- January 3, 2021

APPENDIX 1 Some Mathematical Prerequisites A1.1 Calculus We know that if f (x) = xn where x > 0 and n ∈ R, then f ′(x) = nxn−1. Now suppose f (x) = ax where a > 0 is a constant. Then f (x) = ex ln a. Hence f ′(x) = (ln a)ex ln a = (ln a)ax. An alternative approach: taking logs of f (x) = ax gives ln f (x) = x ln a. Then differentiate to get 1 f (x) f ′(x) = ln a and so f ′(x) = (ln a)ax. The corresponding integration result is displayed below—equation(A1.2a). A1.2 Summary. If f (x) = ax where a > 0 is a constant, then f ′(x) = (ln a) ax∫ ax dx = ax ln a + constant (A1.2a) A1.3 Linear Interpolation Suppose f : R → R is a continuous function. Suppose further that a and b are real numbers with a < b and the numerical values of f (a) and f (b) are known. We require the value of f (c) where a < c < b and it is not possible to calculate the value of f (c) directly. Example 0.3a. The table of the standard normal distribution function, Φ, gives Φ(1.02) = 0.8461 and Φ(1.03) = 0.8485. What is the value of Φ(1.025)? Solution. It seems sensible to give the value which is half way between Φ(1.02) = 0.8461 and Φ(1.03) = 0.8485 and that is 0.8473. We can find an approximate value for f (c) by approximating f with a straight line. If the two values of a and b are close and f is continuous, then this approximation should be reasonable. We are now assuming f is a straight line; hence its gradient is constant. This implies f (c)− f (a) c− a = f (b)− f (a) b− a and hence f (c) = f (a) + c− a b− a [f (b)− f (a)] (A1.3a) Using the gradient after c instead of before c gives the following alternative formulation: f (b)− f (c) b− c = f (b)− f (a) b− a and hence f (c) = f (b)− b− c b− a [f (b)− f (a)] (A1.3b) Both approaches will, of course, give the same approximation. Example 0.3b. Find the value of x with Φ(x) = 0.5095. ST334 Actuarial Methods c©R.J. Reed Sep 27, 2016(9:51) Some Mathematical Prerequisites Page 143 Page 144 Mathematical Prerequisites Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed Solution. From tables we have Φ(0.02) = 0.5080 and Φ(0.03) = 0.5120. The problem can be expressed in the form of the general solution given above if we set f = Φ−1. We have f (0.5080) = 0.02 and f (0.5120) = 0.03 and we require the value of f (0.5095). Using equation (A1.3a) with a = 0.5080 and b = 0.5120 gives f (0.5095) = 0.02 + 0.0015 0.0040 × 0.01 = 0.02375 Using equation (A1.3b) gives f (0.5095) = 0.03− 0.0025 0.0040 × 0.01 = 0.02375 However, it is better to argue from first principles as follows. We know Φ(0.02) = 0.5080 and Φ(0.03) = 0.5120 and we require x with Φ(x) = 0.5095. Using linear interpolation implies Φ(x)− Φ(0.02) Φ(0.03)− Φ(0.02) = x− 0.02 0.03− 0.02 and hence x = 0.02 + 0.01× Φ(x)− Φ(0.02) Φ(0.03)− Φ(0.02) = 0.02 + 0.01× 0.5095− 0.5080 0.5120− 0.5080 which gives the same answer as before. Note that the actual answer is 0.02382. APPENDIX 2 Answers Answers to Exercises: Chapter 1 Section 3 on page 8 (exs1-1.tex) 1. (a) Using c1 = c0(1 + ni) gives ni = 0.2 with n = 2/12. Hence i = 1.2 or 120% p.a. (b) Using c1 = c0(1 + ni) with c1 = 5100, c0 = 5000 and i = 0.06 gives n = 1/3 or 4 months. 2. Under ACT/360 basis, the return over n days is c0 × 0.095 × n/360. Hence if i denotes the yield under ACT/365, then c0 × 0.095× n/360 = c0 × i× n/365. Hence i = 365× 0.095/360 = 0.0963 or 9.63%. 3. 1000× 1.0112 = 1126.82503 and so the answer to (a) is £126.82 assuming the issuer of the investment rounds in its favour, which in this case means down. Also, 1000× 1.0124 = 1269.7346 and so the answer to (b) is £142.91. 4. [ (1 + 0.1)1/12 − 1 ]× 12 = 0.0957 or 9.57%. 5. The amount is 1000× 1.0112 = 1126.83 assuming the lender rounds in its favour, which in this case means up. 6. Denote the effective rate by i. Using( 1 + 0.1× 91 365 ) = (1 + i)91/365 gives i = ( 1 + 0.1× 91 365 )365/91 − 1 = 0.10382 or 10.38%. 7. £1000× 1.065/2 = £1156.82 8. Let i denote the effective annual interest rate. Then 975(1 + i)1/2 = 1000. Hence i = 79/1521 = 0.05194 or 5.19%. 9. (a) 4% p.a. payable every 6 months. (b) (1 + 0.02)2 − 1 = 0.0404 or 4.04%. (c) Using (1 + iep)4 = 1.022 gives iep = 1.021/2 − 1 = 0.00995 or 1.00% to two decimal places. 10. (a) It is equivalent to an effective rate of 3% per 6 months which is equivalent to an effective rate of 1.032−1 = 0.0609 p.a. So answer is 6.09% p.a. (b) Equivalent to an effective rate of 1.061/2 − 1 = 0.029563 for 6 months. Equivalent to a nominal interest rate of 2× 0.02963 = 0.059126, or 5.91% p.a. payable every 6 months. 11. Equivalent to an effective interest rate of 2% per quarter, which is equivalent to an effective interest rate of 1.022−1, or 4.04% per 6 months, which is equivalent to nominal 8.08% p.a. payable every 6 months. 12. (a) Using 1.013 = 1.0303, 1.016 = 1.0615, 1.0112 = 1.12682 and 1.0124 = 1.2697 gives (i) 3.03%; (ii) 6.15%; (iii) 12.68%; (iv) 26.97%. (b) (i) 12%; (ii) 4× 3.03 = 12.12%; (iii) 12.30%; (iv) 12.68%; (v) 13.485%. 13. The equation 1,500 ( 1 + 91i1 365 ) = 1540 leads to i1 = 40 1,500 × 365 91 = 145 1,365 = 0.1069597 and this is 10.70%. Using (1 + i2)91/365 = 1,540 1,500 leads to i2 = ( 77 75 )365/91 = 0.111331 and this is 11.13%. 14. (a) The effective interest rate is 0.025 every 6 months. Let i = 1.025 and let x be the desired answer. Hence: x = 300 1.0253 + 150 1.0257 + 500 1.0259 = 805.1338 or £805.13. (b) The effective interest rate per month is i1 = 1.0251/6 − 1. We want the smallest t with 950 (1 + i1)t > 300 (1 + i1)24 + 150 (1 + i1)48 + 500 (1 + i1)60 = 300 1.0254 + 150 1.0258 + 500 1.02510 where the present time is used as the focal point. This gives (1+ i1)t < 1.209427 or t < 46.2032. So required answer is after 46 months. 15. The initial £50,000 leads to payments of £2,500 at the ends of years 1, 2 and 3. Hence we have: Year Assets at start of year Interest at end of year 1 £50,000 at 5% p.a. £2,500 £50,000 at 5% p.a. £2,500 £2,500 at 6% p.a. £1502 } total = £2,650 £50,000 at 5% p.a. £2,500 £2,500 at 6% p.a. £150 £2,650 at 5.5% p.a. £145.75 3 } total = £2,795.75 Overall the amount returned at the end of year 3 is £50,000 + £2,500 + £2,650 + £2,795.75 = £57,945.75. ST334 Actuarial Methods c©R.J. Reed Sep 27, 2016(9:51) Answers Page 145 Page 146 Answers 1.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 16. (a) The interest rate is an effective 1.5% per 3 months. Hence the discount factor for 6 months is ν1 = 1/1.0152 and the discount factor for 1 year is ν = ν21 = 1/1.015 4. Denote the original loan by ` and the value of the annual payments by y. Then ` = 800ν1(1− ν201 ) 1− ν1 = yν(1− ν10) 1− ν = yν21 (1− ν201 ) 1− ν21 Hence y = 800(1 + ν1)/ν1 = 800(1 + 1.0152) = 1624.18. (b) y = x [ 1 + (1 + i)2 ] . 17. (a) £2,000/1.073 = £1632.60 (b) £2,000/(1 + 0.07× 182/365) = £1,932.55 18. Denote the answer by i. Then 0.45/360 = i/365. Hence i = 0.45625 or 4.56% p.a. 19. (a) 600 = 500(1 + i)4. Hence 500(1 + i)6 = 500× (6/5)3/2 = 657.27. (b) (i) 5,750 = 5,500(1 + i) and so 1 + i = 23/22. Hence 5,500(1 + i)2 = 6011.37 (ii) 5500 × (1 + i)9/12 = 5686.45. However, a lender may use a simple interest calculation in this situation: this gives 5500× (1 + 9i/12) = 5687.50. 20. (a) The amount c0 grows to c0erk/365 after k days. Hence the equivalent simple annual interest rate is 365 k (erk/365 − 1) (b) Invert the previous relation to get (365/k) ln(1 + ki/365). 21. (a) (i) Using ( 1 + i(m) m )m = 1 + i = e0.12 with m = 365/7 gives i(m) = 3657 (e 0.12×7/365 − 1) = 0.1201 or 12.01% per annum compounded every 7 days. (ii) m = 12. Hence i(m) = 12(e0.12/12 − 1) = 0.1206 or nominal 12.06% per annum compounded monthly. (b) (i) £1000e0.12×7/365 = £1002.30; (ii) £1000e0.12/12 = £1010.05. 22. For t ∈ (1, x) we have 1 < t1/n < x1/n. Dividing by t and then integrating this inequality over t between 1 and x gives: ∫ x 1 dt t < ∫ x 1 dt t1−1/n < x1/n ∫ x 1 dt t Hence lnx < n(x1/n − 1) < x1/n lnx. Now x > 0; hence x1/n → 1 as n→∞. Hence result. 23. If i = 0 then f (m) = 0 for all m; hence result. Otherwise, f ′(m) = (1+ i)1/m [ 1− ln(1 + i)/m]−1 and f ′′(m) = (1+ i)1/m( ln(1+ i) )2/m3 > 0 for allm ∈ (0,∞). Hence the function f ′ is strictly increasing. Now limm→∞ f ′(m) = 0. Hence f ′(m) < 0 for all m ∈ (0,∞). Hence f is a decreasing function on (0,∞). 24. A(p) = 1+pip gives the second equality. Using limp↓0 ip = limm↑∞ i(m) = limm↑∞m [ (1 + i)1/m − 1] = ln(1+i) = δ by using question 22. 25. Define g : [0,∞)→ R by g(x) = (1 + x)t − 1− tx. Suppose t > 1. Then g(0) = 0 and g′(x) = t [ (1 + x)t−1 − 1] > 0 for all x > 0. Hence for t > 1 and i > 0 we have (1 + i)t > 1 + ti. Similarly for 0 < t < 1 we have g′(x) < 0 for x > 0; hence for 0 < t < 1 and i > 0 we have (1 + i)t < 1 + ti. 26. (a) 1,000 × 1.1422/365 = 1,116.4949121 or £1,116.49. (b) Interpolate between A(1) = 1000 × 1.1 = 1100 and A(2) = 1000 × 1.12 = 1210. Now A(2) − A(1) = 110. We want A(1) + [A(2)−A(1)] 57/365 = 1,117.18. (c) Because 1 + it > (1 + i)t for 0 < t < 1 by question 25. 27. First investor: £91 grows to £93.90p in 30 days. Let i1 denote the effective interest rate per annum. Then (1 + i1)30/365 = 93.90/91. Hence i1 = 0.4647. Second investor: £93.90 grows to £100 in 60 days. Let i2 denote the effective interest rate per annum. Then (1 + i2)60/365 = 100/93.90. Hence i2 = 0.4665. Hence second investor achieved a higher rate of return. 28. (a) Expected rate of interest is 0.3×0.07+0.5×0.08+0.2×0.1 = 0.081. Hence present value is 10,000/(1.08110) = 4589.26. (b) Expected profit is a4589.26(0.3× 1.0710 + 0.5× 1.0810 + 0.2× 1.110)− 10,000 = 42.94 29. Use ( 1 + i(4)/4 )4 = 1 + i. For part (a) we have ( 1 + i(4)/4 )4 = 1.00512. Hence i(4) = 4 [ 1.0053 − 1] = 0.0603 or 6.03%. For part (b) we have ( 1 + i(4)/4 )4 = (1 + 2 × 0.06)1/2 = 1.121/2. Hence i(4) = 4 [1.121/8 − 1] = 0.5707 or 5.707%. 30. First investor: £96 grows to £97.50 in 45 days. Hence 1 + i1 = (97.50/96)365/45. Second investor: £97.50 grows to £100 in 45 days. Hence 1 + i2 = (100/97.50)365/45. As 100/97.50 = 1.0256 > 97.50/96 = 1.0156, the second investor receives a higher rate of interest. 31. Capital gain is 31p. Hence CGT is 9.3p. Hence 769 grows to (800 − 9.3) + 4 = 794.7 in half a year. Hence 1 + i = (794.7/769)2 giving i = 0.067957 or 6.80%. Appendix 2 Sep 27, 2016(9:51) Answers 1.5 Page 147 32. (a)(i) Suppose n days. Then 4000 = 3600 ( 1 + 0.06n365 ) . Hence n = 365/(9 × 0.06) = 18250/27 = 675.9. Hence 676 days are needed. (ii) Now 1.0154t = 10/9. Hence 4t = ln(10/9)/ ln(1.1015). Or t = 0.25 ln(10/9)/ ln(1.1015) years, or 91.25 ln(10/9)/ ln(1.1015) = 645.7 days. So 646 days are needed. (iii) Now 1.00512t = 10/9. Or t = (1/12)× ln(10/9)/ ln(1.0005) years or (365/12)× ln(10/9)/ ln(1.0005) = 642.54 days. So 643 days are needed. (b) For time periods longer than 1 year, compound interest pays more because the investor receives the benefit of interest on interest. 33. (a) We need d with 3,000 ( 1 + 0.04× d/365) = 3,800. Hence d = 7,300/3 = 2,433.33 or 2,434 days. (b) We need t with 3,000 × 1.04t = 3,800. Hence t = ln(19/15)/ ln(1.04). The number of days is 365t = 2199.9066 or 2,200 days. 34. If i denotes the effective interest rate per annum, then 1 + i = 1.032. Let ν = 1/(1 + i) = 1/1.032; let ν1 = ν1/4 = 1/1.031/2; and finally, let ν2 = ν1/12 = 1/1.031/6. Value of payments in first 4 years at time 0 is 120(1+ν +ν2 +ν3) = 120(1− ν4)/(1− ν). Hence value at time t = 12 is 120(1− ν4)× 1.0324/(1− ν). Value of payments in next 4 years at t = 12 is 30(1−ν161 )×1.0316/(1−ν1). Value of payments in final 4 years at t = 12 is 10(1−ν482 )×1.038/(1−ν2). Hence answer is 894.8930 + 691.0424 + 542.8388 = 2128.77. Answers to Exercises: Chapter 1 Section 5 on page 14 (exs1-2.tex) 1. The discount factor is 1/(1 + 0.08× 91/365) = 0.980. 2. The discount factor is 1/(1 + 0.08)3 = 1/1.083 = 0.794 3. Now i = 25/300 and 1 + i = 13/12. Hence c0(1 + i)2 = 3380 giving c0 = 3380× 12/13× 12/13 = 2880. 4. (a) 720(1 + 1/6)2 = 980 (b) Using (1 + i)(1− d) = 1 gives d = 1/7 or 14.29%. 5. c1 = c0/(1− d) = 275/(1− 1/3) = 275× 3/2 = 412.5 6. By using the relation (1 + i)(1− d) = 1; i increases implies d increases and hence ν = 1− d decreases. 7. Let i1 = ki. Then d1 = i1/(1 + i1) and so d1/d = k(1 + i)/(1 + ki) < k. 8. (a) As i− d = i2/(1 + i), it follows that i− d > 0. 9. (a) Using ( 1− d(m)/m)m = 1 − d with m = 1/2 and d(m) = 0.025 gives d = 1 − √0.95 = 0.02532 or 2.53% p.a. (b)(i) Now ( 1− d(4)/4)4 = 1 − d = 1/(1 + i) = 1/1.045. Hence d(4) = 0.0437756 or nominal 4.38% p.a. payable quarterly. (ii) Using ( 1− d(12)/12)12 = 1 − d = 1/(1 + i) = 1/1.045 gives d(12) = 0.043936 or nominal 4.39% p.a. payable monthly. (iii) For this part, m = 1/3. Using ( 1− d(1/3)/(1/3))1/3 = 1 − d = 1/(1 + i) = 1/1.045 gives 1− 3d(1/3) = 1/1.0453 and hence d(1/3) = 0.041234 or nominal 4.12% p.a. payable every 3 years. 10. (1− dep)1/p = 1− d = 1/(1 + i) = 1/(1 + ieq)1/q . 11. (a) (i) (1.1)k/12−1 (ii) (1.1)p−1. (b) The effective discount rate is d = i/(i+1) = 1/11 p.a. and 1−d = 10/11. Then use (1− dep)1/p = 1− d. (i) Now p = k/12. Hence dep = 1− (10/11)k/12. (ii) dep = 1− (10/11)p 12. Use ( 1− d(m)/m)m = 1 − d. Hence 1 − d = (1 − 0.02)4 = 0.984. Hence (1− d(2)/2)2 = 0.984. Hence d(2) = 2(1− 0.982) = 0.0792 or nominal 7.92% payable every 6 months. 13. (a) The accumulated value after k years is c0(1 + i)k. Hence we want (1 + i)k = 2 or k = ln 2/ ln(1 + i). (b) The accumulated value after k years is c0eik. Hence we want eik = 2 or k = ln 2/i. (c) Annual compounding: 69.66, 14.21, 10.24, and 7.27. Continuous compounding: 69.31, 13.86, 9.90, and 6.93. (d) Now na = ln 2/ ln(1 + i) and nb = ln 2/i. Also i− i2/2 < ln(1 + i) < i for i > 0. 14. (a) Now c0(1 + i2) = c0(1 + i1)2. (i) Hence i2 = 2i1 + i21 and so i2 > 2i1. (ii) Now c0 = (1− d2)c2 = (1− d1)2c2. Hence d2 = 2d1 − d21 and so d2 < 2d1. (b) Now ct = c0(1 + it) = c0(1 + i1)t. Hence (1 + i1) = (1 + it)1/t = (1 + is)1/s. Hence 1 + it = (1 + is)t/s. Using t = 2s gives 1 + it = (1 + is)2 and hence it > 2is. Similarly, (1 − ds) = (1 − d1)s and (1 − dt) = (1 − d1)t imply 1− dt = (1− ds)t/s. Using t = 2s gives dt < 2ds. 15. (a) We are given d1 = 0.01 for 1 month. Hence c0 = (1 − d1)c1/12 = (1 − d1)2c2/12 = (1 − d2)c2/12. Hence d2 = 1− (1− d1)2 = 1− 0.992 = 0.0199. (ii) Similarly dk = 1− (1− d1)k = 1− 0.99k (b) 0.983 × 5000 = 4705.96 16. (a) Now ( 1 + i(m)/m )m = 1 + i = eδ . Hence i(m) = m ( eδ/m − 1). Using (1 + i(m)/m)m (1− d(m)/m)m = 1 gives ( 1− d(m)/m)m = e−δ and hence the result. (b) To show i > i(2) > i(3) > · · · see question 23 on page 9 (section 3 of chapter 1). To show d < d(2) < d(3) < · · ·, let g(m) = m(1−e−δ/m). Then g′(m) = 1−(1+δ/m)e−δ/m. But ex > 1 + x for x > 0 and hence g′(m) > 0 for m > 0. Hence result. (c) By exercise 22 on page 9, we have limn→∞ n(x1/n − 1) = lnx. Hence limm→∞ i(m) = δ. From ( 1 + i(m)/m ) ( 1− d(m)/m) = 1 we get 1/d(m) = 1/m + 1/i(m) and hence limm→∞ d(m) = limm→∞ i(m). Page 148 Answers 1.7 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 17. Using c0 = (1− nd)cn gives c0 = ( 1− 90× 0.06/365) cn = 1798cn/1825. Also cn = (1 + i)90/365c0 where i is the required answer. Hence i = ( 1825/1798 )365/90 − 1 = 0.06231 or effective 6.23% p.a. 18. Using c0 = (1 − nd)cn gives c0 = ( 1− 0.045× 28/365) cn = (1− 1.26365 ) cn = 36,374cn/36,500. Also cn = (1+i)28/365c0 where i is the required answer. Hence i = ( 36,500/36,374 )365/28−1 = 0.04611 or effective 4.61% p.a. 19. Using c0 = (1 − nd)cn gives c0 = (1 − 90 × 0.05/365)cn = 721cn/730. Also cn = (1 + i)90/365c0 where i is the effective rate p.a. and (1 + i(2)/2)2 = 1 + i where i(2) is the required answer. Hence i(2) = 2 [ (1 + i)1/2 − 1] = 2 [ (730/721)365/180 − 1] = 0.05094891 or nominal 5.09% p.a. converted 6-monthly. 20. Using c0 = (1− nd)cn gives c0 = (1− 0.02)4c1 = 0.984c1. (a) Now c1 = (1 + i(2)/2)2 where i(2) is the required answer. Hence i(2) = 2(1/0.982 − 1) = 0.08246564 or nominal 8.25% p.a. convertible 6-monthly. (b) Now c0 = (1 − d(12)/12)12c1 where d(12) is the required answer. Hence d(12) = 12(1− 0.981/3) = 0.0805393 or nominal 8.05% p.a. convertible monthly. 21. (i)(a) Now c1 = 1.045c0 and we want d with c0 = (1−d)c1. Hence d = 0.045/1.045 = 0.04306 or 4.306%. (b) We want d(12) with c0 = (1−d(12)/12)12c1. Hence d(12) = 0.04394 or 4.394%. (c) We want i(4) with (1+i(4)/4)4 = 1.045. Hence i(4) = 0.04426 or 4.426%. (d) Now c5 = 1.0455c0 = 1.24618193765c0 and hence 24.618%. (ii) The discount rate d(12) is applied each month to a value which has already been discounted. Hence the rate d(12)/12 must be higher than d/12 in order to obtain the same total amount of discounting. 22. (The question is not clear: you need to assume the treasury bill is also for 91 days.) Now if c0 is the price of the treasury bill, then c0×1.0491/365 = 100. Hence if d is the annual simple rate of discount, then c0 = (1−91d/365)100. Hence d = 365 91 ( 1− 1 1.0491/365 ) 0.0390296 or 3.90%. 23. (i) Suppose d days. Hence $96.50 grows to $98 in d days. Hence 96.5 × 1.04d/365 = 98 giving d/365 = ln(98/96.5)/ ln(1.04) and hence d = 143.54 or 144 days. (ii) The amount $98 grows to $100 in 38 days. Hence 98(1 + 38i/365) = 100 giving i = 0.196026 or 19.603%. (iii) We want i with 98(1 + i)38/365 = 100. Hence i = (100/98)365/38 − 1 = 0.21415980 or 21.416%. 24. (a) We need c0 with c0 × 1.0391/365 = 100. This gives c0 = 100/1.0391/365 = 99.26576. (b) We need d with c0 = (1− 91d/365)100. Hence d = (365/91)(1− 1/1.0391/365) = 0.029450153 or 2.945%. 25. (i) We have c0 = (1 − 0.055/12)12c1. (a) We need δ with c1 = eδc0. Hence eδ = (1 − 0.055/12)−12 giving δ = 0.0551264 or 5.513%. (b) We need i with 1 + i = (1 − 0.055/12)−12 and hence i = 0.0566741 or 5.667%. (c) We need i(12) with (1+ i(12)/12)12 = (1−0.055/12)−12 and hence i(12) = 12 [(1− 0.055/12)−1 − 1] = 0.0552532 or 5.525%. (ii) In (b), the interest is just added at the end of the year; in (c), the interest is added every month and compounded every month. Because the final sum is the same, the rate in (c) must be less than the rate in (b). (iii) We need d with 100 = (1 − d)8 × 159. Hence d = 0.0563187 or 5.632%. (iv) We need i with (1 + i)2 = 1.12. Hence i = 0.0583006 or 5.830%. Answers to Exercises: Chapter 1 Section 7 on page 18 (exs1-4.tex) 1. (a) 1000 exp(0.05/2) = 1025.31 (b) 1000× 1.050.5 = 1024.6951 or £1,024.70 to nearest penny. 2. 1000 exp (∫ 6 1 δ(s) ds ) = 1000 exp (∫ 6 1 (0.04 + 0.01s)ds ) = 1000 exp(0.2 + 0.175) = 1454.99 3. (a) δ(t) = −ν′(t)/ν(t) (b) b. 4. A(t) = eδt, ν(t) = e−δt and δ̂(t) = δ. 5. Differentiating δ̂(t) = 1t ∫ t 0 δ(s) ds gives dδˆ(t) dt = 1 t2 ( tδ(t)− ∫ t 0 δ(s) ds ) for t > 0. As δ(s) ↑ as s ↑, it follows that ∫ t0 δ(s) ds ≤ tδ(t). Hence result. 6. Recall that δ̂(t) = 1t ∫ t 0 δ(s) ds. Hence δˆ(t) = −ln ν(t)/t. Suppose δ̂ ↑ as t ↑. Then for all t > 0 and all α ∈ [0, 1] we have δˆ(αt) ≤ δˆ(t) and so ln ν(αt) ≥ α ln ν(t). It follows that ν(αt) ≥ (ν(t))α. The proof of the converse is similar. 7. Now ν(t) = exp ( − ∫ t0 δ(s) ds) = exp [−0.06(0.7t − 1)/ ln 0.7]. The present value of £100 in 31/2 years’ time is 100× ν(3.5) = 88.696882 or £88.70. 8. (a) Use 1 + pip(t) = A(t + p)/A(t) and A(t) = exp (∫ t 0 δ(u) du ) . (b) Now ∫ t 0 a u du = (at − 1)/ ln a. Hence pip(t) = exp(0.06× 0.7t(1− 0.7p)× ln 107 )− 1. Appendix 2 Sep 27, 2016(9:51) Answers 2.3 Page 149 9. (a) Using cn = c0(1 + ni) gives 1,550 = 1,500 ( 1 + 0.05n365 ) . Hence n = 730/3 = 243.3 days. It takes 244 days. (b) 1,550 = 1,500e0.05n/365. Hence n = 7,300× ln(31/30) = 239.37 days. It takes 240 days. 10. Now 210 = 200 exp (∫ 5 0 (a + bt 2) dt ) = 200 exp(5a+125b/3). Also 230 = 200 exp (∫ 10 0 (a + bt 2) dt ) = 200 exp(10a+ 1000b/3). So we have the 2 simultaneous equations: 5a + 125b/3 = ln(21/20) and 10a + 1000b/3 = ln(23/20). Hence 750b/3 = ln(23× 20/212) = ln(460/441); hence b = (3/750)× ln(460/441) = 0.00016872646 or 0.0001687. Similarly, 30a = ln(218/(207 × 23)) and so a = 0.008351979 or 0.008352. 11. (a)(i) Using c0 ( 1 + 0.05/12 )120 = 100 gives c0 = 60.716 or £60.72. (ii) For this case, c0 = ( 1− 0.05/12)120×100. Hence £60.59. (iii) Using c0e0.5 = 100 gives c0 = 100e−0.5 = 60.6531 or £60.35. (b). Let i denote the equivalent effective annual interest rate. Hence (1 + i)91/365 = 100/98.91. Hence i = 0.04494 or 4.49%. 12. Now ∫ t 0 δ(s) ds equals 0.04t + 0.005t 2 if t < 8 and 0.64 + (t − 8)0.07 = 0.08 + 0.07t if t > 8. Hence A(t) equals e0.04t+0.005t 2 if t < 8 and e0.08+0.07t if t > 8. (b) The present value is 100/A(10) = 100e−0.78 = 45.84. 13. The general formula is A(t2) = A(t1) exp (∫ t2 t1 δ(s) ds ) . So we want 500 exp (∫ 10 2 δ(s) ds ) + 800 exp (∫ 10 9 δ(s) ds ) . Now ∫ 10 9 δ(s) ds = ∫ 10 9 (0.01s − 0.03)ds = 0.065 and ∫ 10 2 δ(s) ds = 0.345. Hence we want 500e 0.345 + 800e0.065 = 1559.72. (b) We want i where 500(1 + i)8 + 800(1 + i) = 1559.72. Now £1 grows to £e0.345 = £1.412 in 8 years which corresponds to 4.4%. So guess 4%: then 500(1 + i)8 + 800(1 + i) = 1516.28. Guess 5%: then 500(1 + i)8 + 800(1 + i) = 1578.73. So answer is 5% to nearest 1%. (It is 4.71%.) 14. (i) Now c0 = (1 − 0.02)4c1 = 0.984c1 and c1 = eδc0. Hence δ = −4 ln 0.98 = 0.0808108. (ii) Now c1 = (1 + i)c0 = (1+i)0.984c1. Hence i = 1/0.984−1 = 0.0841658. (iii) Now c0 = (1−d(12)/12)12c1. Hence d(12) = 12(1−0.981/3) = 0.0805393. 15. Now∫ t 0 δ(s) ds = { 0.05t + 0.01t2 for 0 ≤ t ≤ 5; 0.15t− 0.25 for t > 5 and hence ν(t) = { e−0.05t−0.01t 2 for 0 ≤ t ≤ 5; e0.25−0.15t for t > 5 (i) We want 1,000ν(12) = 1,000e−1.55 = 212.24797383 or £212.25. (ii) We want d with 212.25 = (1− d)121,000 and hence d = 0.12103392 or 12.1%. Answers to Exercises: Chapter 2 Section 3 on page 27 (exs2-1.tex) 1. (a) NPV(a) = 2575.25 and NPV(b) = 2509.90. (b) NPV(a) = 2372.73 and NPV(b) = 2509.09. (c) At time 0, borrow 2400 and take c0 = b0 = 2500. This transforms a = (100, 2500) into c = (2500,−2400(1 + 0.01) + 2500 = 76). (d) At time 0, invest 2400 and take c0 = a0 = 100. This transforms b = (2500, 10) into c = (100, 2400(1 + 0.1) + 10 = 2650). (e) Consider the two cases a0 ≥ b0 and a0 < b0 and use induction on n. If a0 ≥ b0 then set c0 = b0 and invest a0− b0. This converts a into the sequence (b0, (1 + i)(a0− b0) + a1, a2, . . . , an). Let a∗ = ((1 + i)(a0 − b0) + a1, a2, . . . , an) and b∗ = (b1, . . . , bn). Then NPV(a∗)− NPV(b∗) = (1 + i)(a0 − b0) + n−1∑ j=0 aj+1 (1 + i)j − n−1∑ j=0 bj+1 (1 + i)j = (1 + i) [NPV(a)− NPV(b)] ≥ 0 Hence result by induction assumption. Similarly for a0 < b0. 2. The value at time 0 is: NPV = A(0) + ∫ t 0 ρ(s)ν(s) ds where ν(s) = exp [ − ∫ s 0 δ(u) du ] Hence value at time t is: NPV× exp [∫ t 0 δ(u) du ] = A(0) exp [∫ t 0 δ(u) du ] + ∫ t 0 ρ(s) exp [∫ t s δ(u) du ] ds Alternatively: NowA′(t) = δ(t)A(t)+ρ(t). As usual, let ν(t) = e− ∫ t 0 δ(s) ds denote the value at time 0 of the amount 1 at time t. Integrating the equation ν(t)A′(t) = ν(t)δ(t)A(t) + ν(t)ρ(t) gives ν(t)A(t)− ν(0)A(0) = ∫ t 0 ν(s)ρ(s) ds and hence Page 150 Answers 2.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed A(t) = A(0) ν(t) + ∫ t 0 ν(s) ν(t) ρ(s) ds = A(0)e ∫ t 0 δ(s) ds + ∫ t 0 e ∫ t s δ(h) dh ρ(s) ds (b) A(t) = A(0)(1 + i)t + ∫ t 0 (1 + i) t−sρ(s) ds 3. Using equation (2.9b) gives NPV = 4 ∫ 1 0 νu du + 6 ∫ 2 1 νu du + 8 ∫ 3 2 νu du But ∫ νu du = ∫ exp(u ln ν) du = νu/ ln ν. Hence NPV = [ 4(ν − 1) + 6(ν2 − ν) + 8(ν3 − ν2)] / ln ν = [8ν3 − 2ν2 − 2ν − 4] / ln ν. This leads to A(3) = NPV/ν3 = [ 4× 1.043 + 2× 1.042 + 2.08− 8] / ln 1.04 = 18.935. 4. Now ρ = 100 and NPV = ∫ 6 0 ρ(u)ν(u) du = 100 ∫ 6 0 ν(u) du where ν(t) = exp [ − ∫ t0 δ(u)du]. Hence NPV = 100 ∫ 6 0 exp [−0.06t] dt = 100(1 − e−0.36)/0.06. Then use A(12) = NPV × exp [∫ 12 0 δ(u) du ] . where ∫ 12 0 δ(u) du = 0.06× 6 + ∫ 126 (0.05 + 0.0002s2) ds = 0.36 + 0.3 + 0.1008 = 0.7608. Hence A(12) = NPV× exp(0.7608) = 1078.28. 5. Solving 990(1 + i)− 65(1 + i)0.5 − 1076 = 0 gives (1 + i)0.5 = 1.075875 and so i(2) = 2 [(1 + i)0.5 − 1] = 0.15175 or 15.17%. 6. Using equation (2.9a) gives NPV = ∫ 8 0 50ν(u) du. But for 0 ≤ t ≤ 8 we have ν(t) = exp ( − ∫ t0 δ(u) du) = exp(−0.05t). Hence NPV = 50 ∫ 80 exp(−0.05u) du = 50(1− e−0.4)/0.05 = 1,000(1− e−0.4). Value at time t = 15 is A(15) = NPV/ν(15). Now ∫ 15 0 δ(u) du = 0.4+ ∫ 15 8 (0.04+0.0004u 2) du = 0.68+0.0004(153−83)/3 = 0.68+1.1452/3. Hence A(15) = 953.2295 or £953.23. 7. Use equation (2.9a) with ρ(t) = 50 exp(0.01t) for 9 < t < 12. Hence NPV = ∫ 12 9 50 exp(0.01u)ν(u) du where ν(u) = exp (− ∫ u0 δ(s) ds). Now for u > 8 we have ∫ u0 δ(s) ds = 0.16+0.08+0.48−0.01×24+0.06(u−8) = 0.06u. Hence ν(u) = e−0.06u for u > 8. Hence NPV = 50 ∫ 12 9 e −0.05u du = 1,000(e−0.45 − e−0.6) = 1,000e−0.6(e0.15 − 1) = 88.8165. 8. Let ν = 1/1.09 and use units of £1,000. Then NPV = −60 − 25ν2/3 + 50ν22 +∑4j=0 ∫ 4j+64j+2 (4j + 5)νt dt = −60 − 25ν2/3 + 50ν22 + ∑4 j=0(4j + 5)ν 4j+2 ∫ 0 ν t dt = −60− 25ν2/3 + 50ν22 + ν2(5 + 9ν4 + 13ν8 + 17ν12 + 21ν16)x where x = ∫ 4 0 ν t dt = (ν4−1)/ log(ν). Hence NPV = −60−25ν2/3+50ν22+ν2(−5−4ν4−4ν8−4ν12−4ν16+21ν20)/ ln(ν) = 7.154485046 or £7,154.49. 9. (a) Use equation (6.1c). A(10) = 500 exp (∫ 10 0 δ(t) dt ) where ∫ 10 0 δ(t) dt = 0.56− 0.005t 2 2 ∣∣∣8 0 + 0.12 = 0.68− 0.16 = 0.52. Hence A(10) = 500e0.52 = 841.014 or £841.01. (b) Use equation (2.9a) with ρ(u) = 200e0.1u for 10 < u ≤ 18. Also, for u > 8 we have ∫ u 0 δ(s) ds = 0.4 + 0.06(u− 8) = 0.06u− 0.08 and hence ν(u) = exp (−0.06u + 0.08). Hence NPV = ∫ 18 10 ρ(u)ν(u) du = 200 ∫ 18 10 exp(0.1u) exp(−0.06u + 0.08) du = 200e0.08 ∫ 18 10 e −0.04u du = 200e0.08 ×[ e0.72 − e0.4] /0.04 = 5,000 [e0.8 − e0.48] = 3,047.33. 10. Let i denote the effective interest rate per annum; hence 1 + i = 1.024 and ν = 1/(1 + i) = 1/1.024. Let x = ν1/12. Then NPV = 5,000 [∫ 1 0 νu du + ν 12 (1 + x + x2 + · · · + x11) + ν 2 2 (1 + ν1/2) ] = 5,000 [∫ 1 0 eu ln ν du + ν 12 1− x12 1− x + ν2 2 (1 + ν1/2) ] = 5,000 [ eu ln ν ln ν ∣∣∣∣u=1 u=0 + ν 12 1− ν 1− x + ν2 2 (1 + ν1/2) ] = 5,000 [ ν − 1 ln ν + ν 12 1− ν 1− x + ν2 2 (1 + ν1/2) ] = 13,347.39 11. (i) Use equation (6.2a): ∫ 12 0 δ(s) ds = 0.2 + 0.04t 2 ∣∣t=10 t=5 + (0.0025t 2 + 0.0001t3) ∣∣t=12 t=10 = 0.2 + 0.3 + 0.005 × 22 + 0.0001 × 728 = 0.6828. Hence NPV = ν(12) = e−0.6828 = 0.5052. (ii) We want i with (1 + i)12 = e0.6828, or i = exp(0.6828/12)− 1 = 0.0585499501 or 5.85%. (iii) Use equation (2.9a). For 2 < u < 5, ρ(u) = e−0.05u and∫ u 0 δ(s) ds = 0.04u. Hence NPV = ∫ 5 2 e−0.09u du = −e −0.09u 0.09 ∣∣∣∣5 2 = e−0.18 − e−0.45 0.09 = 2.196 Appendix 2 Sep 27, 2016(9:51) Answers 2.3 Page 151 12. (i) Use equation (6.1c). We have 150 = 100 exp (∫ 5 0 δ(s) ds ) = 100 exp ( 25a 2 + 125b 3 ) or 25a2 + 125b 3 = ln ( 3 2 ) . Similarly and 50a+ 1,000b3 = ln ( 23 10 ) . Hence 500b3 = ln ( 23 10 )−4 ln ( 32) = ln ( 2310)− ln ( 8116) = ln ( 184405). Hence b = 3500 × ln ( 184405) = −0.004734. Also 50a = 8 ln ( 32)− ln ( 2310) = ln ( 328052944 ) and hence a = 0.048216. (ii) We want δ with 100e10δ = 230 or δ = (1/10)× ln(23/10) = 0.08329. (iii) Use equation (2.9a) with ρ(u) = 20e0.05u for 0 < u < 10 and ν(u) = e−δu where δ is given in part (ii). Hence NPV = ∫ 10 0 ρ(u)e−δu du = 20 ∫ 10 0 e(0.05−δ)u du = 20 1− e0.5−10δ δ − 0.05 = 200 23 23− 10e0.5 ln(23/10)− 0.5 = 170.1153 where we have used e−10δ = 10/23 and 10δ = ln(23/10) from part (ii). 13. (i) Now ∫ 10 0 δ(s) ds = 0.2 + 0.01 ( t3 3 − t 2 2 )∣∣∣∣t=10 t=5 = 0.2 + 15.25 6 Hence ν(10) = exp (−0.2− 15.25/6) = 0.06446282 or 0.0645. (ii) Now 1 + i = ν(9)/ν(10) = A(9, 10) = exp [∫ 10 9 0.01(s 2 − s) ds ] = exp [ 0.01× 485/6] = 2.24416 and hence i = 1.24416 or 124.16%. (iii) (a) ν(t) = exp ( − ∫ t0 δ(s) ds) = exp (−0.04t) (b) NPV(ρ) = ∫ 50 ρ(u)ν(u) du = ∫ 50 1 du = 5. 