Topic 3 - Interest Rates

Question 1

What is an Effective Interest Rate?

Answer 1

Interest paid once per unit

time at end of the period

Question 2

What is an Effective Discount Rate?

Answer 2

Interest paid once per unit

time at start of period

Question 3

When are nominal rates used?

Answer 3

When interest is paid more (or

less) frequently than once per unit time

Question 4

What is the standard notation used for a Nominal Interest Rate payable p times per period?

Answer 4

i(p)

Question 5

What does the notation i(p) represent?

Answer 5

A Nominal Interest Rate payable p times per period

Question 6

What does the notation i(12) represent?

Answer 6

Interest payable monthly

Question 7

Q) What would Interest payable quarterly be i(?)

Answer 7

i(4)

Question 8

Q) Express monthly effective interest of 2% as a nominal annual interest rate convertible monthly

Answer 8

i(12)=?
We know that monthly effective interest is 2% (ie i(12)/12 = 2%)
Therefore, i(12) = 12 x 2% = 24% (nominal annual interest rate convertible monthly)
[i(p) = p x i(p)/p]

Question 9

Show the relationships between i and i(p) determined by compounding our pthly effective rate of interest

Answer 9

1+i = (1 + i(p)/p)^p
Rearranging gives: i(p) = p[(1+i)^1/p -1]
NOTE i ≠ i(p)

Question 10

Q) If i(12)=24% pa calculate i

Answer 10

1+i = (1+i(12)/12)^12
1+i = (1+0.24/12)^12
1+i = (1.02)^12
i = 26.8% pa

Question 11

Question
Find the nominal annual interest rate convertible
quarterly, equivalent to an annual effective interest rate of 4% pa

Find the annual effective interest rate, equivalent to
nominal interest rate of 15% pa convertible four-monthly

Answer 11

Solution
Find the nominal annual interest rate convertible quarterly equivalent to an annual effective interest rate of 4% pa
i(4) = 4[(1.04)^1/4 -1) = 3.9414%

Find the annual effective interest rate equivalent to nominal interest rate of 15% pa convertible four-monthly
1+i = (1 + i(3)/3)^3
i = (1 + 0.15/3)^3 -1 = 15.7625%

Question 12

Q) Find accumulated value of £500 after 3 years at a rate of 10% pa convertible half-yearly

Answer 12

Method 1 –calculate i
If i(2) = 10% i = (1+10%/2)^2 -1 = 10.25%
£500 x 1.1025^3 = £670.05

Method 2 –Simplify using smaller time unit
i(2)/2 = Effective rate over 6 months = 5%
Using 6 months as time unit, can perform calculation
£500 x 1.05^6= £670.05

Question 13

Q) Calculate PV of a payment of £1,000 due in 5 years time at an interest rate of 10% pa convertible monthly using both annual and smaller units of time?

Answer 13

Method 1 –calculate i
If i(12) = 10%, i = (1+0.1/12)^12 –1 = 10.4713%
1000 x 1.104713^-5 = £607.79

Method 2 –Simplify using smaller time unit
i(12)/12 = Effective rate over 1 month = 0.8333%
1000 x 1.008333^-(12 x 5) = £607.80 (rounding)

Question 14

What is the standard notation used to represent a Nominal Discount Rate payable p times per period?

Answer 14

d(p)

Question 15

What does the notation d(p) refer to?

Answer 15

A Nominal Discount Rate payable p times per period

Question 16

Show the relationships between d and d(p) determined by compounding our pthly effective rate of discount

Answer 16

1-d = (1-d(p)/p)^p

Rearranging gives: d(p) = p[1-(1-d)^1/p]

Question 17

Using d = 1-v and v = 1/(1+i), derive a relationship for 1+i
in terms of d(p)

Answer 17

Method 1
d(p) = p[1-(1-d)^1/p]
d(p) = p[1-v^1/p]
d(p) = p[1-(1+i)^-1/p]
d(p)/p= 1 -(1+i)^-1/p
1+i = (1-d(p)/p)^-p

Method 2
1-d = (1-d(p)/p)^p
v = (1-d(p)/p)^p
1+i = (1-d(p)/p)^-p

Question 18

Find accumulated value of £500 after 3 years at nominal discount rate of 10% pa convertible half-yearly

Answer 18

1+i = (1-0.1/2)^-2 = 1.10803

£500 x 1.10803^3 = 680.19

Question 19

Calculate PV of a payment of £1,000 due in 5 years time

at a discount rate of 10% pa convertible monthly

Answer 19

1+i = (1 - 0.1/12)^-12 = 1.105634
1000 x 1.105634^-5
£605.26

Question 20

What is the Force of Interest?

