1. The cash flow is as follows:
Time 0 1/m 2/m · · · 1 (m + 1)/m · · · (nm − 1)/m n = nm/m
Cash flow, c −P fr/m fr/m · · · fr/m − t1f r f r/m · · · f r/m C + f r/m− t1f r Hence i is a solution of the equation
P = f r m
∑nm k=1
νk/m+ νnC− t1f r
∑n k=1
νk
= f ra(m)n ,i+ Cνn− t1f ran
Appendix Nov 4, 2014(9:33) Answers 5.2 Page 163 2. Define the function f by
f (n, i) = (1− t1)gCa(m)n ,i+ Cνn= C +[
4. (a) Redemption is at the discretion of the bond issuer. A higher yield means that money is more expensive. This is good news for the purchaser of the bond (low price P and high coupons) but bad news for a bond issuer. The current yield on bonds is 10% p.a. and the coupons on the original bond are only 8% p.a. Hence, the bond issuer can make a profit by purchasing a new bond to pay the coupons on the existing bond rather than redeeming the existing bond.
The bond issuer should retain the current bond—so the answer to part (a) is later.
(b) Use equation 1.5c. We assume no tax. Hence P (n, i) = C + (g− i)Can ,i = C + (0.08− 0.06)Can ,i = C[
1 + 0.02an ,i]
. Hence n↗ implies i ↗. Hence later redemption implies means a higher yield.
5. The cash flow is as follows (where t1= 8/365):
Time 0 t1 t1+1/2 t1+ 1 t1+ 3/2 t1+ 2 · · · t1+13/2 t1+ 7
Cash flow, c −P 0 4 4 4 4 · · · 4 104
Because the bond is ex-dividend, the payment at time t1is zero. The purchase price at time t1should be 8a(2)7 +100ν7. Using tables gives
Alternatively, if x = 1/1.060.5, we have
P = 1
6. Now each coupon after tax is 350× 0.75 = 262.50. If the bond is redeemed at the final possible date, the cash flow is
Time 1/7/91 1/10/91 1/4/92 . . . 1/10/2009 1/4/2010
Payment no. 0 1 2 . . . 37 38
Cash flow −P 262.50 262.50 . . . 262.50 10,262.50
Now i(2)= 0.0591 and (1− t1)g = 0.75× 0.07 = 0.0525. As i(2) > (1− t1)g, assume bond will be redeemed at the latest possible date (and then yield will be at least 6% whatever the redemption date). Let ν = 1/1.06. Hence
P = 262.50 purchaser should price the bond on the assumption it will be redeemed at the earliest possible date of 1/4/04 (and then yield will be at least 5% whatever the redemption date). Hence the cash flow for the second purchaser is as follows:
Time 1/4/99 1/10/99 1/4/00 . . . 1/10/2003 1/4/2004
Payment no. 0 1 2 . . . 9 10
Cash flow −P 262.50 262.50 . . . 262.50 10,262.50
Let ν = 1/1.05. Hence
Page 164 Answers 5.2 Nov 4, 2014(9:33) ST334 Actuarial Methods c⃝R.J. Reed 7.
Time (years) 0 1 2 . . . 19 20
Cash flow −96 4 4 . . . 4 104
Hence 96 = 4a20 + 100ν20and so 24 = a20 + 25ν20. Using method (d) in paragraph 4.3 gives i ≈[
4 + (100− 96)/20]
/96 = 0.044. If i = 0.04, RHS = 25.001; if i = 0.045, RHS = 23.374. Using interpolation gives (x−0.04)/(0.045−0.04) = (f(x)−f(0.04))/(f(0.045)−f(0.04)) and hence x = 0.04 + 0.005(24− 25.001)/(23.374 − 25.001) = 0.043. So answer is 4.3%.
8. Now i = 0.08 and so i(2)= 2[1.081/2−1] = 0.078461. Also (1−t1)g = 0.77×95/110 = 0.665. Hence i(2) > (1−t1)g and so CGT is payable. Hence P is given by
P = 0.77× 9.5a(2)20 ,0.08+ 110ν20− 0.34(110 − P )ν20
= 0.77× 9.5a(2)20 ,0.08+ (72.6 + 0.34P )ν20 and so
P (1− 0.34ν20) = 0.77× 9.5a(2)20 ,0.08+ 72.6ν20 which gives P = 95.79.
