110 = 0
...
071/4 − 1] = 0
...
Hence
i(4) > (1 − t1 )g and so CGT is paid
...
2(110 − P )
i
88 + 0
...
4 (4) a 10 ,0
...
8 × 8a(4) ,0
...
0710
i
1
...
131 and so answer is £101
...

(iii) Running yield is 100(1 − t1 )r/P2 = 0
...
131 = 0
...
382%
...
Using method (d) in paragraph 4
...
092
...
09 gives RHS = 97
...
Trying i = 0
...
275
...
09+0
...
09))/(f (0
...
09)) = 0
...
005×(96
...
877)/(96
...
877) = 0
...
3% approximately
...
(i) Now (1 − t1 )g = 0
...
0613 and i(2) = 2[1
...
059
...
The price P1 is given by
P1 = 0
...
06 + 110ν 13 = 112
...
21
...
9 × 9/110 = 0
...
081/2 − 1] = 0
...
Hence i(2) > (1 − t1 )g and so there

Appendix

Jan 28, 2016(12:36)

Answers 5
...
The price P2 is given by
110 − 0
...
5 + 0
...
9 × 9a(2) ,0
...
1 (2) a 11 ,0
...
0811
i
1
...
45468 or £105
...

(b) The cash ﬂow for the ﬁrst investor is as follows (note there is no capital gain):
Time
0
0
...
5
2
Cash ﬂow before tax
−P1
4
...
5
4
...
5 + P2
Cash ﬂow after tax
−P1
3
...
375
3
...
375 + P2
[
]
Then P1 = 6
...
Using method (d) in paragraph 4
...
75 + (P2 − P1 )/2 /P1 = 0
...

2 ,i
Let f (i) = P1 − 6
...
Then f (0
...
203 and f (0
...
854
...
03)/0
...
01/(1
...
203) and hence i = 0
...
So the answer is 3% to the nearest 1%
...
Now i = 0
...
061/4 − 1) = 0
...
Also (1 − t1 )g = 0
...
08/105 = 0
...
Hence
i(4) > (1 − t1 )g and the purchaser should value on the assumption that redemption takes place at the latest possible
time
...
So we want P with
105 − 0
...
5 + 0
...
8a(4) ,0
...
8 (4) a 20 ,0
...
06
i
1
...
37 or £87,370
...
8 × 8/105 = 0
...
058695
...
The
value of the coupons and redemption payment at 12 months after the time of original issue is
i
x = 0
...
8 × 8a(4) ,0
...
6 + 6
...
06 + 105ν 14 = 108
...
8 × 2 is for the coupon at time 12 months
...
7997 and so the price
is £107,799
...

(b) Again, there is no CGT
...
8 × 2 + 0
...
06 + 105ν 19 = 1
...
4 (4) a 19 ,0
...
0619
1/6
and hence P = ν x = 108
...
72
...
70
...

√
19
...
3 and g = 8/100 and hence (1 − t1 )g = 0
...
06 − 1) = 0
...
Hence
i(2) > (1 − t1 )g
...

Time
1/5/11 1/7/11 1/1/12 1/7/12 · · · 1/1/17 1/7/17 · · · 1/7/21
1/1/22
Before tax
−P
4
4
4
···
4
4
···
4
104
After tax
−P
2
...
8
2
...
8
2
...
8
2
...
25P
Let ν = 1/1
...
Then

(
)
P = 1
...
6a(2) ,0
...
25P )ν 11
11

and hence
P =

1
...
6a(2) ,0
...
061/3 × 75ν 11
11

= 99
...
061/3 × 0
...
31887
...
07 − 1) = 0
...
Hence (1 − t1 )g > i(2)
...

Time
1/4/13
1/7/13
1/1/14
1/7/14
1/1/15
1/7/15
1/1/16
1/7/16
1/7/17
−P1
4
4
4
4
4
4
4
104
Let ν1 = 1/1
...
Then
or £105
...

Time
Before Tax
After tax

(
)
4
P1 = 1
...
624983612
4 ,0
...
31887
−99
...
8

1/1/12
4
2
...
8

1/1/13
4
2
...
625
104
...
081/3 × (2
...
081/2 + 2
...
08 + 2
...
083/2 + 2
...
082 ) + 104
...
0823/12 = 100
...

Also 1
...
8/1
...
8/1
...
8/1
...
8/1
...
0485/1
...
568
...

Page 172 Answers 5
...
J
...
Now (1 − t1 )g = 0
...
04861 and i(m) = i(4) = 4(1
...
04909
...
Let P denote the price per £100
...
75 × 7a(4) ,0
...
35(108 − P )] /1
...
Hence
[
]
0
...
2
P 1−
= 5
...
05 +
leading to P = 107
...
38
...
0520
1
...
(i) Let ν = 1/1
...
Then P1 = 5a 18 ,0
...
9988 or £125
...

(ii) Now let ν = 1/1
...
Then P2 = 5a 13 ,0
...

(iii) Let ν = 1/(1 + i) where i is the required gross annual rate of return
...

This implies ν = 1 and i = 0
...

(iv) The value of P1 would be larger
...
This implies i < 0
...
(i) Deﬁne i by 1 + i = 1
...
Let i′ = 0
...

Time
0
1/2
2/2
3/2
···
19/2
20/2
Cash ﬂow
−P
2
2
2
···
2
2
So P = 4a(2) ,i + 100/(1 + i)10 = 2a 20 ,0
...
01520 = 108
...
58
...
5a 20 ,0
...
01520 and hence the price 91 days later is
P × 1
...
745153968 or £100
...

23
...
Let P denote the price of bond B then the expected return is
[0
...
2(100 − P ) + 0
...
4(106 − P )] /P = [77
...
4 − P ] /P = 5/101 which gives P = 77
...
749057 or e73
...

(ii) Let i′ denote the gross redemption yield—the gross redemption yield always ignores the possibility of default
...
749(1 + i′ ) = 106 giving 1 + i′ = 1
...
73%
...

Answers to Exercises: Chapter 5 Section 5 on page 91

(exs5-2
...
If the payments are linked exactly to the inﬂation index, then the real yield will be the same whatever happens to
inﬂation
...

2
...
01) = (1 − 0
...
Hence 1 + iR = 1
...
98 and so
iR = 3/98 = 0
...
061%
...
Let α = 1
...
08
...
15
P =5
αk ν k/2 =
=
=
= 557
...
93
...
03
1 − αν
(1 + i)
1
...
Let α = 1
...
07
...
05
P =2
αk ν k/2 =
=
=
= 217
...
025
1 − αν
(1 + i)
1
...
90
...
(i) Issued by a government; coupons paid by government—hence virtually no risk of default in most cases
...
hence they provide protection
against inﬂation
...
If inﬂation is higher over these months,
then the payments will not keep up with inﬂation
...
(a) If i denotes the money rate of return, then 10(1 + i) = 11
...
11 or 11%
...
07143 or 7
...

(c) Using (1 + 0
...
6 × 0
...
005067, or
−0
...

[
]
[
]
7
...
04 and ν = 1/1
...
Then P = 12 ν 2/12 + αν 14/12 + α2 ν 26/12 + · · · = 12ν 1/6 1 + αν + α2 ν 2 + · · · =
12ν 1/6 / [1 − αν] = 428ν 1/6 = 423
...
The cash ﬂow is as follows:
Time
0
1
2
3
4
5

...
08)
800(1
...
07)
800(1
...
07)α
800(1
...
07)α2

...
05
...
07
...
48ν 3
= 41876
...
08)ν 2 +
800(1
...
07)αk−3 ν k = 800ν + 864ν 2 +
1 − αν
k=3

Giving an answer of £418
...

Appendix

Jan 28, 2016(12:36)

9
...
0
−20

Answers 5
...
2

2
2
282
...
5

2×245
...
2

2×245
...
2

22×245
...
5

2 × 245
...
0ν 2 22 × 245
...
2
282
...
5
268
...
2
305
...
0 to 305
...
Setting (1 + j)3 = 305
...
0 gives j = 0
...
General result (1 + iR )(1 + j) = 1 + iM where iR is the real rate, j is the inﬂation rate and iM is the
money rate, leads to iR = 0
...
(Or use approximation ν k ≈ 1 − ki
...
022 gives ν = 1/1
...
978
...
021 gives ν = 1/1
...
032
...
021)/(0
...
021) = (f (x) − f (0
...
022) − f (0
...
021 +
0
...
032)/(19
...
032) = 0
...
16%
...
Effective real yield per annum,iR , is given by 1 + iR = 1
...
Cash ﬂow is:
Time
0
10/12
22/12
34/12
Cash ﬂow
0
5
5 × 1
...
032
Inﬂation Index
1
1
...
0222/12
1
...
01252 , we have
∞
∞
∑
∑ 1
...
02β)1/6
5 × 1
...
02
1
...
03
1
...
02β)
R
n=0
n=0
Hence x = 321
...
217
...
Let νm = 1/(1 + iM ) where iM is the money rate of return
...
Using the general result that (1 + iR )(1 + e) = 1 + iM where iR is the real rate, e is the inﬂation
rate and iM is the money rate, we get iR = (d + g − e)/(1 + e) as required
...
Let α = 1
...
05
...
75
1
...
75
3
...
75
α1
...
75
α3
...
75+k 10ν 0
...
75
1
250 =
= 0
...
75+k
k
α
α
α
1
...
75 1 − 1
...
03
k=0

...

...
05ν + 1
...
25 ν 0
...
03 = 26
...
030
...
75 − 25
...
5 + 1
...
25 (1 + i)0
...
75i
We need an initial approximation
...
25 by 1 + 0
...
This leads to i ≈ 0
...
030
...
75 − 0
...
030
...
059
...
25 = 1 + 0
...
059
...
05)
...
05 + 1/25 = 0
...
The approximation used
implies that iM is greater than 0
...

Now use the general result that (1 + iR )(1 + e) = 1 + iM where iR is the real rate, e is the inﬂation rate and iM is the
money rate
...
09/1
...
058
...
058
...
058 then 0
...
030
...
25 − 25
...
0282
...
059 then 0
...
030
...
25 − 25
...
0027
...
0595 then 0
...
030
...
25 − 25
...
010
...
059 and 0
...
059)/(0
...
059) = (f (x) − f (0
...
0595) −
f (0
...
059 + 0
...
0027)/(−0
...
0027) = 0
...
91%
...
(i) Next expected dividend is d1 ; rate of dividend growth is g; let ν = 1/(1 + i)
...

(ii) Analyst I has g = 0; hence 750 = 35/i and i = 35/750 = 7/150 = 0
...
667%
...
1; hence 750 = 35/(i − 0
...
1 + 0
...
14667 or 14
...

Page 174 Answers 5
...
J
...
See answer to exercise 12
...
015 and β = 1
...
The cash ﬂow is as follows:
Time (in years)
0
0
...
25
2
...
25
Cash ﬂow, unadjusted
−125
5
5β
5β 2
5β 3
0
...
25
2
...
25

...

...
25+k
k=0

α0
...
25 ∑ β k ν k
5ν 0
...
25
k
α
α
1
...
25 1 − 1
...
015
k=0

and this leads to
0 = 26ν + 1
...
75 ν 0
...
375
or, multiplying by 1 + i,
0 = 0
...
0150
...
75 − 25
...
Using the ﬁrst method
means we approximate (1 + i)0
...
75i
...
625 + 1
...
75 (1 + 0
...
375i and hence
(
) (
0
...
75
i = 0
...
015
/ 25
...
75 × 1
...
0664
...

If i = 0
...
625 + 1
...
75 (1 + i)0
...
375i = 0
...
If i = 0
...
625 + 1
...
75 (1 + i)0
...
375i = −0
...
Linear interpolation between 0
...
067 gives (x − 0
...
067 − 0
...
0664)/(f (0
...
0664)) and hence x = 0
...
0006(0 − 0
...
0135 − 0
...
0665 or
6
...

15
...

...

...
5(1 + i)1/3
=
1/2 (1 + g)1/2
1−ν
(1 + i)1/2 − 1
...
041/2 = 0
...
This leads to i = 0
...

As in the answer to exercise 12, the value of i will be larger than this approximation
...
096, then 18(1 + i)1/2 − 18 × 1
...
5(1 + i)1/3 = −0
...

If i = 0
...
041/2 − 0
...
0059
...
6% and 10%, or 10% to nearest 1%
...
(i) Are a share in the ownership of a company
...
From point of view of investor:
receive dividends and potential growth in share price
...
No ﬁxed redemption time
...
Shareholders can vote in AGM of company
...
Let α = 1
...
03 and ν = 1/(1 + i) where
i = 0
...
Then the cash ﬂow is as follows:
Time (in years)
0
0
...
5
2
...
5

...