14. (i) If t ≤ 8, ∫ t0 δ(s) ds = ∫ t0 (0.03 + 0.01s) ds = 0.03t + 0.005t2. At t = 8 this has value 0.56. Hence for t > 8 we have ∫ t 0 δ(s) ds = 0.56 + (t − 8)0.05 = 0.05t + 0.16. Hence ν(t) = exp(−0.03t − 0.005t2) if 0 ≤ t ≤ 8 and ν(t) = exp(−0.16 − 0.05t) if t > 8. (ii) (a) We want 500ν(15) = 500 exp(−0.16 − 0.75) = 500 exp(−0.91) = 201.26. (b) let d(4) be the required quantity. Now 201.26 grows to 500 in 15 years. Hence 500 [ 1− d(4)/4]60 = 201.26 and d(4) = 0.0602096 or 6.02%. (iii) We want ∫ 14 10 ρ(t)δ(t) dt = ∫ 14 10 10e −0.02te−0.16−0.05t dt = 10e−0.16 ∫ 14 10 e −0.07t dt = 10 [ e−0.86 − e−1.14] /0.07 = 14.763. 15. (i) If t ≤ 10, ∫ t0 δ(s) ds = ∫ t0 (0.04 + 0.01s) ds = 0.04t + 0.005t2. At t = 10 this has value 0.9. Hence for t > 10 we have ∫ t 0 δ(s) ds = 0.9 + (t − 10)0.05 = 0.05t + 0.4. Hence ν(t) = exp(−0.04t − 0.005t2) if 0 ≤ t ≤ 10 and ν(t) = exp(−0.4 − 0.05t) if t > 10. (ii)(a) We want 1000ν(15) = 1000 exp(−0.4 − 0.75) = 1000 exp(−1.15) = 316.636769 or £316.64. (b) 1000(1− d)15 = 316.64. Hence d = 1− exp(−1.15/15) = 0.07380 or 7.38%. (iii) We want ∫ 15 10 ρ(t)δ(t) dt = ∫ 15 10 20e −0.01te−0.4−0.05t dt = 20e−0.4 ∫ 15 10 e −0.06t dt = 20e−0.4 [ e−0.6 − e−0.9] /0.06 = 1000(e−1 − e−1.3)/3 = 31.78255. 16. (i) For t > 6 we have ∫ t 0 δ(s) ds = 0.6− 0.09 + 0.07(t− 6) = 0.09 + 0.07t Hence ν(7) = exp(−0.58) and ν(10) = exp(−0.79). Answer is 100/ν(10) + 50ν(7)/ν(10) = 100 exp(0.79) + 50 exp(0.21) = 282.0235456 or £282.02. (ii) NPV = ∫ 15 12 ρ(u)ν(u) du = 50 ∫ 15 12 exp(−0.09− 0.02t) dt = 2500(e−0.33 − e−0.39) = 104.667147 or £104.67. 17. (i) Now the value at time t = 0 of £1,000 at time t = 10 is 1,000ν(10) and this has value 1,000ν(10)/ν(5) at time t = 5. Recall ν(t) = exp ( − ∫ t0 δ(s) ds). Hence we want 1,000 exp(− ∫ 105 δ(s) ds). Now ∫ 10 5 δ(s) ds = ∫ 7 5 (0.1− 0.01s) ds + ∫ 10 7 (0.01s− 0.04) ds = 0.2− 0.12 + 0.51/2− 0.12 = 0.215. Hence answer is 1,000 exp(−0.215) = 806.541440 or £806.54. (ii) The equivalent effective rate of interest per annum is given by 806.54(1 + i)5 = 1,000. Hence we want i(12) with 806.54(1 + i(12)/12)60 = 1,000. Hence i(12) = 0.043077 or 4.3%. (iii) Value at time t = 0 is ∫ ρ(s)δ(s) ds = ∫ 4 0 100e 0.02se−0.06s ds = 100 ∫ 4 0 e −0.04s ds = 2500(1 − e−0.16). We want the value at time t = 12 which is 2500(1− e−0.16)/ν(12). Now ∫ 120 δ(s) ds = 0.24 + 0.3− 0.165 + 0.475− 0.2 = 0.65 and hence ν(12) = exp(−0.65). Hence required answer is 2500(1− e−0.16)× exp(0.65) = 708.061523 or £708.06. 18. (i) Now ∫ 10 0 δ(s) ds = 0.2 + 0.003 t3 3 ∣∣∣5 0 + 0.03 + 0.03 t 2 2 ∣∣∣8 5 + 0.04 = 0.98. Hence NPV = 1,000ν(10) = 1,000 exp−0.98 = 375.311099 or £375.31. (ii) Now 1,000(1− d(4)/4)40 = 375.31. Hence d(4) = 0.09680953 or 9.68%. (iii) Now NPV = ∫ 18 10 ρ(u)ν(u) du = ∫ 18 10 100e 0.01uν(u) du. For u > 10 we have ∫ u 0 δ(u) du = 0.02u + 0.78 and so ν(u) = e−0.02u−0.78. Hence NPV = 100e−0.78 ∫ 18 10 e −0.01u du = 10,000e−0.78(e−0.1 − e−0.18) = 318.900257 or £318.90. Page 152 Answers 2.6 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 19. (i) If 0 ≤ t ≤ 9 then ∫ t0 δ(s) ds = 0.03t + 0.005t2. Hence ∫ 90 δ(s) ds = 0.27 + 0.405 = 0.675. Hence for t > 9 we have ∫ t 0 δ(s) ds = 0.675 + 0.06(t− 9) = 0.135 + 0.06t. Hence ν(t) = e−0.03t−0.005t 2 for 0 ≤ t ≤ 9 and ν(t) = e−0.135−0.06t for t > 9. (ii)(a) We need 5,000ν(15) = 5,000e−0.135−0.9 = 5,000e−1.035 = 1776.131905 or £1,776.13. (b) We need δ with 1,776.13e15δ = 5,000. Hence δ = ln(5,000/1,776.13)/15 = 0.06900007148965 or 0.069. (iii) Now NPV = ∫ 15 11 ρ(s)ν(s) ds = ∫ 15 11 100e −0.02se−0.135−0.06s ds = 100 ∫ 15 11 e −0.135−0.08s ds = 100e−0.135(e−0.88 − e−1.2)/0.08 = 124.055318 or 124.055. 20. Now ∫ t 0 δ(s) ds = 0.03t + 0.0025t 2 for t < 3 and 0.0975 + 0.005t for t > 3. Hence NPV = 5,000 ∫ 6 3 ν(u) du = 5,000 ∫ 6 3 e −0.0975−0.005u du = 5,000e−0.0975 [ e−0.015 − e−0.03] /0.005 = 1,000,000 (e−0.1125 − e−0.1275). This has value 13303.9312743 or £13,303.93. (ii) We need i with 1 + i = eδ where 13,303.93 = 5,000 ∫ 6 3 e −δu du. Hence 13,303.93 = 5,000 [ e−3δ − e−6δ] /δ. Definef (i) = 13,303.93− 5,000 [e−3δ − e−6δ] /δ where δ = ln(1 + i). Using the approximation ex ≈ 1 + x + x2/2 shows [ e−3δ − e−6δ] /δ is approximately 3−14δ. This leads to i ≈ 0.025. Now f (0.025) = −121.65 and f (0.03) = 167.84. Linear interpolation gives i ≈ 0.0271 or 2.7%. (Actually 2.70844%.) (iii) 300A(50) = 300 exp(0.0975 + 0.005× 50) = 300e0.3475 = 424.657293 or £424.66. Answers to Exercises: Chapter 2 Section 6 on page 35 (exs2-2.tex) 1. The following function does more than the question requires. It also returns the interest rate and NPV just before the sign changes (if one exists). “irr” n numr tmp NPV signs signchanges if(length(signchanges) == 0) noroot else { noroot firstchange } results if(noroot) { list(results) } else list(table = results, interest = r[firstchange], NPV = NPV[firstchange]) } 2. Applying the bond yield to the cash flow of project A gives: −50 + 8/1.07 + 8/1.072 + · · · + 8/1.078 = −2.23. This suggests that project A is not profitable. 3. Tke IRR is the solution for r of the equation m∑ j=0 cj (1 + i)tj = n∑ j=0 bj (1 + i)t ′ j Define the function g : (−1,∞)→ R by g(r) = n∑ j=0 bj (1 + r)t ′ j −tm − m∑ j=0 cj(1 + r)tm−tj Then g is continuous with limr→−1 g(r) = ∞, limr→∞ g(r) = −∞ and g(r) ↓↓ as r ↑↑ for r ∈ (−1,∞). Hence there exists a unque r∗ ∈ (−1,∞) with g(r∗) = 0 and then m∑ j=0 cj (1 + r∗)tj = n∑ j=0 bj (1 + r∗)t ′ j Appendix 2 Sep 27, 2016(9:51) Answers 2.6 Page 153 4. Use units of £1,000. We have: Time 0 8 9 10 11 Cash flow for project A (£ million) -1,000 1,700 0 0 0 Cash flow for project B (£ million) -1,000 1,000 321 229 245 (i) Let i′ denote the required effective interest rate per annum and let ν = 1/(1+i′). Then 1,700ν8 = 1,000ν8+321ν9+ 229ν10+245ν11 or 700(1+i)3−321(1+i)2−229(1+i)−245 = 0. So let f (i) = 700(1+i)3−321(1+i)2−229(1+i)−245. Now the return on project A is 7%. So try f (0.07) = −0.0128 and f (0.071) = 1.4774767. Interpolation gives i′ = 0.07 + 0.001× 0.0128/(0.0128 + 1.4774767) = 0.7001 or 7.00%. (ii) The cash is received earlier with project A than with project B. Hence an interest rate higher than i′ favours project A; an interest rate lower than i′ favours project B. In particular, if i = 0.06 then NPVA = −1,000 + 1,700/1.068 = 66.6010 and NPVB = −1,000 + 1,000/1.068 + 321/1.069 + 229/1.0610 + 245/1.0611 = 74.3471. 5. The annual effective interest rate is i = 1.032 − 1 = 0.609; so let ν = 1/1.0609. Here are the discrete cash flows: Time 1 Jan 2001 1 August 2001 31 Dec 2011 Time in years 0 7/12 11 Cash flow (£ million) -2 -1.5 3 We also have the continuous cash flow of 0.3 for t ∈ (1, 2), 0.4 for t ∈ (2, 3), . . . , 1.2 for t ∈ (10, 11). Hence NPV(c, ρ) = −2− 1.5ν7/12 + 3ν11 + 0.3 ∫ 2 1 νt dt + · · · + 1.2 ∫ 11 10 νt dt = −2− 1.5ν7/12 + 3ν11 + 12∑ k=3 0.1k ∫ k−1 t=k−2 νt dt = −2− 1.5ν7/12 + 3ν11 + 12∑ k=3 0.1k [ νk−1 − νk−2 ln ν ] = −2− 1.5ν7/12 + 3ν11 + 0.1 ln ν [−3ν − ν2 − ν3 − · · · − ν10 + 12ν11] = −2− 1.5ν7/12 + 3ν11 + 0.1 (1− ν) ln ν [−3ν + 2ν2 + 13ν11 − 12ν12] = −1.883476 + 4.9922 = 3.10875427438 or £3,108,754. 6. Let ν = 1/1.09. The NPV of outgoings (in £ million) is: NPVO = 9 + 12 ∫ 2 1 νu du = 9 + 12νu ln ν ∣∣∣∣2 u=1 = 9 + 12(ν2 − ν) ln ν = 9− 12× 900 1092 ln ν = 9 + 10,800 11,881 ln 1.09 Hence NPVO = 19.5481398596. Now let x = √ v = 1/1.091/2. Incomings (in £ million): Time 1/1/06 1/7/06 1/1/07 1/7/07 1/1/08 1/7/08 1/1/09 · · · 1/1/21 Time in years 0 1/2 1 1.5 2 2.5 3 · · · 15 Cash flow (£ million) 0 0 0 0 0 2.5 2.5 · · · 2.5 After k payments, NPVI = 2.5[x5 + x6 + · · · + xk+4] = 2.5×5(1− xk)/(1− x). Setting NPVI = NPVO gives k∗ = ln [ 1− NPVO(1− x) 2.5×5 ]/ ln(x) = 12.2 or the 13th payment. This gives a DPP of 8.5 years after 1/1/06; that means the payment on 1 July 2014. (ii) Incoming payments are received later than the outgoings. Hence, increasing the interest rate will reduce NPVI more than NPVO. Hence the DPP will increase. 7. (i) Let ν = 1/1.1. Then NPVO = 10 ∫ 2 0 ν t dt = 10 ν t ln ν ∣∣∣2 0 = 10(v 2−1) ln ν . Now NPVI = 8ν3 + 7.5ν4 + 7ν5 + 6.5ν6 + · · · + 4ν11 + 3.5ν12 + 3ν13 + 2ν14 + ν15 = x + 2ν14 + ν15 where 2(1− ν)x = 16ν3 − (ν4 + · · · + ν13)− 6ν14 = ν3(16− 17ν − 5ν11 + 6ν12)/(1− ν). Hence −NPVO + NPVI = 14.5916944289 or £14,591,694. (ii) Incomings are received later than outgoings. Hence increasing i will decrease NPVI more than NPVO. Hence the IRR is more than 10%. (It is 23.7%.) 8. Let α = 1.05 and ν = 1/1.09. The cash flows are as follows: Time 1/1/00 1/7/00 1/1/01 1/1/02 31/12/02 1/1/03 31/12/03 Cash flow −18 −10 −5 −45 60.5 −45α 60.5α Time 1/1/04 31/12/04 1/1/05 31/12/05 1/1/06 31/12/06 1/1/07 31/12/07 Cash flow −45α2 60.5α2 −45α3 60.5α3 −45α4 60.5α4 −45α5 60.5α5 Page 154 Answers 2.6 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed After k years production, we have NPVO = 18 + 10ν1/2 + 5ν + 45ν2[1 + αν + · · · + (αν)k−1] = 18 + 10ν1/2 + 5ν + 45ν2[1− (αν)k]/[1− αν]. Also NPVI = 60.5ν3[1 + αν + · · · + (αν)k−1] = 60.5ν3[1− (αν)k]/[1− αν]. Equating NPVO = NPVI gives k = ln [ 1− (18 + 10ν1/2 + 5ν)(1− αν)/(60.5ν3 − 45ν2)]/ ln(αν) = 3.832792 or 4 years production. This gives a date of 31/12/05 which is 6 years after the start of the project. (ii) Incomings are received later than outgoings. Hence decreasing i will increase NPVI more than NPVO. Hence the DPP will decrease. 9. Use units of £1,000. Let α = 1.06 and ν = 1/1.07. (i) Then NPVO = 180(1 + ν + ν2). We also have the continuous cash flow of 25αt for t ∈ (0, 25). Hence NPVI = ∫ 25 0 25αtνt dt = 25 (αν)25 − 1 ln(αν) Hence NPV = NPVI − NPVO = 51.61779346 or £51,617.80. (ii) We want twith NPVO = ∫ t 0 25(αν) u du = 25[(αν)t−1]/ ln(αν). Hence t = ln [NPVO ln(αν)/25 + 1] / ln(αν) = ln [ 7.2(1 + ν + ν2) ln(αν) + 1 ] / ln(αν) = 22.4204905995 or 22.42 years. (iii) We want to find α1 with NPVO = NPVI and so 180(1 + ν + ν2) = 25 (α1ν)25 − 1 ln(α1ν) Hence ν25α251 − 7.2(1 + ν + ν2) ln(να1) − 1 = 0. If α1 = 1.05 then LHS = 0.00541105; if α1 = 1.06 then LHS =−0.01938711. Linear interpolation gives α1 = 1.05 + 0.01f (1.05)/(f (1.05) − f (1.06)) = 1.052182 or 5.2%. Then f (1.052) = −0.00266205. Linear interpolation gives α1 = 1.05 + 0.002f (1.05)/(f (1.05)− f (1.052)) = 1.05134051 or 5.1%. In fact it is 5.13%.) 10. (i) The existence of the DPP just shows the project is profitable. It does not show how profitable the project is. (ii) For project A we have NPV = −1 + 3.5 1.0410 = 1.3644746 or £1,364,475. For project B, let ν = 1/1.04. Then we have NPV = −1 + 9∑ j=0 ∫ j+1 j (0.08 + 0.01j)νt dt = −1 + 9∑ j=0 (0.08 + 0.01j) ∫ j+1 j νtdt = −1 + 9∑ j=0 (0.08 + 0.01j)νj ν − 1 log ν = 0.0073097664 or £7,309.77. (iii) DPP of project A is 10 (no payments are received before 10). DPP of project B is less than 10. (iv) Project A is preferable because it has the higher NPV (even though the DPP is larger). 11. Let α = 1.04. Then we have (in £million): Date 1/1/14 1/7/14 1/1/15 1/1/16 1/1/17 1/1/18 1/1/19 1/1/20 1/1/21 31/12/21 Cash flow -19 -9 -5 -57 −57α −57α2 −57α3 −57α4 −57α5 75.6 75.6α 75.6α2 75.6α3 75.6α4 75.6α5 Let ν = 1/1.09. Assume production runs for k years where 1 ≤ k ≤ 6 or 31/12/16 to 31/12/21. Then NPV = −19− 9ν1/2 − 5ν − 57 k∑ j=1 αj−1νj+1 + 75.6 k∑ j=1 αj−1νj+2 = −19− 9ν1/2 − 5ν + (75.6ν3 − 57ν2) k∑ j=1 (αν)j−1 = −19− 9ν1/2 − 5ν + (75.6ν3 − 57ν2)1− (αν) k 1− αν giving (αν)k ≤ 1− (19 + 9ν 1/2 + 5ν)(1− αν) 75.6ν3 − 57ν2 = 0.857959 which gives k = 4 years of production or 31/12/19. (ii) If we decrease the interest rate from 9%, then ν increases. The DPP will be shorter. In the extreme, if the interest rate is 0% and so ν = 1, it would take just 2 years to recover the initial costs and so the DPP would be 2 years of production. At the other extreme, if the interest rate was very high, it would never be possible to recover the costs of borrowing the initial amounts which are invested in the project. Appendix 2 Sep 27, 2016(9:51) Answers 2.8 Page 155 12. (i) (a) The DPP is the smallest time such that the accumulated value of the project becomes positive. Equivalently, it is the smallest time t such that the NPV of payments up to time t is positive. (b) The payback period is the time when the net cash flow (incomings less outgoings) is positive. The DPP and payback period are measures of the time it takes for a project to become profitable. (ii) The payback period ignores discounting—and hence ignores the effect of interest. It is therefore an inferior measure and should not be used. Drawbacks of the DPP: • It shows when the project becomes profitable using a forecast interest rate, but it does not show how large (or small) the profit is. • Suppose the DPP of project A is less than the DPP of project B; it is possible that the NPV of project A is less than the NPV of project B—this can occur if there are is a large late cash inflow in project B. (iii) Take time 0 to be 1 January 2004. Let ν = 1/1.1. Then NPV of making bid = 0.2 NPV of building costs = ∫ 7 2 νu du NPV of running costs = ∫ 7.25 7 νu du Total NPV of costs = 0.2 + ∫ 7.25 2 νu du = 0.2 + ν7.25 − ν2 ln ν = 3.613811 NPV of TV rights = 0.3 ∫ 7.25 7 νu du = 0.3 ν7.25 − ν7 ln ν = 0.0380320 Now for the other revenue: Year 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 Time in years 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5 Cash flow 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.8 0.6 0.4 0.2 This has NPV: NPV = ν0.5 10 [ 1 + 2ν + 3ν2 + · · · + 12ν11 − ν8 − 4ν9 − 7ν10 − 10ν11] = ν0.5 10 [ 1 + 2ν + 3ν2 + · · · + 12ν11 − ν8(1 + 4ν + 7ν2 + 10ν3)] = ν0.5 10 [ 1− ν12 (1− ν)2 − 12ν12 1− ν − ν 8 ( 1− 10ν4 1− ν + 3ν 1− ν3 (1− ν)2 )] = ν0.5 10 [ 1 + 2ν12 − 3ν9 (1− ν)2 − 2ν12 + ν8 1− ν ] = 3.052965 Hence we need C with Cν11 = 3.613811− 3.052965− 0.038032 leading to C = 1.49165 or £149.165 million. 13. Use units of £1,000. Let ν = 1/1.08. For the discrete payments we have (in thousands) NPVO = 105 + 105ν1/2 + 105ν + 200ν15. Let N1 denote the net present value of the continuous income for years 2, 3 and 4. Then N1 = 20 ∫ 2 1 νt dt + 23 ∫ 3 2 νt dt + 26 ∫ 4 3 νt dt = 20(ν2 − ν) ln ν + 23(ν3 − ν2) ln ν + 26(ν4 − ν3) ln ν = ν − 1 ln ν [ 20ν + 23ν2 + 26ν3 ] For k ∈ {5, 6, . . . , 30}, let N(k) denote the net present value of the continuous income for years 5, . . . , k. Let α = 1.03. Then N(k) = k∑ j=5 29αj−5 ∫ j j−1 νt dt = 29(ν − 1) ln ν k∑ j=5 αj−5νj−1 = ν − 1 ln ν 29ν4 1− (αν)k−4 1− αν Let NPV(k) = −NPVO + N1 + N(k). (i) We need NPV(30) = −NPVO + N1 + N(30) = 4.30097626. Answer is £4,300.98 (ii) Now NPV(15) = −NPVO + NPV1 + NPV(15) = −129.57469246 and even if we remove the £200,000 payment (which has net present value equal to £63,048.34) from NPVO, this quantity is still negative. Also NPV(2) < NPV(3) < · · · < NPV(15) and so these are all negative. (iii) NPV(29) = −1.97144191 < 0. Hence the DPP occurs in the final year. Answers to Exercises: Chapter 2 Section 8 on page 39 (exs2-3.tex) 1. We use equation (7.4a) which is: (1 + i)t = ft1−0 f0 ft2−0 ft1 · · · ftn−0 ftn−1 ft ftn Now t = 3, t1 = 0.5, etc. Hence 1.113 = 460 400 × 550 510 × 500 550 × 600 540 × 650 600 × X 710 = 460 400 × 500 510 × 650 540 × X 710 and hence X = 715.50062864 or £715,500,629. Page 156 Answers 2.8 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 2. Using equation (7.4a) gives (1 + i)3 = 80 120 × 200 110 × 200 210 = 800 693 leading to i = 0.049024 or 4.90%. 3. The TWRR, i is given by (1 + i)3 = 212 180 261 237 230 261 273 248 295 273 309 311 = 212 180 230 237 295 248 309 311 = 1.35086 Hence i = 0.10544 or 10.54%. 4. Use units of £1,000. Time, t 1 Jan 2001 1 Nov 2001 1 May 2002 31 Dec 2002 Fund value at time t− 0 137 173 205 Net cash flow at time t 20 48 Fund value at time t + 0 120 157 221 Hence the TWRR, i, is given by (1 + i)2 = 137 120 × 173 157 × 205 221 = 4,858,705 4,163,640 Hence i = 0.080248519 or 8.02%. (ii) The disadvantage of the MWRR is that it is sensitive to the amounts and timings of cash flows—and the invest- ment manager has no control over these. The disadvantage of the TWRR is that it requires a lot more information—it requires knowledge of all cash flows and the the value of the fund at the times of all cash flows. 5. Use units of £1,000,000. Let x denote the required amount. Then Time, t 31 Dec 2001 31 Dec 2003 31 Dec 2004 Fund value at time t− 0 x 4.2 Net cash flow at time t 1.44 Fund value at time t + 0 2.2 x + 1.44 Hence the TWRR, i, is given by (1 + i)3 = x 2.2 × 4.2 x + 1.44 = 21x 11(x + 1.44) The MWRR, i, is given by 4.2 = 2.2(1 + i)3 + 1.44(1 + i) or 55i3 + 165i2 + 201i− 14 = 0. Using the approximation (1 + i)3 ≈ 1 + 3i leads to i ≈ 0.07 and the true value will be less than this approximation. Let f (i) = 55i3 + 165i2 + 201i− 14. Then f (0.06) = −1.33412 and f (0.07) = 0.897365. Using linear interpolation gives (0.07− i)/0.01 = 0.897365/(0.897365 + 1.33412) and hence i = 0.66. Using this value in the formula for the TWRR gives x = 11×1.44(1+i)3/ [21− 11(1 + i)3] = 2.5000 or £2.5 million. 6. Use units of £1,000. Time, t 1 Jul 2003 1 Jul 2004 1 July 2005 1 Jul 2006 Fund value at time t− 0 24 32 38 Net cash flow at time t 5 8 Fund value at time t + 0 21 29 40 (i) The MWRR, i, is given by 21(1 + i)3 + 5(1 + i)2 + 8(1 + i) = 38 leading to 21i3 + 68i2 + 81i − 4 = 0. Using the approximation (1 + i)3 ≈ 1 + 3i and (1 + i)2 ≈ 1 + 2i leads to i = 0.049 and the true value will be less than this. Let f (i) = 21i3 + 68i2 + 81i− 4. Then f (0.049) = 0.1347386 and f (0.04) = −0.649856. Using linear interpolation gives: (0.049− i)/0.009 = 0.1347386/(0.1347386 + 0.649856) leading to i = 0.04745 or 4.75%. (ii) The TWRR, i, is given by (1 + i)3 = 24 21 × 32 29 × 38 40 = 1,216 1,015 leading to i = 0.062078 or 6.21%. (iii) The return in the final year is poor (40 decreases to 38), whilst the returns in the first two years are good. The MWRR is biased towards the return in the final year because the size of the fund is much larger in the final year. 7. Use units of £1,000,000. Time, t 1 Jan 2003 1 Jan 2004 1 July 2004 1 Jan 2005 Fund value at time t− 0 450 500 800 Net cash flow at time t 40 100 Fund value at time t + 0 600 490 600 (i) (a) The TWRR, i, is given by (1 + i)2 = 450 600 × 500 490 × 800 600 = 50 49 leading to i = 1.01015254 or 1.015%. (i) (b) Effectively we have the following information: Appendix 2 Sep 27, 2016(9:51) Answers 2.8 Page 157 Time, t 1 Jan 2003 1 Jan 2004 1 July 2004 1 Jan 2005 Fund value at time t− 0 450 800 Net cash flow at time t 40 100 Fund value at time t + 0 600 490 600 For year 1 we have 1 + i1 = 450/600 = 3/4. For year 2, we have i2 with 490(1 + i2) + 100(1 + i2)1/2 = 800, or 49(1 + i2) + 10(1 + i2)1/2− 80 = 0. This quadratic in (1 + i2)1/2 leads to (1 + i2)1/2 = (−10± √ 100 + 320× 49)/98 = (−5±√3945/49) = 1.17978. Hence 1 + i2 = 1.39188. Then (1 + i)2 = (1 + i1)(1 + i2) = 0.75× 1.39188 and hence i = 0.0217197, or 2.17%. (ii) The fund performs well in the period 1 July 2004 to 31 Dec 2004, but not so well in the period 1 Jan 2004 to 1 July 2004. The TWRR for the year is based on the returns in the first 6 months and in the second 6 months, and ignores the fact that there are differing amounts in the fund in these two periods. The LIRR for the year is based on a single return for the whole year—and there is more invested in the fund in the final 6 months. 8. Use units of £1,000,000. Time, t 1 Jan 2000 1 Jan 2001 1 July 2001 31 Dec 2001 Fund value at time t− 0 43 49 53 Net cash flow at time t 4 2 Fund value at time t + 0 40 47 51 Between 1/1/2000 and 1/1/2001 the fund grows from 40 to 43; between 1/1/2001 and 1/7/2001 the fund grows from 47 to 49; and between 1/7/2001 and 1/1/2002 the fund grows from 51 to 53. Hence the TWRR is given by (1 + i)2 = (43/40)× (49/47)× (53/51). Hence i = 0.07921 and the TWRR is 7.92%. For (b), this is equivalent to only having the following information. Time, t 1 Jan 2000 1 Jan 2001 1 July 2001 31 Dec 2001 Fund value at time t− 0 43 53 Net cash flow at time t 4 2 Fund value at time t + 0 40 47 LIRR: first year. 40 grows to 43. Hence i1 = 3/40 = 0.075. LIRR: second year. 47 at time 0 and 2 at time 0.5 grow to 53 at time 1. Hence 47(1 + i2) + 2(1 + i2)0.5 = 53. If x = (1+ i2)0.5 then we get the quadratic 47x2 +2x−53 = 0. Hence x = (−1+2 √ 623)/47 and 1+ i2 = x2 = 1.083368. Hence the LIRR is given by (1+i)2 = 1.075×1.083368 and so it is 7.92%. It is slightly less than the value of TWRR. (ii) The 2 values TWRR and LIRR will be equal if the same intervals are used. In this case, the 3 intervals 1/1/2000 to 1/1/2001, 1/1/2001 to 1/7/2001 and 1/7/2001 to 1/1/2002 would have to be used for the calculation of the LIRR. Also, if the rate of growth in both subintervals is the same, then TWRR will be equal to LIRR. In this case, the growth in the first 6 months is 49/47 and is 53/51 in the second 6 months—these are very similar quantities and so there is little difference between the values of TWRR and LIRR. 9. We have the following information. Time, t 1 Jan 2004 1 July 2004 1 Jan 2005 1 Jan 2006 31 Dec 2006 Fund value at time t− 0 12.5× 1.05 = 13.125 20.9085 29.7225525 38.854229 Net cash flow at time t 12.5 6.6 7.0 8.0 Fund value at time t + 0 12.5 19.725 27.9085 37.7225525 (i) LIRR is given by (1 + i)3 = 1.05× 1.06× 1.065× 1.03. Hence i = 0.06879398 or 6.8794%. (ii) TWRR is given by (1 + i)3 = 13.125 12.5 20.9085 19.725 29.7225525 27.9085 38.854229 37.7225525 = 1.05× 1.06× 1.065× 1.03 This gives i = 0.06879398 or 6.8794%. (iii) MWRR is solution of 12.5(1 + i)3 + 6.6(1 + i)2.5 + 7(1 + i)2 + 8(1 + i) = 38.854229 If f (i) denotes the left hand side, then f (0.06) = 38.86789. Then f (0.055) = 38.454467. Interpolating: (i − 0.055)/(38.854229 − 38.454467) = 0.005/(38.86789 − 38.454467) leading to i = 0.0598 or 5.98%. (In fact, 5.98353%.) (iv) The MWRR is lower because the fund has less money at the beginning when rates are higher. The values of TWRR and LIRR are equal because the value of the fund is known at the times of all cash flows. 10. We have the following information: Time, t 1 Jan 2012 30 September 2012 31 December 2012 Fund value at time t− 0 1.9 0.8 Net cash flow at time t −0.9 Fund value at time t + 0 1.3 1.0 Page 158 Answers 3.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed (i) TWRR is given by 1 + i = 1.9 1.3 × 0.8 1.0 = 19× 8 130 = 76 65 Hence TWRR = 0.16923 or 16.92%. (ii) The MWRR is the solution of 1.3(1+i)−0.9(1+i)1/4 = 0.8 or 13i+5−9(1+i)1/4. So let f (i) = 13i+5−9(1+i)1/4. Then f (0.3) = 8.9−9×1.31/4 = −0.71011; f (0.4) = 10.2−9×1.41/4 = 0.90030 and f (0.35) = 9.55−9×1.351/4 = −0.15121. By linear interpolation we have i− 0.35 = 0.15121× (0.05/1.05151) and so i = 0.35719 or 36% to the nearest 1%. (iii) The fund does well before the large withdrawal. It performs poorly after the large withdrawal when the size of the fund is small—this leads to a higher MWRR. 11. (i) (2.9/2.3) × (4.2/4.4) = 1.203557 or 20.36%. (ii) We want i with 2.3(1 + i) + 1.5(1 + i)8/12 = 4.2. So let f (i) = 23(1 + i) + 15(1 + i)2/3 − 42. Using the approximation (1 + i)2/3 ≈ 1 + 2i/3 gives i ≈ 4/33 = 0.12. Then f (0.12) = −0.062803 and f (0.13) = 0.263347. Linear interpolation gives i = 0.12 + 0.01× 0.062803/(0.263347 + 0.062803) = 0.121926 or 12.2%. (iii) The fund performs well from January to May when the amount in the fund is (relatively) small and performs badly from May to December—when the amount in the fund is large. Hence the MWRR will be less than the TWRR, because the TWRR ignores the amount in the fund. 12. (a) Ignoring inflation we have Time, t 1/4/2001 1/4/2002 1/4/2003 1/4/2006 Fund value at time t− 0 0 1000× 9080 1000 ( 1 + 9080 ) 98 90 1000 [ 130 66 + 473 360 160 98 ] 0 = 1125 = 2313.89 = 4114.82 Net cash flow at time t 1000 1000 0 Fund value at time t + 0 1000 1000 ( 1 + 9080 ) 1000 ( 1 + 9080 ) 98 90 Hence the real TWRR is given by (1 + i1)5 = 98 80 × 130 66 + 473 360 160 98( 1 + 9080 ) 98 90 × 1 1.22 = 1048 561 × 1 1.22 = 1.7856 This gives i1 = 0.12294104322 or 12.294% per annum. (b) The equation for the real MWRR is 1000 + 1000ν 1.02 = 1000 [ 130 66 + 473 360 160 98 ] ν5 1.22 = 4114.82 ν5 1.22 or 19961ν5 − 5802.17647059ν − 5918.22 = 0. Substituting ν = 1/1.12294 gives LHS > 0. Hence MWRR > TWRR. (It is 12.50%.) 13. Use units of £1 million. Time, t 1 January 2010 1 January 2011 1 July 2012 Fund value at time t− 0 120 140 600 Net cash flow at time t 0 200 Fund value at time t + 0 120 340 (i) Let i denote the annual time weighted rate of return. Then (1 + i)5/2 = 140 120 × 600 340 = 35 17 Hence i = 0.3348967 or 33.49%. (ii) The rate of growth in the first year is 16.67% but 46% in the last 18 months. Hence the rate of growth is higher when the fund contains more money; hence the money weighted rate of return will be higher than the time weighted rate of return. (It is 38.87%.) 14. (i) MWRR affected by amounts and timings of cash flows which are not under the control of the fund manager. The equation for the MWRR may not have a unique or, indeed, any solution. But, TWRR requires knowledge of all the cash flows and the fund values at all the cash flow dates. (ii) Denote answer by i. Then (1 + i)2 = (45/41)× (72/19) leading to i = 0.17745183 or 17.75%. Answers to Exercises: Chapter 3 Section 3 on page 51 (exs3-1.tex) 1. (a) and (b) Use sn = ∑n−1 j=0 (1 + i) j , sn+1 = ∑n j=0(1 + i) j and s¨n = ∑n j=1(1 + i) j . (c) If i = 0, a¨n − an−1 = n− (n− 1) = 1. If i 6= 0, a¨n − an−1 = (1 + i)(1− νn)/i− (1− νn−1)/i = 1. (d) If i = 0, ian + νn = 0 + 1 = 1. If i 6= 0, ian + νn = (1− νn) + νn = 1. (e) Use (d) and ian = da¨n . (f) If i = 0, LHS = RHS = 1. If i 6= 0, use equation 2.2a. (g) Use (f) and isn = ds¨n Interpretations: (b) (Value of n+ 1 payments at time n+ 1) = (value of first n payments at time n+ 1) + 1(for payment at time n + 1). (c) (value of n payments at time 1) = (value of n − 1 payments at time 1)+1(payment at time 1). (d) Value at time 0 of loan of size 1, interest rate i with n annual interest payments. (e) As for (d), but interest payments in advance. (f) and (g) correspond to (d) and (e) but valued at time n. Appendix 2 Sep 27, 2016(9:51) Answers 3.3 Page 159 2. Now δ = 0.06; hence 1 + i = eδ and i = 0.0618. Either, (Ia)10 = a10 − 10ν10 δ = ia10 ,i/δ − 10ν10 δ = ia10 ,i δ2 − 10ν 10 δ = 33.86 or, from first principles: (Ia)10 = ∫ 10 0 tνt dt = tνt log ν ∣∣∣∣10 0 − ∫ 10 0 νt log ν dt = 10ν10 log ν − ν t (log ν)2 ∣∣∣∣10 0 = −10ν 10 δ + 1− ν10 δ2 = 33.86 3. Let i denote the annual effective interest rate. Then 1 + i = 1.062. Let x = 1/(1 + i)1/4 = 1/ √ 1.06. NPV = 40(x + x2 + · · · + x16) = 40×1− x 16 1− x = 40√ 1.06− 1 ( 1− 1 1.068 ) = 504.1267 Alternatively, use 2.3bwith i(4) = 4× (√1.06− 1) = 0.118252. Hence NPV = 160a(4)4 ,0.1236 = 160× ( 0.1236/i(4) )× a4 ,0.1236 = 167.236a4 ,0.1236 = 504.13 4. s¨(12)12 = (1 + i) n+1/ma(m)n = 1.0712+1/12a (12) 12 = 1.07 12+1/12 × (i/i(12))× a12 = 18.5597 5. s¨(4)20 = (1 + i) 20a¨(4)20 = (1 + i) 20.25 × (i/i(4))× a20 = 1.07520.25 × 1.027701× 10.194491 = 45.316 6. The cash flow is as follows: Time 0 1 . . . 6 7 Cash flow, c 200 190 . . . 140 130 210 210 . . . 210 210 −10 −20 . . . −70 −80 Hence NPV(c) = 210a¨8 − 10(Ia¨)8 . (ii) Using a¨n = (1 + i)an , (Ia¨)8 = (1 + i)(Ia)8 and (Ia)8 = (an − nνn+1)/(1−ν) gives NPV(c) = 210(1 + i)a8 −10(1 + i)(Ia)8 = 1.07 [ 210a8 − 10(a8 − 8ν9)/(1− ν) ] = 1076.82. 7. We want smallest integer n with 100 ≤ 20an,0.15. Using tables shows that n = 10 is the least such integer. 8. Time 0 1 2 3 4 5 6 7 8 9 Cash flow, c 10 12 14 16 18 20 22 24 26 28 10a¨10 10 10 10 10 10 10 10 10 10 10 2(Ia)9 0 2 4 6 8 10 12 14 16 18 8a¨10 8 8 8 8 8 8 8 8 8 8 2(Ia¨)10 2 4 6 8 10 12 14 16 18 20 NPV(c) = ∑9 t=0 8ν t + 2 ∑9 t=0(t + 1)ν t. Answer is (II) and (III) only. 9. Let x denote a revised payment. Then NPV = 600a(4)n,i = 12xa¨ (12) n,i. Hence x = 50 a(4)n,i a¨(12)n,i = 50 (1 + i)1/12 a(4)n,i a(12)n,i = 50 (1 + i)1/12 i(12) i(4) = 50 eδ/12 12(eδ/12 − 1) 4(eδ/4 − 1) = 150 1− e−δ/12 eδ/4 − 1 = 49.17243 So the answer is £49.17. 