9. (i) Credit risk: most governments will not default. Low volatility risk. Income stream may be volatile relative to inflation. (ii) The cash flow is as follows:
Time (years) 0 1 2 3 4 5 6 7 8 9 10
Cash flow, before tax −P 8 8 8 8 58 4 4 4 4 54
Let ν = 1/1.06. Then P = 8× 0.7a5 + 50ν5+ 4× 0.7a5ν5+ 50ν10 = (5.6 + 2.8ν5)a5 + 50ν5(1 + ν5) = 97.6855.
Or, P = 0.7× 4(a5 + a10) + 50ν5(1 + ν5) = 97.6855.
10. Now i(4)= 4[(1 + i)1/4−1] = 4[1.041/4−1] = 0.0394. Also (1−t1)g = 0.8×5/103 = 0.0388. Hence i(4)> (1−t1)g and so CGT is payable and the investor should assume latest possible redemption.
Using units of £1,000 we have P = 5× 0.8a(4)20+ 103ν20− 0.25(103 − P )ν20= 4a(4)20+ 77.25ν20+ 0.25P ν20. Hence (1− 0.25ν20)P = 4a(4)20 + 77.25ν20leading to P = 102.07201 or £102,072.01.
11. (i) Now t1= 0.25, i(2)= 2(1.051/2− 1) = 0.04939, and (1 − t1)g = 0.75× 7/110 = 0.0477. Hence i(2)> (1− t1)g.
Using the formula P (n, i) = C +[
(1− t1)g− i(m)]
Ca(m)n ,ishows there is a capital gain.
(ii) Because i(2) > (1− t1)g, it follows that the loan is least valuable to the investor if repayment is made by the borrower at the latest possible date.
(iii) Assuming the loan is redeemed at the latest possible date, P = 0.75×3.5[x+x2+· · ·+x30]+[110−0.3(110−P )]x30 where x = 1/1.051/2. Hence P = 2.625x(1− x30)/(1− x) + [77 + 0.3P ]x30. Hence (1− 0.3x30)P = 2.625x(1− x30)/(1− x) + 77x30giving P = 107.75380 or £107,753.80.
Or, P = (5.25a(2)15 + 77ν15)/(1− 0.3ν15) = 107.75383 or £107,753.83.
12. (i) The term “gross redemption yield” means before tax. Let ν = 1/1.05, then P = 4a(2)15 ,0.05+ 100ν15 = 4× 1.012348× 10.379658 + 100/1.0515 = 90.1330.
(ii) (a) Now i(2) = 2[1.051/2− 1] = 0.049390 and (1 − t1)g = 0.75× 4/100 = 0.03. Hence i(2) > (1− t1)g and CGT is payable. Now let ν = 1/1.06; then P = 4× 0.75a(2)7 ,0.06 + 100ν7 − (100 − P ) × 0.4ν7. Hence (1− 0.4ν7)P = 3a(2)7 ,0.06+ 60ν7leading to P = 77.5203.
(b) The cash flow was as follows:
Time (years) 0 0.5 1 1.5 2 2.5 3 3.5 · · · 7.5 8
Cash flow before tax −90.1330 2 2 2 2 2 2 2 · · · 2 2 + 77.5203
Cash flow after tax −90.1330 1.5 1.5 1.5 1.5 1.5 1.5 1.5 · · · 1.5 1.5 + 77.5203 Let i denote the required effective rate of return per annum and let x = 1/(1 + i)1/2. Then 90.133 = 3a8 ,i+ 77.5203ν8 and we need to solve for i.
Using method (d) in paragraph 4.3 gives i≈[
3 + (77.5203− 90.133)/8]
/90.133 = 0.016.
Define f (i) = 1.5x(1− x16)/(1− x) + 77.5203x16where x = 1/(1 + i)1/2. Then f (0.015) = 91.3573 and f (0.02) = 88.2486. Using linear interpolation gives (x− 0.015)/(90.133 − 91.3537) = (0.02 − 0.015)/(88.2486 − 91.3537) and hence x = 0.015 + 0.005× 1.2207/3.1051 = 0.016966. Hence the return is 1.7% to the nearest 0.1%. (More precisely, it is 1.694%.)
13. (i) Assuming £100 nominal is purchased, then price is
P1= 0.75× 10a(2)10 ,0.08+ 110
1.0810 = 102.26 Hence he paid £102.26 per £100 nominal.
(ii) Now (1− t1)g = 0.6× 10/110 = 6/110 = 0.054 and i(2) = 2[(1 + i)1/2− 1] = 2[1.061/2− 1] = 0.059. Hence
Appendix Nov 4, 2014(9:33) Answers 5.2 Page 165 i(2)> (1− t1)g and so the investor is liable for CGT and the price is given by
P2 = 0.6× 10a(2)2 ,0.06+110− 0.4(110 − P )
1.062 = 6a(2)2 ,0.06+66 + 0.4P 1.062 and hence P2 = 108.544.
