Topic 3 - Interest Rates Flashcards
What is an Effective Interest Rate?
Interest paid once per unit
time at end of the period
What is an Effective Discount Rate?
Interest paid once per unit
time at start of period
When are nominal rates used?
When interest is paid more (or
less) frequently than once per unit time
What is the standard notation used for a Nominal Interest Rate payable p times per period?
i(p)
What does the notation i(p) represent?
A Nominal Interest Rate payable p times per period
What does the notation i(12) represent?
Interest payable monthly
Q) What would Interest payable quarterly be i(?)
i(4)
Q) Express monthly effective interest of 2% as a nominal annual interest rate convertible monthly
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]
Show the relationships between i and i(p) determined by compounding our pthly effective rate of interest
1+i = (1 + i(p)/p)^p
Rearranging gives: i(p) = p[(1+i)^1/p -1]
NOTE i ≠ i(p)
Q) If i(12)=24% pa calculate i
1+i = (1+i(12)/12)^12 1+i = (1+0.24/12)^12 1+i = (1.02)^12 i = 26.8% pa
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
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%
Q) Find accumulated value of £500 after 3 years at a rate of 10% pa convertible half-yearly
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
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?
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)
What is the standard notation used to represent a Nominal Discount Rate payable p times per period?
d(p)
What does the notation d(p) refer to?
A Nominal Discount Rate payable p times per period
Show the relationships between d and d(p) determined by compounding our pthly effective rate of discount
1-d = (1-d(p)/p)^p
Rearranging gives: d(p) = p[1-(1-d)^1/p]
Using d = 1-v and v = 1/(1+i), derive a relationship for 1+i
in terms of d(p)
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
Find accumulated value of £500 after 3 years at nominal discount rate of 10% pa convertible half-yearly
1+i = (1-0.1/2)^-2 = 1.10803
£500 x 1.10803^3 = 680.19
Calculate PV of a payment of £1,000 due in 5 years time
at a discount rate of 10% pa convertible monthly
1+i = (1 - 0.1/12)^-12 = 1.105634
1000 x 1.105634^-5
£605.26
What is the Force of Interest?
The nominal rate of interest per unit time convertible continuously
What symbol is used to represent force of interest?
δ (Delta)
State the relationship between i and δ
1+i = e^δ
So δ= ln(1+i)
Find accumulated value of £500 after 3 years at δ of
10% pa?
500e^(0.1 x 3)
£674.93
Calculate PV of a payment of £1,000 due in 5 years time at a δ of 10% pa
1,000e^(-0.1 x 5)
£606.53
Is Force of Interest δ always constant over time?
Force of Interest can also be assumed to be a function of time δ(t)= etc
Q) Find accumulated value of £500 after 3 years at
δ(t) = 0.04 + 0.02t
500exp[3∫0(0.04+ 0.02t dt)]
500exp[(0.04t + 0.01t^2)3,0]
500e^0.21
A(0,3) = £616.84
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
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
State the value of δ in terms of i, v, and d
ln(1+i)
- ln v
- ln(1-d)
State the value of i in terms of δ, v, and d
e^(δ)-1
v^(-1) -1
(1-d)^(-1) -1
State the value of v in terms of δ, i, and d
e^-δ
1/(1+i)
1-d
State the value of d in terms of δ, i, and v
1-e^(-δ)
i/(1+i)
1-v
What is inflation?
A measure of the increase in costs
What affect does inflation have on purchasing power of money?
Inflation erodes the purchasing power of money as time elapses
Say at t=0 a good costs £4 and at t=1 this increased to £4.25, what is the rate of inflation?
Rate of inflation = 1 + j = £4.25/£4, j = 6.25% pa
Say at t=0 a good costs £4 and at t=1 this increased to £4.25. Discuss the change in purchasing power.
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
Allowing for inflation in our calculations introduces what aspect
The idea of real interest rates (So far we have looked at money rates of interest which make
no allowance for inflation)
Q) Find real rate of interest, if £1 is invested at 5% pa for 1 year and inflation is 3% pa
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%
Give the formula for real interest rate
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]
Q) Find real rate of interest, if £1 is invested at 5% pa for 1 year and inflation is -2% pa
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%
What are the two types of force of interest?
Dependent and independent of time