Inﬂation index
1
α0
...
5
α2
...
5

...
5
k=0

and hence

αk+0
...
5
1
dν 0
...
5 d(1 + i)0
...
5
=
=
0
...
02 × 1
...
03
=
= 0
...
050
...
020
...
5 Page 175

17
...
0625
...
06255 ) = 2390
...
13
...
Let a = 1
...
045 × 1
...
065 × 1
...
Then money interest earned is £10,000a
...
6
...
6 and so
iR = (240
...
6)1/5 − 1 = 0
...
61%
...
Redemption proceeds in January 1998 are 10,000 × (274
...
6) × 1
...
Hence, if iM denotes
the effective money rate of return per annum, then (1 + iM )5 = (274
...
6) × 1
...

Of course, we want the real rate of return, iR where (1 + iR )5 = (274
...
6) × 1
...
6
...
0 × 240
...
6 × 275
...
0275 − 1 = 0
...
84%
...
The cash ﬂows are as follows (where x is given in part (i)):
Date
1/93
1/94
1/95
1/96
1/97
Cash ﬂow
−10,000
x
x
x
x
Inﬂation index
240
250
264
...
6
270
...
6

Hence the real rate of return, iR , satisﬁes
(
)
240 2
240 3
240 4
240 5
240
10,000 = x
νR +
ν +
ν +
ν +
ν
250
264
...
6 R 270
...
6 R
(
)
νR
ν2
ν3
ν4
1
+
+ R + R + R
= 240xνR
250 264
...
6 270
...
6
Question only asks which is the greatest real rate of return
...
0284
...
0284 gives RHS ≈ 9967
...
84% and the answer is
Investment B
...
(i) Now g = 6/100 = 0
...
25
...
08 > (1 − t1 )g and so there is a capital gain
...
75
1
...
75
9
...
5
4
...
5
104
...
25(100 − P )
= 79
...
25P
Let x = 1/1
...
Hence:
P = 4
...
75 + x1
...
75 ] + (75 + 0
...
75
= 4
...
75 [1 + x + · · · + x9 ] + (75 + 0
...
75
and hence

(
P (1 − 0
...
75

0
...
5x

1 − x10
1−x

)
+ 75x9
...
06
...
08 = 1
...
4854 or 4
...

19
...
Let α = 1
...
The cash ﬂow is as follows:
Date
0
0
...
5
Payments, gross
−99
4
4
4

···
···

25
4 + 110

The tax payments on the coupons will be 2 at time 1
...
25,
...
25
...
3 at time 25
...

25
0
...
Then 99 = 8a(2) ,iM + 110νM − 2νM a 25 ,iM −
25
[ i
]
0
...
25
25
25
...
3νM = 8 i(2) − 2νM a 25 ,iM + 110νM − 3
...

We need an approximate answer for iM
...
7 after
tax
...
3 gives iM ≈ (6 + 7
...
0637 or 6
...

(d)
]
i
25
25
...
25
If iM = 0
...
3νM = 103
...

[ i
]
25
25
...
25
If iM = 0
...
3νM = 97
...

Using linear interpolation gives iM = 0
...
005 × (99 − 103
...
241 − 103
...
0636
...
0636/1
...
032 or 3
...

(ii) If tax was collected later, then the real rate would increase
...
See answer to exercise 12
...
03 and β = 1
...
The cash ﬂow is as follows:
Time (in years)
0
0
...
667
2
...
667
Cash ﬂow, unadjusted
−21
...
1
1
...
1β 2
1
...
667
α1
...
667
α3
...

...

Page 176 Answers 5
...
J
...
5 =

∞
∑ 1
...
667+k
k=0

α0
...
1ν 0
...
1ν 0
...
667
αk
1
...
667 1 − 1
...
03
k=0

and this leads to
0 = 22
...
1 × 1
...
333 ν 0
...
145
or, multiplying by 1 + i,
0 = 0
...
1 × 1
...
333 (1 + i)0
...
145i
For an initial approximation, either of the methods in the answer to exercise 12 can be used
...
333) by 1 + 0
...
This leads to 0 = 0
...
1 × 1
...
333 (1 + 0
...
145i and
+
(
hence i = 0
...
1 × 1
...
333 / 22
...
333 × 1
...
030
...
070, rounded down
...

If i = 0
...
43 + 1
...
030
...
333 − 22
...
0160
...
071, then 0
...
0150
...
75 − 25
...
00575
...
07 and 0
...
07)/(0
...
07) =
(f (x) − f (0
...
071) − f (0
...
07 + 0
...
016)/(−0
...
016) = 0
...
07%
...
The cash ﬂows were as follows:
Date
1/6/2000
1/12/2000
1/6/2001
1/12/2001
1/6/2002
Payments, not indexed
−94
1
...
5
1
...
5 + 100
Payments, indexed
−94
1
...
5 × 107/100 1
...
5 + 100) × 113/100
−94
1
...
605
1
...
695 + 113
(ii)(a) Before indexing, £94 grows to £113
...
Hence the
CGT is £0
...
489
...
Then 94 = 0
...
53ν 0
...
605ν + 1
...
5 + 1
...
489)ν 2
...
59ν 0
...
815ν + 4
...
5 + 451
...
Transforming into an equation in i
gives 376(1 + i)2 = 4
...
5 + 4
...
995(1 + i)0
...
129 or 376i2 + 747
...
944 − 4
...
5 − 4
...
5 = 0
...
5 ≈ 1 + 0
...
8i − 89
...
The usual formula for
the solution of a quadratic gives i = 0
...

If i = 0
...
185i − 79
...
59(1 + i)1
...
995(1 + i)0
...
8344
...
12 then 376i2 + 747
...
944 − 4
...
5 − 4
...
5 = 4
...

Linear interpolation gives i = 0
...
01 × (3
...
40588 + 3
...
1147 or 11
...

22
...
The cash ﬂow is as follows:
Date
0
1
Payments, not indexed
−25
10
Retail Price Index
170
...
3
Payments, indexed to time t = 0
−25
10 × 170
...
3

2
10
191
...
7/191
...
9
10 × 170
...
9

(i)(a) Let iR denote the real rate of return and let νR = 1/(1 + iR )
...
3
191
...
9
183
...
0
200
...
7465 = 47
...
037
...
037 then 25(1 + iR )3 − 183
...
0 (1 + iR ) − 200
...
0998
...
03 then 25(1 + iR )3 − 183
...
0 (1 + iR ) − 200
...
2636
...
03 + 0
...
2636)/(0
...
2636) = 0
...
51%
...
Then 25 = 10a 3 ,iM
...
095 = 2
...
1 = 2
...
Linear interpolation gives iM = 0
...
005×(2
...
508907)/(2
...
508907) = 0
...
7%
...
9/170
...
0558 or 5
...
The
answers to part (i) suggest 1 + e = (1 + iM )/(1 + iR ) = 1
...
0351 = 1
...
98%
...

23
...

(ii) Let α = 1
...
09
...
02
25 × 1
...
04
25 × 1
...
04α
25 × 1
...
04α2
Hence

NPV = 25
...
52ν 2 + 26
...
52α2 ν 4 + · · · =
(iii) The cash ﬂow is as follows:

25
...
52ν 2 25
...
03 + 26
...
40
1
...
09 × 0
...
02
1
...
5 Page 177
2
25 × 1
...
04 + 900
1
...
035

Hence
25
...
3925ν 2
+
or 874
...
3925ν + 926
...
03
1
...
957195 and hence i = 0
...
47%
...
(i) Let νM = 1/(1 + iM ) denote the money values
...
5a(2) M + 100ν 9
9 ,i

[
]
We need to ﬁnd iM
...
3 gives iM ≈ 7
...
063
...
065
...
5 × 1
...
656104 + 100 × 0
...
4545
...
07
...
5 × 1
...
515232 + 100 × 0
...
0983
...
065)/0
...
4545)/(104
...
4545) and hence iM = 0
...

Using the general result that (1 + iR )(1 + e) = 1 + iM where iR is the real rate, e is the inﬂation rate and iM is the
money rate, we get iR = 1
...
025 − 1 giving 3
...

(ii) let α0 = 159
...
5 and α = 1
...
The cash ﬂow is as follows:
Date
9/10/97
8/4/98
8/10/98
···
8/4/06
9/10/06
1/
Time
0
1
···
8 1/2
9
2
Cash ﬂow
0
α0
α0 α
···
α0 α16
α0 α17
Hence if iM denotes the money rate of return with νM = 1/(1 + iM ), we have
18
∑
k/2
9
NPV =
α0 αk−1 νM + 100α0 α17 νM
k=1

Now (1 + iR )(1 + e) = 1 + iM gives νR = (1 + e)νM = α2 νM and so
18
∑ α0 k/2
α0 9 α0
9
9
NPV =
ν
+ 100 νR =
(2a(2) R + 100νR ) = 2
...
2668(2a(2) R + 100νR ) or
9 ,i
iR
9
9
44
...
3 gives i ≈ 1 + (50 − 44
...
7767 = 0
...

Using tables, iR = 0
...
1649 and iR = 0
...
3602
Interpolating gives (x − 0
...
035 − 0
...
03))/(f (0
...
03))
...
03 +
0
...
7767 − 46
...
3602 − 46
...
034 or 3
...

25
...

Usually the values of the coupon and redemption payments are linked to some inﬂation index, with a time lag
...

The security of government payments tends to lead to a low volatility of return and a low expected return compared
to other investments
...
0250
...
015, the 6-month discount factor
...
2
−P

1 Jan 05 · · · 1 Jan 09

1 July 09

···
1
· · · 113
...
8α
110

1 + 100
113
...
8

1
113
...
8α2

113
...
8×113
...
8α
110
0
...
8α
110
14/6

113
...
8×113
...
8α2/6 113
...
8α
α2

113
...
8×113
...
8α
···
α11

101×113
...
8+100)×113
...
8α
α12

Hence
0
...
8α10 x11 100
...
8α11 x12
0
...
8x 0
...
8αx2 0
...
8α2 x3
+
+
+ ··· +
+
P =
2
3
110α
110α
110α
110α11
110α12
2
3
11
0
...
8x 0
...
8x
0
...
8x
0
...
8x
100
...
8x12
=
+
+
+ ··· +
+
110α
110α
110α
110α
110α
]
113
...
8(1 + x + x2 + · · · + x11 ) + 100x11
110α

Page 178 Answers 5
...
J
...
8x
1 − x12
113
...
8
+ 100x11 =
0
...
383
110α
1−x
110α
0
...
(i) The coupon is £3/2 and this must be adjusted by the inﬂation index at 7/99 (eight months before 3/00)
...
7
×
= 1
...
5
(ii) Let x = 1
...
4/110
...
The
cash ﬂow is as follows:
Time (in years)
0
0
...
5
2
2
...
7
Inﬂation factor
αx4/12
αx10/12
αx16/12
αx22/12
110
...
7 1
...
5αx10/12 1
...
5αx22/12
2 110
...
7 1/2
3/2
5/2
5/2
2
111 =
νM + αx4/12 νM + αx10/12 νM + αx16/12 νM + αx22/12 νM + 100αx22/12 νM
2 110
...
Recall (1 + iR )(1 + e) = 1 + iM where iR is the real rate, e is the inﬂation
rate and iM is the money rate
...
Hence
(
)
3 126
...
5
1/2

Let γ = νR and deﬁne i1 by γ = 1/(1 + i1 )
...
7 −1/2
111 =
x
γ + αx−8/12 γ 2 + αx−8/12 γ 3 + αx−8/12 γ 4 + αx−8/12 γ 5 + 100αx−8/12 γ 5
2 110
...
7 −1/2
3
=
x
− αx−8/12 γ + αx−8/12 a 5 ,i1 + 100αx−8/12 γ 5
2 110
...
0017γ + 1
...
3186γ 5
This must be solved numerically; so we need a rough starting value
...
03 corresponds to i1 =
1
...
014
...
015, then RHS = 112
...
020, then RHS = 109
...
015)/(0
...
015) = (f (x) − f (0
...
020) − f (0
...
015 + 0
...
320)/(109
...
320) = 0
...
Hence νR − 1 = 1/γ 2 − 1 = (1 + i1 )2 − 1 = 0
...
53%
...
5 means that γ must increase
...

27
...
071/2 and β = 206/200
...

Inﬂation index
Payments before adjustment
Payments after adjustment

5/97
0
200
−P
−P

11/97
1
206
2
2β

5/98
2
206α
2
2βα

11/98
3
206α2
2
2βα2

5/99
4
206α3
2
2βα3

···
···
···
···
···

Let iM denote the effective money rate of return and νM = 1/(1 + iM )
...

Hence (i + iR )(1
...
Also νR = 1
...

P =

30
∑

2β ∑ k/2 100β 15
ν
νR +
α
α R
k=1
(
)
β
100
=
2a 30 ,0
...
01530
30

k/2

15
2βαk−1 νM + 100βα29 νM =

k=1

2β
100β
1
P =
a 30 ,0
...
01530
(b) Hence P = 111
...

(ii) Now let γ = 1
...
The cash ﬂow is now as follows (where P = 111
...

