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EECM 3714

Lecture 7: Unit 7

Financial Mathematics
Renshaw, Ch
...

2
...

4
...

6
...


Arithmetic and geometric series
Compounding
Discounting
Investment appraisal
Bonds and interest rates
Loan repayments
Annuities

OBJECTIVES
• Use the geometric series to find the rate at which non-renewable resources are exhausted
• Find the present and future values of payments, deposits, receipts, etc
...
g
...

• The difference between consecutive terms in the sequence is constant
• Consider the sequence 𝐴, 𝐴 + 𝑑, 𝐴 + 2𝑑, 𝐴 + 3𝑑, … − 𝐴 = the initial value/starting value, 𝑑 is the
common difference (note: difference between each term in the sequence is 𝑑)

• Arithmetic series: the sum of the terms of an arithmetic progression
• To get the nth term in the sequence, use 𝐴 + 𝑛 − 1 𝑑
• The sum of the first 𝑛 terms of an arithmetic sequence is
• 𝑛 = 𝑆𝑛 and is equal to

𝑛
2

2𝐴 + 𝑛 − 1 𝑑

• Finance applications: (i) simple interest (interest not reinvested/recapitalised); (ii) investment
paying out a fixed amount every period (e
...
fixed-coupon bond)
• See example 10
...
g
...
e
...
4, p
...
5, p
...
) are exhausted
...
3 on p
...
5 million kWh (kilowatt
hours)
...
6% each year
a)

For each forecast, calculate the electricity consumption in 2019 and in 2024

b) For each forecast, calculate the electricity consumption from 2019 to 2024
...


SOLUTION: APPLICATION OF ARITHMETIC & GEOMETIRC SERIES
• Note: F1 = arithmetic progression (𝑑 = 75000); F2 = geometric progression (𝑅 = 1
...

• Note: 75000/7500000 = 0
...
5, then d must be 0
...
5
...
5 + 3 − 1 0
...
65
• While in 2024, electricity consumption will be
𝑎 + 𝑛 − 1 𝑑 = 7
...
075 = 8
...
5 1
...
5903

While in 2024, electricity consumption will be

𝐴𝑅 𝑛−1 = 7
...
006

8−1

= 7
...

• Since 2019 is now 1𝑠𝑡 period, for 2019: 𝑛 = 1, and
• Electricity consumption in 2019 is a for arithmetic (𝐹1) and A in geometric (𝐹2)
• This implies 𝑎 = 7
...
5903
...


𝐹1𝑏 → Under 𝐹1, total consumption from 2019 to 2024 is
𝑆𝑛 =

𝑛
2

6

2𝑎 + 𝑛 − 1 𝑑 = 2 2 × 7
...
075

𝑆𝑛 = 47
...
5903 1
...
006−1

= 46
...
g
...

• Suppose that the total remaining quantity of a resource is B
• Suppose that current consumption/extraction is A
• Also suppose that resource is extracted/consumed at rate r
• Given this information, we can use the formula for the geometric series to determine how long it
will take to exhaust the resource
...

• ∴ 𝑆𝑛 = 𝐵 = 𝐴

𝑅 𝑛 −1
𝑅−1

• To solve for n, take logs and collect terms

EXAMPLE: COPPER EXHAUSTION
• Geologists estimate that the total remaining copper reserves are 250 million tons
...
After how many years will the reserves of copper be exhausted?

• Solution
• 𝑟 = 0
...
02
• And 𝐴 = 20 and 𝐵 = 250
• The sum of the first n terms is

• 𝐴×
• =

𝑅 𝑛 −1
𝑅−1

20
0
...
02𝑛 −1
1
...
02𝑛 −1
0
...
02𝑛 − 1 = 1000 1
...
02𝑛 − 1 = 250

• 1
...
25
• ⟹ 1
...
25
...
02

𝑛

= l𝑜𝑔 1
...
02 = l𝑜𝑔 1
...
27

l𝑜𝑔 1
...
02

SIMPLE & COMPOUND INTEREST
• Principal = initial value of loan or investment

• Simple interest = interest earned only on principal
• Compound interest = interest earned on principal and previous periods’ interest
• Interest rates are usually quoted as simple annual interest rates = nominal rate
• Interest rate used in this chapter’s calculations = proportionate rate of interest = 𝑝Τ100, where p = the percentage rate of
interest (i
...
if 𝑝 = 6%, then 𝑟 = 0
...
e
...
e
...
a
...
1 × 100 × 10 = 200
• If your interest is compounded annually, then the value of the investment after 10 years will be
100 1 + 0
...
37
• Do Examples 10
...
12, p
...
Find the value of the investment after two years
if interest is compounded
• Annually
• Semi-annually (half-yearly)
• Quarterly
• Monthly
• Weekly
• Daily
• Continuously

SOLUTION: VALUE OF R1000 IN 2 YEARS
Recall the discrete compound growth formula: 𝑦 = 𝑎 1 + 𝑟Τ𝑛
• Annual compounding: 𝑦 = 1000 1 + 0
...
06Τ2

2(2)

• Quarterly: 𝑦 = 1000 1 + 0
...
06Τ12

12(2)

= 1123
...
51

= 1126
...
16

52(2)

= 1127
...
49

• Weekly: 𝑦 = 1000 1 + 0
...
06Τ365

2

𝑛𝑥

• Continuous: 𝑦 = 𝑎𝑒 𝑟𝑥 = 1000𝑒 0
...
50

As frequency of compounding increases, value of investment increases
...
Hence, investment increase
...