10. Using k2 = (k − 1)2 + 2k − 1 we can decompose the given cash flow c into c1 + c2 + c3 + c4 as follows: Time 0 1 2 · · · k · · · n n + 1 Cash flow, c 0 12 22 · · · k2 · · · n2 0 Cash flow, c1 0 0 12 · · · (k − 1)2 · · · (n− 1)2 n2 Cash flow, c2 0 0 0 · · · 0 · · · 0 −n2 Cash flow, c3 0 2 4 · · · 2k · · · 2n 0 Cash flow, c4 0 −1 −1 · · · −1 · · · −1 0 If x denotes the required answer, then x = νx− n2νn+1 + 2(Ia)n − an . 11. The cash flow is as follows: 0 1 2 3 4 5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 Let i1 = 0.04. Then A(12) = 100 [ (1 + i1)22 + (1 + i1)20 + (1 + i1)18 + (1 + i1)16 + (1 + i1)14 + (1 + i1)12 ] + 100 [ (1 + i1)11 + (1 + i1)10 + · · · + 1 ] = 100(1 + i1)12 [ (1 + i1)10 + (1 + i1)8 + · · · + (1 + i1)2 + 1 ] + 100s12 ,0.04 = 100(1 + i1)12 (1 + i1) + 1 s12 ,0.04 + 100s12 ,0.04 = 100 ( 1.0412 2.04 + 1 ) s12 ,0.04 = 2681.835 OR, let x = 1/1.04. Then NPV = 100[x2 + x4 + · · · + x12 + x12a12 ] = 100[x2(1 + x2 + · · · + x10) + x12a12 ]. Hence NPV = 100[ x 1 + x (x + x2 + · · · + x12) + x12a12 ] = 100a12 x + x12 + x13 1 + x = 100a12 1.0412 1.0412 + 2.04 2.04 Page 160 Answers 3.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 12. Let α = 1.0056 and β = 1.03. Thus α is the 6-month growth factor for t ∈ [0, 15] and β is the 6-month growth factor for t ∈ (15, 25]. Clearly there are 50 payments in total. The first 30 payments are made at times 0, 0.5, 1, . . . , 14.5. Their accumulated value at time 15 is c0 = 400(α30 + α29 + · · · + α); and their accumulated value at time 25 is β20c0. The last 20 payments are made at times 15, 15.5, 16, . . . , 24.5. The accumulated value of these 20 payments at time 25 is c1 = 400(β20 + β19 + · · · + β). Hence A(25) = β20c0 + c1 = 400 [ β20α α30 − 1 α− 1 + β β20 − 1 β − 1 ] = 46,702.66 13. (i) Using first principles is probably the easiest method. We have the following cash flow: Time 0 1/12 2/12 3/12 · · · 64/12 65/12 66/12 Cash flow 1/12 1/12 1/12 1/12 · · · 1/12 1/12 0 Let α = 1.131/12. Then A(66/12) = 1 12 (α66 + α65 + · · · + α) = α 12 α66 − 1 α− 1 = 1.131/12 12 1.1311/2 − 1 1.131/12 − 1 = 7.882864 Or, i(12) = 12(1.131/12 − 1) and s¨(12)5.5 = 1.135.5a¨(12)5.5 where a¨(12)5.5 = 1.131/12a(12)5.5 = 1.131/12(1 − 1/1.135.5)/i(12). Hence answer is 7.88286. Or, use (1 − d(12)/12)12 = (1/1.13) to get d(12) = 0.121597 and s¨(12)5.5 = is5.5/d(12) = ((1 + i)5.5 − 1)/d(12) = (1.135.5 − 1)/d(12) = 7.88286 (ii) It is the accumulation after 51/2 years of an annuity with payments of size 1/12 each month in advance at an effective interest rate of 13% per annum. 14. The first 2 results are in the notes. Then use the first result and s(m)n = (1 + i)na (m) n and sn = (1 + i)nan to get the third result. For the third result, use s¨n = (1 + i)sn and d(1 + i) = i. 15. (a) Just use an = ∑n j=1 ν j . (b) We want x with 10,000 = xa120 = x(a100 + ν100a20 ). Then use tables to get x = 143.47. Or work from first principles. 16. Time 0 1 2 3 4 5 Cash flow 0 1 1 1 1 1 s5 = A(5) = 1 + exp( ∫ 5 4 0.02t dt) + exp( ∫ 5 3 0.02t dt) + exp( ∫ 5 2 0.02t dt) + exp( ∫ 5 1 0.02t dt). Now ∫ 5 k 0.02t dt = 0.01(25− k2). Hence s5 = 1 + e0.09 + e0.16 + e0.21 + e0.24 = 5.773. Or, ν(t) = exp(− ∫ t0 δ(s) ds) = exp(− ∫ t0 0.02s ds) = exp(−0.01t2). Hence a5 = ν(1) + ν(2) + ν(3) + ν(4) + ν(5) = e−0.01 + e−0.04 + e−0.09 + e−0.16 + e−0.25. Finally, s5 = a5/ν(5) = a5 e0.25 = 5.773 17. (a) NPV(c) = ∑∞ i=0(1 + g) k/(1 + i)k+1 = 1/(i− g) (b) 10,000/0.06 = 166,667 18. Just use limm→∞ i(m) = δ and limm→∞ d(m) = δ. See example 4.5c for this last limit. 19. If i = 0, then isn = 0 = ds¨n . If i 6= 0, then isn = (1 + i)n − 1 = ds¨n . If i = 0, then ian = 0 = da¨n . If i 6= 0, then ian = 1− νn = da¨n . The other results are in exercise 14. 20. LHS = a¨(m)n,i = (1 + i)1/ma (m) n,i = (1 + i (m)/m)a(m)n,i = (1 + i (m)/m)(i/i(m))an,i = RHS 21. If i = 0 then LHS = n = RHS. If i 6= 0, then RHS = 1− νnm1 jm = 1− νn i(m) = LHS where j = i(m) m and ν1 = 1 1 + j 22. Let ν = 1/(1 + i) denote the discount factor. Using ν = 1− d = (1− dep)m where dep = d(m)/m gives a¨(m)∞ = 1 m ( 1 + ν1/m + ν2/m + · · · ) = 1 m 1 1− ν1/m = 1 mdep = 1 d(m) Using (1 + iep)(1− dep) = 1 gives iep = dep/(1− dep) and so i(m) = d(m)/(1− dep) = d(m)/v1/m. Hence a(m)∞ = 1 m (ν1/m + ν2/m + · · ·) = ν1/m a¨(m)∞ = ν1/m d(m) = 1 i(m) (b) follows immediately from the two tables of cash flows. 23. (a) Algebraically: NPV(c) = k∑ j=1 νjr = νr 1− νkr 1− νr = i (1 + i)r − 1akr = akr sr Or: recall that sr denotes the accumulated value at time r of the sequence of r equal payments of 1 at times 1, 2,. . . r. Hence the single payment of 1 at time r is equivalent to the r payments of 1/sr at times 1, 2, . . . ,r. Hence our annuity is equivalent to kr equal payments of 1/sr at times 1, 2,. . . , kr. (b) Set r = 1/m and k = nm (and hence kr = n) in previous result. Appendix 2 Sep 27, 2016(9:51) Answers 3.3 Page 161 24. In the following x = ν1/m and y = ν1/q . (I (q)a¨)(m)∞ = 1 mq [ (1 + ν1/m + · · · + ν1/q−1/m) + 2(ν1/q + ν1/q+1/m + · · · + ν2/q−1/m) + · · · ] = 1 mq [ (1 + x + · · · + xm/q−1)(1 + 2y + 3y2 + · · ·) ] = 1 mq 1− xm/q 1− x 1 (1− y)2 = 1 mq 1− ν1/q 1− ν1/m 1 (1− ν1/q)2 = [ 1 m 1 1− ν1/m ] [ 1 q 1 1− ν1/q ] = a¨(m)∞ a¨ (q) ∞ = 1 d(m) 1 d(q) 25. Here we show the first equality. The second is done in exercise 35. (Ia¯)n = ∫ n 0 ρ(t)ν(t) dt = n∑ j=1 ∫ j j−1 jνt dt = n∑ j=1 j(νj − νj−1) log ν = 1 + ν + ν2 + · · · + νn−1 − nνn δ = a¨n − nνn δ 26. (a) Just use k|a(m)n = νka (m) n and ν1/ma¨ (m) n = a (m) n . (b) If i = 0, then RHS = (n + k) − k = n = LHS. If i 6= 0, RHS = (1 − νn+k)/i(m) − (1 − νk)/i(m) = νk(1− νn)/i(m) = LHS. (c) Just use a¨(m) n+k = (1 + i) 1/ma(m) n+k and part (b). 27. Use the following decomposition: Time 0 1 2 · · · n Cash flow, c 0 n n− 1 · · · 1 (n + 1)an 0 n + 1 n + 1 · · · n + 1 −(Ia)n 0 −1 −2 · · · −n 28. This decomposition gives (II): 0 1 2 3 4 5 6 7 8 9 . . . 20 0 200 180 160 140 120 100 80 60 60 . . . 60 0 200 200 200 200 200 200 200 200 200 . . . 200 0 0 -20 -40 -60 -80 -100 -120 -140 0 . . . 0 0 0 0 0 0 0 0 0 0 -140 . . . -140 This decomposition gives (III): 0 1 2 3 4 5 6 7 8 9 . . . 20 0 200 180 160 140 120 100 80 60 60 . . . 60 0 60 60 60 60 60 60 60 60 60 . . . 60 0 140 120 100 80 60 40 20 0 0 . . . 0 The decomposition in (I) is incorrect: 0 1 2 3 4 5 6 7 8 9 10 11 . . . 20 0 200 180 160 140 120 100 80 60 60 60 60 . . . 60 0 200 180 160 140 120 100 80 60 40 20 0 . . . 0 0 0 0 0 0 0 0 0 0 −60 −40 −20 . . . 0 60 60 60 60 60 60 60 60 60 60 60 60 . . . 60 −60 −60 −60 −60 −60 −60 −60 −60 0 0 0 0 . . . 0 29. (i) We need i with 250 = 64a¨(4)5 ,i − 7a¨5 ,i = 64a(4)5 ν1/4 − 7a5 ν = a5 [ 64(1 + i)1/4 i i(4) − 7(1 + i) ] Using tables for a5 , (1 + i)1/4 and i/i(4) gives RHS = 253.87 for i = 0.05 and RHS = 248.371 for i = 0.06. Trying i = 0.055 gives RHS = 251.096. Using interpolation between i = 0.055 and i = 0.06 gives (x−0.055)/(0.06−0.055) = (f (x)−f (0.055))/(f (0.06)− f (0.055)) and hence x = 0.055 + 0.005(250− 251.096)/(248.371− 251.096) = 0.057. Hence i = 0.057 or 5.7%. (ii) Using the general result that (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflation rate and iM is the money rate, we get iR = (i− e)/(1 + e) = (0.057− 0.02)/1.02 = 0.0363 of 3.63%. 30. First option: 456 = 240a¨(12)2 ,i . Using a (m) n,i = i d(m) an gives 240a (12) 2 ,0.05 = 240 × 1.026881 × 1.859410 = 458.254. Hence i > 0.05. Second option: 456 = 246a(12)2 ,i . But 246a (12) 2 ,0.05 = 246× 1.022715× 1.859410 = 467.805. Hence i > 0.05. Both deals are unacceptable. Page 162 Answers 3.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 31. Let α = 1.03 and let ν = 1/1.04. First, note that∫ k k−1 νt dt = νt ln ν ∣∣∣∣k t=k−1 = νk−1 ν − 1 ln ν = iνk δ by using ν = 1 1 + i and ln ν = −δ. NPV of costs to companies plus NPV of costs to consumers is 60 [∫ 1 0 νt dt + ∫ 2 1 ανt dt + · · · + ∫ 20 19 α19νt dt ] = 60i δ [ ν + αν2 + · · · + α19ν20] = 60iν δ 1− (αν)20 1− αν NPV of costs to financial advisers is 60 ∫ 1 0 νt dt + 19 ∫ 2 1 νt dt + 18 ∫ 3 2 νt dt + · · · + ∫ 20 19 νt dt = i δ [ 60ν + 19ν2 + 18ν3 + · · · + ν20] NPV of benefits to consumers is 30 ∫ 1 0 νt dt + 33 ∫ 2 1 νt dt + · · · + 87 ∫ 20 19 νt dt = i δ [ 30ν + 33ν2 + · · · + 87ν20] NPV of benefits to companies is 12 ∫ 20 0 νt dt = 12 νt ln ν ∣∣∣∣20 t=0 = 12 1− ν20 δ = 12i δ a20 Hence NPV = i δ [ 12a20 − 40ν + (10ν + 14ν2 + 18ν3 + · · · + 86ν20)− 60ν 1− (αν)20 1− αν ] Let S = 10ν + 14ν2 + 18ν3 + · · · + 86ν20. Then (1− ν)S = 10ν − 86ν21 + 4ν2 1− ν 19 1− ν and so S = 10− 86ν20 i + 4 1− ν19 i2 = 86ia20 − 76 + 4a19 i Hence NPV = i δ [ 98a20 − 40ν + 4a19 − 76 i − 60ν 1− (αν) 20 1− αν ] = 1.019869× [ 98× 13.590326− 40 1.04 + 4× 13.133939− 76 0.04 − 60 1.04 1− (1.03/1.04)20 1− 1.03/1.04 ] = −354.41 Net cost is £354.41 million. 32. Let ν = 1/1.04 and x = ν1/4. For the first alternative we have NPV = 300(1 + x + x2 + · · · + x39) = 300(1 − ν10)/(1−x) = 9975.208909 or £9,975.21. For the second alternative we have NPV = 2,520(ν2 +ν4 +ν6 +ν8 +ν10) = 2,520ν2(1− ν10)/(1− ν2) = 10019.341845 or £10,019.34. Hence the first alternative is preferable for the borrower. 33. (i) Suppose α(t) = ∫ t 0 δ(s) ds. For 0 ≤ t ≤ 4 we have α(t) = 0.08t. Note that α(4) = 0.32. For 4 < t ≤ 9 we have α(t) = 0.32 + 0.12(t− 4)− 0.01t2/2 + 0.08 = −0.08 + 0.12t− 0.005t2. Note that α(9) = 0.595. For t > 9 we have α(t) = 0.595 + 0.05(t− 9) = 0.145 + 0.05t. Hence ν(t) = exp ( − ∫ t 0 δ(s) ds ) = { exp(−0.08t) for 0 ≤ t ≤ 4; exp(0.08− 0.12t + 0.005t2) for 4 < t ≤ 9; exp(−0.145− 0.05t) for t > 9. We have ρ(t) = 100 exp(0.03t) for 10 < t < 12 and NPV(ρ) = ∫ ρ(u)ν(u) du = ∫ 12 10 100 exp(0.03u) exp(−0.145 − 0.05u) du = 100 exp(−0.145) ∫ 1210 exp(−0.02u) du = 5,000(exp(−0.345)−exp(−0.385)) = 138.8485863 or 138.849. (iii) We want 1,000(e−0.08 + e−0.16 + e−0.24) = 2561.88799642 or £2,561.89. 34. For bond redeemed after 10 years: Time 0 1/2 2/2 · · · 19/2 20/2 Cash flow after tax (in e) −100 3/2 3/2 · · · 3/2 3/2 + 100 The return is 1.5% after 6 months. Hence the effective rate of return is 1.0152 − 1 or 3.0225% p.a. For bond redeemed after 20 years: Time 0 1/2 2/2 · · · 19/2 20/2 21/2 · · · 39/2 40/2 Cash flow after tax (in e), c0 −100 3/2 3/2 · · · 3/2 3/2 21/8 · · · 21/8 21/8 + 130 c1 −100 21/8 21/8 · · · 21/8 21/8 21/8 · · · 21/8 21/8 + 130 c2 9/8 9/8 · · · 9/8 9/8 0 · · · 0 0 Note that c0 = c1 − c2. If i denotes the rate of return, then we need i with 100 = 21 4 a(2)20 ,i + 130ν 20 − 9 4 a(2)10 ,i Appendix 2 Sep 27, 2016(9:51) Answers 3.3 Page 163 So let f (i) = 100− 21 8 x 1− x40 1− x − 130x 40 + 9 8 x 1− x20 1− x = 100− 130x 40 − x 8(1− x) ( 12− 21x40 + 9x20) where x = 1/(1 + i)1/2. Spreading the capital gain of 30 over 20 years gives 1.5% p.a. So the return is greater than 3.0225 + 1.5. So try i = 0.045; this gives f (i) = −4.953363. The value i = 0.05 gives f (i) = 2.358334. Linear interpolation gives i ≈ 0.045 + 0.005 × 4.953363/(2.358334 + 4.953363) = 0.048387 or 4.83% p.a. (Actually 4.83377%.) (i) 4.83% p.a. (ii) 3.0225% p.a. (iii) Now 0.5(4.83377 + 3.0225) = 3.928 gives 3.928% p.a. (iv) The amount returned per e100 is 1 2 ( 3a(2)10 ,i + 100ν 10 ) + 1 2 ( 3a(2)10 ,i + 21 4 ν10a(2)10 ,i + 130ν 20 ) = ( 3 + 21 8 ν10 ) a(2)10 ,i + 50ν 10 + 65ν20 = ( 3 + 21 8 x20 ) x(1− x20) 2(1− x) + 50x 20 + 65x40 If i = 0.03928, the right hand side is 103.426 > 100; hence the answer is greater than 3.928% p.a. (It is actually 4.21189% p.a.) 35. (i) Using ln ν = −δ and an = ∫ n 0 ν t dt = (1− νn)/δ gives (Ia)n = ∫ n 0 ρ(t)ν(t) dt = ∫ n 0 tνt dt = nνn ln ν − ∫ n 0 νt ln ν dt = nνn ln ν + 1− νn (ln ν)2 = −nν n δ + 1− νn δ2 = an − nνn δ (ii) Amounts are in thousands. If ρ denotes revenue stream, then ρ(t) = 0 for t ∈ (0, 1); ρ(t) = 250(t − 1) for t ∈ (1, 11); ρ(t) = 2,500− 125(t− 11) for t ∈ (11, 29). Hence present value of revenue stream is NPV(ρ) = ∫ 29 1 ρ(u)νu du = ∫ 11 1 250(u− 1)νu du + ∫ 29 11 ρ(u)νu du = 250ν ∫ 10 0 uνu du + ν11 ∫ 18 0 (2500− 125u)νu du = 250ν(Ia)10 + ν11(2500a18 − 125(Ia)18 ) Using a18 = (1− ν18)/δ = ia18 ,i/δ = 0.2a18 ,0.2/ ln(1.02) = 5.2788 gives NPV(ρ) = 4761.094. Present value of opening cost and additional costs is 500a1 + 200a29 = 1548.470 Hence answer is 4761.094− 1548.470 = 3212.624 or £3,212,624. 36. The cash flow is as follows: Time 0 1/12 2/12 3/12 4/12 5/12 6/12 7/12 8/12 9/12 · · · 24 1112 25 Costs −40 −3 −3 −3 −3 −3 −3 Rental Income 1 1 1 1 · · · 1 0 Maintenance − 212 − 212 − 212 · · · − 212 0 Sale 60 After nmonths, where n > 8, we have NPV = − [40 + 3(x + x2 + · · · + x6)]+(x6+x7+· · ·+xn)− 16 (x7+x8+· · ·+xn). Hence NPV = −40− 3×1− x 6 1− x + x 6 1− xn−5 1− x − x7 6 1− xn−6 1− x = −40 + 37x + 176 x7 + x6 − 56xn+1 1− x Hence NPV ≥ 0 when xn+1 ≤ 65 [ 37x + 176 x 7 + x6 − 40], or when n + 1 ≥ ln [ 65 (37x + 176 x7 + x6 − 40)] / lnx = 112.55 where the inequality has been reversed because lnx < 0. Hence the DPP is 112 months, or 9 years and 4 months. (b) The costs occur mainly in the first 6 months and the income is received later. If the interest rate decreases, then the NPV of the costs increases and the NPV of the income also increases. As the income is received later in time than when the costs are incurred, the decrease in the interest rate will have a larger effect on the income. Hence the DPP will increase. If the interest rate was 0, then the NPV of the costs is -58 and the DPP would be month 76. 37. (i) We need k with 14a(2) k ≥ 120; that means we need k with ak ≥ 60i(2)/7i = 8.42646. Tables shows that k = 14 is sufficient. Checking k = 13.5: a13.5 = (1− ν13.5)/i = 8.554839. Hence the answer is k = 13.5 years. (ii) The NPV of the first 13.5 years of payments is 14a(2)13.5 = 14 i i(2) a13.5 . So profit is 14 i i(2) a13.5 ,0.07 − 120 at 7%, which becomes 1.0713.51.0511.5 [ 14 i i(2) a13.5 − 120 ] at time t = 25. The value of the payments in the final 11.5 years at time t = 13.5 is 14a(2)11.5 ,0.05 = 7.987092. The accumulated value of these payments at time t = 25 is 1.0511.5 × 14a(2)11.5 ,0.05 = 213.3234. So the total profit at time t = 25 is 221.3105. Page 164 Answers 3.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 38. We use units of £10,000. Let ν = 1/1.15 and x = ν1/4. NPV of costs: 500 + 350 ∫ 2 0 νt dt = 500 + 350 νt ln ν ∣∣∣∣2 t=0 = 500 + 350 1− ν2 ln 1.15 = 1110.679238 NPV of benefits if sold after 5 years: NPV = 50 4 (x8 + x9 + x10 + x11) + 55 4 (x12 + x13 + x14 + x15) + 60 4 (x16 + x17 + x18 + x19) + 2050x20 = ν2 4 (50 + 55ν + 60ν2)(1 + x + x2 + x3) + 2050v5 = ν2 4 (50 + 55ν + 60ν2) 1− ν 1− x + 2050v 5 = 1 4× 1.155 (50× 1.15 2 + 55× 1.15 + 60) 0.15 1− x + 2050 1.155 = 1122.03783 NPV of benefits if sold after 3 years: NPV = ν2 4 (50) 1− ν 1− x + 1650v 3 = 1 4× 1.153 × 50× 0.15 1− x + 1650 1.153 = 1120.80588 NPV of benefits if sold after 4 years: NPV = ν2 4 (50 + 55ν) 1− ν 1− x + 1800v 4 = 1 4× 1.154 (50× 1.15 + 55)× 0.15 1− x + 1800 1.154 = 1099.40298 (i) NPV criterion suggests selling after 5 years. (ii) DPP criterion suggests selling after 3 years. (iii) NPV of benefits if sold after 5 years (where x = 1/1.1751/4 and ν = 1/1.175): NPV = 1 4× 1.1756 (50× 1.175 3 + 55× 1.1752 + 60× 1.175 + 65) 0.175 1− x + X 1.1756 and this must equal NPV(costs) = 500 + 350 1− ν2 ln 1.175 and hence X = 1.1756 ( 500 + 350 1− ν2 ln 1.175 ) − (50× 1.175 3 + 55× 1.1752 + 60× 1.175 + 65) 4 0.175 1− x = 2566.51624 (iv) May be periods when property is unoccupied and rental income is reduced. Rental income may not increase at projected rate. Tax and expenses (on rent and proceeds of sale). Maintenance and repair costs will increase as sale is delayed. Forecast sale price is very high (£25,665,160)—this may not be achieved. 39. (i) (a) The equation NPV = 0 when considered as an equation for i. The NPV is the net present value of a sequence of cash flows. (b) The DPP is the first time that the accumulated value of a project becomes positive. (ii) Inflows: NPV(in) = ∫ 6 3 νt dt + 1.9 ∫ 9 6 νt dt + 2.5 ∫ 15 9 νt dt + 8ν15 = ν6 − ν3 ln ν + 1.9 ν9 − ν6 ln ν + 2.5 ν15 − ν9 ln ν + 8ν15 = ν3 − 1 ln ν [ ν3 + 1.9ν6 + 2.5ν9(ν3 + 1) ] + 8ν15 Outflows (where x = ν1/4 and α = 1/05): NPV(out) = 1.5 ∫ 3 0 νt dt + 0.3 4 [x12 + x13 + · · · + x59] + ν3[1 + αν + · · · + α11ν11] = 1.5 ν3 − 1 ln ν + 0.075ν3 1− ν12 1− x + ν 3 1− (αν)12 1− αν Using 9% gives NPV(in) = 12.62517 and NPV(out) = 13.32338. Hence the IRR is less than an effective 9% per annum. Now use 7% per annum on the first 12 years. NPV(in) = ∫ 6 3 νt dt + 1.9 ∫ 9 6 νt dt + 2.5 ∫ 12 9 νt dt = ν6 − ν3 ln ν + 1.9 ν9 − ν6 ln ν + 2.5 ν12 − ν9 ln ν = ν3 − 1 ln ν ν3 [ 1 + 1.9ν3 + 2.5ν6 ] = 9.34598 NPV(out) = 1.5 ∫ 3 0 νt dt + 0.3 4 [x12 + x13 + · · · + x47] + ν3[1 + αν + · · · + α8ν8] = 1.5 ν3 − 1 ln ν + 0.075ν3 1− ν9 1− x + ν 3 1− (αν)9 1− αν = 12.55811 Hence the DPP is more than 12 years. Appendix 2 Sep 27, 2016(9:51) Answers 3.3 Page 165 40. (i) Let α = 1.04 and ν = 1/1.11. Use units of £1,000. The net present value of the continuous income is 24∑ j=0 ∫ j+1 j 80αjνs ds = 24∑ j=0 80αj νs ln ν ∣∣∣∣s=j+1 s=j = 80 ln ν 24∑ j=0 αj(νj+1 − νj) = 80 ln ν (ν − 1)(1 + αν + α2ν2 + · · · + α24ν24) = 80 ln ν (ν − 1)1− α 25ν25 1− αν = 968.241851 The net present value of the discrete payments is 500 + (100ν + 110ν2 + 120ν3 + 130ν4 + 140ν5) + 300ν15− 700ν25 = 947.005379. Or, 500 + 90a5 ,0.11 + 10(Ia)5 ,0.11 + 300ν15 − 700ν25 where (Ia)5 ,0.11 = (a5 ,0.11 − 5ν6)/(1− ν). Hence the overall NPV of the project is 21.236472 or £21,236.47. (ii) The net present value of the outgoings is £998,531.04; the net present value of the income is £1019,767.51. If the interest rate increases, then ν decreases and so the net present value of the outgoings and the income will both decrease. However, over half of the outgoings is the payment of £500,000 at time t = 0 which is unaffected by a change in ν. Hence the income will decrease by more than the outgoings and so the net present value of the project will decrease. For example, if ν = 1/1.12 then the NPV of the project is -£45,257.82. 41. Let ν = 1/1.08 and α = 1.012. In the first year, the total income is £(40,000× 365); hence the pension fund receives £(400 × 365) over the year. This amount is received continuously over the 365 days of the year. Measuring time in years gives the net present value of the income from year 1 to be∫ 1 0 ρ(t)ν(t) dt = ∫ 1 0 (400× 365)νt dt = 400× 365ν − 1 ln ν In the second year we have∫ 2 1 ρ(t)ν(t) dt = ∫ 2 1 (500× 1.1× 365)νt dt = 550× 365ν ν − 1 ln ν Proceeding in this way gives the present value of the income is 365 ν − 1 ln ν [ 400 + 550ν + 550αν2 + 550α2ν3 + · · · + 550α18ν19] = 365ν − 1 ln ν [ 400 + 550ν 1− α19ν19 1− αν ] which is 2275323.80446 or £2,275,323.80. Hence the NPV is £275,323.80. 42. (i) Let i denote the internal rate of return and let ν = 1/(1 + i). Then NPV of outgoings is 1,309,500 + 12,000a(4)25 ,i. The NPV of the incomings is 100,000a(4)5 ,i [ 1 + 1.055ν5 + 1.0510ν10 + 1.0515ν15 + 1.0520ν20 ] = 100,000a(4)5 ,i 1− 1.0525ν25 1− 1.055ν5 Hence i satisfies 13,095 + 120a(4)25 ,i = 1,000a (4) 5 ,i 1− 1.0525ν25 1− 1.055ν5 At 9%, the left hand side is 13095 + 120 × 1.033144 × 9.822580 = 14313. At 9%, the right hand side is 1000 × 1.033144× 3.889651× (1− 1.0525/1.0925)/(1− 1.055/1.095) = 14313. (ii) The NPV of the outgoings is 1,000,000. The NPV of the incomings is 85,000a¨(4)20 ,i + 90,000a¨ (4) 5 ,i/(1 + i) 20. Hence the internal rate of return, i, satisfies f (i) = 1,000− 85a¨(4)20 ,i − 90a¨(4)5 ,i/(1 + i)20 = 0. Now if there was no increase in rent, i would satisfy 200/17 = a¨(4)25 ,i = (1 + i) 1/4 i i(4) a25 ,i which suggests i ≈ 0.08. Now f (0.08) = 43.217460361. Also f (0.07) = −39.0524598472. Using linear interpolation gives i = 0.07+0.01×39.0524598472/(43.217460361+ 39.0524598472) = 0.0747 and hence the answer is 7.5%. (iii) The answers to parts (i) and (ii) suggest project A is preferable, but: the outlay for project A is much larger than the outlay for project B—more money may have to be borrowed and interest rates will change over the period of 25 years; the assumption of no increase in maintenance costs over 25 years for project A is questionable; the assumption of the rise in rents made for both projects A and B is questionable—-an attempt should be made to assess how likely each assumption will be met. 43. Let ν = 1/1.12, γ = ν1/4 and suppose α denotes the required compounding factor. The cash flow is as follows (units of $10,000): Time 0 1/2 4/4 5/4 6/4 7/4 8/4 · · · 23/4 24/4 Cash flow −400 −90 9 9 9 9 9α · · · 9α4 680 Hence 400 + 90ν1/2 = (1 + γ + γ2 + γ3)(9ν + 9αν2 + 9α2ν3 + 9α3ν4 + 9α4ν5) + 680ν6 = (1 + γ + γ2 + γ3)9ν(1 +αν + α2ν2 + α3ν3 + α4ν4) + 680ν6. Hence 1− α5ν5 1− αν = 1− ν1/4 9ν(1− ν) (400 + 90ν 1/2 − 680ν6) = 4.55965787 The left hand side is a¨5 ,β where 1/(1 + β) = αν. Using a¨n = an−1 + 1 gives a4 ,β = 3.55965787. Using tables gives 0.045 < β < 0.05. Linear interpolation gives β = 0.0483515334 and hence α = 1.068344 or 6.83%. Page 166 Answers 3.6 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 44. Use units of £1,000. (i)(a) For project B we have Time 0 1 2 · · · 19 20 Cash flow—out -1,000 Cash flow—in 60 60 60 · · · 60 1,000 Now 162/3× 60 = 1,000. Payments are in advance—so the payback period is 16 years. (i)(b) Let x = 1.005. Project A costs: Time 0 1 2 · · · 19 20 Initial cost −1,000 Further costs −10 −10x −10x2 · · · −10x19 0 Project A income: Time 0 12/12 13/12 · · · 59/12 60/12 · · · 71/12 72/12 · · · 239/12 240/12 Cash flow 0 5 5 · · · 5 5x · · · 5x 5x2 · · · 5x15 2,000 The income each year for project A is less than the income in the same year from project B. Hence the payback period for project A will be longer. (ii)(a) The time when the present value of the incomings is equal or greater than the present value of the outgoings. (b) Project B: we need n with 60(1 + ν + · · · + νn) ≥ 1,000 or an+1 ,0.01 ≥ 50/(3×1.01). Using tables gives n = 18. (c) Again, the NPV of the income from project A in any year is less than that from project B in the same year; hence the DPP for project A will be longer. (iii) If i denotes the IRR, then i satisfies 1,000 = 60a¨20 ,i + 1,000ν20. This leads to 1,000 = 60(1 − ν20)/(1 − ν) + 1,000ν20 and hence ν = 94/100 and hence i = 6/94 = 0.063829787 or 6.383%. (iv) We work out the NPV of project A using the interest rate of 6.383% or i = 6/94 and ν = 94/100. For the costs we have NPV = 1,000 + 10(1− x20ν20)/(1− xν) = 1122.868517. For the income we have NPV = 60a(12)4 ν 11/12 + 60xa(12)1 ν 59/12 + 60x2a(12)1 ,0.05ν 71/12 + · · · + 60x15a(12)1 ν227/12 + 2,000ν20 = 60a(12)4 ν 11/12 + 60a(12)1 xν 59/12 [1 + xν + · · · + x14ν14] + 2,000ν20 = 60 i i(12) [ a4ν 11/12 + a1xν 59/12 1− x15ν15 1− xν ] + 2,000ν20 = 60 i 12 [ (1 + i)1/12 − 1] [ 1− ν4 i ν11/12 + xν71/12 1− x15ν15 1− xν ] + 2,000ν20 = 1227.154537 Hence NPV of project A at 6.383% is 1227.154537−1122.868517 = 104.286. Hence the IRR of project A is greater than 6.383%. (v) The IRR of project A is greater but the DPP is longer. But the DPP ignores the large final payment for project A. So the final decision will depend on other factors such as (a) comparison of NPV using various interest rate scenarios; (b) whether the capital would be useful for another project; etc. Answers to Exercises: Chapter 3 Section 6 on page 62 (exs3-2.tex) 1. Total repayment is 3 × £2,000 = £6,000. Hence original loan is £6,000 − £750 = £5,250. Hence the flat rate is 750/(3× 5250) = 0.0476 or 4.8%. 2. The cash flow is Time 0 1 2 Cash Flow −1000 703.84 703.84 This leads to 1000 = 703.84(ν + ν2). The usual formula for the solution of a quadratic leads to ν = 0.792586 and hence i = 0.261693 or 26.17%. 3. Let x denote a payment. Then 5,000 = xa12 ,0.08. Using the prospective method, the capital outstanding after the 4th payment is y = xa8 ,0.08. Hence the interest element of the 5th payment is 0.08y = 0.08 × 5000a8 ,0.08/a12 ,0.08 = 305.02. 4. The monthly repayment, x, is given by 4,000 = x × 12a(12)5 ,0.15. Using a(m)n = ii(m)an , see equation 2.3b, and tables gives x = 4,000/(12× 1.067016× 3.352155) = 93.193 or £93.93. The flat rate of interest is (60x− 4000)/20000 = 0.0796 or 7.96%. 5. Let i denotes the effective annual rate of interest and let j denote the effective rate of interest over 1 month. Let ν = 1/(1 + i); hence ν1/12 = 1/(1 + j). Then 5000 = 130× 12a(12)4 = 130(ν1/12 + ν2/12 + · · · + ν48/12) = 130a48 ,j We need to solve a48 ,j = 5000/130 = 38.4615 for j. Tables give a48 ,0.005 = 42.580318 and a48 ,0.01 = 37.973959. Interpolating gives (x − 0.005)/(0.01 − 0.005) = (f (x) − f (0.005))/(f (0.01) − f (0.005)). Hence x = 0.005 + 0.005(38.4615− 42.580318)/(37.973959− 42.580318) = 0.00947. Hence ν1/12 ≈ 1/1.00947 and so i ≈ 0.1197. Hence APR is 11.9% (nearest 0.1% rounded down). Appendix 2 Sep 27, 2016(9:51) Answers 3.6 Page 167 6. Suppose the monthly repayment is x. Then 15,000 = 12xa¨(12)2 ,0.1236 = x(1 + α + α 2 + · · · + α23) = x(1− α24)/(1− α) where α12 = 1/1.1236. Hence x = 697.2717 which would be rounded up to £697.28. Hence flat rate is (24x − 15,000)/30,000 = 0.057824 or 5.782%. 7. We need x with 100,000 = 12xa(12)20 ,j where j is the effective rate of interest per annum corresponding to a nominal 6% per annum convertible monthly. Now 6% per annum convertible monthly is 0.5% effective per month; so let ν = 1/1.005. Hence 100,000 =∑240 k=1 xν k = xa240 ,0.005 = xν(1− ν240)/(1− ν). Hence x = 716.43 or £716.43. (ii) After 6 years, 168 repayments are left. Hence capital outstanding is y = x(ν +ν2 + · · ·+ν168) = xν(1−ν168)/(1− ν) = x(1 − ν168)/0.005 = 81298.319786 or £81,298.32. Let ν1 = 1/(1 + 0.055/12) and let z denote the new repayment. Then y = z(ν1 + ν21 + · · · + ν1681 ) = zν1(1− ν1681 )/(1− ν1). Hence z = y(1− ν1)/ν1(1− ν1681 ) = 0.694960 or £694.96. 8. The annuity pays £10,000 per annum. Also i = 0.1. The capital outstanding after 5 years of payments is (prospective loan calculation) x = 10,000a(12)15 ,0.1 = 10,000 × 1.045045 × a15 ,0.1 = 79,486.9588 or £79,487. Hence the interest component of the first instalment of the 6th year is x× (1.11/12 − 1) = 633.84. (ii) NPV of annuity over the whole 20 years is 10,000a(12)20 ,0.1 = 10,000 × 1.045045a20 ,0.1 = 88,970.57 or £88,971. Hence capital repayment over the first 5 years is 88,971 − 79,487.34. Total payment of the annuity over the first 5 years is £50,000. Hence total interest payment over the first 5 years is 50,000− (88,971−79,487.34) = 40,516.34. 9. Let x denote the quarterly payment. Then 20,000 = 4xa(4)20 ,0.1 = 4x×1.036756×a20 ,0.1 = 4x×1.036756×8.513564. Hence x = 566.48. (ii) Loan outstanding after 24th payment is (prospective loan calculation) 4xa(4)14 ,0.1 = 4x i i(4) a14 ,0.1. Hence the interest element is [ 1.11/4 − 1] × 4x i i(4) a14 ,0.1 = xia14 ,0.1 = 417.31. Hence the capital element is x − 417.31 = 149.17. 10. (i) We need x with 100,000 = 12xa(12)25 ,0.05 = 12x i i(12) a25 . Hence x = 100,000/(12 × 1.022715 × 14.093945) = 578.138 or £578.14. (ii) (a) Using the prospective method, it is 12xa(12)13 ,0.05 = 12×578.14×1.022715×9.393573 = 66649.9311 or £66,649.93. (b) Interest portion of 145th payment is 66,649.93× (1.051/12 − 1) = 271.54. Hence capital repayment is 578.14− 271.54 = 306.60. 11. (a) The annual interest payment is £5,000 × 0.1 = £500. (b) The annual sinking fund repayment, x is given by xs10 ,0.08 = 5,000. Hence x = 345.15. (c) The total of (a) and (b) is 845.15. The amortisation repayment, y is given by ya10 ,0.1 = 5,000. Hence y = 813.73 12. (i) 50,000 = xa5 ,0.1 = x × 3.790787 and hence x = 13189.8732 or £13,189.87. Total interest = 5x − 50000 = 15949.35. (ii) Capital outstanding at the beginning of the third year = the value at that time of all outstanding repayments = xa3 ,0.1 = 32,801.25. We now want 32,801.25 = 4ya (4) 5 ,0.12 = 4y i i(4) a5 ,0.12 = 4y × 1.043938 × 3.604776. Hence y = 2179.1012 and so the quarterly payment is £2179.10. (b) After first repayment, capital outstanding is z = y(ν1/4 + ν2/4 + · · ·+ ν19/4) = yν1/4(1− ν19/4)/(1− ν1/4). Hence interest paid is z [ (1 + i)1/4 − 1] = z(1− ν1/4)/ν1/4 = y(1− ν19/4) = 907.09 using ν = 1/1.12. 