Inﬂation index 8 months ago
Payments before adjustment
Payments after adjustment

5/97
0
200
−P
−P

11/97
1
206
2
2β

5/98
2
206γ
2
2βγ

Answers 5
...
97
...
The RPI at 5/12, 15 years later, is 206γ 1/3 γ 30
...
53 × γ 30 = 231
...
Hence
there is no capital gains tax
...
Hence
(
)
(
)
β
100
1
...
53 =
2a 30 ,iR +
=√
γ
(1 + iR )30
(1 + iR )30
1
...
3 gives iR ≈ 1 + (50 − 55
...
478 = 0
...

if iR = 0
...
004
...
02, then RHS = 50
...

Interpolating gives (x − 0
...
02 − 0
...
015))/(f (0
...
015)) and so x = 0
...
005(55
...
004)/(50
...
004) = 0
...
54%
...
478 = a 30 ,iR +

28
...
06
...

...
Hence α = (65 × 1
...
0132353 or 1
...

(ii) Let α = 1
...
Then 260 = 12(αν + α2 ν 2 + · · · + α12 ν 12 ) + 500ν 12
...
03(1 + i′ ) = 1 + i
...
03/(1 + i) = αν then
260 = 12x(1 − x12 )/(1 − x) + 500x12 /1
...
Let f (i′ ) = 260 − 12x(1 − x12 )/(1 − x) − 500x12 /1
...
We have
f (0
...
888284 and f (0
...
977237
...
06 − 0
...
06)/(f (0
...
06)) = 0
...
6% per annum
...
61%
...
(i) The price is given by
P = 0
...
05
1
...
053
0
...
05
63
+ 0
...
3 ×
+ ··· =
=
= 7
...
09
1
...
09
1
...
05
8

or £7
...

(ii)(a) The return from the share is higher; hence P will be higher
...

However, often the rate of inﬂation is just (incorrectly) added—hence 1
...
05 + I and 1
...
09 + I where I is the increase in the inﬂation rate
...
3 × (1
...

0
...

30
...
03, β = 1
...
06 and ν] = 1/1
...

(i) Required answer is
[
]
3500 αν + αβν 2 (1 + γν + γ 2 ν 2 + · · ·) = 3500 αν + αβν 2 /(1 − γν) = 178581
...
81
...

1/1/2012
31/12/2012
31/12/2013
31/12/2014
Date:
Cash ﬂow (in pounds):
-1,720
35α
35αβ
35αβγ + 1,800
Inﬂation Index:
110
...
3
113
...
8
Equation for ν is
35α
35αβ 2 35αβγ + 1,800 3
1,720
=
ν+
ν +
ν or 172 = 3
...
67824647ν 2 + 177
...

110
...
3
113
...
8
So let f (i) = 3
...
67824647ν 2 + 177
...
Now the dividend in 2012
gives a real rate of return of (35α/1720) × (110/112
...
Over the three years, there is also some capital gain
...
03
...
03) = −2
...
Try 2
...
025) = 0
...
So the required
value lies satisﬁes 0
...
03
...
025 + 0
...
114020/(2
...
114020) = 0
...
52%
...
2

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

31
...
022 = 1
...
0404
...
04041/12
...
Then we need 1
...
040425
10,000a(12) =
ν
=
= 158422
...
04041/12 − 1
20
or £158,422
...
30/1
...
69872255 or £59,707
...

(ii) Now x grows to 1
...
Adjusting for inﬂation, x grows to 1
...
If j
denotes the annual effective real return, then (1 + j)20 = 1
...
0055002 or 0
...

(
)
(iii) Capital gain is x(1
...
25x(1
...
0520 + 1 /4
...
0520 + 1 /4
...
002977411
or -0
...

(iv) The tax is paid on the money capital gain
...

32
...
Let i denote the required rate
...
Hence i = (100 × 220)/(95 × 222)
− 1 = 0
...
5%
...
Investment A
...
6 million per year
...

Investment B
...
110 million
...
110 − 1
...
4(1 =
...
6×1
...
4
...
6×1
...
4
which leads to iB = 0
...
94%
...
Capital gains are 1
...
0410
...
6×1
...
4×1
...
So if iC denotes
the net rate of return, we have (1 + iC )10 = 0
...
110 + 0
...
0410 which leads to iC = 0
...
95%
...
Clearly, net rate for Investment C is greater than that for
Investment B because the capital gains are less and hence the tax is less
...

34
...
06
...
5ν + 6ν 1/2 − 175 = 0
or 175i − 6(1 + i)1/2 − 10
...
Approximating (1 + i)1/2 by 1 + i/2 gives i = 0
...
Trying i = 0
...
1535349 and trying i = 0
...
0185989
...
09589 or
9
...
Hence if iR denotes the real rate, then 1 + iR = 1
...
04 leading to iR = 0
...
37%
...
Let f1,2 denote the required value
...
08(1 + f1,2 )2 = 1
...
Hence f1,2
10
...

2
...
Then 1
...
0553
...
0036%
...
tex)

√
= 1
...
08 − 1 = 0
...
0553 /1
...
060036 or

3
...
Hence we
want:
Time
0
1
2
3
Cash Flow
−1
r
r
1+r
Hence
r
1+r
r
+
+
1=
1 + y1 (1 + y2 )2 (1 + y3 )2
(
)
1
1
1
1
=r
+
+
+
1
...
06 × 1
...
06 × 1
...
07
1
...
065 × 1
...
478%
...
06 × 1
...
07 − 1
= 0
...
065 × 1
...
07 + 1

4
...
Then
)
(
1
1
1
1
+
+
+
1=r
2
2
1
...
065
1
...
066
1
...
0662
and hence
r=

1
...
0662 − 1
= 0
...
0662 + 1
...
0662 /1
...
395%
...
2 Page 181

5
...
04 = 0
...
(ii) Ft = limk→0 Ft,k = limk→0 ln(1 + ft,k ) where Ft,k is the forward
force of interest of an investment that starts at time t and lasts for k years
...
Let yk denote the k-year spot rate of interest
...
15%” implies
]
1
1
1
+
1 = 0
...
4 =
+
1 + y1 (1 + y2 )2
Thus if a = 1/(1 + y1 ) and b = 1/(1 + y2 )2 we have the 2 simultaneous equations:
1 = 0
...
0415b
105
...
0415/0
...
4 − 8/0
...
92192
...
0415/1
...
0415 − 105
...
9596
...
0421, or 4
...
0415, or 4
...

7
...
0225 + 0
...
, 5
...
But (1 + y5 )5 = (1 + y3 )3 (1 + f3,2 )2
...
0355 /1
...
Hence f3,2 = 0
...
255%
...
Then
[
]
1
1
1
1
1
1 = iP
+
+
+
+
1 + y1 (1 + y2 )2 (1 + y3 )3 (1 + y4 )4
(1 + y4 )4
and hence 1 = 3
...
8799 and hence iP = 0
...
23%
...
23
3
...
23
100 + 3
...
5
3
...
5
100 + 3
...
0323
...
23
3
...
23
103
...
23
3
...
23
103
...
Then
3
...
5
3
...
5
3
...
5
3
...
5
PB =
+
+
+
=
+
+
+
1 + iB (1 + iB )2 (1 + iB )3 (1 + iB )4 1 + y1 (1 + y2 )2 (1 + y3 )3 (1 + y4 )4
Now
[
] [
]
3
...
5
3
...
5
3
...
5
3
...
5
+
+
+
−
+
+
+
1 + y1 (1 + y2 )2 (1 + y3 )3 (1 + y4 )4
1 + iP (1 + iP )2 (1 + iP )3 (1 + iP )4
[
]
[
]
100
100
1
1
3
...
23 (1 + iP )4
(1 + y4 )4
(1 + y4 )4
(1 + iP )4
[
][
]
3
...
23
(1 + iP )
(1 + y4 )4
Hence
[
]
3
...
5
3
...
5
PB >
+
+
+
1 + iP (1 + iP )2 (1 + iP )3 (1 + iP )4
and hence iB < iP
...
(i) The n-year spot rate of interest is the per annum yield to maturity of a zero coupon bond with n years to maturity
...
Then 1
...
063 (1 + f3,2 )2 gives f3,2 = 1
...
063 − 1 = 0
...
79%
...
Then
(
)
1
1
1
1
1
1
1
1=r
+
+
+
+
+
+
2
3
4
5
6
1
...
05
1
...
07
1
...
08
1
...
086 − 1
= 0
...
086 (1/1
...
052 + 1/1
...
074 + 1/1
...
708%
...
32 + 0
...
421909
...
36 + 0
...
4005) = 1
...
(c) We need exp( 8 δ(t) dt) = A(9) = exp(0
...
352) =
A(8)
1
...

(ii)(a) We need y8 with (1 + y8 )8 = A(8) = exp(0
...
044) − 1 = 0
...
4982%
...
4005) and hence y9 = exp(0
...
045505
or 4
...
(c) We want f8,1 where (1 + f8,1 )(1 + y8 )8 = (1 + y9 )9 and hence f8,1 = exp(0
...
352) =
exp(0
...
0496954
...
0496954 or 4
...

9
...
2

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

10
...
a
...

The usual approximation is i ≈ (7 + 7/3)/98 = 0
...
If i = 0
...
798
...
1 then
RHS = 96
...

Using linear interpolation, (0
...
798 = −0
...
502
...
09319 or 9
...

(ii) For the 1-year bond, ν1 = 98/112 = 7/8 and y1 = 1/7
...
286%
...
9057 and y2 = 0
...
41%
...
91619 and y3 = 0
...
148%
...
Let ft,r denote the forward interest rate p
...
for an agreement starting at time t for r years
...

(i) (a) (1 + f0,2 )2 = (1 + f0,1 )(1 + f1,1 )
...
0567
...

Hence f0,2 = 0
...
(c) (1 + f0,3 )3 = (1 + f0,1 )(1 + f1,2 )2 = (1118/1058) × (1140/1000)
...
06403
...

Time
0
1
2
3
Cash Flow, c
−100
100r
100r
100 + 100r
Hence
r
r
1+r
+
1=
+
2
1 + f0,1 (1 + f0,2 )
(1 + f0,3 )3
Hence r = 0
...

[

]
1058 1000 1058 × 1000
1058 × 1000
+
+
and so 1 = r
+
1118 1118 1118 × 1140
1118 × 1140

12
...
041(1 + f1,1 ) = 1
...
0430 or 4
...

If f2,1 denotes the implied one-year forward rate applicable at time 2, then (1 + y2 )2 (1 + f2,1 ) = (1 + y3 )3 and hence
1
...
0433 and f2,1 = 0
...
50%
...
24
2
1 + y1 (1 + y2 )
(1 + y3 )3
(b) The 2-year par yield is the coupon rate that causes the bond price to be equal to its face value, assuming the bond
is redeemed at par
...
Then the cash ﬂow is as follows:
P =

Time
Cash ﬂow
This gives

0
−100

1
p
100y2

2
p
100y2 + 100

(

)
(
)
1
1
1
1
1
1
p
1=
+
+
= y2
+
+
2
2
2
1 + y1 (1 + y2 )
(1 + y2 )
1
...
042
1
...
042 − 1)/(1
...
041 + 1) = 0
...
198%
...
(i)(a)
5
105
5
+
+
= 101
...
04 1
...
045 1
...
045 × 1
...
6 = 5a 3 ,i + 100ν 3
...
3) gives i ≈ (5 − 1
...
6 = 0
...

If i = 0
...
375
...
04, then right hand side = 102
...

Hence i − 0
...
6 − 101
...
04 − 0
...
785 − 101
...
Hence i = 0
...
42%
...
A higher coupon means more weight on
the coupon at times 1 and 2 when the rate is lower
...

P =

14
...
The usual approximation (method (d) in paragraph 4
...
076
...
5%, RHS = 6×2
...
4961 = 96
...
At 8%, RHS = 6×2
...
3832 = 94
...
Interpolating
gives (x − 0
...
075 − 0
...
099)/(96
...
375) and hence x = 0
...
54%
...
Hence ν(1) = 96/106 = 0
...

Two year bond: 96 = 6ν(1) + 106ν(2)
...
100)/1062 = 0
...

Appendix

Jan 28, 2016(12:36)

Answers 6
...
Gives ν(3) = 96
...
80603
...
Hence f0,1 = 10/96 = 0
...
42%
...
100) = (106/96)(1 + f1,2 ) and so f1,2 = 6/100 = 0
...

Using (1 + y3 )3 = (1 + y2 )2 (1 + f2,3 ) gives 1063 /(96
...
100))(1 + f2,3 ) and so f2,3 = 6/100 = 0
...

15
...
0309278 or 3
...
Two-year spot rate: (1 + y2 )2 =
100/93 and so y2 = 0
...
6952%
...
0435320 or
4
...
Four-year spot rate: (1 + y4 )4 = 100/83 and so y4 = 0
...
7684%
...
44
P =
100
100
100
100
Let i denote the rate of return
...
44 = 4a 4 ,i + 100ν 4
...
3)
gives i ≈ (4 + (100 − 97
...
44 = 0
...