(iii) We need to find i with P1 = 7.5a(2)8 ,i+ P2/(1 + i)8. Using method (d) in paragraph 4.3 gives i≈[
7.5 + (P2− P1)/8]
/P1= 0.081.
Trying i = 0.08 gives RHS = 102.588. Trying i = 0.085 gives RHS = 99.689. Interpolating gives i ≈ 0.08 + 0.005(f (i)− f(0.08))/(f(0.085) − f(0.08)) = 0.08 + 0.005 × (102.26 − 102.588)/(99.689 − 102.588) = 0.0806 or 8.1% approximately.
14. (i) The price is P = 5a20 ,0.06+ 100/1.0620= 88.53.
(ii) (a) Suppose the second investor pays P2. Now (1−t1)g = 0.7×8/100 = 0.056 and i = 0.065. Hence i > (1−t1)g and CGT is payable. Hence
P2= 5× 0.7 × a10 ,0.065+ [100− 0.3(100 − P2)/1.06510
= 3.5a10 ,0.065+ [70 + 0.3P2]/1.06510 and so
P2= 3.5a10 ,0.065+ 70ν10
1− 0.3ν10 = 74.33 (b) We need i where 88.53 = 5a10 ,i+ 74.33/(1 + i)10. Using method (d) in paragraph 4.3 gives i ≈ [
5 + (74.33− 88.53)/10]
/88.53 = 0.0404. So try 4.5%: rhs = 87.4267. So must be lower: try 4%: rhs = 90.7692. Linear interpolation gives: (i− 0.045)/0.005 = (88.53 − 87.4267)/(87.4267− 90.7692). Hence i = 0.04335 or 4.335%.
15. (i) Now t1 = 0.4 and g = 8/100 = 0.08. Also i(2) = 2(1.051/2− 1) = 0.04939. Hence g(1 − t1) = 0.08× 0.6 = 0.048 < i(2). So assume redemption at the latest time of 2011. Also CGT is payable. Let ν = 1/1.05 and β = 0.6.
Time 1/1/01 1/7/01 1/1/02 . . . 1/7/2010 1/1/2011
Cash flow −P 4β 4β . . . 4β 4β + 100− 0.3(100 − P )
Hence, if x = ν1/2= 1/1.051/2,
P = 0.6× 8a(2)10 ,0.05+70 + 0.3P 1.0510 and so
P (1− 0.3x20) = 2.4x1− x20
1− x + 70x20 and so P = 98.668
(ii) In this case, g = 0.08,t1 = 0 and i(2) = 2(1.071/2− 1) = 0.06882. And so g(1 − t1) = 0.08 > i(2). So assume redemption at the earliest time of 2006. Let ν = 1/1.07 and x = ν1/2= 1/1.071/2.
Time 1/1/03 1/7/03 1/1/04 1/7/04 1/1/05 1/7/2005 1/1/2006
Cash flow −P2 4 4 4 4 4 104
Hence P2 = 8a(2)3 ,0.07+ 100ν3 = 4x(1− x6)/(1− x) + 100x6= 102.99.
(iii)
Time 1/1/01 1/7/01 1/1/02 1/7/02 1/1/03
Cash flow −P 2.4 2.4 2.4 2.4 + P2− 0.3(P2− P ) = 2.4 + 101.69 Hence 98.67 = 4.8a(2)2 ,i+ 101.69ν2= 2.4(x + x2+ x3+ x4) + 101.69x4where x = 1/(1 + i)1/2. Using method (d) in paragraph 4.3 gives i≈[
4.8 + (101.69− 98.67)/2]
/98.67 = 0.064.
Try i = 0.06. Then 98.67− 4.8a(2)2 ,i− 101.69ν2=−0.764.
Try i = 0.065. Then 98.67− 4.8a(2)2 ,i− 101.69ν2= 0.1353.
Linear interpolation gives (i− 0.06)/0.005 = (0 + 0.764)/(0.1353 + 0.764) and so i = 0.064248.
Hence i(2)= 2× (1.0642481/2− 1) = 0.0632.
16. We work in multiples of 100,000. The price per £100 nominal is P1= 8a(4)15 ,0.09+ 110
1.0915 = 8 i
i(4)a15 ,0.09+ 110
1.0915 = 96.821947 and hence £9,682,194.70.
