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EECM 3714

Lecture 7: Unit 7

Financial Mathematics
Renshaw, Ch
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2
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4
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6
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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
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fixed-coupon bond)
• See example 10
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g
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e
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4, p
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5, p
...
) are exhausted
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3 on p
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5 million kWh (kilowatt
hours)
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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
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SOLUTION: APPLICATION OF ARITHMETIC & GEOMETIRC SERIES
• Note: F1 = arithmetic progression (𝑑 = 75000); F2 = geometric progression (𝑅 = 1
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• Note: 75000/7500000 = 0
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5, then d must be 0
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5
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5 + 3 − 1 0
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65
• While in 2024, electricity consumption will be
𝑎 + 𝑛 − 1 𝑑 = 7
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075 = 8
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5 1
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5903

While in 2024, electricity consumption will be

𝐴𝑅 𝑛−1 = 7
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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
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5903
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𝐹1𝑏 → Under 𝐹1, total consumption from 2019 to 2024 is
𝑆𝑛 =

𝑛
2

6

2𝑎 + 𝑛 − 1 𝑑 = 2 2 × 7
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075

𝑆𝑛 = 47
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5903 1
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006−1

= 46
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g
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• 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
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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
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02𝑛 −1
1
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02𝑛 −1
0
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02𝑛 − 1 = 1000 1
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02𝑛 − 1 = 250

• 1
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25
• ⟹ 1
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25
...
02

𝑛

= l𝑜𝑔 1
...
02 = l𝑜𝑔 1
...
27

l𝑜𝑔 1
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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
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if 𝑝 = 6%, then 𝑟 = 0
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e
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e
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a
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1 × 100 × 10 = 200
• If your interest is compounded annually, then the value of the investment after 10 years will be
100 1 + 0
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37
• Do Examples 10
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12, p
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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
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06Τ2

2(2)

• Quarterly: 𝑦 = 1000 1 + 0
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06Τ12

12(2)

= 1123
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51

= 1126
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16

52(2)

= 1127
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49

• Weekly: 𝑦 = 1000 1 + 0
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06Τ365

2

𝑛𝑥

• Continuous: 𝑦 = 𝑎𝑒 𝑟𝑥 = 1000𝑒 0
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50

As frequency of compounding increases, value of investment increases
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Hence, investment increase
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• 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
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19 (p
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04
or 4% and is paid semi-anual
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04 2
2

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

• E
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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)
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e
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• 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
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Also find the effective rate of interest in each case
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06

−15

= 41726
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• Monthly compounding: 𝑎 = 𝑦(1 + 𝑟Τ𝑛)−𝑛𝑥
𝑎 = 100000(1 +

0
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24
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06(15) = 40656
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• The effective interest rates are
• Annual = 𝑟 = 0
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06Τ
12

• Continuously = 𝑒 𝑟 − 1 = 𝑒 0
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0618

− 1 = 0
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• 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
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on your rate of time
preference)

• Do Example 10
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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
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22 (p
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e
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• 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
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a

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