Try i = 0
...
44 = −0
...
Try i = 0
...
44 = 0
...

Using linear interpolation gives (i−0
...
005 = (0−0
...
985996−0
...
04719
or 4
...

(ii) The rate of return from the bond is a weighted average of the rates for 1 year, 2 years, 3 years and 4 years
...

(iii) The liquidity preference theory asserts that investors prefer short term investments over long term ones
...

16
...
06 = 100
...
06
...
3% per annum effective:
Time
0
1
2
Cash ﬂow
−P
6
106
Hence P = 6/1
...
0632 = 6/(1 + f0,1 ) + 106/(1 + f0,2 )2 = 6/1
...
Hence i2 = f0,2 =
0
...

The gross redemption yield from a 3-year bond with a 6% annual coupon is 6
...
Hence
P = 6/1
...
0662 + 106/1
...
06 + 6/1
...
066247 or 6
...

Summary: i1 = 0
...
0630905 and i3 = 0
...

(b) We already know that f0,1 = 0
...
Now (1 + f0,1 )(1 + f1,1 ) = (1 + i2 )2 ; hence f1,1 = 0
...

Now (1 + f0,1 )(1 + f1,1 )(1 + f2,1 ) = (1 + i3 )3 ; hence f2,1 = 0
...

Summary: f0,1 = 0
...
06619 and f2,1 = 0
...

√
(ii) Spot rates are geometric averages of forward rates: for example, 1 + i2 = (1 + f0,1 )(1 + f1,1 ), etc
...

17
...
If interest rates are expected to fall, then
long term bonds will become more attractive, and conversely
...
Market segmentation: rates reﬂect supply and demand for bonds of different lengths
...
1, f1,1 = 0
...
08 and f3,1 = 0
...

Gross redemption yield from a 1-year zero coupon bond: i = 0
...

Gross redemption yield from a 3-year zero coupon bond: i = (1
...
09 × 1
...
08997
...
1 × 1
...
08 × 1
...
08194
...
1 × 1
...
08 × 1
...
07595
...

(c) The gross redemption yield of a 5-year coupon-paying bond can be regarded as a weighted average of the gross
redemption yields for zero-coupon bonds of shorter terms—see part (c) of example(1
...
And the early coupons
attract higher rates
...

18
...
08 − 0
...
1n
...
08 − 0
...
9 )9 = 0
...
(ii) We
have (1 + f7,4 )4 (1 + y7 )7 = (1 + y11 )11 and hence (1 + f7,4 )4 = (1
...
04e−1
...
08 − 0
...
7 )7 leading to
f7,4 = 0
...
8243%
...
2

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

leading to
[
] /[
]
1
1
1
1
r = 1−
+
+
(1 + y3 )3
1 + y1 (1 + y2 )2 (1 + y3 )3
[
] /[
]
1
1
1
1
= 1−
+
+
(1
...
04e−0
...
08 − 0
...
1 (1
...
04e−0
...
08 − 0
...
3 )3
= 0
...
0157%
...
(i) Now (1 + f2,2 )2 = (1 + f2,1 )(1 + f3,1 )
...
052 = 1
...
052 /1
...
0550239 or 5
...
(ii) One year: y1 = 0
...
Two year: y2 = 1
...
0425 − 1 = 0
...
125%
...
04 × 1
...
045)1/3 − 1 = 0
...
25%
...
04 × 1
...
052 )1/4 − 1 = 0
...
56%
...
46813
1
...
04 × 1
...
Hence
3
3
3
103
100
94
...
46813
...
045) = 3 × 3
...
8561 − 94
...
150548
and f (0
...
545951 + 82
...
46813 = −1
...
Interpolating gives (i − 0
...
150548) =
0
...
560077 − 0
...
04544 or 4
...
(iv) We have
3
3
103
3
+
+
+
1 + i (1 + i)2 (1 + i)3 (1 + i)4
3
3
3
103
=
+
+
+
1 + f0,1 (1 + f0,1 )(1 + f1,1 ) (1 + f0,1 )(1 + f1,1 )(1 + f2,1 ) (1 + f0,1 )(1 + f1,1 )(1 + f2,1 )(1 + f3,1 )
It follows that
min{f0,1 , f1,1 , f2,1 , f3,1 } < i < max{f0,1 , f1,1 , f2,1 , f3,1 }
The gross redemption yield, i, is a complicated average of the 1-year forward rates
...
(i) The spot rates are i1 = a − b and i2 = a − 2b
...
061 and f11 = 0
...
Hence a − b = 0
...
Hence b = 1
...
061 × 1
...
00199812 and a = 1
...
061 × 1
...
05900188
...
Using the fact that the 4-year par yield is 0
...
07
1 = 0
...
07071303
...
08
P = 0
...
954501698798
1 + a − b (1 + a − 2b)2 (1 + a − 3b)3
(1 + i4 )4
or £0
...

21
...
Then 103 = 6ν + 6ν 2 + 111ν 3
...
7/103 = 0
...
So try i = 0
...
Then 6ν + 6ν 2 +
111ν 3 − 103 = −0
...
Try i = 0
...
1980964
...
06 + 0
...
1980964/(1
...
1849923) = 0
...
43%
...
0776699 or 7
...
The 2 year spot rate,
1
i2 , is given by 103 = 6/(1 + i1 ) + 111/(1 + i2 )2 and hence i2 = 111/ 103 × 111 − 618 − 1 = 0
...
736%
...
Hence i3 = 0
...
394%
...
Now f01 = i1 = 0
...
06736 and f03 = i3 = 0
...

Now (1 + i1 )(1 + f12 ) = (1 + i2 )2 and hence f12 = 0
...
714%
...
05714286 or 5
...

Now (1 + i1 )(1 + f13 )2 = (1 + i3 )3 and hence f13 = 0
...
714%
...

22
...
06 − 0
...
1n
...
844762
...
Then P = 3/(1 + i) + 3/(1 + i)2 + 103/(1 + i)3
...
16 over 3 years
...
4 = 4
...
So try i = 0
...

Let f (i) = P − 3ν − 3ν 2 − 103ν 3 where ν = 1/(1 + i)
...
044) = −0
...
046) = 0
...
Using
linear interpolation gives i − 0
...
044)) = 0
...
046) − f (0
...
0451 or 4
...

(ii) The 4-year par yield, r satisﬁes 1 = r/(1 + y1 ) + r/(1 + y2 )2 + r/(1 + y3 )3 + (1 + r)/(1 + y4 )4
...
046424 or 4
...

Appendix

Jan 28, 2016(12:36)

Answers 6
...
(i) and (ii)(a) See answer to question 9 on page 167
...
Then 1
...
022
...
3611881 or 36
...

24
...
Let ν = 1/(1 + i)
...
The usual approximation gives i = 0
...
08
...
08, then f (i) = −0
...
085 then f (i) = 1
...
Linear interpolation gives i = 0
...
286336 × 0
...
323695 + 0
...
080889 or 8
...

(ii) Let yt denote the t-year spot rate of interest
...
Hence y1 = 12/97 = 0
...

For y2 we have
6
109
6 × 97
109
97 =
+
=
+
2
1 + y1 (1 + y2 )
109
(1 + y2 )2
Hence y2 = 0
...
049%
...
Now 0 δ(s) ds = 0
...
001t2
...
35 + 0
...
399) = 1
...
4903
...
30 + 0
...
336) =
1
...
3993 (c) exp(0
...
336) = exp(0
...
065027 or 1
...

(ii)(a) If yt denotes the required spot rate of interest, then (1 + yt )7 = exp(0
...
Hence yt = exp(0
...
058656 or 5
...

(b) exp(0
...
057598 or 5
...

(c) We want f6,1 where exp(0
...
336)(1 + f6,1 )
...
063) − 1 = 0
...
503%
...
Hence the interest rate in the seventh year,
which is the answer to part (ii)(c), will be higher that the average interest rate over the ﬁrst seven years, which is the
answer to part (ii)(a)
...
01t+0
...
05t−0
...
Hence the present value is
∫ 10
∫ 10
10
e−0
...
06t dt = 30
= 500(e−0
...
6 ) = 143
...
229
...
06 3
3
3
26
...
01 for t = 1/2; y3/2 = 0
...
05 for t = 5/2; y7/2 = 0
...
09 for t = 9/2
...
011/2 + 1/1
...
055/2 + 1/1
...
099/2 =
4
...

(ii) Liquidity preference: investors prefer short term investments and hence longer term bonds must offer higher
yields as an inducement
...
Market segmentation theory:
there is less demand for long-term bonds
...
099/2 = 1
...
Hence f7/2 = 0
...
3%
...

Time
Cash ﬂow

0
−P

0
...
5
5

(exs6-2
...
05 and x = 1/(1 + j) = 20/21
...
Now P = 5a 20 ,j + 100x20 = 100
...
5(Ia) 20 ,j + 100 × 10x20
...
5 × (a 20 ,j − 20x21 ) × 21 + 1000x20 = 654
...
Hence answer
is 6
...

Or from ﬁrst principles:
]
1 [
2
...
5
−
+ 1000x20
100
(1 − x)2
1−x
)
]
(
)
1 [ (
=
2
...
5 1 − x20
100
2
...
4

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

Hence

∑n
2
tk
− ( k=1 tk ctk ν tk )2
k=1 tk ctk ν
∑n
( k=1 ctk ν tk )2
[∑
( ∑n
)2 ]
n
2
tk
2
k=1 tk ctk ν
k=1 tk ctk νtk
∑n
−ν
− ∑n
tk
tk
k=1 ctk ν
k=1 ctk ν

d
dM (i) = −ν 2
di
=

∑n

k=1 ctk ν

tk

∑n

= −ν 2 σ 2
and this quantity is clearly negative
...
First the price:

∫

20

ν t dt = 10

P = 10
0

Hence
10
dM (i) =
P

∫
0

20

[
10
tν t dt =
P

tν t
ln ν

νt
ln ν

∫

20

−
0

0

20

20

=
0

10(ν 20 − 1)
ln ν

]
[
10 20ν 20
νt
νt
dt =
−
ln ν
P
ln ν
(ln ν)2

20

]

0

] 1 − ν 20 − 20ν 20 ln 1
...
70586
2
P (ln ν)
(1 − ν 20 ) ln 1
...
706 years
...
(a) We must have (1) VA (i0 ) = VL (i0 ), the net present values are the same; (2) dA (i0 ) = dL (i0 ), the effective durations
are the same; and (3) cA (i0 ) > cL (i0 ), the convexity of the assets is greater than the convexity of the liabilities
...

(c) Now P (i) = ν + ν 2 + · · · = ν/(1 − ν) = 1/i
...
Let ν = 1/1
...
We need to check
n
n
n
n
n
n
∑
∑
∑
∑
∑
∑
a tk ν tk =
ltk ν tk
t k a tk ν tk =
tk ltk ν)tk
t2 atk ν tk >
t2 ltk ν tk
k
k
k=1

k=1

k=1

k=1

k=1

k=1

First equality: 12
...
946ν 24 = 15ν 13 + 10ν 25
...
9757
...
425ν 12 + 24 × 12
...
Both sides equal 108
...

Third relation: lhs = 144 × 12
...
946ν 24 = 1886
...
73
...

6
...
Unsecured
...
Usually one annual coupon
...

(ii)(a) Now 97 = ra 20 ,0
...
0520
...
0520 )/a 20 ,0
...
75927
...
05 +
= 13
...
05k
1
...
0520
k=1

So the duration is 13
...

7
...
07
...
For the assets we have VA = 7
...
834ν 25 = 12
...

For the liabilities we have VL = 10ν 10 + 20ν 15 = 12
...

Second condition: the durations are the same
...
404ν 2 + 25 × 31
...
569
(1 + i)tk
For the liabilities we have
∑ tk ℓt
k
= 10 × 10ν 10 + 15 × 20ν 15 = 159
...

2
25
10
15
(ii) Let ν1 = 1/1
...
Then proﬁt is 7
...
834ν1 − 10ν1 − 20ν1 = 0
...

(iii) The assets are clearly more spread out than the liabilities; hence all three of Redington’s are satisﬁed and the
company is immunised against small changes in the interest rate
...
(i) Let ν = 1/1
...
The discounted mean term is
8 × 87
...
5ν 19 700 + 2992
...
07061477
dM (i) =
P
(1 + i)tk
87
...
5ν 19
87
...
5ν 11
or 13
...
The convexity is
1 ∑ tk (tk + 1)ctk 8 × 9 × 87
...
5ν 21 6300ν 2 + 59850ν 13
c(i) =
=
=
= 186
...
5ν 8 + 157
...
5 + 157
...
4 Page 187

or 186
...

(ii) We need equal present values
...
5ν 8 + 157
...
85ν 4 + Xν n
...
4763142
...
Hence we need 8 × 87
...
5ν 19 = 4 × 66
...
This second
equation is 700ν 8 + 2992
...
4ν 4 + nXν n
...
85938
...
710827 and hence X = 43
...
07n = 216
...