• If interest is compounded more than once a year, the effective interest rate exceeds the nominal
interest rate
• The effective interest rate is given by 𝐸𝐴𝑅 = 1 +

𝑟 𝑛
𝑛

−1

• Note that if 𝑛 = 1, then 𝐸𝐴𝑅 = 𝑟
• With continuous compounding, the effective rate is 𝑒 𝑟 − 1
• Do Examples 10
...
19 (p
...
04
or 4% and is paid semi-anual
...
04 2
2

EAR= 1 +
− 1 = 0
...
0404∗ 100 = 4
...
) or the amount needed today to produce a specified series of future receipts/payments

• E
...
what is the value of 𝑅1000 in 3 years’ time today?
• Present value: 𝑅1000 today is worth more than 𝑅1000 in a year’s time, because of interest that
could have been earned on the 𝑅1000 (or, because of inflation, less goods and services can be
bought with the 𝑅1000 in a year than in the present)
...
e
...


• So, we know the future value, 𝑦, and want to find the initial value, 𝑎
...
To find PV:
• Given 𝑦 = 𝑎(1 + 𝑟)𝑥 , by simple algebra we get:
• 𝑎=
• 𝑎=
• 𝑎=

𝑦
= 𝑦 1 + 𝑟 −𝑥 , if interest is compounded annually, or
𝑥
1+𝑟
𝑦
𝑟Τ −𝑛𝑥 , if interest is compounded more than
=
𝑦
1
+
𝑛
1+𝑟Τ𝑛 𝑛𝑥
𝑦
= 𝑦𝑒 −𝑟𝑥 , if interest is compounded continuously
𝑟𝑥
𝑒

once a year, or

• Here 𝑎 is said to be the present discounted value (PV) of 𝑦
• It is what we would need to invest now in order to reach any given value, 𝑦, in x years’ time, given
r
...
Also find the effective rate of interest in each case
...
06

−15

= 41726
...


• Monthly compounding: 𝑎 = 𝑦(1 + 𝑟Τ𝑛)−𝑛𝑥
𝑎 = 100000(1 +

0
...
24
...
06(15) = 40656
...

• The effective interest rates are
• Annual = 𝑟 = 0
...
06Τ
12

• Continuously = 𝑒 𝑟 − 1 = 𝑒 0
...
0618

− 1 = 0
...


• Always ask yourself what the present value of a future payment is before deciding on whether to
take the money now or later!
• Your choice of a now or y later then depends on how impatient you are (i
...
on your rate of time
preference)

• Do Example 10
...
340)

INVESTMENT APPRAISAL – THE PV OF A FUTURE INCOME SERIES
• Suppose that we will receive income payments 𝑎 from an investment for n years
• Also suppose that the interest rate r remains constant over the period under consideration
• We can then find the present value of this stream of payments as follows:
𝑛
𝑎1
𝑎2
𝑎𝑛
𝑎𝑖
𝑃𝑉 =
+
+ ⋯+
=෍
2
𝑛
1+𝑟
1+𝑟
1+𝑟
1+𝑟
𝑖=1

• Do examples 10
...
22 (p
...
e
...

• When deciding between multiple projects, choose the project with the highest positive NPV

THE IRR
• The IRR is the rate which yields the same return as the project after the same number of years
• Alternatively, the IRR is the rate which makes the PV of all payments zero
• Thus, the IRR is the rate of return required to increase the investment from 𝑎 to 𝑦 in 𝑥 years
• Rule: invest in project if IRR exceeds market interest rate (𝐼𝑅𝑅 > 𝑟 - otherwise you could get a
better return by just investing your money at the market interest rate)

EXAMPLE
• A project requiring an initial outlay of 𝑅15 000 will produce a return of 𝑅20 000 in 3 years
...
a
...
05)−3
...
75
...
75 − 15000 = 2276
...
3, no
...
321)

1+𝐼𝑅𝑅

• ∴ 20000 = 15000 1 + 𝐼𝑅𝑅
20000
1 + 𝐼𝑅𝑅 3 =

3

⟹

15000

• 1 + 𝐼𝑅𝑅 =
20000
15000

1Τ
3

20000
15000

1Τ
3

∴ 𝐼𝑅𝑅 =

− 1 = 0
...
05

• Because the IRR exceeds the market rate,
invest in the project
• When comparing two projects, choose one
with highest NPV and/or IRR

THE PRESENT VALUE OF AN INFINITE SERIES
• Recall that 𝑃𝑉 =

𝑎1
1+𝑟

+

𝑎2
1+𝑟 2

+ ⋯+

𝑎𝑛
1+𝑟 𝑛

= σ𝑛𝑖=1

𝑎𝑖
1+𝑟 𝑖

• Suppose that the annual payments are constant (i
...
𝑎𝑖 = 𝑎) and that the annual interest rate is
also constant (i
...