13. (i) We want x with 800,000 = xa10 ,0.08 and hence x = 800,000/6.710081 = 119,223.598 or £119,223.60. The total interest paid is 10x− 800,000 = 392,236. (ii) Capital outstanding at start of year 8 is xa3 ,0.08 = 119,223.60 × 2.577097 = 307250.7819. (a) We need y with 307250.78 = 4ya(4)5 ,0.12 = 4y i i(4) a5 ,0.12 = 4y × 1.043938 × 3.604776. Hence y = 20,411.7393 or £20,411.74. (b) Interest payment is 307,250.78× [1.121/4 − 1] = 8,829.57. Hence capital repayment is 20,411.74− 8,829.57 = 11,582.17. 14. The cash flow is as follows: Time 0 1 2 3 . . . 19 20 Cash Flow −c 100 110 120 . . . 280 290 0 90 90 90 . . . 90 90 0 10 20 30 . . . 190 200 Hence c = 90a20 ,0.08 + 10(Ia)20 ,0.08 = 90a20 ,0.08 + 10(a¨20 ,0.08 − 20ν20)/0.08 = 90a20 ,0.08 + 10(1.08a20 ,0.08 − 20ν20)/0.08 = 225a20 ,0.08 − 2500ν20 = 1672.71 (ii) Using the prospective method, the capital outstanding after the 5th payment is equal to the value of all future repayments: hence l5 = 140a15 ,0.08 + 10(Ia)15 ,0.08 = 140a15 ,0.08 + 10(a¨15 ,0.08 − 15ν15)/0.08 = 140a15 ,0.08 + 10(1.08a15 ,0.08 − 15ν15)/0.08 = 275a15 ,0.08 − 1875ν15 = 1762.78 Hence the 6th repayment of 150 consists of interest 1762.78× 0.08 = 141.02 and capital 8.98. Hence the 7th repayment of 160 consists of interest (1762.78− 8.98)× 0.08 = 140.30 and capital 19.70. (iii) The last payment is £290. Let l19 denote the capital outstanding after the 19th payment. Hence the 20th payment Page 168 Answers 3.6 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed must consist of an interest payment of il19 and a capital repayment of l19. Hence (1+i)l19 = 290. Hence l19 = 268.519 and 20th payment consists of an interest payment of 21.48 and a capital payment of 268.52. 15. Cash flow: Time 1/1/80 1/2/80 1/3/80 1/4/80 . . . 1/12/09 1/1/10 Loan −100,000 Interest payments 0 500 500 500 · · · 500 500 Investment premiums 1060/12 1060/12 1060/12 1060/12 · · · 1060/12 0 (a) NPV of investment premiums at 1/1/80 is 1060a¨(12)30 ,0.07 = 1060 i i(12) a¨30 ,0.07 = 1060 i i(12) 1.071/12a30 ,0.07. Accumulated value at time 1/1/10 is 1.0730NPV = 1060 i i(12) 1.071/12s30 ,0.07 = 1060 × 1.031691 × 1.071/12 × 94.460786 = 103885.6856 or £103,885.69. (b) Using effective 4% per annum gives 1060 i i(12) 1.041/12s30 ,0.04 = 1060 × 1.018204 × 1.041/12 × 56.084938 = 60730.42958 or £60,730.43. This implies a shortfall of £39,269.57. (c) Revised premiums leads to additional accumulated value of 3940 i i(12) 1.041/12s10 ,0.04 = 48322.86400. This leads to a surplus of £48,322.86-£39,269.57=£9,053.29. Or, total accumulated value is 1060 i i(12) 1.041/12s20 ,0.04 × 1.0410 + 5000 i i(12) 1.041/12s10 ,0.04 = i i(12) 1.041/12 [ 1060s20 ,0.04 × 1.0410 + 5000s10 ,0.04 ] =1.018204× 1.041/12 [1060× 29.778079× 1.0410 + 5000× 12.006107] = 109053.2949 as before. (d) The interest rate on the investment is less than the rate on the loan. Hence the investor should have paid off the loan rather than investing in the policy. (e) We assume monthly payments paid in arrears. The effective interest rate is iwith 1+i = 1.00512 and ν = 1/(1+i). Let ν1 = ν1/12. We need 12x where 100,000 = xa(12)30 ,i = x [ ν1 + ν21 + · · · + ν3601 ] = xν1(1 − ν3601 )/(1 − ν1). Hence 12x = 12 × 100,000× i/(1− ν30) = 6000/(1− ν30) = 7194.61 If we assume annual payments in arrears, we need x with 100,000 = xa30 ,i = xν(1 − ν30)/(1 − ν) = x(1 − ν30)/(1.00512 − 1). Hence x = 100,000(1.00512 − 1)/(1− 1/1.005360) = 7395.79. 16. (i) See derivation of equation 2.5b. (ii) The cash flow is as follows: Time 0 1 2 3 . . . 20 Cash flow, c 0 10 12 14 . . . 48 (a) Then NPV(c) = 10ν + 12ν2 + 14ν3 + · · · + 48ν20 = 402.372. or use NPV(c) = 8(ν + ν2 + · · · + ν20) + 2(ν + 2ν2 + · · · + 20ν20) = 8a20 + 2(Ia)20 (b) The remaining cash flows are Time 15 16 17 18 19 20 Cash flow, c 0 40 42 44 46 48 Thus we want 40ν + 42ν2 + 44ν3 + 46ν4 + 48ν5 = 200.966. It is also 38a5 + 2(Ia)5 . (c) Now 0.03× 200.966 = 6.029. Hence the capital repayment is 40− 6.029 = 33.97. 17. Time 0 1 2 3 4 · · · 14 15 Cash flow -c 50 48 46 44 · · · 24 22 Let ν = 1/1.06. Hence c = 50ν + 48ν2 + · · · + 24ν14 + 22ν15 = 2v [25 + 24ν + · · · + 12ν13 + 11ν14] = 2νs where s = 25 + 24ν + · · · + 12ν13 + 11ν14. Then (1− ν)s = 25− 11ν15 − [ν + ν2 + · · · + ν14]. Hence c = 2 i [ 25− 11ν15 − 1− ν 14 i ] = 100 3 [ 25− 11ν15 − 50 3 (1− ν14) ] = 370.5033 Hence size of loan is 370.5033. (ii) First payment: interest component is 370.5033× 0.06 = 22.230; hence capital repayment is 50− 22.23 = 27.77. Remaining loan outstanding is 342.7333. Second payment: interest component is 342.7333×0.06 = 20.564; hence capital repayment is 48−20.564 = 27.436. (iii) Using the prospective method: loan outstanding is 24ν + 22ν2 = 118,600/(53× 53) = 42.2214. Hence interest payment of 14th instalment is 42.2214× 0.06 = 2.5333 and capital repayment is 24− 42.2214× 0.06 = 21.468. Appendix 2 Sep 27, 2016(9:51) Answers 3.6 Page 169 18. The amount, x, of each monthly payment is given by 80,000 = 12xa(12)25 ,0.08 = 12x i i(12) a25 ,0.08 Hence x = 602.732. (a) Interest at the end of the first month is 80000(1.081/12− 1) = 514.722. Hence the capital repaid is x− 514.722 = 88.01. (b) Using the prospective method, the loan outstanding after the 228th payment is 12xa(12)6 ,0.08. Hence the total interest paid during the last 6 years is 72x−12xa(12)6 ,0.08 = 12x(6−a(12)6 ,0.08) = 12x(6−ia6 ,0.08/i12)) = 8751.45. (c) The capital outstanding, c is given by c×1.081/12 = 602.732; hence c = 598.88 and the interest paid is 602.73−598.88 = 3.85. (ii) The total annual payments increase and the total interest paid increases. 19. (i) Let x denote the monthly payment which is paid in advance. Then 100,000 = 12xa¨(12)25 ,0.06 = 12x×1.061/12 ii(12) × a25 ,0.06. Hence x = 100,000/(12× 1.061/12 × 1.027211× 12.783356) = 631.5466 or £631.55. Time 0 1/12 2/12 · · · 23/12 24/12 · · · 299/12 300/12 = 25 Loan 100,000 Repayments x x x · · · x x · · · x 0 (ii) Use the prospective method. The outstanding loan at time 23/12 after 24 payments is 12xa(12)23 ,0.06 = 12x i i(12) × a23 ,0.06 = 95779.6067 or £95,779.61. Hence, interest component of next payment (the 25th) is 95,779.61 ×[ 1.061/12 − 1] = 466.212 and the capital repayment is 631.55− 466.21 = 165.34. (iii) (a) After 10 years at time 120/12, the loan outstanding is 12xa¨(12)15 ,0.06. Let y denote the new monthly pay- ment. Then 12ya¨(12)15 ,0.02 = 12xa¨ (12) 15 ,0.06 or y = xa¨ (12) 15 ,0.06/a¨ (12) 15 ,0.02 = x × [ 1.061/12/1.021/12 ] × a(12)15 ,0.06/a(12)15 ,0.02 = 487.474900 or £487.47. (b) The reduction in instalments is x−y = 631.55−487.47 = 144.08 paid at times t = 120/12, 121/12, . . . , 299/12. We need the accumulated value of this cash flow at time t = 300/12 = 25. The value at time t = 119/12 is 12ya(12)15 ,0.02. Hence value at time t = 120/12 is 1.02 1/12 × 12ya(12)15 ,0.02 and the value at time t = 300/12 = 25 is 1.02181/12 × 12ya(12)15 ,0.02. In the usual way, a(12)15 ,0.02 = ii(12)a15 ,0.02 = 1.009134× 12.849264. hence the value at time t = 120/12 is 22455.810 and the value at time t = 300/12 is 30222.563. 20. (i) Now NPV = 1000 [ a10 ,0.05 + a10 ,0.07/1.0510 ] = 12033.6051 or £12,033.61. (ii) First note that 439.52/8790.48 ≈ 0.05; hence x ≤ 10. Loan outstanding at time t = k where 1 ≤ k ≤ 10 is 1000 [ a10 ,0.07/1.0510−k + a10−k ,0.05 ] . Setting this equal to 8790.48 gives 8.79048 = 7.023582α10−k + 1−α 10−k 0.05 = 7.023582α10−k + 20(1 − α10−k) where α = 1/1.05. Hence 11.20952 = 12.976418α10−k; hence 10 − k = log(11.20952/12.976418) log(1/1.05) = 3. Hence k = 7 = x− 1 and so x = 8. (iii) Interest rate now remains at 5%. Value of annuity over 20 years is 1000a20 ,0.05 = 12,462.21, over 19 years is 1000a19 ,0.05 = 12,085.321 and over 18 years is 11,689.587. Hence we need 18 payments of £1,000 with a reduced final payment. Denote this payment by x. Then NPV of the payments is 11,869.587 + x/1.0519 and we want this to equal 12,033.61 which implies a final payment of £869.32. Originally, total interest paid was £20,000-£12,033.61. Under the new scheme, interest paid is £18,000+869.32 - £12,033.61. The reduction is £2,000-£869.32=£1,130.68. 21. The cash flow is Time 0 1 2 · · · 19 20 Cash Flow −c 6000 5800 · · · 2400 2200 So c = 6000ν + 5800ν2 + · · · + 2400ν19 + 2200ν20 = 200x where x = 30ν + 29ν2 + · · · + 12ν19 + 11ν20 and (1− ν)x = 30ν − ν2(1− ν19)/(1− ν)− 11ν21 and so x = 42415.88. (ii) Cash flow after payment 7 is: Time 7 8 9 · · · 19 20 Cash Flow 0 4600 4400 · · · 2400 2200 Hence l7 = 4600ν + 4400ν2 + · · ·+ 2400ν12 + 2200ν13 = 200νy where y = 23 + 22ν + · · ·+ 12ν11 + 11ν12 = 27225.13. year loan outstanding repayment interest due capital repaid loan outstanding before repayment after repayment 8 l7 = 27225.13 x8 = 4600 il7 = 2450.26 x8 − il7 = 2149.74 l8 = 25075.39 9 l8 = 25075.39 x9 = 4400 il8 = 2256.79 x9 − il8 = 2143.21 l9 = 22932.18 (iii)After the ninth payment, the capital outstanding is l9 = 22932.18 and ν = 1/1.07. Time 9 10 11 · · · 19 20 Cash Flow −l9 x x− 200 · · · x− 1800 x− 2000 22932.18 = x(ν +ν2 + · · ·+ν11)− (200ν2 +400ν3 + · · ·+2000ν11) and so xν(1−ν11)/(1−ν) = 22932.18+200ν2(1+ 2ν + · · · + 10ν9) leading to x = 3924.09. Page 170 Answers 3.6 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 22. Let α = 1.05, ν = 1/1.06, ν12 = ν1/12 and ν3 = ν1/3 = ν412. The cash flow is as follows: Time 1/9/98 1/7/99 1/11/99 1/3/00 1/7/00 1/11/00 · · · 1/11/03 1/3/04 Cash Flow −c 1000 1000α 1000α2 1000α3 1000α4 · · · 1000α13 1000α14 Hence c = 1000 [ ν1012 + αν 14 12 + α 2ν1812 + α 3ν2212 + α 4ν2612 + · · · + α13ν6212 + α14ν6612 ] = 1000ν1012 [ 1 + αν3 + α2ν23 + α 3ν33 + α 4ν43 + · · · + α13ν133 + α14ν143 ] = 1000ν5/6 1− α15ν5 1− αν1/3 and this is 17691.768 or £17,692. (ii) Loan outstanding on 1/7/99 is 17,692 × 1.0610/12 = 18,572.277 or £18,572.28. Hence interest repaid in first instalment is 18,572.28− 17,692 = 880.28 and so the capital repaid is 1000− 880.28 = 119.72. (iii) The seventh repayment is 1000α6 on 1/7/01. Capital outstanding at 1/3/01 (just after the 6th payment is made) is 1000α6ν412 [ 1 + αν412 + · · · + α8ν3212 ] = 1000α6ν1/3 1− α9ν3 1− αν1/3 = 13341.566 or £13,341.57. Hence interest in 7th payment is 13341.57 × [1.061/3 − 1] = 261.67 and the capital repaid is 1000α6 − 261.67 = 1,078.43. 23. Let i denote effective interest rate per annum. Then 1 + i = 1.024. Let ν = 1/(1 + i)1/12 = 1/1.021/3. Let α = ν60 = 1/1.0220. Then 10,000 = x(ν+ν2 + · · ·+ν60)+(x+10)(ν61 + · · ·+ν120)+(x+20)(ν121 + · · ·+ν180) which in turn equals[ (1− ν60)/(1− ν)] [10ν61 + 20ν121 + x(ν + ν61 + ν121)] = [(1− α)/(1− ν)] ν [10α + 20α2 + x(1 + α + α2)]. It follows that x(1 + α + α2) = 10,000(1 − ν)/ [(1− α)ν] − 10α − 20α2 and so x = 87.8345. hence the initial payment will be set at £87.84. (ii) Interest at 1 May 1994 is 10,000(1.021/3 − 1). Hence capital repayment is 87.84− 10,000(1.021/3 − 1) = 21.61. (iii) Loan outstanding=(value at time k of original loan)-(accumulated value at time k of repayments). First term is 10,000 × 1.0232. Let β = 1.021/3. Hence second term is x(β95 + β94 + · · · + β36) + (x + 10)(β35 + · · · + β + 1) = x(β96 − 1)/(β − 1) + 10(β36 − 1)/(β − 1) Hence answer is £6,708.31. (iv) Let revised amount be y. Then 6,708.31 = y(ν12 + ν24 + ν36 + ν48) = yν12(1 + ν12 + ν24 + ν36) = yν12(1 − ν48)/(1− ν12). Hence y == 2,036.35. 24. Now (Ia)n = ν + 2ν2 + · · · + nνn = ν(1 + 2ν + · · · + (n − 1)νn−1) = ν1−ν [ 1−νn 1−ν − nνn ] = 1i [ a¨n − nνn ] where a¨n is the present value of the annuity-due, a. (ii) Let ν = 1/1.08. Measuring in hundreds, x = 30ν + 32ν2 + 34ν3 + · · · + 58ν15. Hence (1 − ν)x = 30ν + 2ν2 + · · · + 2ν15 − 58ν16 = 30ν − 58ν16 + 2ν2(1− ν14)/(1− ν). Hence x = 35,255.57. Or use x = 28a15 + 2(Ia)15 . (iii) Let y denote loan outstanding at start of year 9. Then y = 46ν + 48ν2 + 50ν3 + · · · + 58ν7. As above, we have y = 11−ν [ 46ν − 58ν8 + 2ν2(1−ν6)1−ν ] = 267.5415. Start of year 9: loan outstanding = £26,754.15. Repayment = £4,600. Interest = £26,754.15 × 0.08 = £2,140.33. Capital repayment is £2,459.67. Start of year 10: loan outstanding = £24,294.48. Repayment = £4,800. Interest = £24,294.48 × 0.08 = £1,943.56. Leaving loan outstanding = £21,438.04. (iv) Loan outstanding is now £21,438.04. Working in hundreds, we have: Time using original origin 10 11 12 13 14 15 Time using new origin 0 1 2 3 4 5 Cash Flow −214.3804 x x + 2 x + 4 x + 6 x + 8 Hence 214.3804 = xν + (x + 2)ν2 + (x + 4)ν3 + (x + 6)ν4 + (x + 8)ν5 = xν 1− ν5 1− ν + 2ν 2 [1 + 2ν + 3ν2 + 4ν3] = x 1− ν5 i + 2ν2 [ 1 + 2ν + 3ν2 + 4ν3 ] and hence x = ( 214.3804− 2ν2 [1 + 2ν + 3ν2 + 4ν3])× i 1− ν5 = 47.12587 Hence the eleventh payment is £4,712.59. 25. First we find the monthly repayments. Let x denote the monthly repayment for loan A. Then (60x − 10000)/50000 = 0.10715. Hence x = 255.95833. Hence the repayment is £255.96. Let y denote the monthly repayment for loan B. Then 15,000 = 12y [ a(12)2 ,0.12 + 1 1.122 a(12)3 ,0.10 ] Appendix 2 Sep 27, 2016(9:51) Answers 3.6 Page 171 Using the formula a(m)n = ii(m)an and tables gives y = 324.4303 or £324.43. Hence answer to (i) is £600−£255.96− £324.43 = £19.61. (ii) First we find the effective interest rate per annum for loan A. We have 10,000 = 12 × 255.96 × a(12)5 ,i . Hence a(12)5 ,i = 3.255717 which leads to i = 0.2 or 20%. Hence capital outstanding under loan A is 12 × 255.96a(12)3 ,0.2 = 12× 255.96× 1.088651× 2.106481 = 7043.679 or £7,043.68. Capital outstanding under loan B is 12 × 324.43a(12)3 ,0.10 = 12 × 324.43 × 1.045045 × 2.486852 = 10117.8255 or £10,117.83. So interest paid in 25th month is 7,043.68 × [1.21/12 − 1] + 10,117.83 × [1.101/12 − 1] = 188.5160 or £188.52. Hence, capital repayment is £255.96 + £324.43− £188.52 = £391.87. (iii) Total capital outstanding is £7,043.68 + £10,117.83 = £17,161.51. New monthly payment is £(255.96 + 324.43)/2 = £290.20. Hence we need iwith 17161.51 = 12×290.20a(12)10 ,i; hence we need iwith a(12)10 ,i = 4.9280697. Using tables gives a(12)10 ,0.2 = 1.088651 × 4.192472 = 4.5641388 and a(12)10 ,0.15 = 1.067016 × 5.018769 = 5.355107. Using linear interpolation gives i = 0.15 + 0.05 × (5.355107 − 4.9280697)/(5.355107 − 4.5641388) = 0.17699 or an effective rate of 17.7% per annum. Hence interest paid in 25th month is 17161.51 × [1.1771/12 − 1] = 234.6557 or £234.66. Capital repaid is 290.20− 234.66 = 55.54 or £55.54. (iv) The new loan from Freeloans has a higher interest rate than loan B and a lower interest rate than loan A. So the student should investigate the possibility of using Freeloans to repay loan A only. Under the new arrangements, the repayments have been halved. Hence the student could afford to pay off the loan more quickly by maintaining the same total repayments, if such a loan term could be negotiated with Freeloans. 26. Product A. We have 100,000 = 7,095.25a25 ,i and hence a25 ,i = 14.0939361. Hence i = 0.05 or 5% p.a. Product B. 27. (i)(a) The flat rate of interest is xR − l0 nl0 = 4800− 2000 2× 2000 = 0.7 The flat rate of interest is 70%. (b) The flat rate of interest is not a good measure because it take no account of the fact that the loan outstanding decreases each month. (ii) The consumers’ association case. Time 0 1/12 2/12 3/12 · · · 20/12 21/12 22/12 23/12 Cash Flow −2,000 + 200 200 200 200 · · · 200 200 200 200 Hence 2000 = 2400a¨(12)2 ,i = 2400 i d(12) a2 ,i or 5 = 6 i d(12) a2 ,i Now if i = 2, then a2 ,i = (1− 1/9)/2 = 4/9 and (1− d(12)/12)12 = 1/3 and hence d(12) = 1.049823. Hence 6 i d(12) a2 ,i = 6 2 1.049823 4 9 = 5.08 > 5 Hence the claim by the consumers’ association is correct. The banks’ case. Time 0 1/12 · · · 11/12 12/12 · · · 17/12 18/12 · · · 23/12 Receipts −2,000 + 200 200 · · · 200 120 · · · 120 60 · · · 60 Costs −60 −60 · · · −60 −60 · · · −60 −60 · · · −60 c1 60 60 · · · 60 60 · · · 60 60 · · · 60 c2 60 60 · · · 60 60 · · · 60 c3 80 80 · · · 80 Hence 2000 = c3 + c2 = 960a (12) 1 ,i + 720a (12) 1.5 ,i = i d(12) [ 960a1 ,i + 720a1.5 ,i ] Now 1.01463× 1.025 = 1.40. At 4%, i d(12) [ 960a1 ,i + 720a1.5 ,i ] = 1.021537 [ 960× 0.961538 + 7201− ν 1.5 0.04 ] = 1993.51706 < 2000 Hence the banks’ case is also correct. 28. Work in units of £100. Let ν = 1/1.04. Time 0 1 2 3 4 · · · 11 12 13 · · · 19 20 −L 50 48 46 44 · · · 30 28 26 · · · 14 12 c1 52 52 52 52 · · · 52 52 52 · · · 52 52 c2 2 4 6 8 · · · 22 24 26 · · · 38 40 From first principles: L = 50ν + 48ν2 + · · · + 14ν19 + 12ν20 Page 172 Answers 3.6 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed L(ν − 1) = 12ν21 + 2ν2 1− ν 19 1− ν − 50ν and hence L = 2ν(25− 6ν20) 1− ν − 2ν2(1− ν19) (1− ν)2 = 456.386946 Alternatively L = c1 − c2 = 52a20 − 2(Ia)20 = 52a20 − 2 a20 − 20ν21 1− ν = 456.386946 Giving the answer £45,638.69. (ii) The 12th instalment is £2,800. Immediately after the 11th instalment has been paid at t = 11, the capital outstand- ing is the sum of all future repayments which is L1 = 28ν + 26ν2 + 24ν3 + · · · + 14ν8 + 12ν9. Hence L1 = 30a9 − 2(Ia)9 = 30a9 − 2 a9 − 9ν10 1− ν = 152.586727 Hence the capital outstanding at t = 11 is £15,258.67, and the interest component of the 12th instalment is 15,258.67× 0.04 = 610.35. The capital repayment is therefore £2,800− £610.35 = £2,189.65. (iii) The loan outstanding just after the 12th instalment is paid is £15,258.67− £2,189.65 = £13,069.02. So we want k and x where x < 28 and 130.6902 = 28ak + xν k+1 Hence k = 5 and xν6 = 130.6902− 124.651016 = 6.039184. Hence x = 7.64149437. We have established that the remaining term of the revised loan is 6 years with 5 repayments of £2,800 and one final repayment of £764.15. (iii) The total of all repayments is 50 + 48 + 46 + · · · + 28 + 5× 28 + 7.6415 = 615.6415. Hence the total interest paid is 615.6415− 456.386946 = 159.254554 or £15,925.46. 29. (i) Denote the annual payment by x. Then 500,000 = xa10 ,0.09 which gives x = 77910.041327 or £77,910.04. The total amount of interest is 10× £77,910.04− £500,000 = £279,100.40. (ii)(a) Loan outstanding at t = 7 is value at t = 7 of the remaining 3 instalments. This is xa3 ,0.09. Let y denote the amount of the new quarterly payment. Then 4ya4 ,0.12 = xa3 ,0.09 giving y = 77,910.04 × 2.531295/(4 × 1.043938 × 3.037349) or £15,549.16. (ii)(b) Let ν denote the quarterly discount factor. Hence ν = 1/1.121/4. Capital outstanding after first quarterly instalment is y(ν + · · · + ν15); capital outstanding after second quarterly instalment is y(ν + · · · + ν14). Hence capital repayment in second quarterly instalment is yν15 and so interest content is y(1− ν15) = 15,549.16(1− 1/1.1215/4) = 5383.411819 or £5,383.41. 30. (i) Let ν = 1/1.04. The cash flow of the repayments is as follows: Time 0 1 2 . . . 14 15 Cash Flow, c 0 400 380 . . . 140 120 c1 0 420 420 . . . 420 420 c2 0 20 40 . . . 280 300 So we want c = c1 − c2 = 420a15 ,0.04 − 20(Ia)15 ,0.04 = 420a15 ,0.04 − 20 a15 ,0.04 − 15ν16 1− ν = 3052.64507035 or £3,052.65. (ii) Interest component is £3,052.65× 0.04 = £122.11. So capital repayment is £400− £122.11 = £277.89. (iii) The beginning of the ninth year is just after the payment of £260 has been made. So the capital outstanding is 260a7 ,0.04 − 20(Ia)7 ,0.04 = 260a15 ,0.04 − 20 a7 ,0.04 − 7ν8 1− ν = 1099.178046207 or £1,099.18. So new loan is £549.59. Let x denote the size of the new first repayment. Time 0 1 2 . . . 9 10 Cash Flow, c −549.59 x x + 2 . . . x + 16 x + 18 c1 0 x− 2 x− 2 . . . x− 2 x− 2 c2 0 2 4 . . . 18 20 Hence 549.59 = (x− 2)a10 ,0.08 + 2(Ia)10 ,0.08 leading to x = 2 + (549.59− 2× 32.686907)/6.710081 = 74.162495 or £74.16. 31. (i) Time 0 1 2 3 . . . 9 10 Cash Flow, c 0 00 300 400 . . . 1,000 1,100 c1 0 100 100 100 · · · 100 100 c2 0 100 200 300 . . . 900 1,000 Appendix 2 Sep 27, 2016(9:51) Answers 4.6 Page 173 Now c = c1 + c2 = 100 [ a10 ,0.06 + (Ia)10 ,0.06 ] = 100 [ 7.360087 + 7.360087− 10ν11 1− ν ] = 4432.249451415 or £4,432.25. (ii) Immediately after the sixth repayment (at time t = 6 + 0), the loan outstanding is 800ν + 900ν2 + 1000ν3 + 1100ν4 = 700a4 ,0.06 + 100(Ia)4 ,0.06 = 3266.64. Hence the interest component of the seventh repayment is £3266.64× 0.06 = £196.00 and the capital component is £800− £196 = £604. (iii) The loan outstanding after the seventh repayment is £3266.64− £604 = £2662.64. Let x denote the new annual payment. Then 2662.64 = xa8 ,0.08. Hence x = 463.3386576 or £463.34. 32. (i) After 1 month. amount owing is 300,000×1.0851/12. Hence monthly interest payment is 300,000× (1.0851/12− 1) = 2,046.45 leading to a total interest payment over the first 15 years of 180× 2,046.45 = 368361 or £368,361.00. Similarly, for the remaining 10 years, the total interest paid is 40× 3,090.66 = 123,626.40. Total is £491,987.40. (ii) Let i denote the effective annual rate of interest. Then 1 + i = 1.0452. Let ν = 1/(1 + i) and x = ν1/12. Let z denote the monthly contribution to the savings account. Then the NPV of the 180 monthly contributions is z(1 + x + x2 + · · · + x179) = z(1 − x180)/(1 − x) = z(1 − ν15)/(1 − x). We want this to have value £150,000 after 15 years. Hence 1.04530z(1 − ν15)/(1 − x) = 150,000 and z = 150,000(1 − x)/(1.04530 − 1) = 399.369219 or £399.37. (iii) Value of savings after 15 years is 399.37 × (1.115 − 1)/(1 − 1/1.11/12) = 160395.465880 or £160,395.46. So outstanding loan is £139,604.54. Let y denote the new monthly repayment. Then 139,604.54 = 12ya(12)10 ,0.07. Hence y = 139604.54/(12× 1.031691× 7.023582) = 1605.498842 or £1,605.50. 33. (i) In the notes. (ii) Clearly 30,000 = X(Ia)60 ,0.0125. Easier from first principles: 30,000 = X(ν + 2ν2 + · · · + 60ν60) = Xν [ (1− ν60)/(1− ν)2 − 60ν60/(1− ν)]. Hence X = 26.6222 or £26.62. (iii) Let i denote the effective rate of interest per annum and let ν = 1/(1 + j) = 1/(1 + i)1/12. Then 30,000 = 958.32ν36a60 ,j and hence ν36a60 ,j = 31.3047834. Using tables shows LHS = 31.4201977 when j = 0.01. So try j = 0.011; then LHS=29.5097845. Linear interpolation gives j = 0.010060 and hence i = 0.0128 or 12.8%. (iv) The total repayments are 1,830X = 48,714.60 under the first schedule and 57,499.20 under the second schedule. So the second schedule has a lower interest rate but the total repayments are larger—the reason for this is that the payments do not start for 3 years during which time the outstanding loan grows larger. 34. Let i denote the yield obtained by purchasing this loan for the price P . Then P = gCa(m)n,i + Cαν n = gC 1− νn i(m) + Cανn = g αi(m) [ C ′ −K] +K where C ′ = Cα and K = C ′νn. 35. Use units of £1,000. Now K = value at time 0 of the instalments = 1.05× 20(ν4 + ν8 + ν12 + ν16 + ν20) = 21ν4(1− ν20)/(1− ν4) where ν = 1/1.08. Also C ′ = 100× 1.05 = 105. Finally, after allowing for tax, g = 0.1× 0.8 = 0.08 and i(m) = 0.08(2) = 2(1.081/2 − 1). Hence P = 103.2856015262 or £103,285.60. Answers to Exercises: Chapter 4 Section 6 on page 79 (exs4-1.tex) 1. (a) £1,000,000 ( 1 + 0.05× 90/365) = 1,012,328.8; (b) £1,012,328.8/(1 + 0.045× 67/365) = £1,004,035.2 (c) Price on 4 May is £1,012,328.8/(1 + 0.04× 37/365). Hence we want[ price on 4 May price on 4 April − 1 ] 365 30 = [ 1 + 0.045× 67/365 1 + 0.04× 37/365 − 1 ] 365 30 = 0.050960034 or 5.1%. 2. (a) A futures contract is a legally binding contract to buy or sell an agreed quantity of an asset at an agreed price at an agreed time in the future. An option is a contract that gives the buyer the option of buying an agreed quantity of an asset at an agreed price at an agreed time in the future. (b) Convertibles can be converted into ordinary shares at a fixed price at a fixed date in the future. Hence they give the owner the option to purchase a specified quantity of ordinary shares at a specified price at a specified time in the future. 3. Suppose party A has a loan at a variable rate. Party A pays a series of fixed payments to party B for a fixed term. Party B, the other party to the interest rate swap, pays a series of variable payments—the interest payments on the loan. 4. Preference shares usually pay a fixed dividend whilst ordinary shares do not. Preference shares rank above ordinary shares for payments of dividends and if the company is wound up. Preference shares usually have a lower return to reflect the lower risk. Preference shares usually do not have voting rights; ordinary shares do. 5. Short term—less than one year. Sold at a discount. Used to fund short term spending of a government. In the UK, denominated in sterling, but also in euros. Sold at a discount and redeemed at par. Secure and marketable. Often used as a benchmark risk-free short term investment. Page 174 Answers 5.2 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 6. (i) Issued by the government. The term “gilt-edged market” refers to the market in government bonds. Coupon and redemption payments linked to an inflation index. (ii) Most bonds are linked to an inflation index but there is a time lag. When the bond is redeemed, the inflation is usually calculated on the basis of the inflation index at some earlier reference date. If inflation increases over the period of the bond, then the payments will not keep up with inflation. 7. (a) Market risk. Interest rates may move. If the current rate of 5.5% decreases, then the company receives less but has to pay out the same each month. If the interest rate moves above 6%, the other party would have to pay out more than it receives. Credit risk: the other party defaults on the payments. (b) At the current time, the company does not face a credit risk, because it is paying 6% but only receiving 5.5%. 8. Issued by companies. Entitle holders to share in profits (the dividend) after interest on loans, etc have been paid. They have lowest ranking. Higher expected returns than most other asset classes—but the risk of capital loss is also higher. Size of dividend is variable. Possible capital gain from increase in price of share. Marketability depends on size of company. Voting rights in proportion to number held. 9. (a) Issued by large companies, banks and governments. Usually unsecured. Regular interest payments and redeemed at par. Issued in currency other than issuer’s home currency and outside the issuer’s home country. Yields depend on issuer, size of issue and time to redemption. Yields typically slightly lower than for conventional loan stocks from the same issuer. Usually a bearer security. Negotiable with an active secondary market. (b) Certificates issued by banks and building societies stating that money has been deposited. Usually last for 28 days to 6 months. Interest paid on maturity. Security and marketability depends on issuing bank. Active secondary market. 10. (i) Purchase the security at time 0. This is a negative cash flow. Receive periodic coupon payments (usually every 6 months or every year) and a capital repayment on maturity. The coupons (and possibly the capital payment) are indexed by a price index such as CPI or RPI. Often they are linked to the value of the index some time previously so that there is no delay in making the payments. (ii) Purchase the share at time 0; this is a negative cash flow. Receive periodic dividends, usually every 6 months or every year. Dividends are not fixed and depend on the success of the company. Dividends cease if company fails and then share may have no value. Receive value of share when it is sold—the value may be much higher than the purchase price. 11. See notes on page 74 in section 4.4. 