∑
Now we check the∑ condition: for the assets t2 atk ν tk = 16 × 66
...
52023
...
82353
...

k
9
...
The zero-coupon bond must have the same present value
...
Hence its spread is less than the spread of the liabilities
and so immunisation is not possible
...
07
...
43ν 15 + 7
...
5550 and present value of liabilities
is VL = 4ν 19 + 6ν 21 = 2
...
Equal to third decimal place
...
So we need: 15×3
...
12ν 25 = 19×4ν 19 +21×6ν 21
...
44420
and rhs = 51
...
Equal to second decimal place
...
For the assets
t k a tk ν tk =
∑ 2
tk
2
19
2
21
2
15
2
25
15 × 3
...
12ν = 1099
...
For the liabilities tk ℓtk ν = 19 × 4ν + 21 × 6ν = 1038
...

This implies convexity of assets is greater than convexity of liabilities and hence we have immunisation
...
(i) Let ν = 1/1
...

Present value of liabilities is VL = 3ν 3 + 5ν 5 + 9ν 9 + 11ν 11 = 15
...
Present value of assets is VA = a 5 ,0
...
992710 + Rν n
...
01169
...
For the liabilities, tk ℓtk ν tk = 9ν 3 +25ν 5 +81ν 9 +121ν 11 = 116
...
For the assets,
∑
tk atk ν tk = (Ia) 5 ,0
...
Hence nRν n = 116
...
08 = 116
...
08 − 5ν 6 )/(1 − ν) =
105
...

Hence n = 9
...
01169 × 1
...
9718
...
For the assets
∑
∑ 2
∑5
tk atk ν tk = k=1 k 2 ν k + n2 Rν n = 40
...
474
...
031
...

3
5
9
11
(ii)(a) Let ν1 = 1/1
...
For the liabilities we have VL = 3ν1 + 5ν1 + 9ν1 + 11ν1 = 21
...
For the assets we have
n
9
...
03 +Rν = 4
...
9718ν
= 21
...
This is a deﬁcit of 0
...
(b) Immunisation
only protects against small changes in the interest rate
...
(i)(a) Let ν = 1/1
...
Present value of liabilities is VL = a 20 ,0
...
5ν 20 a 20 ,0
...
5ν 20 )a 20 ,0
...
Duration
of liabilities is
] 0
...
06 + 0
...
06
1 [
dM (i) =
(Ia) 20 ,0
...
5(21ν 21 + 22ν 22 + · · · + 40ν 40 ) =
VL
(1 + 0
...
06
Using (Ia) n = (a n − nν n+1 )/(1 − ν) and tables gives
a 40 ,0
...
06 − 20ν 21
= 11
...
06
or 11
...

(b) Present value of assets for £100 nominal is VA = 10a 15 ,0
...
Hence duration of assets is
dM (i) =

10(Ia) 15 ,0
...
06 + 150ν 15
=
= 9
...
06 + 100ν 15
a 15 ,0
...
35236 years
...
Part (i) shows that the
duration of assets invested in the government bonds will always be too short
...
The duration of the liabilities is greater than the duration of the assets and
the duration is a measure of the rate of change of the present value with respect to i
...

12
...
07
...
07 (a 10 ,0
...
946
P
a 10 ,0
...
07
(b) ∑
There are 3 conditions to check:
∑n
n
(I) k=1 ltk ν tk = 7ν 5 + 8ν 8 = 9
...
07 + Xν n = 7
...

Hence we want Xν n = 2
...

∑n
(II) k=1 tk ltk ν tk = 7 × 5ν 5 + 8 × 8ν 8 = 62
...
4
∑n

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

= (ν + 2ν 2 + · · · + 10ν 10 ) + nXν n = (Ia) 10 + nXν n = (a 10 ,0
...

Hence n = 10
...
32695 million
...
7612
k
∑n 2
tk
= (ν +∑ ν 2 + · · · + 102 ν 10 ) + n2 Xν n = 228
...
9673
22
k=1 tk atk ν
∑n 2
n
tk
As k=1 tk atk ν > k=1 t2 ltk ν tk , we have Redington immunisation
...
(Ia) n is the present value (time 0) of the following increasing annuity:
Time
0
1
2
···
n
Cash Flow, c
0
1
2
···
n
Hence:
(Ia) n = ν + 2ν 2 + · · · + nν n
k=1 tk atk ν

tk

(1 − ν)(Ia) n = ν + ν 2 + · · · + ν n − nν n+1 = ν a n − nν n+1
¨
n
= ν(a n − nν )
¨
and so
(Ia) n =
(ii) The cash ﬂow is as follows:
Time
Cash Flow, c
Now

] a n − nν n a n − nν n+1
ν [
¨
a n − nν n =
¨
=
1−ν
i
1−ν
0
−P

0
...
03

= 5a 20 ,0
...
755
Hence

n
]
1 ∑ tk ctk
1 [
=
5(0
...
5ν 3 + · · · + 10ν 20 ) + 10 × 100ν 20
tk
P
(1 + i)
P
k=1
[
]
]
a − 20ν 21
1
1 [
2
...
739
=
= 6
...
The duration is an average of the times
of payments weighted by their sizes: decreasing the coupon to £8 means that the ﬁnal payment has relatively more
weight and so the duration will be longer
...
(i) Let ν = 1/1
...
Ignoring tax we would have P = 4a(2) ,0
...
But we have CGT at the rate of 25% and
20
income tax of an annual amount of 1 at the end of each year
...
06 − a 20 ,0
...
25(110 − P )] ν 20
20
and hence
4a(2) ,0
...
06 + 82
...
014782 − 1) × 11
...
5ν 20
= 65
...
25ν 20
1 − 0
...
Hence the price is given by
P = 4a(2) ,0
...
06 + 110ν 20
20
P =

=

1
...
469921 + 110ν 20 = 69
...
5
1
1
...
5
3
Cash Flow—coupons
0
2
2
2
2
2
2
Cash Flow—redemption value
0
0
0
0
0
0
0
Cash Flow—tax
0
0
−1
0
−1
0
−1
Let x = ν 1/2
...
5(x + 2x2 + · · · + 40x40 ) + 20 × 110x40 − (Ia) 20 ,0
...
06
= x + 2x2 + · · · + 40x40 + 587
...
2700 = 976
...
5
2
0
0

20
2
110
−1

Appendix

Jan 28, 2016(12:36)
dM (i) =

Answers 6
...
09822
= 14
...
38648

or 14
...

15
...

(i) Let ν = 1/1
...
The present value of the liabilities is VL = 160a 15 ,0
...
93610
...
07 + 10 × 200ν 10 16(Ia) 15 ,0
...
96971
VL
160a 15 ,0
...
07 + 20ν 10
or 6
...

(iii) Let A denote the nominal amount of security A purchased and B denote the nominal amount of security B
purchased
...
08a 8 ,0
...
03a 25 ,0
...
08(Ia) 8 ,0
...
03(Ia) 25 ,0
...
05971302A + 0
...
93610
6
...
97613131B = 10865
...
63689237 × 0
...
05971302 × 7
...
90926094
...
33253 × 0
...
93610 × 7
...
74879459/∆ = 1351
...
63689237 × 1558
...
05971302 × 10865
...
64324/∆ = 237
...
So we need
£1,351,272
...
00 of B
...
Hence Redington immunisation should hold
...
Let ν = 1/1
...
Liabilities:
Time
0
Cash Flow
0

5
3,000

10
5,000

15
7,000

20
9,000

25
11,000

(i) NPV = 1000(3ν 5 + 5ν 10 + 7ν 15 + 9ν 20 + 11ν 25 ) = 1000x[3 + 5x + 7x2 + 9x3 + 11x4 ] = 11,570
...

(ii) dM (i) = 5000x[3 + 10x + 21x2 + 36x3 + 55x4 ]/NPV = 171,353/NPV = 14
...

(iii) The present value of £100 nominal of A is 5a 26 ,0
...
348443
...
07 + 100ν 32 = 62
...

Equating the present values gives one equation: 76
...
06033405B = 11,570
...

For dM (i) for the assets, we proceed as follows:
[
]
a 26 ,0
...
07 + 100 × 26ν = 5
+ 2600ν 26 = 1031
...
07 − 32ν 33
=4
+ 3200ν 32 = 930
...
07 + 100 × 32ν

32

Hence the second equation is 1031
...
6057861B = 171,353
...
9746 and B = 163
...
Hence we need £1,897
...

17
...
06, α = 1
...
03ν
...
1ν 7 + 6 × 1
...
16 ν 12 +
(6 × 1
...
03k ν 12+k )
k=1

= 6ν 6 (1 + α + α2 + · · · + α6 ) + 6ν 6 α6

∞
∑
k=1

[

1
...
544
1−α
1−β

and for Boring plc we have
P = 4ν(1 + 1
...
0052 ν 2 + · · ·) =

4ν
= 72
...
005ν

(ii) Now let ν = 1/1
...
For Cyber plc we have:
]
[
β
1 − α7
+ α6
= 151
...
16%
...
538 Drop is 15
...

1 − 1
...
Hence its duration is larger and hence its dependence on the interest rate
is larger
...
4

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

18
...
08 + 1000ν 11 )/100 = a 10 ,0
...
710081
...
247 × ν 8
...
247 ) = 7
...
71018
...
247 × 8
...
247 /(100,000ν 7
...
247 × 8
...
240
...
247 years is 100,000ν 7
...
337
...

Hence NPV of assets = NPVof liabilities = 57,250
...

Volatility of assets = volatility of liabilities, by (ii) (a) and (b)
Convexity of assets = 60
...
24 by (ii)(b)
...

(iii) Liabilities: 100,000 to be paid in 7
...

Assets: 57,250
...
0857
...
685
...
337(0
...
085 + ν 10 ) = 55,372
...
Hence proﬁt (in 1/3/1997 values) is 6
...

19
...
07
...
5
10
10
...
5
2500

Value of this annuity at time 9
...
07
...
07
= 20ν 15 + 5ν 9
...
017204 × 11
...
41568
If Vl denotes the volatility of liabilities, then
VL × PL = 20 × 15ν 16 + 2
...
5ν 11
...
5ν 35
...
5 a(2) ,0
...
5
25

= 300ν 16 + 1
...
5 (20ν 0
...
0 ) = 300ν 16 + 1
...
5

50
∑
(k + 19)ν k/2
k=1

Let x = ν

1/2

...
25x
(k + 19)xk
32

21

k=1

(
32

21

16

21

= 300x + 1
...
25x

1 − x50 x(1 − x50 ) 50x51
19x
+
−
1−x
(1 − x)2
1−x

1 − ν 25 x(1 − ν 25 ) 50x51
19x
+
−
1−x
(1 − x)2
1−x

)
)

= 101
...
25x21 (450
...
73726 − 267
...
69543
Assets: suppose invest in yY nominal of bond Y and it matures at time tY
...
41568
tY
(i) Hence yY ν = 25
...

(ii) Volatility of assets:
25 × 10ν 11 + tY yY ν tY +1 = VL × PL = 651
...
922
...
182 years and yY = 115
...

(iii) For Redington immunisation, have 3 conditions:
• value of assets = value of liabilities
• volatility of assets = volatility of liabilities
• convexity of assets > convexity of liabilities
First two conditions are satisﬁed
...
4 Page 191

20
...

(ii) Present value of liabilities is:
PL = 60a 9 ,0
...
176
1
...
07 + 10 × 750ν 10
PL
Let yA and yB denote nominal amounts in bonds A and B respectively
...
406 or £115,406 and yA = 656
...

(iii) For Redington immunisation, have 3 conditions:
• value of assets = value of liabilities
• volatility of assets = volatility of liabilities
• convexity of assets > convexity of liabilities
First two conditions are satisﬁed
...
(i)(a) Present value of liabilities is VL = 100a 5 ,0
...
9477
...
05
...
90 years
...

Then present value is yA ν + yB ν 5 = VL
...
Hence (yA ν + 5yB ν 5 ) =
VL dM (i) = 100(a 5 − 5ν 6 )/(1 − ν) = 1256
...
Hence 4yB ν 5 and so yB = 262
...
Hence yA = 238
...

(iii) (a) Convexity of assets: (2yA ν 3 + 30νB ν 7 )/(yA ν + yB ν 5 ) = 13
...
89 years
...
Hence Redington’s immunisation should have been achieved
...
9477 = 12
...
Hence convexity of assets
is greater than convexity of liabilities
...
For proof of ﬁrst formula, see exercise 10 in section 3
...

Liabilities (in 1,000s):
Time
0
1
2
3
···
19
20
Cash ﬂow
0
100
105
110
···
190
195
(a) Present value: NPV = 95a 20 ,0
...
07 = 95a 20 ,0
...
07 − 20ν 21 )/(1 − ν) = 1446
...