Answer 20

The nominal rate of interest per unit time convertible continuously

Question 21

What symbol is used to represent force of interest?

Answer 21

δ (Delta)

Question 22

State the relationship between i and δ

Answer 22

1+i = e^δ

So δ= ln(1+i)

Question 23

Find accumulated value of £500 after 3 years at δ of

10% pa?

Answer 23

500e^(0.1 x 3)

£674.93

Question 24

Calculate PV of a payment of £1,000 due in 5 years time at a δ of 10% pa

Answer 24

1,000e^(-0.1 x 5)

£606.53

Question 25

Is Force of Interest δ always constant over time?

Answer 25

Force of Interest can also be assumed to be a function of time δ(t)= etc

Question 26

Q) Find accumulated value of £500 after 3 years at

δ(t) = 0.04 + 0.02t

Answer 26

500exp[3∫0(0.04+ 0.02t dt)]
500exp[(0.04t + 0.01t^2)3,0]
500e^0.21
A(0,3) = £616.84

Question 27

Find accumulated value at t = 10 of an investment of
£100 at t = 0, if Force of Interest is..
δ(t) = 0.04 0≤t<5
0.1−0.01t 5≤t

Answer 27

Hint cannot calc A(0,10) but can calc A(0,5) x A(5,10)
100 x A(0,5) x A(5,10)
100 x exp[5∫0(0.04dt)] x exp[10∫5(0.1−0.01tdt)]
100e^0.2e^(0.5-0.375)
£138.40

Question 28

State the value of δ in terms of i, v, and d

Answer 28

ln(1+i)

• ln v
• ln(1-d)

Question 29

State the value of i in terms of δ, v, and d

Answer 29

e^(δ)-1
v^(-1) -1
(1-d)^(-1) -1

Question 30

State the value of v in terms of δ, i, and d

Answer 30

e^-δ
1/(1+i)
1-d

Question 31

State the value of d in terms of δ, i, and v

Answer 31

1-e^(-δ)
i/(1+i)
1-v

Question 32

What is inflation?

Answer 32

A measure of the increase in costs

Question 33

What affect does inflation have on purchasing power of money?

Answer 33

Inflation erodes the purchasing power of money as time elapses

Question 34

Say at t=0 a good costs £4 and at t=1 this increased to £4.25, what is the rate of inflation?

Answer 34

Rate of inflation = 1 + j = £4.25/£4, j = 6.25% pa

Question 35

Say at t=0 a good costs £4 and at t=1 this increased to £4.25. Discuss the change in purchasing power.

Answer 35

If you had £100 you could buy 25 units at t =0 and now 23 units at t = 1, so purchasing power eroded as time evolves

Question 36

Allowing for inflation in our calculations introduces what aspect

Answer 36

The idea of real interest rates (So far we have looked at money rates of interest which make
no allowance for inflation)

Question 37

Q) Find real rate of interest, if £1 is invested at 5% pa for 1 year and inflation is 3% pa

Answer 37

At end of year £1 accumulates to £1.05
Allowing for inflation the £1.05 is worth
1.05/1.03 = 1.0194
Real rate of interest is 1.94%

Question 38

Give the formula for real interest rate

Answer 38

1+i’= (1+i)/(1+j)
i = effective interest rate
j = rate of inflation
i’= real interest rate 
[Formula still valid during deflation i.e. j is negative]

Question 39

Q) Find real rate of interest, if £1 is invested at 5% pa for 1 year and inflation is -2% pa

Answer 39

At end of year £1 accumulates to £1.05
Allowing for inflation the £1.05 is worth
1.05/(1-0.02) = 1.0714
Real rate of interest is 7.14%

Question 40

What are the two types of force of interest?

Answer 40

Dependent and independent of time