12. The running yield equals income divided by price. Some reasons: property is a less flexible investment; investing in property incurs high expenses (buying, selling and valuation expenses); difficult to value a property; commercial rents often fixed for 3 to 5 years but dividends can change every year. Answers to Exercises: Chapter 5 Section 2 on page 86 (exs5-1.tex) 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− t1fr fr/m · · · fr/m C + fr/m− t1fr Hence i is a solution of the equation P = fr m nm∑ k=1 νk/m + νnC − t1fr n∑ k=1 νk = fra(m)n,i + Cν n − t1fran 2. Define the function f by f (n, i) = (1− t1)gCa(m)n,i + Cνn = C + [ (1− t1)g − i(m) ] Ca(m)n,i Clearly f (n, i)↘ as i↗. For bond A we have P = f (n1, i1) and for bond B we have P = f (n2, i2). (a) If P = C, then i(m)1 = (1− t1)g = i(m)2 . Hence i1 = i2. (b) If P < C then i(m)1 > (1 − t1)g. Also f (n, i1) ↘ as n ↗. Hence f (n2, i1) < f (n1, i1) = P = f (n2, i2). But f (n, i)↘ as i↗. Hence i2 < i1. (c) If P > C then i(m)1 < (1 − t1)g. Also f (n, i1) ↗ as n ↗. Hence f (n2, i1) > f (n1, i1) = P = f (n2, i2). Using f (n, i)↘ as i↗ shows that i2 > i1. Appendix 2 Sep 27, 2016(9:51) Answers 5.2 Page 175 3. The cash flow is: Time 0 t0 t0 + 1/m t0 + 2/m t0 + 3/m · · · Cash flow −P (1− t1)fr/m (1− t1)fr/m (1− t1)fr/m (1− t1)fr/m · · · P = (1− t1)fr m ( νt0 + νt0+1/m + νt0+2/m + · · · ) = (1− t1)frνt0 m 1 1− ν1/m = (1− t1)frν t0 a¨(m)∞ Recall (1 − d(m)/m)m = 1 − d = ν and hence d(m) = m(1 − ν1/m). Hence an alternative expression is P = (1− t1)frνt0/d(m). 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 t1 is zero. The purchase price at time t1 should be 8a (2) 7 +100ν 7. Using tables gives P = 1 1.068/365 [ 8a(2)7 + 100ν 7 ] = 8× 1.014782× 5.582381 + 66.5057 1.068/365 = 111.682 Alternatively, if x = 1/1.060.5, we have P = 1 1.068/365 4 [ x + x2 + · · · + x13 + x14 + 25×14] = 1 1.068/365 4x [ 1− x14 1− x + 25x 13 ] = 111.682 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 37∑ k=0 ν0.25+0.5k + 10000ν18.75 = 262.50ν0.25 1− ν19 1− ν1/2 + 10000ν 18.75 = 9385.46 (ii) Now i(2) = 2 ( 1.051/2 − 1) = 0.0494 and (1 − t1)g = 0.75 × 0.07 = 0.0525. As i(2) < (1 − t1)g, the second 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 P = 262.50 10∑ k=1 νk/2 + 10000ν5 = 262.50ν0.5 1− ν5 1− ν1/2 + 10000ν 5 = 10136.30 7. Time (years) 0 1 2 . . . 19 20 Cash flow −96 4 4 . . . 4 104 Hence 96 = 4a20 + 100ν20 and 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%. Page 176 Answers 5.2 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 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 + 0.25Pν20. Hence (1− 0.25ν20)P = 4a(4)20 + 77.25ν20 leading 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,i shows 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 − 0.3×30)P = 2.625x(1 − x30)/(1− x) + 77×30 giving P = 107.75380 or £107,753.80. Or, P = (5.25a(2)15 + 77ν 15)/(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 + 100ν 15 = 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 + 60ν7 leading 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) + 77.5203×16 where 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 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. Appendix 2 Sep 27, 2016(9:51) Answers 5.2 Page 177 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]/1.065 10 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− x 20 1− x + 70x 20 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 + 100ν 3 = 4x(1− x6)/(1− x) + 100×6 = 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 + 101.69ν 2 = 2.4(x + x2 + x3 + x4) + 101.69×4 where 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 = 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 P1 is 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 Page 178 Answers 5.2 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed is CGT for the second investor. The price P2 is 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 + P2ν 2. 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 + 105ν 14 = 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 )ν 11 ) 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 + 100ν 4 1 ) = 105.624983612 or £105.6250. (iii) We have 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%. Appendix 2 Sep 27, 2016(9:51) Answers 5.5 Page 179 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 P1 would 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. Answers to Exercises: Chapter 5 Section 5 on page 93 (exs5-2.tex) 1. If the payments are linked exactly to the inflation index, then the real yield will be the same whatever happens to inflation. However, the delay of 8 months before allowance is made for inflation means that the investor will be worse off. 2. Using (1 + iM ) = (1 + q)(1 + iR) gives (1 + 0.01) = (1 − 0.02)(1 + iR). Hence 1 + iR = 1.01/0.98 and so iR = 3/98 = 0.3061 or 3.061%. 3. Let α = 1.03 and ν = 1/1.08. Then P = 5 ∞∑ k=1 αkνk/2 = 5αν1/2 1− αν1/2 = 5α (1 + i)1/2 − α = 5.15 1.081/2 − 1.03 = 557.93 Hence £557.93. 4. Let α = 1.025 and ν = 1/1.07. Then P = 2 ∞∑ k=1 αkνk/2 = 2αν1/2 1− αν1/2 = 2α (1 + i)1/2 − α = 2.05 1.071/2 − 1.025 = 217.90 Hence £217.90. 5. (i) Issued by a government; coupons paid by government—hence virtually no risk of default in most cases. Coupon and redemption values linked to a specified inflation index, usually with a time lag. hence they provide protection against inflation. (ii) Usually, payments linked to value of inflation index some months earlier. If inflation is higher over these months, then the payments will not keep up with inflation. 6. (a) If i denotes the money rate of return, then 10(1 + i) = 11.1 and hence i = 0.11 or 11%. (b) If q denotes the rate of inflation, then 112(1 + q) = 120 and hence q = 8/112 = 0.07143 or 7.143%. (c) Using (1 + 0.6i) = (1 + q)(1 + iR) gives 1 + iR = (1 + 0.6 × 0.11) × 112/120 and hence iR = −0.005067, or −0.5067%. 7. Let α = 1.04 and ν = 1/1.07. Then P = 12 [ ν2/12 + αν14/12 + α2ν26/12 + · · ·] = 12ν1/6 [1 + αν + α2ν2 + · · ·] = 12ν1/6/ [1− αν] = 428ν1/6 = 423.20 8. The cash flow is as follows: Time 0 1 2 3 4 5 . . . Cash flow 0 800 800(1.08) 800(1.08)(1.07) 800(1.08)(1.07)α 800(1.08)(1.07)α2 . . . where α = 1.05. Let ν = 1/1.07. Hence required value is NPV = 800ν + 800(1.08)ν2 + ∞∑ k=3 800(1.08)(1.07)αk−3νk = 800ν + 864ν2 + 924.48ν3 1− αν = 41876.15 Giving an answer of £418.76. Page 180 Answers 5.5 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 9. Cash flow is as follows: Time 0 1 2 3 Cash flow, unadjusted −20 2 2 22 Inflation index, Q(t) 245.0 268.2 282.2 305.5 Cash flow, adjusted −20 2×245.0268.2 2×245.0282.2 22×245.0305.5 Hence 20 = 2× 245.0ν 268.2 + 2× 245.0ν2 282.2 + 22× 245.0ν3 305.5 = 490ν 268.2 + 490ν2 282.2 + 5390ν3 305.5 Inflation index goes from 245.0 to 305.5 in 3 years. Setting (1 + j)3 = 305.5/245.0 gives j = 0.076 as the average rate of inflation. General result (1 + iR)(1 + j) = 1 + iM where iR is the real rate, j is the inflation rate and iM is the money rate, leads to iR = 0.022 as a first estimate of the real rate of return. (Or use approximation νk ≈ 1− ki.) Trying i = 0.022 gives ν = 1/1.022 and RHS = 19.978. Trying i = 0.021 gives ν = 1/1.022 and RHS = 20.032. Interpolating gives (x− 0.021)/(0.022− 0.021) = (f (x)− f (0.021)/(f (0.022)− f (0.021)) and hence x = 0.021 + 0.001(20− 20.032)/(19.978− 20.032) = 0.0216 or 2.16%. 10. Effective real yield per annum,iR, is given by 1 + iR = 1.01252. Cash flow is: Time 0 10/12 22/12 34/12 · · · Cash flow 0 5 5× 1.03 5× 1.032 · · · Inflation Index 1 1.0210/12 1.0222/12 1.0234/12 · · · Hence, if β = 1 + iR = 1.01252, we have x = ∞∑ n=0 5× 1.03n 1.02n+10/12(1 + iR)n+10/12 = 5 (1.02β)10/12 ∞∑ n=0 1.03n 1.02nβn = 5 (1.02β)1/6 1.02β − 1.03 Hence x = 321.68 or £3.217. 11. Let νm = 1/(1 + iM ) where iM is the money rate of return. Hence 1 = ∞∑ k=1 d(1 + g)k−1νk = dν 1− (1 + g)ν = d iM − g Hence iM = d + g. Using the general result that (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflation rate and iM is the money rate, we get iR = (d + g − e)/(1 + e) as required. 12. Let α = 1.03 and β = 1.05. The cash flow is as follows: Time (in years) 0 0.75 1.75 2.75 3.75 . . . Cash flow, unadjusted −250 10 10β 10β2 10β3 . . . Inflation factor 1 α0.75 α1.75 α2.75 α3.75 . . . Hence we require: 250 = ∞∑ k=0 10βkν0.75+k α0.75+k = 10ν0.75 α0.75 ∞∑ k=0 βkνk αk = 10ν0.75 1.030.75 1 1− 1.05ν/1.03 and this leads to 0 = 25× 1.05ν + 1.030.25ν0.75 − 25× 1.03 = 26.25ν + 1.030.25ν0.75 − 25.75 or, multiplying by 1 + i, 0 = 0.5 + 1.030.25(1 + i)0.25 − 25.75i We need an initial approximation. Either, approximate (1 + i)0.25 by 1 + 0.25i. This leads to i ≈ (0.5 + 1.030.25) / (25.75− 0.25× 1.030.25) = 0.059. In fact, (1 + i)0.25 = 1 + 0.25i + α where α > 0; hence i is greater than the approximation 0.059. Or, ignoring inflation and assuming the next dividend occurs in 12 months’ time gives 250 = 10ν + 10βν2 + · · · = 10ν/(1 − βν) = 10/(1 + iM − β) = 10/(iM − 0.05). Hence iM ≈ 0.05 + 1/25 = 0.09. The approximation used implies that iM is greater than 0.09. Now use the general result that (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflation rate and iM is the money rate. This gives 1 + iR = 1.09/1.03 and hence iR ≈ 0.058. So iR is slightly more than 0.058. If i = 0.058 then 0.5 + 1.030.25(1 + i)0.25 − 25.75i = 0.0282. If i = 0.059 then 0.5 + 1.030.25(1 + i)0.25 − 25.75i = 0.0027. If i = 0.0595 then 0.5 + 1.030.25(1 + i)0.25 − 25.75i = −0.010. Linear interpolation between 0.059 and 0.0595 gives (x−0.059)/(0.0595−0.059) = (f (x)−f (0.059)/(f (0.0595)− f (0.059)) and hence x = 0.059 + 0.0005(0− 0.0027)/(−0.010− 0.0027) = 0.0591 or 5.91%. 13. (i) Next expected dividend is d1; rate of dividend growth is g; let ν = 1/(1 + i). Then P = ∑∞ k=1 d1(1 + g) k−1νk = d1ν/(1− (1 + g)ν) = d1/(i− g). (ii) Analyst I has g = 0; hence 750 = 35/i and i = 35/750 = 7/150 = 0.04667 or 4.667%. Analyst II has g = 0.1; hence 750 = 35/(i− 0.1) leading to i = 0.1 + 0.04667 = 0.14667 or 14.667%. Appendix 2 Sep 27, 2016(9:51) Answers 5.5 Page 181 14. See answer to exercise 12. Let α = 1.015 and β = 1.04. The cash flow is as follows: Time (in years) 0 0.25 1.25 2.25 3.25 . . . Cash flow, unadjusted −125 5 5β 5β2 5β3 . . . Inflation factor 1 α0.25 α1.25 α2.25 α3.25 . . . Hence we require: 125 = ∞∑ k=0 5βkν0.25+k α0.25+k = 5ν0.25 α0.25 ∞∑ k=0 βkνk αk = 5ν0.25 1.0150.25 1 1− 1.04ν/1.015 and this leads to 0 = 26ν + 1.0150.75ν0.25 − 25.375 or, multiplying by 1 + i, 0 = 0.625 + 1.0150.75(1 + i)0.75 − 25.375i For an initial approximation, either of the methods in the answer to exercise 12 can be used. Using the first method means we approximate (1 + i)0.75 by 1 + 0.75i. This leads to 0 = 0.625 + 1.0150.75(1 + 0.75i) − 25.375i and hence i = ( 0.625 + 1.0150.75 ) / ( 25.375− 0.75× 1.0150.75) = 0.0664. As in the answer to exercise 12, the value of i will be larger than this approximation. If i = 0.0664, then 0.625 + 1.0150.75(1 + i)0.75 − 25.375i = 0.00128. If i = 0.067, then 0.625 + 1.0150.75(1 + i)0.75 − 25.375i = −0.0135. Linear interpolation between 0.0664 and 0.067 gives (x− 0.0664)/(0.067− 0.0664) = (f (x)− f (0.0664)/(f (0.067)− f (0.0664)) and hence x = 0.0664 + 0.0006(0− 0.00128)/(−0.0135− 0.00128) = 0.0665 or 6.65%. 15. (i) The cash flow sequence is: Time (in years) 0 4/12 10/12 16/12 . . . Cash flow −P d1 d1(1 + g)1/2 d1(1 + g) . . . Hence P = ∞∑ k=0 d1(1 + g)k/2 (1 + i)(4+6k)/12 = d1 (1 + i)4/12 ∞∑ k=0 ( 1 + g 1 + i )k/2 = d1(1 + i)1/6 (1 + i)1/2 − (1 + g)1/2 (ii) The cash flow sequence is Time (in years) 0 2/12 8/12 14/12 20/12 . . . Cash flow −£18 d1 d1(1 + g)1/2 d1(1 + g) d1(1 + g)3/2 . . . Hence 18 = d1ν 2/12 1− ν1/2(1 + g)1/2 = 0.5(1 + i)1/3 (1 + i)1/2 − 1.041/2 or 18(1 + i)1/2 − 18× 1.041/2 = 0.5(1 + i)1/3 Initial approximation: replace (1 + i)1/2 by 1 + i/2 and (1 + i)1/3 by 1 + i/3. This leads to i = 0.096, rounding down. As in the answer to exercise 12, the value of i will be larger than this approximation. If i = 0.096, then 18(1 + i)1/2 − 18× 1.041/2 − 0.5(1 + i)1/3 = −0.0278. If i = 0.1, then 18(1 + i)1/2 − 18× 1.041/2 − 0.5(1 + i)1/3 = 0.0059. Hence answer lies between 9.6% and 10%, or 10% to nearest 1%. 16. (i) Are a share in the ownership of a company. Can be readily bought and sold. From point of view of investor: receive dividends and potential growth in share price. Size of dividends is not guaranteed. No fixed redemption time. Shareholders last in line to receive payment if company is liquidated. Shareholders can vote in AGM of company. (ii) Denote the current price by P and the next dividend by d. Let α = 1.02, β = 1.03 and ν = 1/(1 + i) where i = 0.05. Then the cash flow is as follows: Time (in years) 0 0.5 1.5 2.5 3.5 . . . Cash flow −P d dβ dβ2 dβ3 . . . Inflation index 1 α0.5 α1.5 α2.5 α3.5 . . . Hence P = d ∞∑ k=0 βkνk+0.5 αk+0.5 = dν0.5 α0.5 1 1− βν/α = dν0.5α0.5 α− βν = d(1 + i)0.5α0.5 α(1 + i)− β and hence d P = 1.02× 1.05− 1.03 1.050.5 × 1.020.5 = 0.0396 Page 182 Answers 5.5 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 17. (i) We need x where 10,000 = xa5 ,0.0625. Hence x = 10,000i/(1 − ν5) = 625/(1 − 1/1.06255) = 2390.13 or £2,390.13. (ii) Investment A. Let a = 1.035 × 1.045 × 1.055 × 1.065 × 1.075. Then money interest earned is £10,000a. The January, 1993 value of this amount is £10,000a × 240/275.6. Hence (1 + iR)5 = a × 240/275.6 and so iR = (240.0a/275.6)1/5 − 1 = 0.0261 or 2.61%. Investment B. Redemption proceeds in January 1998 are 10,000 × (274.0/237.6) × 1.02755. Hence, if iM denotes the effective money rate of return per annum, then (1 + iM )5 = (274.0/237.6)× 1.02755. Of course, we want the real rate of return, iR where (1 + iR)5 = (274.0/237.6) × 1.02755 × 240/275.6. Hence iR = (274.0× 240.0)/(237.6× 275.6)1/5 × 1.0275− 1 = 0.0284 or 2.84%. (iii) Investment C. The cash flows are as follows (where x is given in part (i)): Date 1/93 1/94 1/95 1/96 1/97 1/98 Cash flow −10,000 x x x x x Inflation index 240 250 264.4 266.6 270.4 275.6 Hence the real rate of return, iR, satisfies 10,000 = x ( 240 250 νR + 240 264.4 ν2R + 240 266.6 ν3R + 240 270.4 ν4R + 240 275.6 ν5R ) = 240xνR ( 1 250 + νR 264.4 + ν2R 266.6 + ν3R 270.4 + ν4R 275.6 ) Question only asks which is the greatest real rate of return. So we need to decide if iR > 0.0284. Substituting iR = 0.0284 gives RHS ≈ 9967. Hence real yield from Investment C is less than 2.84% and the answer is Investment B. 18. (i) Now g = 6/100 = 0.06 and t1 = 0.25. Hence i = 0.08 > (1− t1)g and so there is a capital gain. The cash flow is as follows: Time (in years) 0 0.75 1.75 · · · 8.75 9.75 Cash flow (before tax) −P 6 6 · · · 6 106 Cash flow (after tax) −P 4.5 4.5 · · · 4.5 104.5− 0.25(100− P ) = 79.5 + 0.25P Let x = 1/1.08. Hence: P = 4.5[x0.75 + x1.75 + · · · + x9.75] + (75 + 0.25P )x9.75 = 4.5×0.75[1 + x + · · · + x9] + (75 + 0.25P )x9.75 and hence P (1− 0.25×9.75) = 4.5×0.75 ( 1− x10 1− x ) + 75×9.75 and so P = 75.06. (ii) Using (1 + iM ) = (1 + iR)(1 + q) gives 1.08 = 1.03(1 + q) and so q = 0.4854 or 4.854%. 19. Work in units of £100. Let α = 1.03. The cash flow is as follows: Date 0 0.5 1 1.5 · · · 25 Payments, gross −99 4 4 4 · · · 4 + 110 The tax payments on the coupons will be 2 at time 1.25, 2 at time 2.25, . . . , and 2 at time 25.25. The CGT will be 3.3 at time 25.25. Let iM denote the money rate of return and let νM = 1/(1 + iM ). Then 99 = 8a (2) 25 ,iM + 110ν25M − 2ν0.25M a25 ,iM − 3.3ν25.25M = [ 8 i i(2) − 2ν0.25M ] a25 ,iM + 110ν 25 M − 3.3ν25.25M . We need an approximate answer for iM . Note that the coupon is 6 per year after tax and the capital gain is 7.7 after tax. Using method (d) in paragraph 4.3 gives iM ≈ (6 + 7.7/25)/99 = 0.0637 or 6.4%. If iM = 0.06, then [ 8 i i(2) − 2ν0.25M ] a25 ,iM + 110ν 25 M − 3.3ν25.25M = 103.453. If iM = 0.065, then [ 8 i i(2) − 2ν0.25M ] a25 ,iM + 110ν 25 M − 3.3ν25.25M = 97.241. Using linear interpolation gives iM = 0.06 + 0.005× (99− 103.453)/(97.241− 103.453) = 0.0636. Now 1 + iM = (1 + q)(1 + iR) and hence iR = 1.0636/1.03− 1 = 0.032 or 3.2%. (ii) If tax was collected later, then the real rate would increase. 20. See answer to exercise 12. Let α = 1.03 and β = 1.05. The cash flow is as follows: Time (in years) 0 0.667 1.667 2.667 3.667 . . . Cash flow, unadjusted −21.5 1.1 1.1β 1.1β2 1.1β3 . . . Inflation factor 1 α0.667 α1.667 α2.667 α3.667 . . . Appendix 2 Sep 27, 2016(9:51) Answers 5.5 Page 183 Hence we require: 21.5 = ∞∑ k=0 1.1βkν0.667+k α0.667+k = 1.1ν0.667 α0.667 ∞∑ k=0 βkνk αk = 1.1ν0.667 1.030.667 1 1− 1.05ν/1.03 and this leads to 0 = 22.575ν + 1.1× 1.030.333ν0.667 − 22.145 or, multiplying by 1 + i, 0 = 0.43 + 1.1× 1.030.333(1 + i)0.333 − 22.145i For an initial approximation, either of the methods in the answer to exercise 12 can be used. Using the first method means we approximate (1 + i)0.333 by 1 + 0.333i. This leads to 0 = 0.43 + 1.1× 1.030.333(1 + 0.333i)− 22.145i and hence i = ( 0.43 + 1.1× 1.030.333) / (22.145− 0.333× 1.1× 1.030.333) = 0.070, rounded down. As in the answer to exercise 12, the value of i will be larger than this approximation. If i = 0.07, then 0.43 + 1.1 × 1.030.333(1 + i)0.333 − 22.145i = 0.0160. If i = 0.071, then 0.625 + 1.0150.75(1 + i)0.75 − 25.375i = −0.00575. Linear interpolation between 0.07 and 0.071 gives (x − 0.07)/(0.071 − 0.07) = (f (x) − f (0.07)/(f (0.071) − f (0.07)) and hence x = 0.07 + 0.001(0 − 0.016)/(−0.00575 − 0.016) = 0.0707 or 7.07%. 21. The cash flows were as follows: Date 1/6/2000 1/12/2000 1/6/2001 1/12/2001 1/6/2002 Payments, not indexed −94 1.5 1.5 1.5 1.5 + 100 Payments, indexed −94 1.5× 102/100 1.5× 107/100 1.5× 111/100 (1.5 + 100)× 113/100 −94 1.53 1.605 1.665 1.695 + 113 (ii)(a) Before indexing, £94 grows to £113. But the purchase price of £94 is indexed to £94 × 118/102. Hence the CGT is £0.35× (113− 94× 118/102) = 1.489. (ii)(b) Let i denote the yield and let ν = 1/(1 + i). Then 94 = 0.75 × (1.53ν0.5 + 1.605ν + 1.665ν1.5 + 1.695ν2) + (113− 1.489)ν2. This leads to 376 = 4.59ν0.5 + 4.815ν + 4.995ν1.5 + 451.129ν2. Transforming into an equation in i gives 376(1 + i)2 = 4.59(1 + i)1.5 + 4.815(1 + i) + 4.995(1 + i)0.5 + 451.129 or 376i2 + 747.185i− 79.944− 4.59(1 + i)1.5 − 4.995(1 + i)0.5 = 0. Using the usual approximation of (1 + i)0.5 ≈ 1 + 0.5i, etc gives 376i2 + 737.8i − 89.5 = 0. The usual formula for the solution of a quadratic gives i = 0.11. If i = 0.11 then 376i2 + 747.185i− 79.944− 4.59(1 + i)1.5 − 4.995(1 + i)0.5 = −3.8344. If i = 0.12 then 376i2 + 747.185i− 79.944− 4.59(1 + i)1.5 − 4.995(1 + i)0.5 = 4.40588. Linear interpolation gives i = 0.11 + 0.01× (3.8344)/(4.40588 + 3.8344) = 0.1147 or 11.47%. 22. Work in units of £1,000. The cash flow is as follows: Date 0 1 2 3 Payments, not indexed −25 10 10 10 Retail Price Index 170.7 183.3 191.0 200.9 Payments, indexed to time t = 0 −25 10× 170.7/183.3 10× 170.7/191.0 10× 170.7/200.9 (i)(a) Let iR denote the real rate of return and let νR = 1/(1 + iR). Then 25 = 1707 183.3 νR + 1707 191.0 ν2R + 1707 200.9 ν3R or 25(1 + iR) 3 − 1707 183.3 (1 + iR)2 − 1707191.0(1 + iR)− 1707 200.9 = 0 Using the usual approximation of (1 + iR)3 ≈ 1 + 3iR, etc gives 1.7465 = 47.5iR and hence iR = 0.037. If i = 0.037 then 25(1 + iR)3 − 1707183.3 (1 + iR)2 − 1707191.0 (1 + iR)− 1707200.9 = 0.0998. If i = 0.03 then 25(1 + iR)3 − 1707183.3 (1 + iR)2 − 1707191.0 (1 + iR)− 1707200.9 = −0.2636. Using linear interpolation gives iR = 0.03 + 0.007× (0 + 0.2636)/(0.0998 + 0.2636) = 0.0351, or 3.51%. (b) Let iM denote the money rate of return. Then 25 = 10a3 ,iM . Using tables shows that a3 ,0.095 = 2.508907 and a3 ,0.1 = 2.486852. Linear interpolation gives iM = 0.095+0.005×(2.5−2.508907)/(2.486852−2.508907) = 0.097 or 9.7%. (ii) Let e denote the average rate of inflation, then (1 + e)3 = 200.9/170.7 and hence e = 0.0558 or 5.58%. The answers to part (i) suggest 1 + e = (1 + iM )/(1 + iR) = 1.097/1.0351 = 1.0598 or 5.98%. They are not equal because the inflation index does not increase by a constant rate. 23. (i) See exercise 16. (ii) Let α = 1.06 and ν = 1/(1 + i) = 1/1.09. The cash flow is as follows: Date 1 2 3 4 5 Dividends 25× 1.02 25× 1.02× 1.04 25× 1.02× 1.04α 25× 1.02× 1.04α2 · · · Hence NPV = 25.5ν + 26.52ν2 + 26.52αν3 + 26.52α2ν4 + · · · = 25.5 1.09 + 26.52ν2 1− αν = 25.5× 0.03 + 26.52 1.09× 0.03 = 834.40 (iii) The cash flow is as follows: Page 184 Answers 5.5 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed Date 0 1 2 Cash flow −820 25× 1.02 25× 1.02× 1.04 + 900 Inflation Index 1 1.03 1.03× 1.035 Hence 820 = 25.5ν 1.03 + 926.3925ν2 1.06605 or 874.161 = 26.3925ν + 926.52ν2 Solving the quadratic give ν = 0.957195 and hence i = 0.04472 or 4.47%. 24. (i) Let νM = 1/(1 + iM ) denote the money values. Then 107 = 7.5a(2)9 ,iM + 100ν 9 We need to find iM . Using method (d) in paragraph 4.3 gives iM ≈ [ 7.5 + (100− 107)/9] 107 = 0.063. Try iM = 0.065. Then RHS = 7.5× 1.015994× 6.656104 + 100× 0.567353 = 107.4545. Try iM = 0.07. Then RHS = 7.5× 1.017204× 6.515232 + 100× 0.543934 = 104.0983. Interpolation gives (iM − 0.065)/0.005 = (107− 107.4545)/(104.0983− 107.4545) and hence iM = 0.0657. Using the general result that (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflation rate and iM is the money rate, we get iR = 1.0657/1.025− 1 giving 3.97%. (ii) let α0 = 159.5/69.5 and α = 1.0251/2. The cash flow is as follows: Date 9/10/97 8/4/98 8/10/98 · · · 8/4/06 9/10/06 Time 0 1/2 1 · · · 81/2 9 Cash flow 0 α0 α0α · · · α0α16 α0α17 Hence if iM denotes the money rate of return with νM = 1/(1 + iM ), we have NPV = 18∑ k=1 α0α k−1νk/2M + 100α0α 17ν9M Now (1 + iR)(1 + e) = 1 + iM gives νR = (1 + e)νM = α2νM and so NPV = 18∑ k=1 α0 α ν k/2 R + 100 α0 α ν9R = α0 α (2a(2)9 ,iR + 100ν 9 R) = 2.2668(2a (2) 9 ,iR + 100ν9R) (iii) We want to find iR with 203 = 2.2668(2a (2) 9 ,iR + 100ν9R) or 44.7767 = a(2)9 ,iR + 50ν 9 R = iR i(2)R a9 ,iR + 50ν 9 R Using method (d) in paragraph 4.3 gives i ≈ [1 + (50− 44.7767)/9] /44.7767 = 0.0353. Using tables, iR = 0.03 gives RHS = 46.1649 and iR = 0.035 gives RHS = 44.3602 Interpolating gives (x − 0.03)/(0.035 − 0.03) = (f (x) − f (0.03))/(f (0.035) − f (0.03)). Hence x = 0.03 + 0.005(44.7767− 46.1649)/(44.3602− 46.1649) = 0.034 or 3.4%. 25. (i) Payments guaranteed by a government. Usually the values of the coupon and redemption payments are linked to some inflation index, with a time lag. Guaranteed real return if held to maturity—except for effect of time lag in indexation. The security of government payments tends to lead to a low volatility of return and a low expected return compared to other investments. (ii) Let α = 1.0250.5, the 6-month inflation factor and let x = 1/1.015, the 6-month discount factor. Cash flow is: Date 1 July 03 1 Jan 04 1 July 04 1 Jan 05 · · · 1 Jan 09 1 July 09 Cash flow, not indexed −P 1 1 1 · · · 1 1 + 100 Inflation index, 8 months earlier 113.2 113.8 113.8α 113.8α2 · · · 113.8α10 113.8α11 Cash flow, indexed, before tax −P 113.8110 113.8α110 113.8α 2 110 · · · 113.8α 10 110 101×113.8 110 Cash flow, indexed, after tax −P 0.8×113.8110 0.8×113.8α110 0.8×113.8α 2 110 · · · 0.8×113.8α 10 110 (0.8+100)×113.8α11 110 Inflation index 113.8α2/6 113.8α8/6 113.8α14/6 113.8α20/6 · · · 113.8α68/6 113.8α74/6 Equivalent Inflation index 1 α α2 α3 · · · α11 α12 Hence P = 0.8× 113.8x 110α + 0.8× 113.8αx2 110α2 + 0.8× 113.8α2×3 110α3 + · · · + 0.8× 113.8α 10×11 110α11 + 100.8× 113.8α11×12 110α12 = 0.8× 113.8x 110α + 0.8× 113.8×2 110α + 0.8× 113.8×3 110α + · · · + 0.8× 113.8x 11 110α + 100.8× 113.8×12 110α = 113.8x 110α [ 0.8(1 + x + x2 + · · · + x11) + 100×11] Appendix 2 Sep 27, 2016(9:51) Answers 5.5 Page 185 = 113.8x 110α [ 0.8 1− x12 1− x + 100x 11 ] = 113.8 110α [ 0.8 1− x12 0.015 + 100×12 ] = 94.383 26. (i) The coupon is £3/2 and this must be adjusted by the inflation index at 7/99 (eight months before 3/00). Hence the answer is 3 2 × 126.7 110.5 = 1.72 (ii) Let x = 1.04 and α = 127.4/110.5 which is the inflation factor at 9/99 (used for 5/00, eight months later). The cash flow is as follows: Time (in years) 0 0.5 1 1.5 2 2.5 Date 9/99 3/00 9/00 3/01 9/01 3/02 Inflation factor 126.7110.5 αx 4/12 αx10/12 αx16/12 αx22/12 Cash flow −111 32 126.7110.5 1.5αx4/12 1.5αx10/12 1.5αx16/12 101.5αx22/12 Hence the yield, i satisfies νM = 1/(1 + i) where ν is given by 111 = 3 2 ( 126.7 110.5 ν 1/2 M + αx 4/12νM + αx10/12ν 3/2 M + αx 16/12ν2M + αx 22/12ν 5/2 M ) + 100αx22/12ν5/2M We want the real effective annual yield. Recall (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflation rate and iM is the money rate.Hence νR = (1 + e)νM = xνM . Hence 111 = 3 2 ( 126.7 110.5 x−1/2ν1/2R + αx −8/12νR + αx−8/12ν 3/2 R + αx −8/12ν2R + αx −8/12ν5/2R ) + 100αx−8/12ν5/2R Let γ = ν1/2R and define i1 by γ = 1/(1 + i1). Then 111 = 3 2 ( 126.7 110.5 x−1/2γ + αx−8/12γ2 + αx−8/12γ3 + αx−8/12γ4 + αx−8/12γ5 ) + 100αx−8/12γ5 = 3 2 ( 126.7 110.5 x−1/2 − αx−8/12 ) γ + 3 2 αx−8/12a5 ,i1 + 100αx −8/12γ5 = 0.0017γ + 1.6848a5 ,i1 + 112.3186γ 5 This must be solved numerically; so we need a rough starting value. Now an answer of 0.03 corresponds to i1 = 1.031/2 − 1 = 0.014. So try i1 = 0.015, then RHS = 112.320 (Use tables!) So try i1 = 0.020, then RHS = 109.673 Hence interpolation gives (x − 0.015)/(0.020 − 0.015) = (f (x) − f (0.015)/(f (0.020) − f (0.015)) and hence x = 0.015 + 0.005(111− 112.320)/(109.673− 112.320) = 0.0175. Hence νR − 1 = 1/γ2 − 1 = (1 + i1)2 − 1 = 0.0353 or 3.53%. (iii) Increasing 110.5 means that γ must increase. Hence answer must decrease. 27. (i) Let α = 1.071/2 and β = 206/200. The cash flow is as follows: Date 5/97 11/97 5/98 11/98 5/99 · · · 11/11 5/12 Serial no. 0 1 2 3 4 · · · 29 30 Inflation index 200 206 206α 206α2 206α3 · · · 206α28 206α29 Payments before adjustment −P 2 2 2 2 · · · 2 102 Payments after adjustment −P 2β 2βα 2βα2 2βα3 · · · 2βα28 102βα29 Let iM denote the effective money rate of return and νM = 1/(1 + iM ). Let iR denote the effective real rate of return and νR = 1/(1 + iR). Hence (i + iR)(1.07) = i + iM and α2νM = νR. Also νR = 1.0152. P = 30∑ k=1 2βαk−1νk/2M + 100βα 29ν15M = 2β α 30∑ k=1 ν k/2 R + 100β α ν15R P = 2β α a30 ,0.015 + 100β α 1 1.01530 = β α ( 2a30 ,0.015 + 100 1.01530 ) (b) Hence P = 111.53. (ii) Now let γ = 1.051/2. The cash flow is now as follows (where P = 111.53): Page 186 Answers 5.5 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed Date 5/97 11/97 5/98 11/98 5/99 · · · 11/11 5/12 Serial no. 0 1 2 3 4 · · · 29 30 Inflation index 8 months ago 200 206 206γ 206γ2 206γ3 · · · 206γ28 206γ29 Payments before adjustment −P 2 2 2 2 · · · 2 102 Payments after adjustment −P 2β 2βγ 2βγ2 2βγ3 · · · 2βγ28 102βγ29 The redemption money is 100βγ29 = 208.97. The RPI at 3/97 is 206 and hence 206γ1/3 at 5/97, the purchase date. The RPI at 5/12, 15 years later, is 206γ1/3γ30. Hence the purchase price revalued according to the RPI to the redemption date is 111.53 × γ30 = 231.86. Hence there is no capital gains tax. Let iR denote the real effective yield per 6 months. Hence 111.53 = β γ ( 2a30 ,iR + 100 (1 + iR)30 ) = 1.03√ 1.05 ( 2a30 ,iR + 100 (1 + iR)30 ) and hence we need to solve 55.478 = a30 ,iR + 50 (1 + iR)30 Using method (d) in paragraph 4.3 gives iR ≈ [ 1 + (50− 55.478)/30] /55.478 = 0.015. if iR = 0.015, then RHS = 56.004. If iR = 0.02, then RHS = 50.0. Interpolating gives (x − 0.015)/(0.02 − 0.015) = (f (x) − f (0.015))/(f (0.02) − f (0.015)) and so x = 0.015 + 0.005(55.478− 56.004)/(50.0− 56.004) = 0.0154 or 1.54%. 28. (i) Let α denote the rate of inflation and let ν = 1/1.06. Then the cash flow is as follows: Time 0 1 2 3 . . . Cash flow in pence −260 12α 12α2 12α3 . . . Hence 260 = 12(αν + α2ν2 + α3ν3 + · · ·) = 12αν/(1− αν). Hence α = (65× 1.06)/68 = 1.0132353 or 1.324%. (ii) Let α = 1.03 and ν = 1/(1 + i) where i is the rate of return. Then 260 = 12(αν + α2ν2 + · · · + α12ν12) + 500ν12. We want the real rate of return, i′, where 1.03(1 + i′) = 1 + i. So if x = 1/(1 + i′) = 1.03/(1 + i) = αν then 260 = 12x(1 − x12)/(1 − x) + 500×12/1.0312. Let f (i′) = 260 − 12x(1 − x12)/(1 − x) − 500×12/1.0312. We have f (0.06) = −14.888284 and f (0.07) = 8.977237. Hence i′ ≈ 0.06 − 0.01f (0.06)/(f (0.07) − f (0.06)) = 0.066238 or 6.6% per annum. (It is 6.61%.) 