∑20
∑20
Duration: dM (i) = k=1 (95 + 5k)kν k /NPV = (95(Ia) 20 ,0
...
889/NPV = 9
...
43 years
...
Hence present value
is xA ν 25 + 0
...
07 + xB ν 12 = 1446
...
Duration is (25xA ν 25 + 0
...
07 + 12xB ν 12 )/1446
...
429
...
Solving (take 25 × ﬁrst equation
minus second equation) gives xB = 1249
...
27
...
Cash ﬂow:
Time
0
1/2
1
3/2
4/2
···
39/2
40/2
Cash ﬂow
0
3
...
75
3
...
75
···
3
...
75
(i)(a) P = 7
...
1 + 110ν 20 = 7
...
1 + 110ν 20 = 7
...
024404 × 8
...
120 = 81
...

20
(b) Let dM (i) denote the volatility
...
75ν ∑ k/2
k
dM (i) × P =
=
3
...
75ν 2a 20 − 40ν
=
+ 2,200ν 21
2
1 − ν 1/2

k=1

and hence dM (i) = 8
...

(ii) Let x denote amount of each liability
...
Then present value of liabilities is
x(ν t1 + ν 10 ) = P = 81
...

Let dL denote volatility of the liabilities
...

M
M
Hence (t1 ν t1 +1 + 10ν 11 )/(ν t1 + ν 10 ) = 8
...
61 shows that this is the required value
...
526
c(i) =
tk +2
(1 + i)
xν t1 + xν 10
k=1

Page 192 Answers 6
...
J
...
In this case, liabilities are very close (9
...
Hence there should be immunisation
...
5a(2) ,0
...
5 × 1
...
36492 + 110/1
...
254
...
5a(2) = 24
...

4 ,0
...
9
...
9 × 1
...
77
...
(i) Let ν = 1/1
...
05 and β = αν = 1
...
03
...
379704
1−β
(ii) Let x denote the nominal amount of the bond
...
04xa 20 ,0
...
04 × 14
...
553676]
1
...
91494 or £9,446
...

(iii) Duration of liabilities:
]
100 [
1 ∑
ctk tk ν tk =
dM (i) =
ν + 2αν 2 + 3α2 ν 3 + · · · + 59α58 ν 59 + 60α59 ν 60
NPV
NPV
[
]
] 100ν 1 − β 60
100ν [
60β 60
2
=
1 + 2β + 3β + · · · + 59β 58 + 60β 59 =
−
= 36
...
1437 years
...
04x(1 + 2ν + 3ν 2 + · · · + 20ν 20 ) + 20xν 20 =
−
+ 20ν 20
0
...
572532
or 14
...

(v) For the liabilities, d(i) = νdM (i) = 35
...
148089
...
5% in the interest rate, then the liabilities change by 35 × 1
...
5% approximately whilst the assets change by
14
...
5 = 21
...
Thus the liabilities would increase by approximately 52
...
2 = 31
...

/
25
...
08
...
08/0
...
818147 − 20/1
...
818147 =
8
...
037 years
...
08 and ν = 1/1
...
Then P = 500ν(1 + αν + · · · + α19 α19 ) =
500ν × 20 and dM (i) = 500ν(1 + 2αν + · · · + 20α19 α19 )/P = (1 + 2 + 3 + · · · + 20)/20 = 10
...
5 years
...
For (b), the later
cash ﬂows are higher than the corresponding cash ﬂows in (a); hence the duration will be higher
...
Let ν = 1/1
...
05
...

1
Outgoings = 0
...
1ν + 0
...
9ν 9 + ν 10 = 0
...
25 +
−
10 (1 − ν)2
1−ν
8
9
27
27
Incomings = 0
...
5ν 8 (1 − ν 20 )/(1 − ν) + 5ν 27
...
379448 or £379,448
...
NPV = 0
...
64ν 10 −4
...
21ν(1−α10 ν 10 )/(1−αν)+5
...
2 =
0
...
Hence the second option has the larger NPV
...
Clearly the discounted mean term of the second option is less than 10 years
...
1(ν + 22 ν 2 + 32 ν 3 + · · · + 102 ν 10 ) + 0
...
The sum over the ﬁrst 10 years (terms up to ν 10 ) is negative
...
(In fact, d1 P = 49
...
)
27
...
08
...

Present value of liabilities is VL = 400ν 10
...

Let x denote nominal amount in the zero-coupon bond and let y denote the nominal amount in the ﬁxed-interest
stock
...
08ya 16 ,0
...
1yν 16
...
The right hand side is 4,000ν 10 and the left hand side is 12xν 12 +
0
...
08 + 1
...
This gives two equations for the two unknowns, x and y:
xν 12 + 0
...
08 + 1
...
08y(Ia) 16 ,0
...
6yν 16 = 4,000ν 10

Appendix

Jan 28, 2016(12:36)

Answers 6
...
96a 16 ,0
...
4ν 16 − 0
...
08 = 320ν 10
and hence y = 63
...
Substituting in the ﬁrst equation gives x = 254
...

(ii) Amount invested in the zero-coupon bond is xν 12 = 101
...
Hence amount in the ﬁxed interest
stock must be 400ν 10 − 40ν 10 − xν 12 = 0
...
08 + 1
...
647069 or £65,647
...
Hence we do have immunisation
...
Let ν = 1/1
...

(i) We have VL = 300ν 15 + 200a 20 ,0
...
20191149 or
£2,058,201
...
(ii) We have
∑
tk ℓtk ν tk 15 × 300ν 15 + 200(ν + 2ν 2 + · · · + 20ν 20 ) 4,500ν 15 + 200(Ia) 20 ,0
...
356894
VL
VL
VL
or 8
...
(iii) Suppose purchase yA of security A and yB of security B
...
09a 12 ,0
...
04a 30 ,0
...
07536078yA + 0
...
We also need tk atk ν tk = 4,500ν 15 + 200(Ia) 20 ,0
...
17420214 where
∑
[
]
[
]
tk atk ν tk = yA 0
...
08 + 12ν 12 + yB 0
...
08 + 30ν 30 = 8
...
56986281yB
...
07536078yA + 0
...
20191149
8
...
56986281yB = 17200
...
56986281 × 2058
...
54968865 × 17200
...
469843361
1
...
56986281 − 8
...
54968865
Similarly, yB = 255
...
So we need £1,783,470 in security A and £255,287 in security B
...
Assuming the ﬁrst two conditions hold
as in part (iii), this reduces to:
∑
∑
t 2 a tk ν tk >
t2 ℓtk ν tk
k
k
∑
29
...
(ii) Recall duration is dM (i) = tk ctk ν tk /P
...
05
...

Three year bond: P = 4ν + 4ν 2 + 104ν 3 and dM (i) = (4ν + 8ν 2 + 312ν 3 )/P = 2
...
88 years
...
05 + 100ν 5 = 95
...
05 + 500ν 5 )/P
...
05 =
(a 5 ,0
...
Hence dM (i) = 4
...
62 years
...
If the coupon
rate increased to 8%, then the coupons would form a larger part of the return from the bond
...

(iv) Let i denote the rate of return and ν = 1/(1 + i)
...
Because 95 = 4 + 4 + 4 + 4 + 79, the rate of return is clearly 0
...
A capital gain of 6 is obtained after 12 years; this is an average of 0
...

So that suggests the rate of return is between 1
...
5% and closer to 1
...
5%
...
03 gives 95 − 4a 4 ,i − ν 4 a 8 ,i − 100ν 12 = 3
...

Trying i = 0
...
89927474
...
025 + 0
...
89927474/(3
...
89927474) = 0
...
6% per annum
...
Tax will be different for the two options
...
Cash obtained from Option 1 could be invested elsewhere to
improve return from Option 1 over the period of 12 years
...
(i) The payback period is the time when the net cash ﬂow is positive—it completely ignores the effect of interest
...
However, it does not look at the size of
the overall proﬁt of the project
...
Then incomings over [0, t] are 0
...
Hence we want t0 with 0
...
6875 years
...
04
...
64 0 0 ν t dt = 0
...
Setting
this equal to 3 gives ν t0 = 3 ln ν/0
...
17975169 or 5
...

(iii) Let α = 1
...
For project A we have NPV = fA (i) = −3 + 0
...
5αν 3/2 + · · · + 0
...
5ν 1/2 (1 − α6 ν 6 )/(1 − αν)
...
64 0 ν t dt
...
5(1 − α6 )/(1 − α) + 3 − 3
...
When i = 0
...
Hence
there is at least one root in (0, 0
...

(iv) Duration of incomings of project A is dA (i) = ν 1/2 (1+3αν +5α2 ν 2 +· · ·+11α5 ν 5 )/(4P ) where P = 0
...
Hence dA (i) = 0
...
16327775 or 3
...

M
∫6
∫6
∫6
∫6
Duration of incomings of project B is dB (i) = 0 0
...
64ν t dt = 0 tν t dt/ 0 ν t dt
...
6

Jan 28, 2016(12:36)

c
ST334 Actuarial Methods ⃝R
...
Reed

is (ν 6 − 1)/ ln ν and the numerator is 6ν 6 / ln ν − (ν 6 − 1)/(ln ν)2
...
882446 or 2
...

M
(v) The duration is a measure of the sensitivity of the price to interest rate changes
...
Hence if the interest rate increases, the
net present value of the incomings of project A will decrease more than the net present value of the incomings of
project B
...
Let ν = 1/1
...
Now VL = 6ν 8 + 11ν 15 and VA = Xν 5 + Y ν 20
...
Solving gives 15Xν 5 = 72ν 8 + 55ν 15 and 15Y ν 20 = 18ν 8 + 110ν 15
...
50877088 and Y = 13
...
(ii) For the third condition, we need to check
tk atk ν tk >
tk ltk ν tk
...
76749 and rhs = 6 × 48ν + 11 × 225ν = 935
...
Hence the /third
condition is satisﬁed
...
Let ν = 1/1
...
Use units of £1 million
...
04 = 197
...

Let x denote the amount invested in the ﬁrst security; then this security has NPV equal to 0
...
04 + xν 10 =
1
...
Hence x = 98
...
081108969 = 91
...
37
...
1ya 5 ,0
...
267109307y
...
96387/1
...
102078055 or £78,102,078
...

(iii) The duration of the liabilities is 10(ν + 2ν 2 + 3ν 3 + · · · + 39ν 39 + 40ν 40 )/197
...
04 /197
...
04 = (a 40 ,0
...
32307934
...
476511
or 15
...

[
]
(iv) Duration of the assets is 0
...
04 + 10xν 10 + 0
...
04 + 5yν 5 /197
...
04 =
(8
...
9922537936 and (Ia) 5 ,0
...
451822 − 5ν 6 )/(1 − nu) = 13
...

This leads to 6
...
23 years
...
Hence if the interest rate falls then
the increase in the NPV of the liabilities will be more than the increase in the NPV of the assets
...

∑n
33
...
07 + 100/1
...
047167 or 114
...
(ii) Now dM (i) = k=1 tk ctk ν tk /114
...
07 + 1000ν 10 /114
...
19880 or 7
...

(iii) Duration is a measure of the position of the
mean of the cash ﬂows
...
(iv)(a) Effective
duration d(i) = dM (i)/(1 + i) = 7
...
07 = 6
...
(b) The effective duration is a measure of the rate of
change of the net present value with i
...
07 to 0
...
728ϵ)114
...

Answers to Exercises: Chapter 6 Section 6 on page 122

(exs6-3
...
Y = ln(1 + it ) ∼ N (µ = 0
...
01)
...
05] = Pr[Y ≤ ln(1
...
06)/0
...
05) − 0
...
01] = Φ(−1
...
12) = 1 − 0
...
1314
2
...
, i12 are i
...
d
...
06 and
standard deviation σ = 0
...

Hence E[A(12)] = 10 (E[1 + i1 ])12 = 10(1 + µ)12 = 10 × 1
...
122
(
)12
Now E[A(12)2 ] = E[100(1 + i1 )2 (1 + i2 )2 · · · (1 + i12 )2 ] = 100 E[(1 + i1 )2 ] = ]
100(σ 2 + 1
...

[ 2
2
2
2 12
24
Hence var[A(12)] = E[A(12) ] − E[A(12)] = 100 (σ + 1
...
06
and so the standard deviation is
√
var[A(12)] = 0
...

3
...
Then the
accumulated value is for £100 nominal is
A = 109 + 4(1 + i3 ) + 4(1 + i2 )(1 + i3 ) + 4(1 + i1 )(1 + i2 )(1 + i3 )
Using independence gives
E[A] = 109 + 4(1 + E[i3 ]) + 4(1 + E[i2 ])(1 + E[i3 ]) + 4(1 + E[i1 ])(1 + E[i2 ])(1 + E[i3 ])
= 109 + 4 × 1
...
045 × 1
...
045 × 1
...
055
= 109 + 4
...
06 + 1
...
055] = 122
...
29
...
Let S10 = (1 + i1 ) · · · (1 + i10 ) where i1 ,
...
i
...
lognormal(µ = 0
...
00064)
...
75, 0
...

