Tutorial 1

Question 1

1) When i = 8%, calculate
i. v
ii. i(4)
iii. i(3)
iv. d
v. d(12)
vi. δ
vii. Real rate of interest if inflation is 3%

Answer 1

v =(1+i)^-1 = 1.08^-1 = 0.926
i(4) = p((1+i)^1/p -1) = 4(1.08^1/4 -1) = 7.77%
i(3) = p((1+i)^1/p -1) = 3(1.08^1/3 -1) = 7.8%
d = i/(1+i) = 0.08/1.08 = 7.41%
d(12) = p(1-(1+i)^-1/p) = 12(1-1.08^-1/12) = 7.67%
δ = ln(1+i) = ln(1.08) = 7.7%
i’ = (1+i)/(1+j) -1 = 1.08/1.03-1 = 4.85%

Question 2

2) A bank account pays 5% pa simple interest. Calculate the accumulated amount at t =5 if £1,000
is paid into the account at t =0.

Answer 2

1000(1+0.05x5) = £1,250

Question 3

3) A rate of interest of 4% pa convertible quarterly is equivalent to what effective annual rate of
discount?

Answer 3

i(4) = 4%, i = (1+0.04/4)^4 -1 = 4.06%, d =i/(1+i) = 0.0406/1.0406 = 3.902%

Question 4

4) Calculate the total present value as at 1 July 2012 of the following payments assuming an
interest rate of 10% convertible monthly

Payment date Amount £
30 September 2012 1,000
30 November 2012 2,500
31 January 2013 -2,000

Answer 4

i(12) = 10%, i = (1+0.10/12)^12 -1 = 10.47% 
PV = 1000v^3/12 + 2500v^5/12 – 2000v^7/12  = 1000(1.1047)^-3/12 + 2500(1.1047)^-5/12 – 2000(1.1047)^-7/12 
PV = £1,487

Question 5

5) Calculate the accumulated value at 31 December 2012 of the following payments assuming a
discount rate of 6% convertible quarterly

Payment date Amount £
1 January 2011 500
30 September 2011 1,500
30 April 2012 750

Answer 5

1+i = (1-d(4)/p)^-4  =(1-0.06/4)^-4 = 1.06232 
A = 500(1.06232)^2 +1500(1.06232)^1.25 +750(1.06232)^8/12 = £2,963

Question 6

6) Calculate the present value at 1 April 2012 of £2,500 payable on 30 November 2013 using a
force of interest of 6% pa

Answer 6

v^1.667 = e^-1.667δ , 2500e^(-1.667x0.06) = £2,262

Question 7

7) δ(t) = 0.05t + 0.001t^2 for all t
i. Calculate the accumulated value at t = 6, of an investment of £1,000 made at t = 0
ii. Calculate the constant annual effective interest rate earned over the six years

Answer 7

Calculate the accumulated value at t = 6, of an investment of £1,000 made at t = 0
A(0,6) = 1000exp[∫0,6(0.05t+0.001t^2 dt)]
A(0,6) = 1000exp[(0.025t^2 + 0.000333t^3)]6,0
A(0,6) = 1000e^(0.9+0.072)
A(0,6) = £2,643.22
Calculate the constant annual effective interest rate earned over the six years
1000(1+i)^6 = 2643.22
1+i= 2.643^1/6
i = 17.6%

Question 8

Question to work through in tutorial

Find accumulated value at t = 8 of an investment of £2,000 at t = 2, if Force of Interest is..
δ(t) = { 0.05 0≤t<4
0.2-0.02t 4≤t

Answer 8

A =2000 x A(2,4) x A(4,8)
A =2000 x exp[∫2,4 (0.05 dt)] x exp[∫4,8 (0.2-0.02t dt)]
A = 2000 x exp[(0.05t)2,4] x exp[(0.2t – 0.01t^2)4,8]
A = 2000e^(0.2-0.1)e^(0.96-0.64)
A = £3,043.92