29. (i) The price is given by P = 0.3× 1.05 1.09 + 0.3× 1.05 2 1.092 + 0.3× 1.05 3 1.093 + · · · = 0.3× 1.05 1.09− 1.05 = 63 8 = 7.875 or £7.875. (ii)(a) The return from the share is higher; hence P will be higher. (b) If profits increase with inflation and dividends are a constant percentage of profits, then the price may stay much the same because the investor will expect the same real rate of return. However, often the rate of inflation is just (incorrectly) added—hence 1.05 becomes 1.05 + I and 1.09 becomes 1.09 + I where I is the increase in the inflation rate. Hence the new P is 0.3× (1.05 + I) 0.04 which is greater than the old P . (c) The risk the share will not actually give the returns specified is higher—hence the price P will be lower. 30. Let α = 1.03, β = 1.05, γ = 1.06 and ν = 1/1.08. (i) Required answer is 3500 [ αν + αβν2(1 + γν + γ2ν2 + · · ·)] = 3500 [αν + αβν2/(1− γν)] = 178581.01852 or £1,785.81. (ii) Let i denote the required real rate of return and let ν = 1/(1 + i). Date: 1/1/2012 31/12/2012 31/12/2013 31/12/2014 Cash flow (in pounds): -1,720 35α 35αβ 35αβγ + 1,800 Inflation Index: 110.0 112.3 113.2 113.8 Equation for ν is 1,720 110.0 = 35α 112.3 ν + 35αβ 113.2 ν2 + 35αβγ + 1,800 113.8 ν3 or 172 = 3.53116652ν + 3.67824647ν2 + 177.86783963ν3. So let f (i) = 3.53116652ν + 3.67824647ν2 + 177.86783963ν3− 172 where ν = 1/(1 + i). Now the dividend in 2012 gives a real rate of return of (35α/1720)× (110/112.3) or 2%. Over the three years, there is also some capital gain. So try i = 0.03. Then f (0.03) = −2.330313; so it is less than 3%. Try 2.5%: f (0.025) = 0.114020. So the required value lies satisfies 0.025 < i < 0.03. Using linear interpolation gives i ≈ 0.025 + 0.005 × 0.114020/(2.330313 + 0.114020) = 0.025233 or 2.52%. Appendix 2 Sep 27, 2016(9:51) Answers 6.2 Page 187 31. Effective interest rate is i where 1 + i = 1.022 = 1.0404; hence i = 0.0404. Let ν = 1/(1 + i)1/12 = 1/1.04041/12. Let x be the required amount. Then we need 1.0520x to equal the cost of the annuity which is 10,000a(12)25 ,i = 2500 3 ν 1− ν300 1− ν = 2500 3 1− 1/1.040425 1.04041/12 − 1 = 158422.3048754 or £158,422.30 and hence x = 158,422.30/1.0520 = 59707.69872255 or £59,707.70. (ii) Now x grows to 1.0520x in 20 years. Adjusting for inflation, x grows to 1.0520x × (143/340) in 20 years. If j denotes the annual effective real return, then (1 + j)20 = 1.0520x× (143/340) and hence j = 0.0055002 or 0.55%. (iii) Capital gain is x(1.0520− 1) and hence the tax is 0.25x(1.0520− 1) and the new payout is x (3× 1.0520 + 1) /4. Let j′ denote the new real rate of return; then (1+j′)20 = (143/340)×(3× 1.0520 + 1) /4. Hence j′ = −0.002977411 or -0.2977%. (iv) The tax is paid on the money capital gain. After deducting the tax, the remaining return is less than inflation which leads to a negative real return. 32. Adjusting for inflation, £95 grows to £100 × 220/222. Let i denote the required rate. Then 95(1 + i)91/365 = 100× 220/222. Hence i = ( (100× 220)/(95× 222) )365/91 − 1 = 0.18463896 or 18.5%. 33. Investment A. After tax, the interest payments are £0.6 million per year. This is a net interest rate of 6%. Investment B. After 10 years, we have 1.110 million. The capital gains are 1.110 − 1. Amount returned after tax is 1, 110−0.4(1 = .110−1) = 0.6×1.110+0.4. So if iB denotes the net rate of return, we have (1+iB)10 = 0.6×1.110+0.4 which leads to iB = 0.06940531 or 6.94%. Investment C. Capital gains are 1.110−1.0410. Hence, amount returned is 0.6×1.110 +0.4×1.0410. So if iC denotes the net rate of return, we have (1 + iC)10 = 0.6× 1.110 + 0.4× 1.0410 which leads to iC = 0.079469 or 7.95%. (ii) Gross rate of return is 10% for all 3 investments. Clearly, net rate for Investment C is greater than that for Investment B because the capital gains are less and hence the tax is less. The net rate for Investment B is greater than that for Investment A because the tax is deferred until the end for Investment B and so greater compounding is achieved. 34. Let i be the effective money yield per annum and let ν = 1/(1+i); let α = 1.06. Then we have 175 = 6ν1/2 +6αν3/2 + 6α2ν5/2 + · · · = 6ν1/2 [1 + αν + (αν)2 + · · ·] = 6ν1/2/(1− αν) which gives the equation 185.5ν + 6ν1/2 − 175 = 0 or 175i − 6(1 + i)1/2 − 10.5 = 0. Approximating (1 + i)1/2 by 1 + i/2 gives i = 0.096. Trying i = 0.095 gives LHS = −0.1535349 and trying i = 0.096 gives LHS = 0.0185989. Linear interpolation gives i = 0.09589 or 9.589%. Hence if iR denotes the real rate, then 1 + iR = 1.09589/1.04 leading to iR = 0.0537 or 5.37%. 35. (i) See §4.1 on page 73. (ii) If dividends are assumed to grow by a constant proportion, g, each year, then the cash flow is: Time 0 1 2 3 . . . Cash flow, before tax −P d d(1 + g) d(1 + g)2 . . . Cash flow, after tax −P (1− t)d (1− t)d(1 + g) (1− t)d(1 + g)2 . . . and so the rate of return r of the share given by: P = ∞∑ k=1 (1− t)d(1 + g)k−1 (1 + r)k = (1− t)d r − g (iii) So d = 6, g = 0.01, r = 0.06 and t = 0.2. Hence P = 96p. (iv) The required return, r would increase; hence P would decrease. (v) Cost on 1 August 2014 is £960. On 1 August 2015, receives dividend income £60 gross or £48 net and sells shares for £1200 with CGT of£60. Hence £960 grows to £1,188. Let r denote the net real rate of return; then 960(1 + r)/123 = 1,188/126. Hence 1 + r = 123× 1,188 126× 960 = 41× 33 14× 80 = 1.208036 or 20.80%. Answers to Exercises: Chapter 6 Section 2 on page 107 (exs6-1.tex) 1. Let f1,2 denote the required value. Then 1.08(1 + f1,2)2 = 1.0953. Hence f1,2 = √ 1.0953/1.08 − 1 = 0.10258 or 10.26%. 2. Let f1,2 denote the answer. Then 1.045(1 + f1,2)2 = 1.0553. Hence f1,2 = √ 1.0553/1.045 − 1 = 0.060036 or 6.0036%. 3. For the par yield, we want the coupon rate, r which ensures that the bond price equals the face value. Hence we want: Time 0 1 2 3 Cash Flow −1 r r 1 + r Hence 1 = r 1 + y1 + r (1 + y2)2 + 1 + r (1 + y3)2 Page 188 Answers 6.2 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed = r ( 1 1.06 + 1 1.06× 1.065 + 1 1.06× 1.065× 1.07 ) + 1 1.06× 1.065× 1.07 and hence r = 1.06× 1.065× 1.07− 1 1.065× 1.07 + 1.07 + 1 = 0.06478 or 6.478%. 4. Let r denote the required answer. Then 1 = r ( 1 1.06 + 1 1.0652 + 1 1.06× 1.0662 ) + 1 1.06× 1.0662 and hence r = 1.06× 1.0662 − 1 1.0662 + 1.06× 1.0662/1.0652 + 1 = 0.06395 or 6.395%. 5. (i) Ft,1 = ln(1 + ft,1) = ln 1.04 = 0.39221. (ii) Ft = limk→0 Ft,k = limk→0 ln(1 + ft,k) where Ft,k is the forward force of interest of an investment that starts at time t and lasts for k years. 6. Let yk denote the k-year spot rate of interest. Then “the two-year par yield at time t = 0 is 4.15%” implies 1 = 0.0415 [ 1 1 + y1 + 1 (1 + y2)2 ] + 1 (1 + y2)2 The other condition implies 105.4 = 8 1 + y1 + 106 (1 + y2)2 Thus if a = 1/(1 + y1) and b = 1/(1 + y2)2 we have the 2 simultaneous equations: 1 = 0.0415a + 1.0415b 105.4 = 8a + 106b This leads to (106 − 8 × 1.0415/0.0415)b = 105.4 − 8/0.0415 and hence b = 0.92192. Similarly, (106 × 0.0415/1.0415− 8)a = 106/1.0415− 105.4 and hence a = 0.9596. So y1 = 1/a− 1 = 0.0421, or 4.21%, and y2 = √ 1/b− 1 = 0.0415, or 4.15%. 7. We are given that yn = 0.0225 + 0.0025n for n = 1, 2, . . . , 5. (a) We want f3,2. But (1 + y5)5 = (1 + y3)3(1 + f3,2)2. Hence (1 + f3,2)2 = 1.0355/1.033. Hence f3,2 = 0.04255 or 4.255%. (b) Let iP denote the 4-year par yield. Then 1 = iP [ 1 1 + y1 + 1 (1 + y2)2 + 1 (1 + y3)3 + 1 (1 + y4)4 ] + 1 (1 + y4)4 and hence 1 = 3.717853iP + 0.8799 and hence iP = 0.0323 or 3.23%. (c) Par yield bond: Time 0 1 2 3 4 Cash flow for par yield bond −100 3.23 3.23 3.23 100 + 3.23 Cash flow for other bond, bond B −PB 3.5 3.5 3.5 100 + 3.5 The gross redemption yield of the par yield bond is clearly the same as the par yield; this is iP = 0.0323. Writing this as an equation: 100 = 3.23 1 + iP + 3.23 (1 + iP )2 + 3.23 (1 + iP )3 + 103.23 (1 + iP )4 = 3.23 1 + y1 + 3.23 (1 + y2)2 + 3.23 (1 + y3)3 + 103.23 (1 + y4)4 Let iB denote the gross redemption yield and PB denote the price of the other bond. Then PB = 3.5 1 + iB + 3.5 (1 + iB)2 + 3.5 (1 + iB)3 + 103.5 (1 + iB)4 = 3.5 1 + y1 + 3.5 (1 + y2)2 + 3.5 (1 + y3)3 + 103.5 (1 + y4)4 Now [ 3.5 1 + y1 + 3.5 (1 + y2)2 + 3.5 (1 + y3)3 + 103.5 (1 + y4)4 ] − [ 3.5 1 + iP + 3.5 (1 + iP )2 + 3.5 (1 + iP )3 + 103.5 (1 + iP )4 ] = 3.5 3.23 [ 100 (1 + iP )4 − 100 (1 + y4)4 ] + 100 [ 1 (1 + y4)4 − 1 (1 + iP )4 ] = [ 3.5× 100 3.23 − 100 ] [ 1 (1 + iP )4 − 1 (1 + y4)4 ] > 0 Hence PB > [ 3.5 1 + iP + 3.5 (1 + iP )2 + 3.5 (1 + iP )3 + 103.5 (1 + iP )4 ] and hence iB < iP . Appendix 2 Sep 27, 2016(9:51) Answers 6.2 Page 189 8. (i) The n-year spot rate of interest is the per annum yield to maturity of a zero coupon bond with n years to maturity. (ii) Let f3,2 denote the answer. Then 1.0755 = 1.063(1 + f3,2)2 gives f3,2 = √ 1.0755/1.063− 1 = 0.09790 or 9.79%. (iii) Let r denote the 6-year par yield. Then 1 = r ( 1 1.04 + 1 1.052 + 1 1.063 + 1 1.074 + 1 1.0755 + 1 1.086 ) + 1 1.086 and hence r = 1.086 − 1 1.086(1/1.04 + 1/1.052 + 1/1.063 + 1/1.074 + 1/1.0755) + 1 = 0.07708 or 7.708%. 9. (i)(a)We have A(8) = exp (∫ 8 0 δ(t) dt ) = exp(0.32 + 0.032) = 1.421909. (b) We have A(9) = exp (∫ 9 0 δ(t) dt ) = exp(0.36 + 0.0405) = exp(0.4005) = 1.492571. (c) We need exp( ∫ 9 8 δ(t) dt) = A(9) A(8) = exp(0.4005)/ exp(0.352) = 1.049695. (ii)(a) We need y8 with (1 + y8)8 = A(8) = exp(0.352) and hence y8 = exp(0.044) − 1 = 0.044982 or 4.4982%. (b) We need y9 with (1 + y9)9 = A(9) = exp(0.4005) and hence y9 = exp(0.0445) − 1 = 0.045505 or 4.5505%. (c) We want f8,1 where (1 + f8,1)(1 + y8)8 = (1 + y9)9 and hence f8,1 = exp(0.4005)/ exp(0.352) = exp(0.0485) = 1.0496954. Hence f8,1 = 0.0496954 or 4.96954%. 10. (i) Let ν = 1/(1 + i) where i is the gross redemption yield p.a. Then 98 = 7ν + 7ν2 + 112ν3. The usual approximation is i ≈ (7 + 7/3)/98 = 0.0952. If i = 0.09 then RHS = 98.798. If i = 0.1 then RHS = 96.296. Using linear interpolation, (0.09− α)/0.798 = −0.01/2.502. Hence α = 0.09319 or 9.319%. (ii) For the 1-year bond, ν1 = 98/112 = 7/8 and y1 = 1/7. Hence 1-year spot rate is y1 = 1/7 or 14.286%. Using the 2-year bond gives 98 = 7ν1 + 112ν22 or 16ν 2 2 = 14 − ν1 which gives ν2 = 0.9057 and y2 = 0.10410 or 10.41%. Using the 3-year bond gives 98 = 7ν1 + 7ν22 + 112ν 3 3 or 16ν 3 3 = 14 − ν1 − ν22 = 15(14 − ν1)/16 = which gives ν3 = 0.91619 and y3 = 0.09148 or 9.148%. 11. Let ft,r denote the forward interest rate p.a. for an agreement starting at time t for r years. We are given (1 + f0,2)2 = 1118/1000; (1 + f1,2)2 = 1140/1000; 1 + f1,1 = 1058/1000. (i) (a) (1 + f0,2)2 = (1 + f0,1)(1 + f1,1). Hence 1 + f0,1 = 1118/1058 and f0,1 = 0.0567. (b) (1 + f0,2)2 = 1118/1000. Hence f0,2 = 0.057355. (c) (1 + f0,3)3 = (1 + f0,1)(1 + f1,2)2 = (1118/1058)× (1140/1000). Hence f0,3 = 0.06403. (ii) Three year par yield is coupon rate r which causes current bond price to be equal to its face value, assuming it is redeemed at par. Time 0 1 2 3 Cash Flow, c −100 100r 100r 100 + 100r Hence 1 = r 1 + f0,1 + r (1 + f0,2)2 + 1 + r (1 + f0,3)3 and so 1 = r [ 1058 1118 + 1000 1118 + 1058× 1000 1118× 1140 ] + 1058× 1000 1118× 1140 Hence r = 0.0636. 12. (i) If f1,1 denotes the implied one-year forward rate applicable at time 1, then (1 + y1)(1 + f1,1) = (1 + y2)2 and hence 1.041(1 + f1,1) = 1.0422 and f1,1 = 0.0430 or 4.30%. If f2,1 denotes the implied one-year forward rate applicable at time 2, then (1 + y2)2(1 + f2,1) = (1 + y3)3 and hence 1.0422(1 + f2,1) = 1.0433 and f2,1 = 0.0450 or 4.50%. (ii) The cash flow is as follows: Time 0 1 2 3 Cash flow −P 3 3 113 Hence P = 3 1 + y1 + 3 (1 + y2)2 + 113 (1 + y3)3 = 105.24 (b) The 2-year par yield is the coupon rate that causes the bond price to be equal to its face value, assuming the bond is redeemed at par. Let yp2 denote the 2-year par yield. Then the cash flow is as follows: Time 0 1 2 Cash flow −100 100yp2 100yp2 + 100 This gives 1 = yp2 ( 1 1 + y1 + 1 (1 + y2)2 ) + 1 (1 + y2)2 = yp2 ( 1 1.041 + 1 1.0422 ) + 1 1.0422 and hence yp2 = (1.042 2 − 1)/(1.0422/1.041 + 1) = 0.04198 or 4.198%. Page 190 Answers 6.2 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 13. (i)(a) P = 5 1.04 + 5 1.04× 1.045 + 105 1.04× 1.045× 1.048 = 101.60 (b) The gross redemption yield is the value for i which satisfies 101.6 = 5a3 ,i + 100ν3. The usual approximation (method (d) in paragraph 4.3) gives i ≈ (5− 1.6/3)/101.6 = 0.044. If i = 0.045, then right hand side = 101.375. If i = 0.04, then right hand side = 102.785. Hence i− 0.45 = (101.6− 101.375)(0.04− 0.045)/(102.785− 101.375). Hence i = 0.0442 or 4.42%. (ii) The gross redemption yield is a weighted average of the forward rates. A higher coupon means more weight on the coupon at times 1 and 2 when the rate is lower. Hence the yield will be lower. 14. (i) For the 3-year bond: Time 0 1 2 3 Cash flow −96 6 6 106 Hence 96 = 6a3 ,i + 100ν3. The usual approximation (method (d) in paragraph 4.3) gives i ≈ (6 + 4/3)/96 = 0.076. At 7.5%,RHS = 6×2.600526+80.4961 = 96.099. At 8%,RHS = 6×2.577097+79.3832 = 94.846. Interpolating gives (x− 0.075)/(0.075− 0.07) = (96− 96.099)/(96.099− 97.375) and hence x = 0.0754 or 7.54%. (ii) One year bond: 96 = ν(1)106. Hence ν(1) = 96/106 = 0.90566. Two year bond: 96 = 6ν(1) + 106ν(2). Hence ν(2) = (96.100)/1062 = 0.85440. Three year bond: 96 = 6ν(1) + 6ν(2) + 106ν3. Gives ν(3) = 96.1002/1063 = 0.80603. (iii) 1 + f0,1 = 1 + y1 = 106/96. Hence f0,1 = 10/96 = 0.1042 or 10.42%. Using (1 + y2)2 = (1 + y1)(1 + f1,2) gives 1062/(96.100) = (106/96)(1 + f1,2) and so f1,2 = 6/100 = 0.06 or 6%. Using (1 + y3)3 = (1 + y2)2(1 + f2,3) gives 1063/(96.1002) = (1062/(96.100))(1 + f2,3) and so f2,3 = 6/100 = 0.06 or 6%. 15. (i)(a) One-year spot rate: 1+y1 = 100/97 and so y1 = 3/97 = 0.0309278 or 3.0928%. Two-year spot rate: (1+y2)2 = 100/93 and so y2 = 0.0369517 or 3.6952%. Three-year spot rate: (1 + y3)3 = 100/88 and so y3 = 0.0435320 or 4.3532%. Four-year spot rate: (1 + y4)4 = 100/83 and so y4 = 0.0476844 or 4.7684%. (b) Now P = 4× 97 100 + 4× 93 100 + 4× 88 100 + 104× 83 100 = 97.44 Let i denote the rate of return. Then 97.44 = 4a4 ,i + 100ν4. The usual approximation (method (d) in paragraph 4.3) gives i ≈ (4 + (100− 97.44)/4)/97.44 = 0.048. Try i = 0.05, then 4a4 ,i + 100ν4 − 97.44 = −0.985996. Try i = 0.045, then 4a4 ,i + 100ν4 − 97.44 = 0.766204. Using linear interpolation gives (i−0.045)/0.005 = (0−0.766204)/(−0.985996−0.766204) leading to i = 0.04719 or 4.72%. (ii) The rate of return from the bond is a weighted average of the rates for 1 year, 2 years, 3 years and 4 years. The 1 year, 2 year and 3 year rates are all less than the 4 year spot rate. (iii) The liquidity preference theory asserts that investors prefer short term investments over long term ones. This is consistent with the answers in part (i)(a): the rates increase as the term increases. 16. The gross redemption yield from a one year bond with a 6% annual coupon is 6% per annum effective: Time 0 1 Cash flow −P 106 Hence P = 106/1.06 = 100. Hence i1 = f0,1 = 0.06. The gross redemption yield from a 2-year bond with a 6% annual coupon is 6.3% per annum effective: Time 0 1 2 Cash flow −P 6 106 Hence P = 6/1.063 + 106/1.0632 = 6/(1 + f0,1) + 106/(1 + f0,2)2 = 6/1.06 + 106/(1 + f0,2)2. Hence i2 = f0,2 = 0.0630905. The gross redemption yield from a 3-year bond with a 6% annual coupon is 6.6% per annum effective. Hence P = 6/1.066 + 6/1.0662 + 106/1.0663 = 6/(1 + f0,1) + 6/(1 + f0,2)2 + 106/(1 + f0,3)2 = 6/1.06 + 6/1.06309052 + 106/(1 + f0,3)3 Hence i3 = f0,3 = 0.066247 or 6.6247%. Summary: i1 = 0.06; i2 = 0.0630905 and i3 = 0.066247. (b) We already know that f0,1 = 0.06. Now (1 + f0,1)(1 + f1,1) = (1 + i2)2; hence f1,1 = 0.06619. Now (1 + f0,1)(1 + f1,1)(1 + f2,1) = (1 + i3)3; hence f2,1 = 0.07259. Summary: f0,1 = 0.06; f1,1 = 0.06619 and f2,1 = 0.07259. (ii) Spot rates are geometric averages of forward rates: for example, 1 + i2 = √ (1 + f0,1)(1 + f1,1), etc. If forward rates are increasing, then the spot rates will increase more slowly—because the geometric average depends on the initial lower rates also. Appendix 2 Sep 27, 2016(9:51) Answers 6.2 Page 191 17. (i)(a) Bond yields are determined by expectations of future interest rates. If interest rates are expected to fall, then long term bonds will become more attractive, and conversely. (b) Liquidity preference: short term investments are preferred over longer term investments—hence longer term bonds should offer higher returns. Market segmenta- tion: rates reflect supply and demand for bonds of different lengths. (ii)(a) We are given y1 = 0.1, f1,1 = 0.09, f2,1 = 0.08 and f3,1 = 0.07. Gross redemption yield from a 1-year zero coupon bond: i = 0.1. Gross redemption yield from a 3-year zero coupon bond: i = (1.1× 1.09× 1.08)1/3 − 1 = 0.08997. Gross redemption yield from a 5-year zero coupon bond: i = (1.1× 1.09× 1.08× 1.072)1/5 − 1 = 0.08194. Gross redemption yield from a 10-year zero coupon bond: i = (1.1× 1.09× 1.08× 1.077)1/10 − 1 = 0.07595. (b) The values decrease. (c) The gross redemption yield of a 5-year coupon-paying bond can be regarded as a weighted average of the gross redemption yields for zero-coupon bonds of shorter terms—see part (c) of example(1.2a). And the early coupons attract higher rates. Hence the gross redemption yields of coupon-paying bonds will decrease with time more slowly than the gross redemption yields of zero coupon bonds. 18. We are given that yn = 0.08− 0.04e−0.1n. (i) P = 1/(1 + y9)9 = 1/(1.08− 0.04e−0.9)9 = 0.57344375. (ii) We have (1 + f7,4)4(1 + y7)7 = (1 + y11)11 and hence (1 + f7,4)4 = (1.08 − 0.04e−1.1)11/(1.08 − 0.04e−0.7)7 leading to f7,4 = 0.078243 or 7.8243%. (iii) We need r with 1 = r 1 + y1 + r (1 + y2)2 + 1 + r (1 + y3)3 leading to r = [ 1− 1 (1 + y3)3 ]/[ 1 1 + y1 + 1 (1 + y2)2 + 1 (1 + y3)3 ] = [ 1− 1 (1.08− 0.04e−0.3)3 ]/[ 1 1.08− 0.04e−0.1 + 1 (1.08− 0.04e−0.2)2 + 1 (1.08− 0.04e−0.3)3 ] = 0.050157 or 5.0157%. 19. (i) Now (1 + f2,2)2 = (1 + f2,1)(1 + f3,1). This implies 1.052 = 1.045(1 + f3,1) and hence f3,1 = 1.052/1.045 − 1 = 0.0550239 or 5.5024%. (ii) One year: y1 = 0.04 or 4%. Two year: y2 = √ 1.04× 1.0425 − 1 = 0.04124925 or 4.125%. Three year: y3 = (1.04× 1.0425× 1.045)1/3 − 1 = 0.042498 or 4.25%. Four year: y4 = (1.04× 1.0425× 1.052)1/4 − 1 = 0.0456155 or 4.56%. (iii) Cash flow from bond is as follows: Time 0 1 2 3 4 Cash flow −P 3 3 3 103 Hence P = 3 1.04 + 3 1.04× 1.0425 + 3 (1 + y3)3 + 103 (1 + y4)4 = 94.46813 Let i denote gross redemption yield. Hence 94.46813 = 3 1 + i + 3 (1 + i)2 + 3 (1 + i)3 + 103 (1 + i)4 = 3a4 ,i + 100 (1 + i)4 Let f (i) = 3a4 ,i + 100/(1 + i)4 − 94.46813. Then f (0.045) = 3 × 3.587526 + 83.8561 − 94.46813 = 0.150548 and f (0.05) = 3 × 3.545951 + 82.2702 − 94.46813 = −1.560077. Interpolating gives (i − 0.045)/(−0.150548) = 0.005/(−1.560077− 0.150548) leading to i = 0.04544 or 4.544%. (iv) We have 3 1 + i + 3 (1 + i)2 + 3 (1 + i)3 + 103 (1 + i)4 = 3 1 + f0,1 + 3 (1 + f0,1)(1 + f1,1) + 3 (1 + f0,1)(1 + f1,1)(1 + f2,1) + 103 (1 + f0,1)(1 + f1,1)(1 + f2,1)(1 + f3,1) It follows that min{f0,1, f1,1, f2,1, f3,1} < i < max{f0,1, f1,1, f2,1, f3,1} The gross redemption yield, i, is a complicated average of the 1-year forward rates. 20. (i) The spot rates are i1 = a − b and i2 = a − 2b. Also f01 = 0.061 and f11 = 0.065. Hence a − b = 0.061 and (1 + f01)(1 + f11) = (1 + a − 2b)2. Hence b = 1.061 − √ 1.061× 1.065 = −0.00199812 and a = 1.122 −√ 1.061× 1.065 = 0.05900188. (ii) First we need i4, the 4-year spot rate. Using the fact that the 4-year par yield is 0.07, we get 1 = 0.07 ( 1 1 + a− b + 1 (1 + a− 2b)2 + 1 (1 + a− 3b)3 ) + 1.07 (1 + i4)4 Hence i4 = 0.07071303. Then P = 0.05 ( 1 1 + a− b + 1 (1 + a− 2b)2 + 1 (1 + a− 3b)3 ) + 1.08 (1 + i4)4 = 0.954501698798 or £0.9545 per £1 nominal. Page 192 Answers 6.2 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 21. (i) Let i denote the gross redemption yield and let ν = 1/(1 + i). Then 103 = 6ν + 6ν2 + 111ν3. Spreading the capital gain of £2 equally over the 3 years, suggests i ≈ 6.7/103 = 0.065. So try i = 0.065. Then 6ν + 6ν2 + 111ν3 − 103 = −0.1849923. Try i = 0.06; then 6ν + 6ν2 + 111ν3 − 103 = 1.1980964. Linear interpolation gives i = 0.06 + 0.005× 1.1980964/(1.1980964 + 0.1849923) = 0.064331 or 6.43%. (ii) The 1 year spot rate, i1 is given by 111 = (1 + i1)103 and hence i1 = 0.0776699 or 7.767%. The 2 year spot rate, i2, is given by 103 = 6/(1 + i1) + 111/(1 + i2)2 and hence i2 = 111/ √ 103× 111− 618− 1 = 0.067357 or 6.736%. The 3-year spot rate, i3 is given by 103 = 6/(1 + i1) + 6/(1 + i2)2 + 111/(1 + i3)3. Hence i3 = 0.0639414 or 6.394%. (iii) To answer to 3 decimal places, we need more than 3 decimal places for the spot rates. Now f01 = i1 = 0.07767, f02 = i2 = 0.06736 and f03 = i3 = 0.06394. Now (1 + i1)(1 + f12) = (1 + i2)2 and hence f12 = 0.05714286 or 5.714%. Now (1 + i2)2(1 + f23) = (1 + i3)3 and hence f23 = 0.05714286 or 5.714%. Now (1 + i1)(1 + f13)2 = (1 + i3)3 and hence f13 = 0.05714286 or or 5.714%. In fact f12 = f23 = f13. 22. (i) Let yn = 0.06− 0.02e−0.1n. Then the price per £100 of the bond is P = 3/(1 + y1) + 3/(1 + y2)2 + 103/(1 + y3)3 = 95.844762. Let i denote the gross redemption yield. Then P = 3/(1 + i) + 3/(1 + i)2 + 103/(1 + i)3. Now the capital gain is approximately £4.16 over 3 years. This suggests a yield of 3 + 1.4 = 4.4% approximately. So try i = 0.044. Let f (i) = P − 3ν − 3ν2 − 103ν3 where ν = 1/(1 + i). Then f (0.044) = −0.299419; f (0.046) = 0.234823. Using linear interpolation gives i− 0.044/(−f (0.044)) = 0.002/(f (0.046)− f (0.044)) and hence i ≈ 0.0451 or 4.5%. (ii) The 4-year par yield, r satisfies 1 = r/(1 + y1) + r/(1 + y2)2 + r/(1 + y3)3 + (1 + r)/(1 + y4)4. Hence r = 1− 1/(1 + y4)4)/(1/(1 + y1) + 1/(1 + y2)2 + 1/(1 + y3)3 + 1/(1 + y4)4) = 0.046424 or 4.64%. 23. (i) and (ii)(a) See answer to question 9 on page 174. (ii) Let x denote the forward rate of interest per annum convertible monthly in the second month. Then 1.01(1 + x/12) = 1.022. Hence x = 0.3611881 or 36.12%. 24. (i) Let i denote the gross redemption yield (which is the same as the yield to maturity or the internal rate of return when taxation is ignored). Let ν = 1/(1 + i). Then 97 = 6ν + 6ν2 + 109ν3 or 97(1 + i)3−6(1 + i)2−6(1 + i)−109 = 0 or f (i) = 97i3 + 285i2 + 273i− 24 = 0. The usual approximation gives i = 0.06 + 6/(3× 100) = 0.08. If i = 0.08, then f (i) = −0.286336 and if i = 0.085 then f (i) = 1.323695. Linear interpolation gives i = 0.08 + 0.286336× 0.005/(1.323695 + 0.286336) = 0.080889 or 8.09%. (ii) Let yt denote the t-year spot rate of interest. Then 97(1 + y1) = 109. Hence y1 = 12/97 = 0.12371 or 12,37%. For y2 we have 97 = 6 1 + y1 + 109 (1 + y2)2 = 6× 97 109 + 109 (1 + y2)2 Hence y2 = 0.09049 or 9.049%. 25. Now ∫ t 0 δ(s) ds = 0.05t + 0.001t 2. (a) exp( ∫ 7 0 δ(s) ds) = exp(0.35 + 0.049) = exp(0.399) = 1.490334 or 1.4903. (b) exp(0.30 + 0.036) = exp(0.336) = 1.399339 or 1.3993 (c) exp(0.399)/ exp(0.336) = exp(0.063) = 1.065027 or 1.0650. (ii)(a) If yt denotes the required spot rate of interest, then (1 + yt)7 = exp(0.399). Hence yt = exp(0.399/7) − 1 = 0.058656 or 5.866%. (b) exp(0.336/6) − 1 = 0.057598 or 5.760%. (c) We want f6,1 where exp(0.399) = exp(0.336)(1 + f6,1). Hence f6,1 = exp(0.063)− 1 = 0.065027 or 6.503%. (iii) The formula for δ(t) shows that the interest rate is increasing with t. Hence the interest rate in the seventh year, which is the answer to part (ii)(c), will be higher that the average interest rate over the first seven years, which is the answer to part (ii)(a). (iv) Now ρ(t) = 30e−0.01t+0.001t 2 for 3 < t < 10 and ν(t) = e−0.05t−0.001t 2 . Hence the present value is∫ 10 3 ρ(t)ν(t) dt = 30 ∫ 10 3 e−0.06t dt = 30 e−0.06t −0.06 ∣∣∣∣10 3 = 500(e−0.18 − e−0.6) = 143.229287659 or 143.229. 26. (i) Spot rates of interest are y1/2 = 0.01 for t = 1/2; y3/2 = 0.03 for t = 3/2; y5/2 = 0.05 for t = 5/2; y7/2 = 0.07 for t = 7/2; and y9/2 = 0.09 for t = 9/2. Hence NPV = 1/1.011/2 + 1/1.033/2 + 1/1.055/2 + 1/1.077/2 + 2/1.099/2 = 4.98307921162 or £4,983,079. (ii) Liquidity preference: investors prefer short term investments and hence longer term bonds must offer higher yields as an inducement. Expectations theory: investors expect interest rates to rise. Market segmentation theory: there is less demand for long-term bonds. (iii) Now (1 + y9/2)9/2 = (1 + y7/2)7/2(1 + f7/2) or 1.099/2 = 1.077/2(1 + f7/2). Hence f7/2 = 0.1629901 or 16.3%. 27. (i) Let ν = 1/(1 + i). Then 101 = 3ν + 3ν2 + 103ν3. Let f (i) = 103ν3 + 3ν2 + 3ν − 101. The value of i is clearly less than 3%. Now f (0.03) = −1 and f (0.025) = 0.428012. Linear interpolation gives i = 0.025 + 0.005 ∗ 0.428012/1.428012 = 0.0264986 or 2.65%. (It is actually 2.64885%.) (ii) One-year bond: 101(1 + i1) = 103. Hence i1 = 2/101 = 0.01980198 or 1.980%. Two-year bond: 101 = 3/(1 + i1) + 103/(1 + i2)2; hence 103/(1 + i2)2 = (101× 103− 303)/103. Hence i2 = 0.0248883 or 2.489%. Three- year bond: 101 = 3/(1 + i1) + 3/(1 + i2)2 + 103/(1 + i3)3 = 303/103 + 3(101 × 103 − 303)/1032 + 103/(1 + i3)3; hence i3 = 0.026589379 or 2.659%. (iii) Now (1 + i2)2(1 + f2,1) = (1 + i3)3. Hence f2,1 = 0.03 exactly or 3%. (iv) See notes: expectations theory, liquidity preference, market segmentation with explanation. Appendix 2 Sep 27, 2016(9:51) Answers 6.4 Page 193 Answers to Exercises: Chapter 6 Section 4 on page 116 (exs6-2.tex) 1. Time 0 0.5 1 · · · 9.5 10 Cash flow −P 5 5 · · · 5 105 Let j = 0.05 and x = 1/(1 + j) = 20/21. Also let 1 + i = (1 + j)2. Now P = 5a20 ,j + 100x20 = 100. Hence dM (i) = ∑n k=1 tkctkν tk/P where the numerator is 2.5(Ia)20 ,j + 100× 10x20. Using the general result that (Ia)n = (an −nνn+1)/(1− ν) implies the numerator equals 2.5× (a20 ,j − 20x21)× 21 + 1000x20 = 654.266. Hence answer is 6.543 years. Or from first principles: dM (i) = 1 100 [ 2.5(x + 2x2 + 3x3 + · · · + 20x20) + 1000x20] = 1 100 [ 2.5 ( x(1− x20) (1− x)2 − 20x21 1− x ) + 1000x20 ] = 1 100 [ 2.5 ( 20× 21(1− x20)− 20× 21x21) + 1000x20] = 10.5 (1− x20) 2. Now dM (i) = ∑n k=1 tkctkν tk∑n k=1 ctkν tk and dν di = −ν2 Hence d di dM (i) = −ν2 ∑n k=1 ctkν tk ∑n k=1 t 2 kctkν tk − (∑nk=1 tkctkνtk )2 ( ∑n k=1 ctkν tk )2 = −ν2 [∑n k=1 t 2 kctkν tk∑n k=1 ctkν tk − (∑n k=1 tkctkνtk∑n k=1 ctkν tk )2] = −ν2σ2 and this quantity is clearly negative. 3. First the price: P = 10 ∫ 20 0 νt dt = 10 νt ln ν ∣∣∣∣20 0 = 10(ν20 − 1) ln ν Hence dM (i) = 10 P ∫ 20 0 tνt dt = 10 P [ tνt ln ν ∣∣∣∣20 0 − ∫ 20 0 νt ln ν dt ] = 10 P [ 20ν20 ln ν − ν t (ln ν)2 ∣∣∣∣20 0 ] = 10 P (ln ν)2 [ 20ν20 ln ν − ν20 + 1] = 1− ν20 − 20ν20 ln 1.04 (1− ν20) ln 1.04 = 8.70586 or 8.706 years. 4. (a) We must have (1) VA(i0) = VL(i0), the net present values are the same; (2) dA(i0) = dL(i0), the effective durations are the same; and (3) cA(i0) > cL(i0), the convexity of the assets is greater than the convexity of the liabilities. (b) We have dM (i) = (1 + i)d(i) where dM is the duration and d is the volatility. (c) Now P (i) = ν + ν2 + · · · = ν/(1− ν) = 1/i. Hence dP (i) di = − 1 i2 and so − 1 P dP (i) di = 1 i 5. Let ν = 1/1.08. We need to check n∑ k=1 atkν tk = n∑ k=1 ltkν tk n∑ k=1 tkatkν tk = n∑ k=1 tkltkν) tk n∑ k=1 t2katkν tk > n∑ k=1 t2kltkν tk First equality: 12.425ν12 + 12.946ν24 = 15ν13 + 10ν25. Both sides equal 6.9757. Second equality: 12× 12.425ν12 + 24× 12.946ν24 = 13× 15ν13 + 25× 10ν25. Both sides equal 108.21. Third relation: lhs = 144×12.425ν12 +576×12.946ν24 = 1886.46 and rhs = 169×15ν13 +625×10ν25 = 1844.73. hence we have the conditions for immunisation against small interest rate changes. 