We want Pr(2000S10 > 4500) = Pr(S10 > 2
...
25) = Pr((ln S10 − 0
...
08 > (ln 2
...
75)/0
...
7616) = 1 − (0
...
16 × 0
...
7769 = 0
...
6 Page 195

5
...
001 and var[1 + i] = 4 × 10−6
...
Using
2
2
2
1 2
E[Y ] = eµ+ 2 σ = 1
...
0012 and hence
2
2
eσ = 1 + 4 × 10−6 /1
...
992 × 10−6
...
001/eσ /2 ) = 0
...
Or, quoting the results
√
of example 5
...
0012 ) = 3
...
0012 / 4 × 10−6 + 1
...
0009975
...
We want j with P[i ≤ j] = 0
...
Hence 0
...
Hence ln(1 + j) = µ − 1
...
002288
...
Now S3 = (1 + i1 )(1 + i2 )(1 + i3 ) where i1 , i2 and i3 are i
...
d
...
08 × 0
...
04 × 0
...
02 × 0
...
0625 and variance σ 2 = (0
...
625) + (0
...
25) + (0
...
125) − µ2 = 0
...
06253 = 1
...

2
Now E[(1 + i1 )2 ] = var[1 + i1 ] + (1 + µ)2 = σ 2 + (1 + µ)2 and hence E[S3 ] = (σ 2 + (1 + µ)2 )3
...
00054375 + 1
...
0625 = 0
...
0456
...
Let R = ln(1 + i) where E[i] = 0
...
0032 = 0
...
Then R ∼ N (µ, σ 2 )
...
0015 =
2
2
2
1 2
E[1 + i] = E[eR ] = eµ+ 2 σ and 0
...
Hence eσ − 1 = 0
...
0152 and so
1
σ 2 = 8
...
Hence µ = ln 1
...
00149439
...
9
...
9 = P[i > j] = P[ln(1 + i) > ln(1 + j)] = P[R > ln(1 + j)] = P[(R − µ)/σ >
(ln(1 + j) − µ)/σ)] = 1 − Φ((ln(1 + j) − µ)/σ)
...
1) = −0
...
23445%
...
(i) Now Sn = (1 + i1 )(1 + i2 ) · · · (1 + in )
...

k=1
2
Now E[(1 + ik )2 ] = 1 + 2j + s2 + j 2
...

(ii)(a) E[S8 ] = (1 + j)8 = 1
...
593848
...
12 + 0
...
062 )8 − 1
...
343646
...
Work in units of £1 million
...
06 and var[Rk ] = 0
...
0064
...
0610 = 1
...

2
2
1 2
(b) Now Rk is lognormal
...
06 = eµ+ 2 σ and 0
...
Using
√
the method of example 5
...
0064/1
...
00567982 and µ = ln(1
...
0064 + 1
...
0554290
...
Hence P[S10 < 0
...
611763] = P[ln(S10 ) < ln(1
...
61173) − 10µ)/ 10σ 2 ) = Φ(−0
...
3733
...
Suppose 1 + it ∼ lognormal(µ, σ 2 ) and hence Z = ln(1 + it ) ∼ N (µ, σ 2 )
...
05 and var[1 + it ] = e2µ+σ (eσ − 1) = 0
...
Hence eσ = 0
...
052 + 1 and so σ 2 = 0
...
0433325
...
04 < it < 0
...
04) < ln(1 + it ) < ln(1
...
07) − µ)/σ) − Φ((ln(1
...
233) − Φ(−0
...
233) − (1 − Φ(0
...
233) + Φ(0
...
5910 + 0
...
0038 + 0
...
9 × 0
...
1077
1

2

2

2

1

2

11
...
Then Y = ln Z ∼ N (µ, σ 2 )
...
Hence eµ+ 2 σ = 1
...
04
...
04/1
...
03434 and µ = ln(1
...
072 + 0
...
05049
...
Hence S15 ∼
lognormal(15µ, 15σ 2 )
...
5] = Pr[ln S15 > ln 2
...
5 − 15µ)/ 15σ 2 ] = 1 − Φ((ln 2
...
221457) = 1 − 0
...
4123
...
221457) − Φ(0
...
23) − Φ(0
...
221457 − 0
...
23 −
0
...
1457 and hence Φ(0
...
22) + 0
...
23) − Φ(0
...
5877
...
We are given that ln(1 + it ) ∼ N (µ = 0
...
0003493) and E[1 + it ] = 1
...
0004
...
05 = 997,500
...

(b) 15% of assets in guaranteed return gives 950,000 × 0
...
05 = 149,625
...
Hence S = 950,000 × 0
...
We want Pr[S < 1,000,000 −
149,625] = Pr[S < 850,375] = Pr[1 + it < 1
...
053096] = Pr[(ln(1 + it ) − µ)/σ <
(ln 1
...
8425) = 1 − Φ(0
...
7995 + 0
...
20
(ii) (a) Variance = 0
...
022 = 260,822,500
...

13
...

(i) A(5) = 425 × 1
...
Using independence gives E[A(5)] =
425×1
...
035 +425×1
...
4581615 or £997,458
...
Also var[A(5)] = 4252 var[S5 ]
...
035 + 0
...
035√5 = 1
...
Hence
)
k=1
10
var[S5 ] = 1
...
035 = 0
...
Hence the standard deviation of A(5) is 425 var[S5 ] = 32
...
21
...
6

c
ST334 Actuarial Methods ⃝R
...
Reed

Jan 28, 2016(12:36)

(ii) The new expectation √ 850 × 1
...
53336 or £1,009,533
...
This is considerably higher
...
486418 or £65,486
...
So there is
greater probability of covering the liability of £1,000,000, but there is also a much higher probability of a large
shortfall
...
Let S2 denote the amount after 2 years
...
03 with probability 1/3;
{
0
...
7;
I1 = 0
...
04 with probability 0
...

0
...

Using the independence of I1 and I2 gives E[S2 ] = 10,000(1 + E[I1 ])(1 + E[I2 ]) = ]
10,000(1 + 0
...
047) =
[
2
10,923
...
(ii) E[S2 ] = 108 E[(1 + I1 )2 ]E[(1 + I2 )2 ] = 108 1 + 2 × 0
...
0061/3 [1 + 2 × 0
...
00223] =
108 × 1
...
Hence var[S2 ] = 108 (1
...
092372 ) = 19,338
...
(i) S10 = 1,000Π10 (1 + Ij )
...
06
...
0610 = 1,790
...

j=1
2
(ii) Now E[(1 + I1 )2 ] = 1 + 0
...
00392 = 1
...
Hence E[S10 ] = 106 × 1
...
12392 − 1
...
60
...
6326
...
Hence E[S10 ] is unchanged
...
Hence the standard
deviation of S10 will become smaller
...

16
...

...
i
...
lognormal(µ = 0
...
006)
...
073
...
18471
giving the answer £595,184
...

(ii) Let x denote the required amount
...
99
...
So we require 0
...
Hence 0
...

√
Hence (ln 600 − ln x − 10µ)/ 10σ 2 = −2
...
326 348 10σ 2 giving £526,771
...

17
...

2
Hence E[Sn ] = (1 + j)n and E[Sn ] = [s2 + (1 + j)2 ]n
...

(ii) Now ln(1 + it ) ∼ N (µ = 0
...
04)
...

Hence Pr[1000S16 > 4250] = Pr[S16 > 4
...
25] = Pr[(ln S16 − 16µ)/(4σ) > (ln 4
...
25 − 16µ)/(4σ)) = 1 − Φ((ln 4
...
28)/0
...
0432436) = 0
...

18
...
, in are i
...
d
...
Hence
2
E[Sn ] = (1 + j)n and E[Sn ] = E[(1 + i1 )2 ]n = [var(1 + i1 ) + E(1 + i1 )2 ]n = [s2 + (1 + j)2 ]n
...

(ii) Suppose j = 0
...
01
...
0001 and Z = ln(1 + i) ∼ N (µ, σ 2 )
...
06 and s2 = 0
...
Hence eσ = 0
...
062 + 1 and so
σ 2 = 0
...
059928 and µ = 0
...

(b) S12 = (1 + i1 ) · · · (1 + i12 )
...
69869, 0
...

(c) Pr[S12 > 2] = Pr[ln S12 > ln 2] = Pr[(ln S12 − 0
...
69869)/0
...
1696) =
Φ(0
...
5636 + 0
...
0039 = 0
...
Let W10 = (1 + i1 ) · · · (1 + i10 ) where i1 , i2 ,
...
i
...
with expectation 0
...
09
...
07 + 0
...
072 = 1
...

(i) Hence we are interested in S10 = 1000W10
...
0710 = 1967
...
Using
√
2
10
E[W10 ] = 1
...
15310 − 1
...
66
...
Now ln(1 + i1 ) has a lognormal distribution
...
Hence eµ+ 2 σ = E[1 + √ ] = 1
...
153
...
153/1
...
00705
...
07 / 1
...
064134
...

We want P[S10 < 0
...
5E[W10 ]] = P[ln(W10 ) < ln(0
...
0710 )] = Φ[(ln(0
...
0710 ) −
√
0
...
0705] = Φ(−2
...
00661
...
64134)/ 0
...
835] = 0
...

20
...
Then value at time 5 is A(5) = xS5 where S5 = R1 R2 R3 R4 R5 and R1 , R2 ,
...
i
...
lognormal with expectation 1
...
02
...
3b gives ln Rk ∼ N (µ, σ 2 )
√
where σ 2 = ln(1 + 0
...
042 ) = 0
...
042 / 0
...
042 ) = 0
...
Hence ln S5 ∼
N (5µ, 5σ 2 ) or ln S5 ∼ N (0
...
091611)
...
01 = P[xS5 < 5,000] = P[ln S5 < ln 5,000 − ln x] = P[(ln S5 − 0
...
091611 <
√
√
(ln 5,000 − ln x − 0
...
091611]
...
01) and hence x =

Appendix

21
...

23
...

Jan 28, 2016(12:36)

Answers 6
...
1) = 8699
...
48
...
The problem is that the returns have
expectation 1
...
02
...

(i) Now Sn = Πn (1+ik ) where i1 ,
...
i
...
with mean j and standard deviation s
...

(b) Using independence again gives E[Sn ] = Πn E[(1 + ik )2 ] =
k=1
k=1
n
2
2
2 n
2
2
Πk=1 (1 + 2j + E[ik ]) = (1 + 2j + s + j )
...

(ii)(a) Now j = E[ik ] = (i1 + i2 )/2
...
Hence var[ik ] = (i2 + i2 )/2 − (i1 + i2 )2 /4 = (i2 + i2 −
2
1
2
1
2
1
k
2i1 i2 )/4 = (i1 − i2 )2 /4
...
5; hence j = 0
...
Also (1 + 2j + s2 + j 2 )25 − (1 + j)50 = 0
...
25
...
25+5
...
5
...
00037738
...
14113724
and i1 − i2 = 2s = 0
...
Hence i1 = 0
...
0511424 or 8
...
11424%
...
Then the accumulated value at time 10 is
[
]
(1 + i)10 − 1
S(10) = 800 (1 + i) + (1 + i)2 + · · · + (1 + i)10 = 800(1 + i)
i
We are given that
{ 0
...
25
{ 8934
...
25
i = 0
...
55 and hence S(10) = 9989
...
55
0
...
2
11826
...
2
Hence E[S(10)] = 10093
...
11
...
2
...
Then S10 = (1+i1 ) · · · (1+i10 )
...
0710 = 1
...

2
(ii) Using independence again gives E[S10 ] = Π10 E[(1 + ik )2 ] = (1 + 2 × 0
...
016 + 0
...
07 + 0
...
072 )10 − 1
...
576097
...

Use units of £1,000
...
06, E[1 + i2 ] =
3
1
7
3
4 × 1
...
05 = 1
...
06 + 10 × 1
...
054
...
06 × 1
...
054 = 95
...
85
...
Using independence again gives E[S3 ] = E[(1+i1 )2 ]E[(1+i2 )2 ]E[(1+i3 )2 ]
...
04 +1
...
08 ); E[(1+i2 ) ] = 4 1
...
05 = 1
...
062 + 10 1
...
111
...
416304978
...
41630498−1
...
435073 or 3,435,073
in units of £2
...
08 × 1
...
06 = 97
...
For any other possibility we have A(3) < 97
...
175
...
Let Y = 1 + it
...
06 and var[Y ] = 0
...
Now ln(Y ) ∼ N (µ, σ 2 )
...
06 and
2µ+σ 2 σ 2
2
σ2
2
2
2
µ
2
e
(e − 1) = 0
...
Hence e = 1 + 0
...
06 leading to σ = 0
...
Also e = 1
...
082 + 1
...
0554290
...
Then S10 = j=1 (1 + ij )
...
Hence
2E[S10 ] = 2 exp(10µ + 10σ 2 /2) = 2 exp(0
...
082 /1
...
5816954 or £3
...