(ii) Now (1− t1)g = 0.8× 8/110 = 6.4/110 = 0.0581 and i(4) = 4[(1 + i)1/4− 1] = 4[1.071/4− 1] = 0.068. Hence i(4)> (1− t1)g and so CGT is paid. Hence the price is given by
P2= 0.8× 8a(4)10 ,0.07+110− 0.2(110 − P )
1.0710 = 6.4 i
i(4)a10 ,0.07+ 88 + 0.2P2 1.0710 which gives P2 = 101.131 and so answer is £101.131.
(iii) Running yield is 100(1− t1)r/P2= 0.8× 8/101.131 = 0.06328 or 6.382%.
(iv) We need i with P1= 8a(4)5 ,i+ P2/(1 + i)5. Using method (d) in paragraph 4.3 gives i≈[
8 + (P2− P1)/5] /P1=
Page 166 Answers 5.2 Nov 4, 2014(9:33) ST334 Actuarial Methods c⃝R.J. Reed 0.092. Trying i = 0.09 gives RHS = 97.877. Trying i = 0.095 gives RHS = 96.275. Interpolating gives i≈ 0.09+0.005(f(i)−f(0.09))/(f(0.095)−f(0.09)) = 0.09+0.005×(96.822−97.877)/(96.275−97.877) = 0.093 or 9.3% approximately.
17. (i) Now (1− t1)g = 0.75× 9/110 = 0.0613 and i(2)= 2[1.061/2− 1] = 0.059. Hence i(2)< (1− t1)g and so there is no CGT. The price P1is given by
P1= 0.75× 9a(2)13 ,0.06+ 110ν13= 112.21121 Hence answer is £112.21.
(ii)(a) Now (1− t1)g = 0.9× 9/110 = 0.0736 and i(2) = 2[1.081/2− 1] = 0.0784. Hence i(2) > (1− t1)g and so there is CGT for the second investor. The price P2is given by
P2= 0.9× 9a(2)11 ,0.08+110− 0.35(110 − P2) 1.0811 = 8.1 i
i(2)a11 ,0.08+ 71.5 + 0.35P2 1.0811 which leads to P2= 105.45468 or £105.45.
(b) The cash flow for the first investor is as follows (note there is no capital gain):
Time 0 0.5 1 1.5 2
Cash flow before tax −P1 4.5 4.5 4.5 4.5 + P2
Cash flow after tax −P1 3.375 3.375 3.375 3.375 + P2
Then P1= 6.75a(2)2 ,i+ P2ν2. Using method (d) in paragraph 4.3 gives i≈[
6.75 + (P2− P1)/2]
/P1= 0.030.
Let f (i) = P1− 6.75a(2)2 ,i− P2ν2. Then f (0.03) = −0.203 and f(0.04) = 1.854. Using linear interpolation gives (i− 0.03)/0.203 = 0.01/(1.854 + 0.203) and hence i = 0.031. So the answer is 3% to the nearest 1%.
18. Now i = 0.06 and hence i(4)= 4(1.061/4− 1) = 0.058695. Also (1 − t1)g = 0.6× 100 × 0.08/105 = 0.04571. Hence i(4) > (1− t1)g and the purchaser should value on the assumption that redemption takes place at the latest possible time. Also, capital gains will be paid. So we want P with
P = 4.8a(4)20 ,0.06+ 105− 0.3(105 − P )
1.0620 = 4.8 i
i(4)a20 ,0.06+73.5 + 0.3P 1.0620 which gives P = 87.37 or £87,370.
(a) Now (1− t1)g = 0.8× 8/105 = 0.060952 and i(4)= 0.058695. Hence i(4)< (1− t1)g and there is no CGT. The value of the coupons and redemption payment at 12 months after the time of original issue is
x = 0.8× 2 + 0.8 × 8a(4)14 ,0.06+ 105ν14= 1.6 + 6.4 i
i(4)a14 ,0.06+ 105ν14= 108.8517
where the term 0.8× 2 is for the coupon at time 12 months. It follows that P = ν1/6x = 107.7997 and so the price is £107,799.70.
(b) Again, there is no CGT. The value of the coupons and redemption payment at 12 months after the time of original issue is
x = 0.8× 2 + 0.8 × 8a(4)19 ,0.06+ 105ν19= 1.6 + 6.4 i
i(4)a19 ,0.06+ 105 1.0619 and hence P = ν1/6x = 108.2467 and the price is £108,246.72.
(ii) Purchaser should pay no more than 107,799.70. If purchaser pays more than this and bond is redeemed early, then yield will be less than 6%.
19. (i) Now t1 = 0.3 and g = 8/100 and hence (1− t1)g = 0.056; also i(2) = 2(√
1.06− 1) = 0.05913. Hence i(2)> (1− t1)g. Assume capital gain and bond redeemed at latest possible time.