6. (i) For medium and long-term investments. Unsecured. Bearer bonds. Usually one annual coupon. Issued in eurocurrency—currency owned by a non-resident of the country where the currency is legal tender. (ii)(a) Now 97 = ra20 ,0.05 + 100/1.0520. Hence r = (97− 100/1.0520)/a20 ,0.05 = 4.75927. (b) We have dM (i) = 1 97 [ r 20∑ k=1 k 1.05k + 100× 20 1.0520 ] = 1 97 [ r(Ia)20 ,0.05 + 100× 20 1.0520 ] = 13.2146691547 So the duration is 13.2147 years. Page 194 Answers 6.4 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 7. (i) Let ν = 1/1.07. First condition: present values are equal. For the assets we have VA = 7.404ν2 + 31.834ν25 = 12.3323. For the liabilities we have VL = 10ν10 + 20ν15 = 12.3324. Second condition: the durations are the same. For the assets we have ∑ tkatk (1 + i)tk = 2× 7.404ν2 + 25× 31.834ν25 = 159.569 For the liabilities we have ∑ tk`tk (1 + i)tk = 10× 10ν10 + 15× 20ν15 = 159.569 Hence the durations (which are these figures multiplied by ν/P ) are equal. (ii) Let ν1 = 1/1.075. Then profit is 7.404ν21 + 31.834ν 25 1 − 10ν101 − 20ν151 = 0.0158. (iii) The assets are clearly more spread out than the liabilities; hence all three of Redington’s are satisfied and the company is immunised against small changes in the interest rate. 8. (i) Let ν = 1/1.07. The discounted mean term is dM (i) = 1 P ∑ tkctk (1 + i)tk = 8× 87.5ν8 + 19× 157.5ν19 87.5ν8 + 157.5ν19 = 700 + 2992.5ν11 87.5 + 157.5ν11 = 13.07061477 or 13.070615 years. The convexity is c(i) = 1 P ∑ tk(tk + 1)ctk (1 + i)tk+2 = 8× 9× 87.5ν10 + 19× 20× 157.5ν21 87.5ν8 + 157.5ν19 = 6300ν2 + 59850ν13 87.5 + 157.5ν11 = 186.8959854 or 186.895984. (ii) We need equal present values. Hence 87.5ν8 + 157.5ν19 = 66.85ν4 +Xνn. Hence Xνn = 43.4763142. We also need equal durations. Hence we need 8 × 87.5ν8 + 19 × 157.5ν19 = 4 × 66.85ν4 + nXνn. This second equation is 700ν8 + 2992.5ν19 = 267.4ν4 + nXνn. Hence nXνn = 1030.85938. Hence n = 23.710827 and hence X = 43.4763142× 1.07n = 216.25511. Now we check the third condition: for the assets ∑ t2katkν tk = 16× 66.85ν4 + n2Xνn = 25258.52023. For the liabilities: ∑ t2k`tkν tk = 18980.82353. Hence immunisation occurs. 9. (i) Present value of liabilities is 4ν19 + 6ν21. The zero-coupon bond must have the same present value. Hence it must have some value £X at time n years where 19 < n < 21. Hence its spread is less than the spread of the liabilities and so immunisation is not possible. (ii) Let ν = 1/1.07. Then present value of assets is VA = 3.43ν15 + 7.12ν25 = 2.5550 and present value of liabilities is VL = 4ν19 + 6ν21 = 2.5551. Equal to third decimal place. We also need equal durations. So we need: 15×3.43ν15 +25×7.12ν25 = 19×4ν19 +21×6ν21. Now lhs = 51.44420 and rhs = 51.44528. Equal to second decimal place. Now we need to check convexity of assets is greater than convexity of liabilities. For the assets ∑ t2katkν tk = 152 × 3.43ν15 + 252 × 7.12ν25 = 1099.6266. For the liabilities∑ t2k`tkνtk = 192 × 4ν19 + 212 × 6ν21 = 1038.3217. This implies convexity of assets is greater than convexity of liabilities and hence we have immunisation. 10. (i) Let ν = 1/1.08. Present value of liabilities is VL = 3ν3 +5ν5 +9ν9 +11ν11 = 15.0044. Present value of assets is VA = a5 ,0.08 +Rνn = 3.992710 +Rνn. Equating these gives Rνn = 11.01169. We also need equal volatilities. For the liabilities, ∑ tk`tkν tk = 9ν3+25ν5+81ν9+121ν11 = 116.5741. For the assets,∑ tkatkν tk = (Ia)5 ,0.08 + nRνn. Hence nRνn = 116.5741 − (Ia)5 ,0.08 = 116.5741 − (a5 ,0.08 − 5ν6)/(1 − ν) = 105.20899. Hence n = 9.5543 and hence R = 11.01169× 1.08n = 22.9718. For immunisation we need the convexity of the assets to be greater than the convexity of the liabilities. For the assets∑ t2katkν tk = ∑5 k=1 k 2νk + n2Rνn = 40.275 + n2Rνn = 1045.474. For the liabilities ∑ t2k`tkν tk = 27ν3 + 125ν5 + 729ν9 + 1331ν11 = 1042.031. This implies convexity of assets is greater than convexity of liabilities and hence we have immunisation. (ii)(a) Let ν1 = 1/1.03. For the liabilities we have VL = 3ν31 + 5ν 5 1 + 9ν 9 1 + 11ν 11 1 = 21.9029. For the assets we have VA = a5 ,0.03+Rνn = 4.579707+22.9718ν9.5543 = 21.8996. This is a deficit of 0.0033 or £3,300. (b) Immunisation only protects against small changes in the interest rate. This is a large change! 11. (i)(a) Let ν = 1/1.06. Present value of liabilities is VL = a20 ,0.06 + 0.5ν20a20 ,0.06 = (1 + 0.5ν20)a20 ,0.06. Duration of liabilities is dM (i) = 1 VL [ (Ia)20 ,0.06 + 0.5(21ν 21 + 22ν22 + · · · + 40ν40)] = 0.5(Ia)40 ,0.06 + 0.5(Ia)20 ,0.06 (1 + 0.5ν20)a20 ,0.06 Using (Ia)n = (an − nνn+1)/(1− ν) and tables gives dM (i) = a40 ,0.06 − 40ν41 + a20 ,0.06 − 20ν21 (1− ν)(2 + ν20)a20 ,0.06 = 11.302655 Appendix 2 Sep 27, 2016(9:51) Answers 6.4 Page 195 or 11.30265 years. (b) Present value of assets for £100 nominal is VA = 10a15 ,0.06 + 100ν15. Hence duration of assets is dM (i) = 10(Ia)15 ,0.06 + 15× 100ν15 10a15 ,0.06 + 100ν15 = (Ia)15 ,0.06 + 150ν15 a15 ,0.06 + 10ν15 = 9.3523637 or 9.35236 years. (ii)(a) For immunisation, the duration of the assets must equal the duration of the liabilities. Part (i) shows that the duration of assets invested in the government bonds will always be too short. (b) If interest rates decrease, then the values of VA and VL would both rise. The duration of the liabilities is greater than the duration of the assets and the duration is a measure of the rate of change of the present value with respect to i. Hence the value of VL would rise by more than the rise in VA. 12. (a) The cash flow (in £1m) is as follows: Time 0 1 2 3 4 5 6 7 8 9 10 Cash flow −P 1 1 1 1 1 1 1 1 1 1 Let ν = 1/1.07. Then dM (i) = ν + 2ν2 + · · · + 10ν10 P = (Ia)10 ,0.07 a10 ,0.07 = (a10 ,0.07 − 10ν11) (1− ν)a10 ,0.07 = 4.946 (b) There are 3 conditions to check: (I) ∑n k=1 ltkν tk = 7ν5 + 8ν8 = 9.646976 and ∑n k=1 atkν tk = a10 ,0.07 +Xνn = 7.023582 +Xνn. Hence we want Xνn = 2.623394. (II) ∑n k=1 tkltkν tk = 7× 5ν5 + 8× 8ν8 = 62.2031∑n k=1 tkatkν tk = (ν + 2ν2 + · · · + 10ν10) + nXνn = (Ia)10 + nXνn = (a10 ,0.07 − 10ν11)/(1− ν) + nXνn. Hence n = 10.46886 years and so X = 5.32695 million. (III) ∑n k=1 t 2 kltkν tk = 7× 25ν5 + 8× 64ν8 = 422.7612∑n k=1 t 2 katkν tk = (ν + 22ν2 + · · · + 102ν10) + n2Xνn = 228.451 + n2Xνn = 515.9673 As ∑n k=1 t 2 katkν tk > ∑n k=1 t 2 kltkν tk , we have Redington immunisation. 13. (Ia)n is the present value (time 0) of the following increasing annuity: Time 0 1 2 · · · n Cash Flow, c 0 1 2 · · · n Hence: (Ia)n = ν + 2ν2 + · · · + nνn (1− ν)(Ia)n = ν + ν2 + · · · + νn − nνn+1 = νa¨n − nνn+1 = ν(a¨n − nνn) and so (Ia)n = ν 1− ν [ a¨n − nνn ] = a¨n − nνn i = an − nνn+1 1− ν (ii) The cash flow is as follows: Time 0 0.5 1 · · · 9 10 Cash Flow, c −P 5 5 · · · 5 105 Now P = 5(ν + ν2 + · · · + ν20) + 100ν20 where ν = 1 1.03 = 5a20 ,0.03 + 100ν 20 = 129.755 Hence dM (i) = 1 P n∑ k=1 tkctk (1 + i)tk = 1 P [ 5(0.5ν + ν2 + 1.5ν3 + · · · + 10ν20) + 10× 100ν20] = 1 P [ 2.5(Ia)20 + 1000ν 20] = 1 2P [ 5 a20 − 20ν21 1− ν + 2000ν 20 ] = 1815.739 2P = 6.997 years (iii) Now suppose the coupon rate is £8 per annum instead of £10 per annum. The duration is an average of the times of payments weighted by their sizes: decreasing the coupon to £8 means that the final payment has relatively more weight and so the duration will be longer. Page 196 Answers 6.4 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 14. (i) Let ν = 1/1.06. Ignoring tax we would have P = 4a(2)20 ,0.06 + 110ν 20. But we have CGT at the rate of 25% and income tax of an annual amount of 1 at the end of each year. Hence the price is given by P = 4a(2)20 ,0.06 − a20 ,0.06 + [110− 0.25(110− P )] ν20 and hence P = 4a(2)20 ,0.06 − a20 ,0.06 + 82.5ν20 1− 0.25ν20 = (4× 1.014782− 1)× 11.469921 + 82.5ν20 1− 0.25ν20 = 65.95296 (ii) Now there is no CGT. Hence the price is given by P = 4a(2)20 ,0.06 − a20 ,0.06 + 110ν20 = (4× 1.014782− 1)× 11.469921 + 110ν20 = 69.38648 We now need to evaluate ∑ tkctkν tk for the following cash flow: Time 0 0.5 1 1.5 2 2.5 3 · · · 19.5 20 Cash Flow—coupons 0 2 2 2 2 2 2 · · · 2 2 Cash Flow—redemption value 0 0 0 0 0 0 0 · · · 0 110 Cash Flow—tax 0 0 −1 0 −1 0 −1 · · · 0 −1 Let x = ν1/2. Then∑ tkctkν tk = 2× 0.5(x + 2×2 + · · · + 40×40) + 20× 110×40 − (Ia)20 ,0.06 = (x + 2×2 + · · · + 40×40) + 2200ν20 − (Ia)20 ,0.06 = x + 2×2 + · · · + 40×40 + 587.2700 = x(1− x40) (1− x)2 − 40×41 1− x + 587.2700 = 976.09822 and hence dM (i) = 976.09822 69.38648 = 14.067556 or 14.0676 years. 15. Use units of £1,000. (i) Let ν = 1/1.07. The present value of the liabilities is VL = 160a15 ,0.07 + 200ν10 = 1558.93610. (ii) Discounted mean term is dM (i) = ∑ tkctkν tk VL = 160(Ia)15 ,0.07 + 10× 200ν10 160a15 ,0.07 + 200ν10 = 16(Ia)15 ,0.07 + 200ν10 16a15 ,0.07 + 20ν10 = 6.96971 or 6.96971 years. (iii) Let A denote the nominal amount of security A purchased and B denote the nominal amount of security B purchased. Then the present value of the assets is VA = A(0.08a8 ,0.08 +ν8)+B(0.03a25 ,0.08 +ν25) and the discounted mean term is A(0.08(Ia)8 ,0.08 + 8ν8) +B(0.03(Ia)25 ,0.08 + 25ν25) VA So we have the equations 1.05971302A + 0.53385667B = 1558.93610 6.63689237A + 7.97613131B = 10865.33253 Discriminant is ∆ = 6.63689237 × 0.53385667 − 1.05971302 × 7.97613131 = −4.90926094. Hence A = (10865.33253 × 0.53385667 − 1558.93610 × 7.97613131)/∆ = −6633.74879459/∆ = 1351.272396 and B = (6.63689237 × 1558.93610 − 1.05971302 × 10865.33253)/∆ = −1167.64324/∆ = 237.8450. So we need £1,351,272.40 of A and £237,845.00 of B. (iv) The assets are more spread out than the liabilities. Hence Redington immunisation should hold. 16. Let ν = 1/1.07 and x = ν5. Liabilities: Time 0 5 10 15 20 25 Cash Flow 0 3,000 5,000 7,000 9,000 11,000 (i) NPV = 1000(3ν5 + 5ν10 + 7ν15 + 9ν20 + 11ν25) = 1000x[3 + 5x + 7×2 + 9×3 + 11×4] = 11,570.339. (ii) dM (i) = 5000x[3 + 10x + 21×2 + 36×3 + 55×4]/NPV = 171,353/NPV = 14.8097 years. (iii) The present value of £100 nominal of A is 5a26 ,0.07 + 100ν26 = 76.348443. Also the present value of £100 nominal of B is 4a32 ,0.07 + 100ν32 = 62.06033405. Equating the present values gives one equation: 76.348443A + 62.06033405B = 11,570.339. For dM (i) for the assets, we proceed as follows: 5(Ia)26 ,0.07 + 100× 26ν26 = 5 [ a26 ,0.07 − 26ν27 1− ν ] + 2600ν26 = 1031.744022 Appendix 2 Sep 27, 2016(9:51) Answers 6.4 Page 197 Similarly: 4(Ia)32 ,0.07 + 100× 32ν32 = 4 [ a32 ,0.07 − 32ν33 1− ν ] + 3200ν32 = 930.6057861 Hence the second equation is 1031.744022A + 930.6057861B = 171,353. Solving these 2 equations gives: A = 18.9746 and B = 163.09. Hence we need £1,897.46 nominal of A and £16,309 nominal of B. 17. (i) Let ν = 1/1.06, α = 1.1ν and β = 1.03ν. Then for Cyber plc we have P = 6ν6 + 6× 1.1ν7 + 6× 1.12ν8 + · · · + 6× 1.16ν12 + ∞∑ k=1 (6× 1.16 × 1.03kν12+k) = 6ν6(1 + α + α2 + · · · + α6) + 6ν6α6 ∞∑ k=1 1.03kνk = 6ν6 [ 1− α7 1− α + α 6 β 1− β ] = 214.544 and for Boring plc we have P = 4ν(1 + 1.005ν + 1.0052ν2 + · · ·) = 4ν 1− 1.005ν = 72.727 (ii) Now let ν = 1/1.07. For Cyber plc we have: P = 6ν6 [ 1− α7 1− α + α 6 β 1− β ] = 151.982 Drop is 29.16%. and for Boring plc we have P = 4ν 1− 1.005ν = 61.538 Drop is 15.38%. (iii) The cash flows for Cyber plc are later. Hence its duration is larger and hence its dependence on the interest rate is larger. 18. (i)(a) The volatility or modified duration is given by d(i) = − 1 P (i) dP di = n∑ k=1 tkCtk (1 + i)tk+1 / n∑ k=1 Ctk (1 + i)tk (b) Convexity is c(i) = 1 P d2P di2 = n∑ k=1 tk(tk + 1)Ctk (1 + i)tk+2 / n∑ k=1 Ctk (1 + i)tk (ii)(a) Using 100ian,i + 100νn = 100 gives volatility is (8ν(Ia)10 ,0.08 + 1000ν11)/100 = a10 ,0.08 = 6.710081. (b) Volatility: 100,000× 7.247× ν8.247/(100,000ν7.247) = 7.247ν = 6.71018. Convexity: 100,000× 7.247× 8.247ν9.247/(100,000ν7.247) = 7.247× 8.247ν2 = 51.240. (c) Present value of liability of £100,000 to be paid in 7.247 years is 100,000ν7.247 = 57,250.337. This is the amount invested in the loan stock. Hence NPV of assets = NPVof liabilities = 57,250.337. Volatility of assets = volatility of liabilities, by (ii) (a) and (b) Convexity of assets = 60.53 and convexity of liabilities = 51.24 by (ii)(b). Hence Redington immunisation. (iii) Liabilities: 100,000 to be paid in 7.247 years. Assets: 57,250.337 of loan stock with 8% coupon NPV of liabilities = 100,000/1.0857.247 = 55,365.685. NPV of assets = 57,250.337(0.08a10 ,0.085 + ν10) = 55,372.14. Hence profit (in 1/3/1997 values) is 6.45. 19. Let ν = 1/1.07. The cash flows for the annuity certain are: Time 0 · · · 9.5 10 10.5 11 · · · 34 34.5 Cash flow 0 · · · 0 2500 2500 2500 · · · 2500 2500 Value of this annuity at time 9.5 is a(2)25 ,0.07. Hence, net present value of liabilities is PL = 20ν15 + 5ν9.5a (2) 25 ,0.07 = 20ν 15 + 5ν9.5 i i(2) a25 ,0.07 = 20ν15 + 5ν9.5 × 1.017204× 11.653583 = 38.41568 If Vl denotes the volatility of liabilities, then VL × PL = 20× 15ν16 + 2.5(10ν11 + 10.5ν11.5 + · · · + 34.5ν35.5) = 300ν16 + 1.25ν10.5(20ν0.5 + 21ν1 + · · · + 69ν25.0) = 300ν16 + 1.25ν10.5 50∑ k=1 (k + 19)νk/2 Let x = ν1/2. Then Page 198 Answers 6.4 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed VL × PL = 300×32 + 1.25×21 50∑ k=1 (k + 19)xk = 300×32 + 1.25×21 ( 19x 1− x50 1− x + x(1− x50) (1− x)2 − 50×51 1− x ) = 300ν16 + 1.25×21 ( 19x 1− ν25 1− x + x(1− ν25) (1− x)2 − 50×51 1− x ) = 101.62038 + 1.25×21(450.45473 + 712.73726− 267.74145) = 651.69543 Assets: suppose invest in yY nominal of bond Y and it matures at time tY . Present value of assets: 25ν10 + yY νtY = PL = 38.41568 (i) Hence yY νtY = 25.70695. (ii) Volatility of assets: 25× 10ν11 + tY yY νtY +1 = VL × PL = 651.69543 and hence tY yY νtY +1 = 532.922. Hence tY = 22.182 years and yY = 115.302. (iii) For Redington immunisation, have 3 conditions: • value of assets = value of liabilities • volatility of assets = volatility of liabilities • convexity of assets > convexity of liabilities First two conditions are satisfied. Remains to check last condition, where convexity is given by: c(i) = n∑ k=1 tk(tk + 1)Ctk (1 + i)tk+2 / P 20. (i)(a) and (b) in notes. (ii) Present value of liabilities is: PL = 60a9 ,0.07 + 750 1.0710 = 772.176 Volatility of liabilities is: VL = 60(Ia)9 ,0.07 + 10× 750ν10 PL Let yA and yB denote nominal amounts in bonds A and B respectively. Then net present value of assets is yAν 5 + yBν20 = PL The volatility of the assets is 5yAν5 + 20yBν20 PL This leads to yB = 115.406 or £115,406 and yA = 656.768 or £656,768. (iii) For Redington immunisation, have 3 conditions: • value of assets = value of liabilities • volatility of assets = volatility of liabilities • convexity of assets > convexity of liabilities First two conditions are satisfied. Remains to check last condition, where convexity is given by: c(i) = n∑ k=1 tk(tk + 1)Ctk (1 + i)tk+2 / P 21. (i)(a) Present value of liabilities is VL = 100a5 ,0.05 = 432.9477. (b) Let ν = 1/1.05. Macaulay duration is dM (i) = 100(Ia)5/VL = 100(a5 − 5ν6)/(1− ν)VL = 2.90 years. (ii) Suppose purchase nominal yA of 1-year bond and yB of 5-year bond. Then present value is yAν + yBν5 = VL. Macaulay duration is (yAν + 5yBν5)/VL = dM (i). Hence (yAν + 5yBν5) = VLdM (i) = 100(a5 − 5ν6)/(1− ν) = 1256.64. Hence 4yBν5 and so yB = 262.8158. Hence yA = 238.3759. (iii) (a) Convexity of assets: (2yAν3 + 30νBν7)/(yAν + yBν5) = 13.89356 or 13.89 years. (b) Assets more spread out than liabilities. Hence Redington’s immunisation should have been achieved. Check of convexity of liabilities: 100(2ν + 6ν2 + 12ν3 + 20ν4 + 30ν5)/432.9477 = 12.08. Hence convexity of assets is greater than convexity of liabilities. 22. For proof of first formula, see exercise 10 in section 3.3. Liabilities (in 1,000s): Time 0 1 2 3 · · · 19 20 Cash flow 0 100 105 110 · · · 190 195 Appendix 2 Sep 27, 2016(9:51) Answers 6.4 Page 199 (a) Present value: NPV = 95a20 ,0.07 + 5(Ia)20 ,0.07 = 95a20 ,0.07 + 5(a20 ,0.07 − 20ν21)/(1− ν) = 1446.947. Duration: dM (i) = ∑20 k=1(95 + 5k)kν k/NPV = (95(Ia)20 ,0.07 + 5 ∑20 k=1 k 2νk)/NPV = 13643.889/NPV = 9.429 or 9.43 years. (b) Suppose xA and xB denote nominal (or face values) of amounts on (A) and (B) respectively. Hence present value is xAν25 + 0.08xBa12 ,0.07 + xBν12 = 1446.947. Duration is (25xAν25 + 0.08xB(Ia)12 ,0.07 + 12xBν12)/1446.947 = 9.429. We have two simultaneous equations for the two unknowns: xA and xB . Solving (take 25 × first equation minus second equation) gives xB = 1249.27 and hence xA = 534.27. 23. Cash flow: Time 0 1/2 1 3/2 4/2 · · · 39/2 40/2 Cash flow 0 3.75 3.75 3.75 3.75 · · · 3.75 113.75 (i)(a) P = 7.5a(2)20 ,0.1 + 110ν 20 = 7.5(i/i(2))a20 ,0.1 + 110ν20 = 7.5× 1.024404× 8.513564 + 110/1.120 = 81.76077. (b) Let dM (i) denote the volatility. Hence dM (i)× P = ∑ tkCtk (1 + i)tk+1 = 40∑ k=1 k 2 3.75νk/2+1 + 20× 110ν21 = 3.75ν 2 40∑ k=1 kνk/2 + 2,200ν21 = 3.75ν 2 2a(2)20 − 40ν41/2 1− ν1/2 + 2,200ν 21 and hence dM (i) = 8.9104. (ii) Let x denote amount of each liability. Let t1 denote time first liability is due. Then present value of liabilities is x(νt1 + ν10) = P = 81.76077. Let dLM denote volatility of the liabilities. Then d L M × P = x(t1νt1+1 + 10ν11). Hence (t1νt1+1 + 10ν11)/(νt1 + ν10) = 8.9104 and substituting t1 = 9.61 shows that this is the required value. (b) Convexity: c(i) = n∑ k=1 tk(tk + 1)Ctk (1 + i)tk+2 / P = t1(t1 + 1)xνt1+2 + 10× 11xν12 xνt1 + xν10 = 87.526 Payments which are more spread out will have greater convexity. In this case, liabilities are very close (9.61 and 10 years) and assets more spread out. Hence there should be immunisation. (iii) Present value one year later (at time t = 1) is 7.5a(2)19 ,0.1 + 110ν 19 = 7.5 × 1.024404 × 8.36492 + 110/1.119 = 82.254. Present value (times t = 1) of interest payments for next four years is 7.5a(2)4 ,0.1 = 24.354. Hence present value (time t = 1) of asset minus interest payments is 57.9. Hence forward price at time 5 (four years later) is 57.9× 1.14 = 84.77. 24. (i) Let ν = 1/1.03, α = 1.05 and β = αν = 1.05/1.03. Then NPV = 100 [ ν + αν2 + α2ν2 + · · · + α59ν60] = 100ν(1− β60) 1− β = 10852.379704 (ii) Let x denote the nominal amount of the bond. Then for the assets NPV = 0.04xa20 ,0.03 + x 1.0320 = x [0.04× 14.877475 + 0.553676] Hence x = 9446.91494 or £9,446.91494m. (iii) Duration of liabilities: dM (i) = 1 NPV ∑ ctktkν tk = 100 NPV [ ν + 2αν2 + 3α2ν3 + · · · + 59α58ν59 + 60α59ν60] = 100ν NPV [ 1 + 2β + 3β2 + · · · + 59β58 + 60β59] = 100ν NPV [ 1− β60 (1− β)2 − 60β60 1− β ] = 36.143707 or 36.1437 years. (iv) Duration of the assets: 1 NPV [ 0.04x(1 + 2ν + 3ν2 + · · · + 20ν20) + 20xν20] = x NPV [ 0.04ν ( 1− ν20 (1− ν)2 − 20ν20 1− ν ) + 20ν20 ] = 14.572532 or 14.57 years. (v) For the liabilities, d(i) = νdM (i) = 35.090977 and for the assets d(i) = 14.148089. Hence if there is a change of 1.5% in the interest rate, then the liabilities change by 35× 1.5 = 52.5% approximately whilst the assets change by 14.15 × 1.5 = 21.2% approximately. Thus the liabilities would increase by approximately 52.5 − 21.2 = 31.3% of their original value more than the change in the value of the assets. Page 200 Answers 6.4 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 25. (i)(a) Now P = 500a20 ,0.08. Hence dM (i) = (Ia)20 / a20 = (1.08/0.08) × (9.818147 − 20/1.0821)/9.818147 = 8.03694757 or 8.037 years. (i)(b) Let α = 1.08 and ν = 1/1.08. Then P = 500ν(1 + αν + · · · + α19α19) = 500ν × 20 and dM (i) = 500ν(1 + 2αν + · · · + 20α19α19)/P = (1 + 2 + 3 + · · · + 20)/20 = 10.5 or 10.5 years. (ii) The duration is a measure of the mean term of the cash flows weighted by their present values. For (b), the later cash flows are higher than the corresponding cash flows in (a); hence the duration will be higher. 26. Let ν = 1/1.07 and α = 1.05. First option. Outgoings = 0.25 + 0.1ν + 0.2ν2 + · · · + 0.9ν9 + ν10 = 0.25 + 1 10 (ν + 2ν2 + · · · + 9ν9 + 10ν10) = 0.25 + 1 10 [ ν(1− ν10) (1− ν)2 − 10ν11 1− ν ] Incomings = 0.5(ν8 + ν9 + · · ·+ ν27) + 5ν27 = 0.5ν8(1− ν20)/(1− ν) + 5ν27. Then NPV = Incomings−Outgoings = 0.379448 or £379,448. Second Option. NPV = 0.21ν[1+αν+α2ν2+· · ·+α9ν9]+5.64ν10−4.2 = 0.21ν(1−α10ν10)/(1−αν)+5.64ν10−4.2 = 0.472592 or £472,592. Hence the second option has the larger NPV. (ii) Recall the discounted mean term or Macaulay duration is mean term of the cash flows weighted by their present values: dM (i) = ∑ tkctkν tk/P . Clearly the discounted mean term of the second option is less than 10 years. The discounted mean term of the first option, d1 is given by d1P = −0.1(ν + 22ν2 + 32ν3 + · · ·+ 102ν10) + 0.5(8ν8 + 9ν9 + · · · + 27ν27) + 5 × 27ν27. The sum over the first 10 years (terms up to ν10) is negative. Hence the discounted mean term must be larger than 10 years. (In fact, d1P = 49.404532.) 27. Let ν = 1/1.08. Use units of £1,000. Present value of liabilities is VL = 400ν10. Hence, amount held in cash is 40ν10. Let x denote nominal amount in the zero-coupon bond and let y denote the nominal amount in the fixed-interest stock. Hence the present value of the assets is VA = 40ν10 + xν12 + 0.08ya16 ,0.08 + 1.1yν16. Second condition is ∑ k tkatkν tk = ∑ k tk`tkν tk . The right hand side is 4,000ν10 and the left hand side is 12xν12 + 0.08y(Ia)16 ,0.08 + 1.1× 16yν16. This gives two equations for the two unknowns, x and y: xν12 + 0.08ya16 ,0.08 + 1.1yν 16 = 360ν10 12xν12 + 0.08y(Ia)16 ,0.08 + 17.6yν 16 = 4,000ν10 Multiplying the first equation by 12 and subtracting the second equation gives y [ 0.96a16 ,0.08 − 4.4ν16 − 0.08(Ia)16 ,0.08 ] = 320ν10 and hence y = 63.785239 or £63,785. Substituting in the first equation gives x = 254.593513 or £254,593. (ii) Amount invested in the zero-coupon bond is xν12 = 101.102587 or £101,103. Hence amount in the fixed interest stock must be 400ν10 − 40ν10 − xν12 = 0.08ya16 ,0.08 + 1.1yν16 = 65.647069 or £65,647. (iii) The assets are more spread out than the liabilities (which are at a single point). Hence we do have immunisation. 28. Let ν = 1/1.08 and work in units of £1,000. (i) We have VL = 300ν15 + 200a20 ,0.08 = 2058.20191149 or £2,058,201.91. (ii) We have dM (i) = ∑ tk`tkν tk VL = 15× 300ν15 + 200(ν + 2ν2 + · · · + 20ν20) VL = 4,500ν15 + 200(Ia)20 ,0.08 VL = 8.356894 or 8.36 years. (iii) Suppose purchase yA of security A and yB of security B. Then VA = yA(0.09a12 ,0.08 + ν12) + yB(0.04a30 ,0.08 + ν30) = 1.07536078yA + 0.54968865yB . We also need ∑ tkatkν tk = 4,500ν15 + 200(Ia)20 ,0.08 = 17200.17420214 where∑ tkatkν tk = yA [ 0.09(Ia)12 ,0.08 + 12ν 12] + yB [0.04(Ia)30 ,0.08 + 30ν30] = 8.56066413yA + 7.56986281yB . So we have two equations in two unknowns: 1.07536078yA + 0.54968865yB = 2058.20191149 8.56066413yA + 7.56986281yB = 17200.17420214 and hence yA = 7.56986281× 2058.20191149− 0.54968865× 17200.17420214 1.07536078× 7.56986281− 8.56066413× 0.54968865 = 1783.469843361 Similarly, yB = 255.2870422689. So we need £1,783,470 in security A and £255,287 in security B. (iv) We need the convexity of the assets to be greater than the convexity of the liabilities. Assuming the first two conditions hold as in part (iii), this reduces to: ∑ t2katkν tk > ∑ t2k`tkν tk 29. (i) See notes. (ii) Recall duration is dM (i) = ∑ tkctkν tk/P . We have ν = 1/1.05. One year bond: P = 104ν and dM (i) = 104ν/P = 1 or 1 year. Three year bond: P = 4ν + 4ν2 + 104ν3 and dM (i) = (4ν + 8ν2 + 312ν3)/P = 2.88437972 or 2.88 years. Five year bond: P = 4a5 ,0.05 + 100ν5 = 95.67052465 and dM (i) = (4(Ia)5 ,0.05 + 500ν5)/P . Now (Ia)5 ,0.05 = (a5 ,0.05 − 5ν6)/(1− nu). Hence dM (i) = 4.62032257 or 4.62 years. Appendix 2 Sep 27, 2016(9:51) Answers 6.4 Page 201 (iii) The duration, dM (i), is the average of the times tk weighted by their present values, ctkν tk/P . If the coupon rate increased to 8%, then the coupons would form a larger part of the return from the bond. Hence the average time of the cashflows would be less. (iv) Let i denote the rate of return and ν = 1/(1 + i). Option 1: we have 95 = 4a4 ,i + 79ν5. Because 95 = 4 + 4 + 4 + 4 + 79, the rate of return is clearly 0. Option 2: the new cashflow is as follows time 0 1 2 3 4 5 6 7 8 9 10 11 12 cashflow -95 4 4 4 4 1 1 1 1 1 1 1 101 Hence 95 = 4a4 ,i + ν4a8 ,i + 100ν12. A capital gain of 6 is obtained after 12 years; this is an average of 0.5 per year. So that suggests the rate of return is between 1.5% and 4.5% and closer to 1.5% than to 4.5%. Trying i = 0.03 gives 95− 4a4 ,i − ν4a8 ,i − 100ν12 = 3.77426736. Trying i = 0.025 gives 95 − 4a4 ,i − ν4a8 ,i − 100ν12 = −0.89927474. Hence required i = 0.025 + 0.005 × 0.89927474/(3.77426736 + 0.89927474) = 0.02596209 or 2.6% per annum. (v) Harder to predict inflation risk for Option 2. Tax will be different for the two options. Risk of further default harder to predict for option with the longer term. Cash obtained from Option 1 could be invested elsewhere to improve return from Option 1 over the period of 12 years. 30. (i) The payback period is the time when the net cash flow is positive—it completely ignores the effect of interest. The DPP is the time, t, when the NPV of payments up to time t is positive. However, it does not look at the size of the overall profit of the project. (ii)(a) Measure time in years. Then incomings over [0, t] are 0.64t. Hence we want t0 with 0.64t0 = 3; hence t0 = 4.6875 years. (b) Let ν = 1/1.04. Then NPV = ∫ ρ(t)ν(t) dt = 0.64 ∫ t0 0 ν t dt = 0.64(νt0 − 1)/ ln ν. Setting this equal to 3 gives νt0 = 3 ln ν/0.64 + 1; taking logs gives t0 = 5.17975169 or 5.18 years. (iii) Let α = 1.1. For project A we have NPV = fA(i) = −3 + 0.5ν1/2 + 0.5αν3/2 + · · · + 0.5α5ν11/2 = −3 + 0.5ν1/2(1− α6ν6)/(1− αν). For project A we have NPV = fB(i) = −3 + 0.64 ∫ 6 0 ν t dt. When i = 0 we have ν = 1 and fA(0)− fB(0) = −3 + 0.5(1− α6)/(1− α) + 3− 3.84 > 0. When i = 0.04 we have fA(0)− fB(0) < 0. Hence there is at least one root in (0, 0.04). (iv) Duration of incomings of projectA is dAM (i) = ν 1/2(1+3αν+5α2ν2 +· · ·+11α5ν5)/(4P ) where P = 0.5ν1/2(1− α6ν6)/(1−αν). Hence dAM (i) = 0.5(1+3αν+5α2ν2 + · · ·+11α5ν5)(1−αν)/(1−α6ν6) = 3.16327775 or 3.16 years. Duration of incomings of project B is dBM (i) = ∫ 6 0 0.64tν t dt/ ∫ 6 0 0.64ν t dt = ∫ 6 0 tν t dt/ ∫ 6 0 ν t dt. The denominator is (ν6 − 1)/ ln ν and the numerator is 6ν6/ ln ν − (ν6 − 1)/(ln ν)2. Hence dBM (i) = 2.882446 or 2.88 years. (v) The duration is a measure of the sensitivity of the price to interest rate changes. The duration of the incomings of project A is greater than the duration of the incomings of project B. Hence if the interest rate increases, the net present value of the incomings of project A will decrease more than the net present value of the incomings of project B. 31. Let ν = 1/1.08. Now VL = 6ν8 + 11ν15 and VA = Xν5 + Y ν20. First two conditions are 6ν8 + 11ν15 = Xν5 + Y ν20 and 48ν8 + 165ν15 = 5Xν5 + 20Y ν20. Solving gives 15Xν5 = 72ν8 + 55ν15 and 15Y ν20 = 18ν8 + 110ν15. Hence X = 5.50877088 and Y = 13.7968767. (ii) For the third condition, we need to check ∑ t2katkν tk > ∑ t2kltkν tk . Now lhs = 25Xν5 + 400Y ν20 = 1277.76749 and rhs = 6 × 48ν8 + 11 × 225ν15 = 935.820659. Hence the /third condition is satisfied. 