Additional Questions Flashcards
- The force of interest δ(t) at time t is given by:
δ(t)={0.05+0.04t, 0≤t<4
0.06, 4≤t
(i) Calculate the present value of a sum of £2,000 payable in 10 years time. (3 Marks)
(ii) Calculate the effective rate of interest per unit time from t = 9 to t = 10. (3 Marks)
(iii) Calculate the discount rate pa convertible monthly implied by part i). (2 Marks)
(Total 8 Marks)
i) Calculate the present value of a sum of £2,000 payable in 10 years time. (3)
v(10) = v(4)xv(4,10) v(10) =e-{∫0,4 (0.05+0.04t) dt +∫4,10 (0.06) dt} (1) So 2000v(10) = 2000e-{(0.05t +0.02t^2)0,4 +(0.06t)4,10} (1) 2000v(10) = 2000e-{(0.52) + (0.6-0.24)} = 2000e^-0.88 = £829.57 (1)
ii) Calculate the effective rate of interest per unit time from t= 9 to t = 10. (3)
A(9,10) = e^(∫9,10 0.06 dt) (1)
A(9,10) = e^[0.06t]9,10 (1)
A(9,10) =e^(0.6-0.54) = 1.0618, i = 6.18% (1)
iii) Calculate the discount rate pa convertible monthly implied by part i). (2)
2000(1+i)^-10 = 829.57 (1)
So i = (829.57/2000)^-1/10 – 1 = 9.2% pa (1/2)
d(12) = 12(1-1.092^-1/12) = 8.77% (1/2)
- (i) If d(4) = 10.3% pa calculate the following:
- i(4)
- δ
- Rate of inflation if real rate of interest is 8.3% pa(4 Marks)
(ii) Evaluate the following based on i above.
- 𝑎̈∞¬(4)
- a10¬(4)
- 𝑠̈15¬(4)
- ӯ10¬
- (lᾱ)5¬
[ӯ = s bar]
(6 Marks)(Total 10 Marks)
Q2 i) If d(4) = 10.3% calculate the following: (4)
- i(4)
i = (1-0.103/4)^-4 -1= 11% (1)
i(4) = 4(1.11^1/4 -1) = 10.57% (1)
- δ
ln(1.11) = 10.436% (1) - Rate of inflation if real rate of interest is 8.3%
(1. 11)/(1+j) -1 = 8.3%
(1. 11/1.083)-1 =2.5% (1)
ii) Evaluate the following based on i above. (6)
- 𝑎̈∞¬(4)
(1/0.103) = 9.709 (1)
- a10¬(4)
(1-1.11^-10)/0.1057 = 6.1288 (1) - 𝑠̈15¬(4)
(1. 11^15)(1-1.11^-15)/0.103 = 36.744 (1) - ӯ10¬
(1. 11^10)(1-1.11^-10)/ln1.11 = 17.626 (1) - (I𝑎̅)5¬
[𝑎̈5¬ – 5v^5]/δ (1)
[(1-1.11^-5)/0.11x1.11^-5(1.11^-5)]/ln1.11 = 10.8776 (1)
- A pensioner approaching retirement has accumulated a pension fund of
£300,000. They have received the following annuities quotes payable from
1 April 2016 ceasing on 31 March 2046. Calculate the initial instalment of
each quote, assume an annual effective rate of interest of i =5% pa.
A pension payable quarterly in advance
A pension payable annually in arrears decreasing at a rate
of 2.5% pa from year two
A pension payable annually in advance that increases by
£150 pa from year two
(Total 9 Marks)
- A pension payable quarterly in advance
300000 = 4P𝑎̈30¬(4) (0.5)
d(4) = 4.85%
300000 = 4P(1-1.05^-30)/0.0485 (0.5)
P = 300000/(4x15.848) =£4,732.46 pq (1) - A pension payable annually in arrears decreasing at a rate of 2.5% pa from year two
300000 = P[v + 1.025^-1v^2+ 1.025^-2v^3.....] (1) 300000 = 1.025P[1.025^-1v + 1.025^-2v^2+ 1.025^-3v^3 ....] (0.5) 300000 = 1.025P𝑎30¬ @j% (1) j = 1.05x1.025 – 1 = 7.625% (0.5) 300000 = 1.025P (1-1.07625^-30)/0.07625 (0.5) P = £25,084.03 (0.5)
- A pension payable annually in advance that increases by £150 pa from year two
300000= [P-150] 𝑎̈30¬ + 150(I𝑎̈)30¬ (1)
300000 = [P-150]x1.05(1-1.05^-30)/0.05
+150x1.05{1.05[1.05(1-1.05^-30)/0.05 -30(1.05^-30)}/0.05 (1)
300000 =[P-150]x16.141 +150x193.195 (0.5)
P = £16,940.83 (0.5)[Total 9]