Time 1/5/11 1/7/11 1/1/12 1/7/12 · · · 1/1/17 1/7/17 · · · 1/7/21 1/1/22
Before tax −P 4 4 4 · · · 4 4 · · · 4 104
After tax −P 2.8 2.8 2.8 · · · 2.8 2.8 · · · 2.8 2.8 +
75 + 0.25P Let ν = 1/1.06. Then
P = 1.061/3 (
5.6a(2)11 ,0.06+ (75 + 0.25P )ν11 )
and hence
P =
1.061/3× 5.6a(2)11 ,0.06+ 1.061/3× 75ν11
1− 1.061/3× 0.25ν11 = 99.3188735468 or £99.31887. (ii) Now t1 = 0, g = 8/100 and i(2)= 2(√
1.07− 1) = 0.0688. Hence (1 − t1)g > i(2). Assume no capital gain and bond is redeemed at earliest time of 1/1/2017.
Time 1/4/13 1/7/13 1/1/14 1/7/14 1/1/15 1/7/15 1/1/16 1/7/16 1/7/17
−P1 4 4 4 4 4 4 4 104
Let ν1= 1/1.07. Then
P1= 1.071/4 (
8a(2)4 ,0.07+ 100ν14 )
= 105.624983612 or £105.6250. (iii) We have
Appendix Nov 4, 2014(9:33) Answers 5.5 Page 167
Time 1/5/11 1/7/11 1/1/12 1/7/12 1/1/13 1/4/13
Before Tax −99.31887 4 4 4 4 105.625
After tax −99.31887 2.8 2.8 2.8 2.8 104.0485
Now 1.081/3× (2.8/1.081/2+ 2.8/1.08 + 2.8/1.083/2+ 2.8/1.082) + 104.0485/1.0823/12= 100.2255.
Also 1.091/3× (2.8/1.091/2+ 2.8/1.09 + 2.8/1.093/2+ 2.8/1.092) + 104.0485/1.0923/12= 98.568.
Hence the net yiell is between 8% and 9%.
20. Now (1− t1)g = 0.75× 7/108 = 0.04861 and i(m)= i(4) = 4(1.051/4− 1) = 0.04909. Hence i(m)> (1− t1)g and so there is a capital gain and assume late redemption. Let P denote the price per £100. Then P = 0.75× 7a(4)20 ,0.05+ [108− 0.35(108 − P )] /1.0520. Hence
P [
1− 0.35 1.0520
]
= 5.25a(4)20 ,0.05+ 70.2
1.0520 leading to P = 107.24537545 and the answer £107,245.38.
21. (i) Let ν = 1/1.03158. Then P1= 5a18 ,0.03158+ 100ν18= 5ν(1− ν18)/(1− ν) + 100ν18= 124.9988 or £125.00.
(ii) Now let ν = 1/1.05. Then P2= 5a13 ,0.05+ 100ν13 = 100.
(iii) Let ν = 1/(1 + i) where i is the required gross annual rate of return. Then 125 = 5(ν + ν2+ ν3+ ν4+ ν5) + 100ν5. This implies ν = 1 and i = 0. Hence the rate of return is 0%.
(iv) The value of P1would be larger. Hence we would have ν > 1 in part (iii). This implies i < 0.
22. (i) Define i by 1 + i = 1.0152. Let i′ = 0.015.
Time 0 1/2 2/2 3/2 · · · 19/2 20/2
Cash flow −P 2 2 2 · · · 2 2
So P = 4a(2)10 ,i+ 100/(1 + i)10 = 2a20 ,0.015+ 100/1.01520= 108.584319822 or £108.58.
(ii) Assuming tax of 25%, the price at time 0 is P = 1.5a20 ,0.015+ 100/1.01520 and hence the price 91 days later is P× 1.015182/365= 100.745153968 or £100.75.
23. (i) The return from bond A is i = 5/101. Let P denote the price of bond B then the expected return is [0.1× (−P ) + 0.2(100 − P ) + 0.3(50 − P ) + 0.4(106 − P )] /P = [77.4 − P ] /P Hence we want [77.4− P ] /P = 5/101 which gives P = 77.4 × 101/106 = 73.749057 or e73.749.
(ii) Let i′ denote the gross redemption yield—the gross redemption yield always ignores the possibility of default.
Hence 73.749(1 + i′) = 106 giving 1 + i′ = 1.437308 and hence 43.73%.
(iii) Bond B has higher risk—hence an investor will require a higher return.