(ii) We want P[2S10 < 0
...
5816954] = P[S10 < 1
...
554290, 0
...
Hence
√
P[S10 < 1
...
432678) − 0
...
0567982) = Φ(−0
...
20692
...
(i) Option A
...
Then E[A1 ] = 100 × 1
...
81 by independence
...
11 + 0
...
0552
...
11 + 0
...
0552 )10 − 1
...
20
...
For this case

5
5
 100 × 1
...
01 with probability 0
...
06 × 1
...
3;
A2 =
 100 × 1
...
065 with probability 0
...
065 × 1
...
3
...
06 ×[0
...
015 +0
...
035 +0
...
065 +0
...
085 ] = £169
...
Also var[A2 ] = 1002 ×
1
...
2×1
...
3×1
...
2×1
...
3×1
...
2×1
...
3×1
...
2×1
...
3×1
...
62
...
Now ln(1 + it ) ∼ N (µ, σ 2 )
...
055 and e2µ+2σ = E[(1 + it )2 ] =
2
1
...
072 + 0
...
Hence eσ = (1
...
072 + 0
...
0552 giving σ 2 = 0
...

√
Also eµ = 1
...
11 + 0
...
0552 giving µ = 0
...

Now ln(A1 /100) ∼ N (10µ, 10σ 2 ) = N (0
...
0439275)
...
15 − 0
...
0372989 or 0
...

P[A1 < 115] = P[ln(A1 /100) < ln 1
...
0439275

Page 198 Answers 7
...
J
...
The smallest possible amount is 100 × 1
...
015 = 140
...
So the
probability is 0
...
There is a
much higher probability of a low return with option A
...
(i) Let A denote the accumulated value
...
By independence, E[A] = Π10 E[1+ij ] =
j=1
E[1 + i1 ] (E[1 + i2 ])9 = 1
...
049 = 1
...
(ii) Now E[(1 + i1 )2 ] = 1 + 0
...
0029 = 1
...
08 + 0
...
08184
...
1029 × 1
...

Hence var[A] = E[A2 ] − (E[(A)])2 = 0
...
0726376 or £72,638
...
Hence the value of E[A] will be unchanged
...

1

2

28
...
Let S20 denote the required answer
...
04 − 1)/(α − 1) = 180498
...
93
...
Hence
20µ+10σ 2
1
...
04
E(S20 ) = e
= e
...
04) where ln S20 ∼ N (20µ, 20σ 2 ) = N (1, 0
...

√
1
...
04/ 0
...
44376854 or 0
...

29
...
04 × 0
...
06 × 0
...
54
...
05420 = 200,000 and hence we get x = 69858
...
26
...
0420 with
probability 0
...
0620 with probability 0
...
0410 × 1
...
49
...
5443921 or £201,336
...
54
...
0420 = 153068
...
05; largest amount is x × 1
...
903673
or £224,044
...
Hence the range is £224,044
...
05 = 70,976
...

30
...
07
...
0715 = 27590
...
32
...
00506
...
14506
...
1450615 and
j
√
hence var[A] = 108 (1
...
0730 ), and so the standard deviation is 104 1
...
0730 = 1263
...
837
...
Hence E[A] is unchanged
...
(b) The mean will be less because the money is invested for a shorter
period
...

31
...
85(1 + i1 )(1 + i2 )(1 + i3 )(1 + i4 )(1 + i5 )
...
85[E(1 + i1 )]5 = 7
...
0555 =
10
...
Now E[(1 + i1 )2 ] = 1 + 2 × 0
...
042 + 0
...
114625
...
852 1
...
05510 and stdev[S5 ] = 7
...
1146255 − 1
...
871061457 or £871,061
...
85 × 1
...
550725284 or £9,550,725
...
The policy of part (i) has an expected proﬁt but there is the possibility of a very large loss
...

∏n
2
32
...
Using independence gives E[Sn ] = t=1 E[1 + it ] = (1 + j)n
...
Hence var[Sn ] = (1 + 2j + j + s ) − (1 + j)2n
...
122 /1
...
013226 and eµ = 1
...
122 + 1
...
032608
...
Hence ln(1 + ( t ) ∼ N (µ, σ 2 ) and (ln(1 + it ) − µ) /σ ∼ N (0, 1)
...
06 < 1 + it < 1
...
08) − µ)/σ − Φ (ln(1
...
06184493 or 0
...
2%
...
So the probability it lies between 6% and 8% will be small; The reason it is so small
is a consequence of the shape of the lognormal distribution
...
01), dlnorm( seq(0,2, by=0
...

Answers to Exercises: Chapter 7 Section 3 on page 136

(exs7-1
...
Now 1 + i = (1 + 0
...
Hence 80 = K/(1 + i)1/4 = K/1
...
0125 = 81
...
We have S0 = 12
...
Then
K
1
...
5
1
...
03
...
03 − 1
...
03 − 1
...
034/6 − 1
...
031/6 = 8
...
02
...
(i) See page 129
...
07×7/12 = 60 − 2e−0
...
07×6/12 which gives K = 60e0
...
28/12
0
...
4418
...
Now £1
...
a
...
20 × 1
...
2147
...
2147
...
3 Page 199

5
...
If the contract speciﬁes that A will buy the asset from B at some time in the future,
then A holds a long forward position and B holds a short forward position
...
Hence Ke−0
...
03/4 − 30e−0
...
02/4 − 30e0
...
62661
...
Let K denote the forward price
...
3
0
...
045
1
...
04
This gives K = 6
...
3 × 1
...
041/2 − 0
...
662588
...
(i) A forward contract is a legally binding contract to buy (or sell) an agreed quantity of an asset at an agreed price
at an agreed time in the future
...
Thus futures contracts are over-the-counter or OTC
...

Settlement of a forward contract occurs entirely at maturity and then the asset is normally delivered by the seller to
the buyer
...

The party agreeing to sell the asset is said to hold a “short forward position” and the buyer is said to hold a “long
forward position
...
Hence K = S0 × 1
...
04
...
82
...
Hence K = S0 × 1
...
04
...
20
...
20 − 134
...
38 or 24
...
043 = 21
...

Or: the value of the contract to its owner who has the right to buy the asset in 3 years’ time for the price 134
...
042 − 134
...
043 = 21
...

8
...

(ii) Let K denote the forward price, α = 1
...
015
...
25 0
...
75
1
1
...
5
1
...
25
2
...
75
3
Dividend 0 0
...
2α2 0
...
2α4 0
...
2α6 0
...
2α8 0
...
2α8 β 2 0
...
2α8 β 4
Hence
Ke−0
...
5 −

8
∑

0
...
05k/4 −

k=1

= 4
...
2

4
∑

0
...
1−0
...
2α8 e−0
...
5 − 0
...
2α8 e−0
...
05/4 = 0
...
05/4 = 1
...
Hence K = 2
...

(
)
(∫
)
15
15
0
...
(i)(a) Accumulation is 100 exp 5 δ(t) dt = 100 exp(0
...
08t + 2
= 100 exp(0
...
4 +
10

0
...
25) exp(0
...
058329
...
25) exp 0
...
003t2
2

14

)
= 100 exp(0
...
32 +

10

0
...
25) exp(0
...
214352
...
5875 − 0
...
145000
...
058329 and hence δ = 0
...

∫t
(ii) Now ν(t) = exp(− 0 δ(u) du)
...
05t)
...
04t
ρ(t)ν(t) dt =
100e0
...
05t dt = 100
e−0
...
2 ) = 453
...
04 0
0
0
0
(iii) Let K denote the forward price
...
5) where ν(t) is given above
...
05) − 7 exp(0
...
204123
...
Use £100 nominal
...

Investor has a short position
...
He receives a
coupon of 2
...
5 and a coupon of 2
...
So the time t = 0 value of these three receipts is
98e−0
...
5e−0
...
5e−0
...
850707
The current price is £95
...
850707 − £95 = £2
...
This was for £100 nominal
...
850707 = £28,507
...

11
...
If investors expect interest rates to fall, then long term bonds will become more attractive; this will cause the yields on long term bonds to fall and lead to a decreasing yield curve
...

Page 200 Answers 7
...
J
...
08 or 8%
...
08 × 1
...
07498837
...
08×1
...
06 and hence i3 = 0
...
Finally, (1+i4 )4 = 1
...
07×1
...
05 and hence i4 = 0
...

(b) The price of the bond is
[
]
1
1
1
1
100
+
+
+
+
= 94
...
To get a ﬁrst approximation, use the usual method of spreading
the capital gain over the coupons; this gives 5 + (100 − P )/4 = 6
...

If i = 0
...
8598
...
07, then 5a 4 ,i + 100/(1 + i)4 − P = −1
...

Using linear interpolation gives i = 0
...
01 × 1
...
8598 + 1
...
06562
...
56%
...
Hence
K
4
= 400 −
2
(1 + i2 )
1 + i1
and hence K = 1
...
07 × (400 − 4/1
...
96
12
...
Then Ke−0
...
5e−0
...
Hence K = 9
...
So
the forward price of £9
...
Hence a risk-free proﬁt can be made as follows
...

Also enter into a forward contract to sell the share in 6 months’ time for £9
...

Time 1 month: receive dividend of £0
...

Time 6 months: dividend has grown to £0
...
03×5/12 ; also owe £10e0
...
70 for the share
...
70 + 0
...
03×5/12 − 10e0
...
70 − K) = £0
...

13
...

(ii) The given information is summarised in the following table:
t=0
t = 1, market up
t = 1, market down
cost of security A
20
25
15
cost of security B
15
20
12

Now 25/20 = 15/12
...
Hence for no
arbitrage, the prices of the shares at time t = 0 must be in the ratio 5 : 4
...

At 15, the price of B is too low: consider selling three A and buying four B at time 0
...

14
...
Then
K1
5
6
= 95 −
−
12
7
1
...
03
1
...
Then
K2
5
6
= 145 −
−
7
2
1
...
03
1
...
In units of money now it is
K2 − K1
= 145 − 95 × 1
...
86896
1
...
0312 1
...
03
1
...
In units of money now it is
145
95 × 1
...
025 − 95 × 1
...
39365
7
7
1
...
02
1
...
0212
15
...
051/2 1
...
06
This leads to K = 900 × 1
...
06/1
...
2773 or 852p
...
3 Page 201

16
...
051/12 1
...
0511/12
Hence K = 1
...
057/12 − 50 × 1
...
84489 or 9
...

1000 =

17
...
At
time T , both lead to same value—ownership of the asset
...

(ii) Let K denote the price of the forward contract
...
02 1
...
022 − 10 × 1
...
88 or £197
...

18
...

(ii) Let t = 0 denote the time four years ago and let
K0 denote the price of the initial forward contract at that time
...
20 at t = 0 with
dividends of £1
...
So we are comparing the following 2 cash ﬂows:
Time
0
1
2
3
4
5
6
7
8
9
c1
−7
...
20
1
...
20
1
...
20
c2
−K0
Hence K0 = 7
...
2a 5 ,0
...
0254
...
Then K1 = 10
...
2a 5 ,0
...
Hence the value of the original contract at t = 4 is K1 − 1
...
45 − 7
...
0254 = 2
...
50
...
03t units of the security over the interval (0, t)
...
2/1
...
2 × 1
...
039 at t = 4
...
45/1
...
Hence the value of the original forward contract at t = 4 is
10
...
035 − 7
...
0254 /1
...
923201 or £2
...

19
...
042
...
Hence K = 7
...
5956
...
Whether or not the share price
increases or decreases between 1/9/12 and 1/7/13 does not affect the forward price and indeed this information is
unknown when the cost of the forward contract is determined
...

20
...
An option gives the right but not the obligation to buy or sell a speciﬁed quantity of
an asset at a speciﬁed price at a speciﬁed time in the future
...
A put option gives the owner the right but not the obligation to sell a speciﬁed asset for
a speciﬁed price at some speciﬁed time in the future
...
50
1
...
10
c2
−K
Assuming no arbitrage, we must have NPV(c1 ) = NPV(c2 )
...
1
1
...
50 +
+
=
1
...
0252 1
...
801306 or £8
...

21
...
Then
10
K
+
1/2
1
...
043/4
[
]
1/4
1/2
Hence K = 1
...
04 − 10 = 175
...
275p
...
(i) See page 129
...
6 on page 134
...
09×0
...
035×0
...
09−0
...
75 = 6
...
25268
...
035×0
...
Hence at time 0, the investor has e−0
...
75 units
of the stock which grow to 1 unit after 9 months
...
30
...
035×0
...
09×0
...
09−0
...
75 = £6
...
Hence he makes
a proﬁt of £6
...
25268
...

C denotes correction and A denotes addition
...
Question 30, exercises 5
...

C: pp 176–177
...
5
...
J

Title: mathmatics for finance
Description: Well arranged short notes, and easy to comprehend and understand.. Especialy actuarial science students

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