• Also suppose that the payments are to be received from now on forever
• The present value of this series of payments is 𝑃𝑉 =

𝑎
𝑟

PERPETUAL BOND PRICES
• Suppose that the government wants to finance its expenditure by selling bonds that promise a
fixed annual payment 𝑎 for all time
...

• The British government actually issued these bonds (called consols) for a long time (in 1751 for
the first time)
...
At which price are these bonds
bought and sold?
• So the price of a perpetual bond can be found as the PV of a perpetual income series, i
...
𝑃𝑏 =
𝑎
𝑟

𝑃𝑉 = , where 𝑎 is known as the coupon (fixed interest payment)
• This means that there is a negative relationship between the interest rate and the price of the
bond (see F10
...
346)

• This is one of the most important relationships in macroeconomics
...
g
...
8, p
...
9, p
...
11
𝑟
𝑃1 = 𝐾
(1 + 𝑟)𝑥 − 1
This formula gives you the capital repaid only
and not interest paid, which is give by rK
...
12, p
...
4

EXAMPLE
Suppose that you borrow 𝑅500 000 to buy a house and that you agree to repay the mortgage in 25
equal annual instalments, at an interest rate of 6% per annum (year)
...

• Find the value of the monthly instalment payment
• How much interest will you pay over the 25 years?
• Also do Example 10
...
4

SOLUTION: ANNUAL INSTALMENTS
• Annual instalment = 𝑎 =

𝑎=
•
•
•
•

0
...
06 −25

𝑟𝐾
1− 1+𝑟 −𝑥

= 39113
...


Total value of instalments after 25 years:
𝑥𝑎 = 25 × 39113
...
9777
Principal = 500 000
Therefore, total interest payments = 977833
...
9777

• Monthly instalment = 𝑎 =
𝑎=

1−

0
...
06Τ12 −(25)(12)

1−

𝑟Τ 𝐾
𝑛
1+𝑟Τ𝑛 −𝑛𝑥

= 3221
...


• Total value of instalments over 25 years = 𝑛𝑥𝑎 = 25 × 12 × 3221
...
1
• Principal = 500 000
• ∴ total interest payments = 966452
...
1

ANNUITY
• Recall that the present value of a stream of income receipts and/or payments is:
• 𝑃𝑉 =

𝑎1
1+𝑟

𝑎2
1+𝑟 2

+

+ ⋯+

𝑎𝑛
1+𝑟 𝑛

= σ𝑛𝑖=1

𝑎𝑖
1+𝑟 𝑖

• Annuity = sequence of equal payments made at fixed points of time over specified period of time
(e
...
𝑅100 per month for 10 years) (𝑎 is constant, i
...
𝑎1 = 𝑎2 = ⋯ = 𝑎𝑛 = 𝑎
• Examples = pensions, retirement annuities, structured savings plans
• The present value of an annuity (with interest compounded annually) is:

• 𝑃𝑉 = 𝑃0 =

𝑎
𝑟

1−

1
1+𝑟 𝑥

• The present value of an annuity (with interest compounded more than once a year) is 𝑃𝑉 = 𝑃0 =
𝑎
1
1
−
Τ
Τ 𝑛𝑥
𝑟 𝑛

1+𝑟 𝑛

• Useful to know this to determine how much we should invest now to pay for a series of equal
regular payments to be received in future (e
...
retirement annuity/pension)

ANNUITY
• The future value of an annuity (with interest compounded annually) is:
• 𝐹𝑉 = 𝑃𝑛 =

𝑎
𝑟

1+𝑟

𝑥

−1

• The future value of an annuity (with interest compounded more than once a year) is:
𝑎

• 𝐹𝑉 = 𝑃𝑛 = 𝑟Τ

𝑛

1 + 𝑟Τ𝑛

𝑛𝑥

−1

• Useful to know this to determine how much the annuity will be worth after x years have expired;
if we know the FV, we can determine the monthly deposit/payment required to reach our
savings/investment goal

EXAMPLE 1
Calculate the present and future values of a deposit of 𝑅1000 per year for 8 years if the interest

rate is 6% p
...

Solution:
• 𝑃𝑉 = 𝑃0 =
𝑃𝑉 =

1000
0
...
06

𝑎
𝑟

1−

1
1+𝑟 𝑥

1
1
...
06

1+𝑟
8

= 6209
...

𝑥

−1

− 1 = 9897
...


EXAMPLE 2
Suppose that a person takes out an annuity that is worth 𝑅40 000 after 10 years, and that the
interest rate is 7
...
a
...
075
𝑛

• Plug in the values:
𝑎
40000 = 0
...
075Τ12

(12)(10)

12

−1
...

• Multiply both sides by 0
...
075Τ12 = 𝑎 1 + 0
...

• Divide both sides by 1 + 0
...
075Τ12)
1+0
...
81

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