32. Let ν = 1/1.04. Use units of £1 million. The NPV of the liabilities is 10a40 ,0.04 = 197.92774 or £197,927,400. Let x denote the amount invested in the first security; then this security has NPV equal to 0.05xa10 ,0.04 + xν10 = 1.081108969x. Hence x = 98.96387/1.081108969 = 91.5392183746 or £91,539,218.37. Let y denote the amount invested in the first security; then this security has NPV equal to 0.1ya5 ,0.04 + yν5 = 1.267109307y. Hence y = 98.96387/1.267109307 = 78.102078055 or £78,102,078.06. (iii) The duration of the liabilities is 10(ν + 2ν2 + 3ν3 + · · · + 39ν39 + 40ν40)/197.92774 = 10(Ia)40 ,0.04/197.92774 where (Ia)40 ,0.04 = (a40 ,0.04 − 40ν41)/(1 − ν) = 306.32307934. Hence the duration of the liabilities is 15.476511 or 15.48 years. (iv) Duration of the assets is [ 0.05x(Ia)10 ,0.04 + 10xν10 + 0.1y(Ia)5 ,0.04 + 5yν5 ] /197.92774 where (Ia)10 ,0.04 = (8.110896 − 10ν11)/(1 − nu) = 41.9922537936 and (Ia)5 ,0.04 = (4.451822 − 5ν6)/(1 − nu) = 13.0064836551. This leads to 6.23034183924 or 6.23 years. (v) The duration of the liabilities is much greater than the duration of the assets. Hence if the interest rate falls then the increase in the NPV of the liabilities will be more than the increase in the NPV of the assets. Hence the fund will make a loss. 33. (i) The price is 9a10 ,0.07 + 100/1.0710 = 114.047167 or 114.047. (ii) Now dM (i) = ∑n k=1 tkctkν tk/114.047 =[ 9(Ia)10 ,0.07 + 1000ν10 ] /114.047 = 7.19880 or 7.199 years. (iii) Duration is a measure of the position of the mean of the cash flows. So if 3%, the mean will be later because of the payment of 100 at the end. (iv)(a) Effective duration d(i) = dM (i)/(1 + i) = 7.1988/1.07 = 6.728 years. (b) The effective duration is a measure of the rate of change of the net present value with i. If the interest rate changes from 0.07 to 0.07 + then the new price will be approximately (1− d(i))P (i) = (1− 6.728)114.047. Page 202 Answers 6.6 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 34. Recall the duration is dM (i) = ∑n k=1 tkctkν tk . For this question, ν = 1/1.04 = 25/26. (i) Now P = 500a12 ,0.04 and dM (i) = 500(Ia)12 ,0.04/500a12 ,0.04 = (9.385074 − 12ν13) × 1.04/0.04 × 1/9.385074 = 56.632809/9.385074 = 6.0343487 or 6.03435 years. (ii) dM (i) = [ 5(Ia)8 ,0.04 + 800ν8 ] / [ 5a8 ,0.04 + 100ν8 ] . Now denominator = 5 × 6.732745 + 100ν8 = 106.7327455 and numerator = 5 × (6.732745 − 8ν9) × 1.04/0.04 + 800ν8 = 729.1188090. Hence dM (i) = 6.8312569 or 6.83126 years. (iii) The duration of the assets is greater than the duration of the liabilities. Now the duration is a measure of the rate of change of the NPV. Hence, if the interest rate increases by a small amount, the NPV of the assets will decrease by a larger amount than the decrease in the NPV of the liabilities; hence the insurance company makes a loss. Answers to Exercises: Chapter 6 Section 6 on page 124 (exs6-3.tex) 1. Y = ln(1 + it) ∼ N (µ = 0.06, σ = 0.01). Hence Pr[it ≤ 0.05] = Pr[Y ≤ ln(1.05)] = Pr[(Y − 0.06)/0.01 ≤ (ln(1.05)− 0.06)/0.01] = Φ(−1.12) = 1− Φ(1.12) = 1− 0.8686 = 0.1314 2. Accumulated value is A(12) = 10(1 + i1)(1 + i2) · · · (1 + i12) where i1, i2, . . . , i12 are i.i.d. with mean µ = 0.06 and standard deviation σ = 0.01. Hence E[A(12)] = 10 (E[1 + i1])12 = 10(1 + µ)12 = 10× 1.0612 = 20.122 Now E[A(12)2] = E[100(1 + i1)2(1 + i2)2 · · · (1 + i12)2] = 100 ( E[(1 + i1)2] )12 = 100(σ2 + 1.062)12. Hence var[A(12)] = E[A(12)2] − E[A(12)]2 = 100 [(σ2 + 1.062)12 − 1.0624] and so the standard deviation is√ var[A(12)] = 0.65775. 3. Let i1, i2 and i3 denote the annual effective rate of interest in years 2004, 2005 and 2006 respectively. Then the accumulated value is for £100 nominal is A = 109 + 4(1 + i3) + 4(1 + i2)(1 + i3) + 4(1 + i1)(1 + i2)(1 + i3) Using independence gives E[A] = 109 + 4(1 + E[i3]) + 4(1 + E[i2])(1 + E[i3]) + 4(1 + E[i1])(1 + E[i2])(1 + E[i3]) = 109 + 4× 1.045 + 4× 1.045× 1.06 + 4× 1.045× 1.06× 1.055 = 109 + 4.18[1 + 1.06 + 1.06× 1.055] = 122.285294 or £122,285.29. 4. Let S10 = (1 + i1) · · · (1 + i10) where i1, . . . , i10 are i.i.d. lognormal(µ = 0.075, σ2 = 0.00064). Hence lnS10 ∼ N (0.75, 0.0064). We want Pr(2000S10 > 4500) = Pr(S10 > 2.25) = Pr(lnS10 > ln 2.25) = Pr((lnS10 − 0.75)/0.08 > (ln 2.25 − 0.75)/0.08) = 1− Φ(0.7616) = 1− (0.7764 + 0.16× 0.003) = 1− 0.7769 = 0.2231 5. Now 1 + i ∼ lognormal(µ, σ2) where E[1 + i] = 1.001 and var[1 + i] = 4× 10−6. We need to find µ and σ2. Using E[Y ] = eµ+ 12σ 2 = 1.001 and var[Y ] = e2µ+σ 2 (eσ 2 − 1) = 4 × 10−6 gives (eσ2 − 1) = 4 × 10−6/1.0012 and hence eσ 2 = 1 + 4× 10−6/1.0012 and σ2 = 3.992× 10−6. Hence µ = ln(1.001/eσ2/2) = 0.0009975. Or, quoting the results of example 5.3b gives σ2 = ln(1 + 4 × 10−6/1.0012) = 3.9920 × 10−6 and µ = ln(1.0012/√4× 10−6 + 1.0012) = 0.0009975. Now Z = ln(1+ i) ∼ N (µ, σ2). We want j with P[i ≤ j] = 0.05. Hence 0.05 = P[1+ i ≤ 1+j] = P[ln(1+ i) ≤ ln(1+ j)] = P[Z ≤ ln(1 + j)] = P[(Z−µ)/σ ≤ (ln(1 + j)−µ)/σ] = Φ((ln(1 + j)−µ)/σ). Hence ln(1 + j) = µ−1.644 854σ giving j = −0.002288. 6. Now S3 = (1 + i1)(1 + i2)(1 + i3) where i1, i2 and i3 are i.i.d. with expectation µ = 0.08 × 0.625 + 0.04 × 0.25 + 0.02× 0.125 = 0.0625 and variance σ2 = (0.082 × 0.625) + (0.042 × 0.25) + (0.022 × 0.125)− µ2 = 0.00054375 Now E[S3] = (1 + µ)3 = 1.06253 = 1.1995. Now E[(1 + i1)2] = var[1 + i1] + (1 + µ)2 = σ2 + (1 + µ)2 and hence E[S23 ] = (σ2 + (1 + µ)2)3. Hence var[S3] = (σ2 + (1 + µ)2)3 − E[S3]2 = (0.00054375 + 1.06252)3 − 1.06256 = 0.0020798 and so σ = 0.0456. 7. Let R = ln(1 + i) where E[i] = 0.0015 and var[i] = 0.0032 = 0.000009. Then R ∼ N (µ, σ2). Hence 1.0015 = E[1 + i] = E[eR] = eµ+ 12σ 2 and 0.0032 = var[i] = var[eR] = e2µ+σ 2 (eσ 2 − 1). Hence eσ2 − 1 = 0.0032/1.0152 and so σ2 = 8.9730× 10−6. Hence µ = ln 1.0015− 12σ2 = 0.00149439. We want j with P[i > j] = 0.9. Hence 0.9 = P[i > j] = P[ln(1+ i) > ln(1+j)] = P[R > ln(1+j)] = P[(R−µ)/σ > (ln(1 + j)− µ)/σ)] = 1− Φ((ln(1 + j)− µ)/σ). Hence ln(1 + j) = µ + σΦ−1(0.1) = −0.0023445 or -0.23445%. 8. (i) Now Sn = (1 + i1)(1 + i2) · · · (1 + in). Using independence gives E[Sn] = Πnk=1 (1 + E[ik]) = (1 + j)n. Now E[(1 + ik)2] = 1 + 2j + s2 + j2. Using independence again gives E[S2n] = (1 + 2j + s2 + j2)n and hence var[Sn] = (1 + 2j + s2 + j2)n − (1 + j)2n. (ii)(a) E[S8] = (1 + j)8 = 1.068 = 1.593848. (b) var[S8] = (1.12 + 0.082 + 0.062)8 − 1.0616 = 0.343646. Appendix 2 Sep 27, 2016(9:51) Answers 6.6 Page 203 9. Work in units of £1 million. (a) Let Rk denote the annual effective rate of return in year k Then E[Rk] = 1.06 and var[Rk] = 0.082 = 0.0064. Then S10 = R1R2 · · ·R10 and E[S10] = 1.0610 = 1.790848. (b) Now Rk is lognormal. Hence lnRk ∼ N (µ, σ2) where 1.06 = eµ+ 12σ2 and 0.0064 = e2µ+σ2 (eσ2 − 1). Using the method of example 5.3b gives σ2 = ln(1 + 0.0064/1.062) = 0.00567982 and µ = ln(1.062/ √ 0.0064 + 1.062) = 0.0554290. Now S10 ∼ lognormal(10µ, 10σ2). Hence P[S10 < 0.9E[S10]) = P[S10 < 1.611763] = P[ln(S10) < ln(1.611763)] = Φ((ln(1.61173)− 10µ)/ √ 10σ2) = Φ(−0.32301) = 0.3733. 10. Suppose 1 + it ∼ lognormal(µ, σ2) and hence Z = ln(1 + it) ∼ N (µ, σ2). Using E[etZ] = etµ+ 12 t2σ2 gives E[1 + it] = eµ+ 1 2σ 2 = 1.05 and var[1 + it] = e2µ+σ 2 (eσ 2 − 1) = 0.112. Hence eσ2 = 0.112/1.052 + 1 and so σ2 = 0.010915 and µ = 0.0433325. (ii) Pr[0.04 < it < 0.07] = Pr[ln(1.04) < ln(1 + it) < ln(1.07)] = Φ((ln(1.07) − µ)/σ) − Φ((ln(1.04) − µ)/σ) = Φ(0.233)−Φ(−0.039) = Φ(0.233)− (1−Φ(0.039)) = Φ(0.233) +Φ(0.039)− 1 = 0.5910 + 0.3× 0.0038 + 0.5120 + 0.9× 0.004− 1 = 0.1077 11. Let Z = 1+it. Then Y = lnZ ∼ N (µ, σ2). Then E[Z] = eµ+ 12σ2 and var[Z] = e2µ+2σ2 (eσ2−1). Hence eµ+ 12σ2 = 1.07 and e2µ+σ 2 (eσ 2 − 1) = 0.04. Hence σ2 = ln(1 + 0.04/1.072) = 0.03434 and µ = ln(1.072/ √ 1.072 + 0.04) = 0.05049. (ii) S15 = (1 + i1)(1 + i2) · · · (1 + i15) and hence lnS15 = ∑15 j=1 ln(1 + ij) ∼ N (15µ, 15σ2). Hence S15 ∼ lognormal(15µ, 15σ2). Pr[S15 > 2.5] = Pr[lnS15 > ln 2.5] = Pr[(lnS15 − 15µ)/ √ 15σ2 > (ln 2.5 − 15µ)/ √ 15σ2] = 1 − Φ((ln 2.5 − 15µ)/ √ 15σ2)) = 1− Φ(0.221457) = 1− 0.5877 = 0.4123. (Note that interpolation has been used: (Φ(0.221457)−Φ(0.22))/(Φ(0.23)−Φ(0.22)) = (0.221457− 0.22)/(0.23− 0.22) = 0.1457 and hence Φ(0.221457) = Φ(0.22) + 0.1457(Φ(0.23)− Φ(0.22)) = 0.5877.) 12. We are given that ln(1 + it) ∼ N (µ = 0.06748, σ2 = 0.0003493) and E[1 + it] = 1.07 and var[1 + it] = 0.0004. (i) (a) 950,000× 1.05 = 997,500. Hence probability = 1. (b) 15% of assets in guaranteed return gives 950,000×0.15×1.05 = 149,625. Let S denote accumulated value from 85% in the risky investment. Hence S = 950,000 × 0.85(1 + it) = 807,500(1 + it). We want Pr[S < 1,000,000 − 149,625] = Pr[S < 850,375] = Pr[1 + it < 1.053096] = Pr[ln(1 + it) < ln 1.053096] = Pr[(ln(1 + it) − µ)/σ < (ln 1.053096− µ)/σ] = Φ(−0.8425) = 1− Φ(0.8425) = 1− (0.7995 + 0.0007) = 0.20 (ii) (a) Variance = 0. (b) Return = 149,625 +S = 149,625 + 807,500(1 + it) which has variance 807,5002var(it) = 807,5002 × 0.022 = 260,822,500. The units of the variance are £2. 13. Use units of £1,000. (i) A(5) = 425 × 1.035 + 425S5 where S5 = (1 + i1)(1 + i2) · · · (1 + i5). Using independence gives E[A(5)] = 425×1.035 +425E[S5] = 425×1.035 +425×1.0355 = 997.4581615 or £997,458.16. Also var[A(5)] = 4252var[S5]. Using independence again gives E[S25 ] = Π5k=1E(1 + ik) 2 = (1 + 2 × 0.035 + 0.032 + 0.0352)5 = 1.4165344. Hence var[S5] = 1.4165344− 1.03510 = 0.0059356. Hence the standard deviation of A(5) is 425 √ var[S5] = 32.7432092 or £32,743.21. (ii) The new expectation is 850 × 1.0355 = 1009.53336 or £1,009,533.36. This is considerably higher. The new standard deviation is 850 √ var[S5] = 65.486418 or £65,486.42 which is twice the previous amount. So there is greater probability of covering the liability of £1,000,000, but there is also a much higher probability of a large shortfall. 14. Let S2 denote the amount after 2 years. Then S2 = 10,000(1 + I1)(1 + I2) where I1 = { 0.03 with probability 1/3; 0.04 with probability 1/3; 0.06 with probability 1/3. and I2 = { 0.05 with probability 0.7; 0.04 with probability 0.3. Using the independence of I1 and I2 gives E[S2] = 10,000(1 + E[I1])(1 + E[I2]) = 10,000(1 + 0.13/3)(1.047) = 10,923.7. (ii) E[S22 ] = 108E[(1 +I1)2]E[(1 +I2)2] = 108 [ 1 + 2× 0.13/3 + 0.0061/3] [1 + 2× 0.047 + 0.00223] = 108 × 1.193465601. Hence var[S2] = 108(1.193465601− 1.092372) = 19,338.41 15. (i) S10 = 1,000Π10j=1(1 + Ij). Now E[1 + I1] = 1.06. Hence E[S10] = 1,000× 1.0610 = 1,790.85. (ii) Now E[(1 + I1)2] = 1 + 0.12 + 0.00392 = 1.12392. Hence E[S210] = 106 × 1.1239210 and so var[S10] = 106 × [1.1239210 − 1.0620] = 9,145.60. Hence the standard deviation is 95.6326. (iii) (a) Clearly E[I1] is unchanged. Hence E[S10] is unchanged. The variance of I1 is smaller. Hence the standard deviation of S10 will become smaller. (b) Clearly both E[S10] and the standard deviation of S10 will become larger. 16. The cash flow is as follows: Time 0 1 2 3 . . . 9 10 Cash Flow (thousands) 150 20 20 20 . . . 20 20 HenceA(10) = 150(1+i1) · · · (1+i10)+20(1+i2) · · · (1+i10)+· · ·+20 where the 1+ij are i.i.d. lognormal(µ = 0.07, σ2 = 0.006). Let α = eµ+ 1 2σ 2 = e0.073. Then E[A(10)] = 150α10 + ∑0 j=9 20α j = 150α10 +20(α10−1)/(α−1) = 595.18471 Page 204 Answers 6.6 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed giving the answer £595,184.71. (ii) Let x denote the required amount. We require P[x(1+i1) · · · (1+i10) ≥ 600] = 0.99. NowZ = (1+i1) · · · (1+i10) ∼ lognormal(10µ, 10σ2) and hence lnZ ∼ N (10µ, 10σ2). So we require 0.99 = P[xZ ≥ 600] = P[lnZ ≥ ln 600 − lnx]. Hence 0.99 = P[(lnZ−10µ)/ √ 10σ2 ≥ (ln 600− lnx−10µ)/ √ 10σ2] where (lnZ−10µ)/ √ 10σ2 ∼ N (0, 1). Hence (ln 600− lnx− 10µ)/ √ 10σ2 = −2.326 348 or lnx = ln 600− 10µ + 2.326 348 √ 10σ2 giving £526,771.15. 17. (i) Sn = (1 + i1) · · · (1 + in) where E[1 + it] = 1 + j, var[1 + it] = s2 and hence E[(1 + it)2] = s2 + (1 + j)2. Hence E[Sn] = (1 + j)n and E[S2n] = [s2 + (1 + j)2]n. Hence var[Sn] = [s2 + (1 + j)2]n − (1 + j)2n. (ii) Now ln(1 + it) ∼ N (µ = 0.08, σ = 0.04). Hence lnS16 ∼ N (16µ, 16σ2). Hence Pr[1000S16 > 4250] = Pr[S16 > 4.25] = Pr[lnS16 > ln 4.25] = Pr[(lnS16 − 16µ)/(4σ) > (ln 4.25 − 16µ)/(4σ)] = 1− Φ((ln 4.25− 16µ)/(4σ)) = 1− Φ((ln 4.25− 1.28)/0.16) = 1− Φ(1.0432436) = 0.1485 by using interpolation on the tables. 18. (i) Sn = (1 + i1) · · · (1 + in) where i1, . . . , in are i.i.d. with mean j and variance s2. Hence E[Sn] = (1 + j)n and E[S2n] = E[(1 + i1)2]n = [var(1 + i1) + E(1 + i1)2]n = [s2 + (1 + j)2]n. Hence var(Sn) = [s2 + (1 + j)2]n − (1 + j)2n. (ii) Suppose j = 0.06 and s = 0.01. Hence s2 = 0.0001 and Z = ln(1 + i) ∼ N (µ, σ2). (a) E[1 + i] = eµ+ 12σ 2 = 1.06 and s2 = 0.0001 = var[1 + i] = e2µ+σ 2 (eσ 2 − 1). Hence eσ2 = 0.0001/1.062 + 1 and so σ2 = 0.00008996 and eµ = 1.059928 and µ = 0.05822. (b) S12 = (1 + i1) · · · (1 + i12). Hence lnS12 ∼ N (12µ, 12σ2) = N (0.69869, 0.0010679). (c) Pr[S12 > 2] = Pr[lnS12 > ln 2] = Pr[(lnS12 − 0.69869)/σ > (ln 2− 0.69869)/0.032679] = 1− Φ(−0.1696) = Φ(0.1696) = 0.5636 + 0.96× 0.0039 = 0.567 19. Let W10 = (1 + i1) · · · (1 + i10) where i1, i2, . . . , i10 are i.i.d. with expectation 0.07 and standard deviation 0.09. Now E[(1 + i1)2] = 1 + 2× 0.07 + 0.092 + 0.072 = 1.153. (i) Hence we are interested in S10 = 1000W10. Now E[S10] = 1000E[W10] = 1000 × 1.0710 = 1967.15. Using E[W 210] = 1.15310 gives the standard deviation of S10 to be 1000 √ var[W10] = 1000 √ 1.15310 − 1.0720 = 531.66. (ii) We need the distribution of S10. Now ln(1 + i1) has a lognormal distribution. Define µ and σ by ln(1 + i1) ∼ N (µ, σ2). Hence eµ+ 1 2σ 2 = E[1 + i1] = 1.07 and e2µ+2σ 2 = E[(1 + i1)2] = 1.153. Hence eσ 2 = 1.153/1.072 and so σ2 = 0.00705. Hence eµ = 1.072/ √ 1.153 and so µ = 0.064134. It follows that ln(W10) ∼ N (10µ, 10σ2). We want P[S10 < 0.5E[S10]] = P[W10 < 0.5E[W10]] = P[ln(W10) < ln(0.5 × 1.0710)] = Φ[(ln(0.5 × 1.0710) − 0.64134)/ √ 0.0705] = Φ(−2.4777974) = 0.00661. (iii) P[1200W10 < 1400] = P[W10 < 7/6] = P[ln(W10) < ln(7/6)] = Φ[(ln(7/6) − 0.64134)/ √ 0.0705] = Φ[−1.835] = 0.0333. 20. Suppose invest £x at time 0. Then value at time 5 is A(5) = xS5 where S5 = R1R2R3R4R5 and R1, R2, . . . ,R5 are i.i.d. lognormal with expectation 1.04 and variance 0.02. Using the method of example 5.3b gives lnRk ∼ N (µ, σ2) where σ2 = ln(1 + 0.02/1.042) = 0.0183222 and µ = ln(1.042/ √ 0.02 + 1.042) = 0.0300596. Hence lnS5 ∼ N (5µ, 5σ2) or lnS5 ∼ N (0.150298, 0.091611). (i) We want x with 0.01 = P[xS5 < 5,000] = P[lnS5 < ln 5,000 − lnx] = P[(lnS5 − 0.150298)/ √ 0.091611 < (ln 5,000 − lnx − 0.150298)/√0.091611]. Hence (ln 5,000 − lnx − 5µ)/ √ 5σ2 = Φ−1(0.01) and hence x = exp [ ln 5,000− 5µ− √ 5σ2Φ−1(0.1) ] = 8699.476176 or £8,699.48. (ii) This amount is substantially more than £5,000, the size of the liability. The problem is that the returns have expectation 1.04, just larger than 1 but the variance is 0.02. Hence the variance is relatively large and there is a substantial probability that the value of the return is less than 1. 21. (i) Now Sn = Πnk=1(1+ik) where i1, . . . , in are i.i.d. with mean j and standard deviation s. (a) Using independence gives E[Sn] = Πnk=1(1 + E[ik]) = (1 + j)n. (b) Using independence again gives E[S2n] = Πnk=1E[(1 + ik)2] = Πnk=1(1 + 2j + E[i2k]) = (1 + 2j + s2 + j2)n. Hence var[Sn] = E[S2n]− E[Sn]2 = (1 + 2j + s2 + j2)n − (1 + j)2n. (ii)(a) Now j = E[ik] = (i1 + i2)/2. Also E[i2k] = (i21 + i22)/2. Hence var[ik] = (i21 + i22)/2− (i1 + i2)2/4 = (i21 + i22 − 2i1i2)/4 = (i1−i2)2/4. (b) Now (1+j)25 = 5.5; hence j = 0.07056862. Also (1+2j+s2 +j2)25− (1+j)50 = 0.52 = 0.25. Hence (1+2j+s2 +j2)25 = 0.25+5.52 = 30.5. Hence s2 = 0.00037738. It follows that i1 +i2 = 2j = 0.14113724 and i1 − i2 = 2s = 0.03885239. Hence i1 = 0.0899948 and i2 = 0.0511424 or 8.99948% and 5.11424%. 22. Suppose the effective interest rate is i per annum. Then the accumulated value at time 10 is S(10) = 800 [ (1 + i) + (1 + i)2 + · · · + (1 + i)10] = 800(1 + i) (1 + i)10 − 1 i We are given that i = { 0.02 with probability 0.25 0.04 with probability 0.55 0.07 with probability 0.2 and hence S(10) = { 8934.97233579 with probability 0.25 9989.0811263 with probability 0.55 11826.8794549 with probability 0.2 Hence E[S(10)] = 10093.1135944 or £10,093.11. (b) This is clearly 0.2. Appendix 2 Sep 27, 2016(9:51) Answers 6.6 Page 205 23. (i) Let ik denote the effective rate of interest per annum in year k. Then S10 = (1+i1) · · · (1+i10). Using independence gives E[S10] = 1.0710 = 1.967151. (ii) Using independence again gives E[S210] = Π10k=1E[(1 + ik) 2] = (1 + 2 × 0.07 + 0.016 + 0.072)10 and var[S10] = E[S210]− E[S10]2 = (1 + 2× 0.07 + 0.016 + 0.072)10 − 1.0720 = 0.576097. (iii) If 1,000 units are invested, then the expectation is 1,000E[S10] and the variance is 1,000,000var[S10]. 24. Use units of £1,000. (i) Accumulated amount is A(3) = 80S3 where S3 = (1 + i1)(1 + i2)(1 + i3) and E[1 + i1] = 1.06, E[1 + i2] = 3 4 × 1.07 + 14 × 1.05 = 1.065 and E[1 + i3] = 710 × 1.06 + 310 × 1.04 = 1.054. Using independence gives E[A(3)] = 80× 1.06× 1.065× 1.054 = 95.18885 or £95,188.85. (ii) Now var[A(3)] = 6,400var[S3]. Using independence again gives E[S23 ] = E[(1+i1)2]E[(1+i2)2]E[(1+i3)2]. Now E[(1+i1)2] = 13 (1.04 2 +1.062 +1.082); E[(1+i2)2] = 34 1.07 2 + 14 1.05 2 = 1.1343, and E[(1+i3)2] = 710 1.06 2 + 310 1.04 2 = 1.111. Hence E[S23 ] = 1.416304978. Hence var[A(3)] = 6,400(1.41630498−1.18986062) = 3.435073 or 3,435,073 in units of £2. (iii) Now 80× 1.08× 1.07× 1.06 = 97.99488. For any other possibility we have A(3) < 97. Hence the probability is 13 × 34 × 710 = 740 = 0.175. 25. Let Y = 1 + it. Then E[Y ] = 1.06 and var[Y ] = 0.082. Now ln(Y ) ∼ N (µ, σ2). Hence eµ+ 12σ2 = 1.06 and e2µ+σ 2 (eσ 2 − 1) = 0.082. Hence eσ2 = 1 + 0.082/1.062 leading to σ2 = 0.00567982. Also eµ = 1.062/ √ 0.082 + 1.062 leading to µ = 0.0554290. Let S10 denote the value at time 10 of £1 at time 0. Then S10 = ∏10 j=1(1 + ij). Hence ln(S10) ∼ N (10u, 10σ2). Hence 2E[S10] = 2 exp(10µ + 10σ2/2) = 2 exp(0.55429)× (1 + 0.082/1.062)5 = 3.5816954 or £3.581695m. (ii) We want P[2S10 < 0.8 × 3.5816954] = P[S10 < 1.432678] where ln(S10 ∼ N (0.554290, 0.0567982). Hence P[S10 < 1.432678] = Φ((ln(1.432678)− 0.554290)/ √ 0.0567982) = Φ(−0.817143) = 0.20692. 26. (i) Option A. Let A1 = 100(1 + i1)(1 + i2) · · · (1 + i10). Then E[A1] = 100 × 1.05510 = £170.81 by independence. Also E[A21] = 1002 ( 1 + 2E[i1] + E[i21] )10 = 1002 ( 1.11 + 0.072 + 0.0552 )10 . Hence var[A1]/1002 = (1.11 + 0.072 + 0.0552)10 − 1.05520 giving the standard deviation of £36.20. Option B. For this case A2 = 100× 1.065 × 1.015 with probability 0.2; 100× 1.065 × 1.035 with probability 0.3; 100× 1.065 × 1.065 with probability 0.2; 100× 1.065 × 1.085 with probability 0.3. Hence E[A2] = 100×1.065×[0.2×1.015 +0.3×1.035 +0.2×1.065 +0.2×1.085] = £169.48. Also var[A2] = 1002× 1.0610×(0.2×1.0110 +0.3×1.0310 +0.2×1.0610 +0.3×1.0810−(0.2×1.015 +0.3×1.035 +0.2×1.065 +0.3×1.085)2 giving a standard deviation of £21.62. (ii) Option A. Now ln(1 + it) ∼ N (µ, σ2). Hence eµ+ 12σ2 = E[1 + it] = 1.055 and e2µ+2σ2 = E[(1 + it)2] = 1.11 + 0.072 + 0.0552. Hence eσ 2 = (1.11 + 0.072 + 0.0552)/1.0552 giving σ2 = 0.00439275. Also eµ = 1.0552/ √ 1.11 + 0.072 + 0.0552 giving µ = 0.0513444. Now ln(A1/100) ∼ N (10µ, 10σ2) = N (0.513444, 0.0439275). We want P[A1 < 115] = P[ln(A1/100) < ln 1.15] = Φ [ ln 1.15− 0.513444√ 0.0439275 ] = 0.0372989 or 0.0373. Option B. . The smallest possible amount is 100 × 1.065 × 1.015 = 140.648853 which is greater than 115. So the probability is 0. (iii) Both options have similar expectations, but option A has a much higher variance and so is riskier. There is a much higher probability of a low return with option A. 27. (i) LetA denote the accumulated value. ThenA = (1+i1)(1+i2) · · · (1+i10). By independence, E[A] = Π10j=1E[1+ij] = E[1 + i1] (E[1 + i2])9 = 1.05 × 1.049 = 1.494477 or £1,494,477. (ii) Now E[(1 + i1)2] = 1 + 0.1 + 0.0029 = 1.1029 and E[(1 + i2)2] = 1 + 0.08 + 0.00184 = 1.08184. Hence E[A2] = E[(1 + i1)2] ( E[(1 + i2)2] )9 = 1.1029× 1.081849. Hence var[A] = E[A2]− (E[(A)])2 = 0.00527622 giving a standard deviation of 0.0726376 or £72,638. (iii) The value of E[i2] is unchanged. Hence the value of E[A] will be unchanged. The value of var[i2] is increased— hence the value of the standard deviation of A will be increased. 28. Let α = E(1 + i) = eµ+ 12σ 2 . Let S20 denote the required answer. Then S20 = 5000(1 + i1) · · · (1 + i20) + 5000(1 + i2) · · · (1 + i20) + · · · + 5000(1 + i20) and hence E(S20) = 5,000(α20 + α19 + · · · + α) = 5,000α(α20 − 1)/(α − 1) = 5,000α(e1.04− 1)/(α− 1) = 180498.925165 or £180,498.93. (ii) For this part S20 = (1 + i1) · · · (1 + i20). Hence E(S20) = e20µ+10σ 2 = e1.04. We want P(S20 > e1.04) = P(lnS20 > 1.04) where lnS20 ∼ N (20µ, 20σ2) = N (1, 0.08). Hence P(S20 > e1.04) = 1− Φ(0.04/ √ 0.08) = 0.44376854 or 0.44. Page 206 Answers 7.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 29. (i) The expected annual interest rate is 0.04 × 0.3 + 0.06 × 0.7 = 0.54. If x denotes the required amount, then x × 1.05420 = 200,000 and hence we get x = 69858.262239 or £69,858.26. (ii) The value is x × 1.0420 with probability 0.09, and x × 1.0620 with probability 0.49 and x × 1.0410 × 1.0610 with probability 0.49. This has expectation 201336.5443921 or £201,336.54 giving the answer £1,336.54. (iii) Smallest amount is x×1.0420 = 153068.050218 or £153,068.05; largest amount is x×1.0620 = 224044.903673 or £224,044.90. Hence the range is £224,044.90− £153,068.05 = 70,976.85. 30. (i) Let A = 10,000(1 + i1)(1 + i2) · · · (1 + i15) where E[ij] = 0.07. Hence E[A] = 10,000 × 1.0715 = 27590.3154 or £27,590.32. (ii) Now E[i2j] = 0.00506. Hence E[(1 + ij)2] = 1.14506. Hence E[A2] = 108 × 1.1450615 and hence var[A] = 108(1.1450615−1.0730), and so the standard deviation is 104√1.1450615 − 1.0730 = 1263.83669 or 1,263.837. (iii) (a) The mean, E[ij], is clearly the same. Hence E[A] is unchanged. However, there is less variability and so the standard deviation of A will be less. (b) The mean will be less because the money is invested for a shorter period. Also the standard deviation will be less because the range of possibilities for A will be less. 31. (i) Let S5 = 7.85(1 + i1)(1 + i2)(1 + i3)(1 + i4)(1 + i5). Hence E[S5] = 7.85[E(1 + i1)]5 = 7.85 × 1.0555 = 10.259636050 or £10,259,636. Now E[(1 + i1)2] = 1 + 2 × 0.055 + 0.042 + 0.0552 = 1.114625. Hence var[S5] = 7.852 [ 1.1146255 − 1.05510] and stdev[S5] = 7.85√1.1146255 − 1.05510 = 0.871061457 or £871,061. (ii) In this case S5 = 7.85 × 1.045 = 9.550725284 or £9,550,725. So there is a certain loss of £449,274 with the policy of part (ii). The policy of part (i) has an expected profit but there is the possibility of a very large loss. Using the approximation of 95% of a distribution within ±2 standard deviations, implies a chance of 1 in 40 of a loss of over £1,400,000. 32. (i) Now Sn = (1 + i1) · · · (1 + in). Using independence gives E[Sn] = ∏n t=1 E[1 + it] = (1 + j) n. Similarly E[S2n] =[ E(1 + it)2 ]n = [ 1 + 2j + s2 + j2 ]n because E[i2t] = var[it] + E[it]2. Hence var[Sn] = (1 + 2j + j2 + s2)n− (1 + j)2n. (ii)(a) See notes: σ2 = ln(1 + 0.122/1.042) = 0.013226 and eµ = 1.042/ √ 0.122 + 1.042 leading to µ = 0.032608. (b) Now 1 + it ∼ lognormal(µ, σ2). Hence ln(1 + it) ∼ N (µ, σ2) and (ln(1 + it)− µ) /σ ∼ N (0, 1). Hence P[1.06 < 1 + it < 1.08] = Φ ( ln(1.08)− µ)/σ) − Φ ((ln(1.06)− µ)/σ) = 0.06184493 or 0.062 or 6.2%. (iii) The expectation of the return is 4%. So the probability it lies between 6% and 8% will be small; The reason it is so small is a consequence of the shape of the lognormal distribution. Try the following plot of the density: plot( seq(0,2, by=0.01), dlnorm( seq(0,2, by=0.01), m=mu,s=sqrt(sigma1)), type="l"). Answers to Exercises: Chapter 7 Section 3 on page 138 (exs7-1.tex) 1. Now 1 + i = (1 + 0.05/4)4. Hence 80 = K/(1 + i)1/4 = K/1.0125 and so K = 80× 1.0125 = 81. 2. We have S0 = 12. Let i denote effective risk-free interest rate per annum and let K denote the forward price. Then K (1 + i)10/12 = 12− 1.5 (1 + i)3/12 − 1.5 (1 + i)6/12 − 1.5 (1 + i)9/12 Now 1 + i = 1.032. Hence K = 12× 1.035/3 − 1.5× 1.037/6 − 1.5× 1.034/6 − 1.5× 1.031/6 = 8.01609 or £8.02. 3. (i) See page 131. (ii) Now Ke−0.07×7/12 = 60 − 2e−0.07×3/12 − 2e−0.07×6/12 which gives K = 60e0.49/12 − 2e0.28/12 − 2e0.07/12 = 58.4418. 4. Now £1.20 invested for 91 days at 5% p.a. effective will accumulate to 1.20 × 1.0591/365 = 1.2147. No arbitrage implies that the forward price to be paid in 91 days must be £1.2147. 5. (i) A forward contract is a legally binding agreement to buy or sell an agreed quantity of an asset at an agreed price at an agreed time in the future. If the contract specifies that A will buy the asset from B at some time in the future, then A holds a long forward position and B holds a short forward position. (ii) Using time 0 values gives Ke−δT = S0e−rT − ce−δt1 . Hence Ke−0.05/4 = 150e−0.03/4 − 30e−0.05/6 and hence K = 150e0.02/4 − 30e0.05/12 = 120.62661. 6. Let K denote the forward price. Hence K 1.045 = 6− 0.3 1.041/2 − 0.3 1.045 This gives K = 6.27− 0.3× 1.045/1.041/2 − 0.3 = 5.662588. 7. (i) A forward contract is a legally binding contract to buy (or sell) an agreed quantity of an asset at an agreed price at an agreed time in the future. The contract is usually tailor-made between two financial institutions or between a financial institution and a client. Thus futures contracts are over-the-counter or OTC. Such contracts are not normally traded on an exchange. Settlement of a forward contract occurs entirely at maturity and then the asset is normally delivered by the seller to the buyer. Forward contracts are subject to credit risk—the risk of default. The party agreeing to sell the asset is said to hold a “short forward position” and the buyer is said to hold a “long forward position.” (ii) Original contract: Price S0 = 96. Hence K = S0 × 1.0410 − 7 × 1.04. Hence the agreed forward price was K = 134.82. Appendix 2 Sep 27, 2016(9:51) Answers 7.3 Page 207 Either: position now is that the price is S0 = 148. Hence K = S0 × 1.043 − 7× 1.04. Hence the new forward price should be K = 159.20. This gives a difference of 159.20− 134.82 = 24.38 or 24.28/1.043 = 21.67 in today’s terms. Or: the value of the contract to its owner who has the right to buy the asset in 3 years’ time for the price 134.82 is 148− 7/1.042 − 134.82/1.043 = 21.67. 8. (i) See page 131. (ii) Let K denote the forward price, α = 1.01 and β = 1.015. The dividends are as follows: Time 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 Dividend 0 0.2α 0.2α2 0.2α3 0.2α4 0.2α5 0.2α6 0.2α7 0.2α8 0.2α8β 0.2α8β2 0.2α8β3 0.2α8β4 Hence Ke−0.15 = 4.5− 8∑ k=1 0.2αke−0.05k/4 − 4∑ k=1 0.2α8βke−0.1−0.05k/4 = 4.5− 0.2 8∑ k=1 xk1 − 0.2α8e−0.1 4∑ k=1 xk2 = 4.5− 0.2x1 1− x81 1− x1 − 0.2α 8e−0.1x2 1− x42 1− x2 where x1 = αe−0.05/4 = 0.99745358 and x2 = βe−0.05/4 = 1.00239147. Hence K = 2.47433598. 9. (i)(a) Accumulation is 100 exp (∫ 15 5 δ(t) dt ) = 100 exp(0.25) exp ( 0.08t + 0.003t 2 2 ∣∣∣15 10 ) = 100 exp(0.25) exp(0.4 + 0.1875) = 100 exp(0.25) exp(0.5875) = 231.058329. (b) Accumulation is 100 exp (∫ 14 5 δ(t) dt ) == 100 exp(0.25) exp ( 0.08t + 0.003t 2 2 ∣∣∣14 10 ) = 100 exp(0.25) exp(0.32 + 0.144) = 100 exp(0.25) exp(0.464) = 204.214352. (c) Accumulation is 100 exp (∫ 15 14 δ(t) dt ) = 100 exp(0.5875− 0.464) = 113.145000. (d) We want δ with 100e10δ = 231.058329 and hence δ = 0.083750. (ii) Now ν(t) = exp(− ∫ t0 δ(u) du). Hence for 0 ≤ t ≤ 5 we have ν(t) = exp(−0.05t). We need∫ 5 0 ρ(t)ν(t) dt = ∫ 5 0 100e0.01te−0.05t dt = 100 ∫ 5 0 e−0.04t dt = 100 e−0.04t −0.04 ∣∣∣∣5 0 = 2500(1− e−0.2) = 453.173117 (iii) Let K denote the forward price. Then Kν(1) = 300 − 7ν(0.5) where ν(t) is given above. Hence K = 300 exp(0.05)− 7 exp(0.025) = 308.204123. 10. Use £100 nominal. Let K denote the forward price. Investor has a short position. This means he must supply the bond at time t = 1 for the price of £98. He receives a coupon of 2.5 at time t = 0.5 and a coupon of 2.5 at time t = 1. So the time t = 0 value of these three receipts is 98e−0.052 + 2.5e−0.023 + 2.5e−0.052 = 97.850707 The current price is £95. Hence the value to the investor is £97.850707− £95 = £2.850707. This was for £100 nom- inal. For £1 million, the value is 10,000× £2.850707 = £28,507.07. 11. (i) This theory asserts that the relative attraction of long and short term bonds depends on the expectations of future movements in interest rates. If investors expect interest rates to fall, then long term bonds will become more attrac- tive; this will cause the yields on long term bonds to fall and lead to a decreasing yield curve. Similarly, if investors expect interest rates to rise, then long term bonds will become less attractive and so lead to an increasing yield curve. (ii)(a) Clearly i1 = 0.08 or 8%. Now (1 + i2)2 = 1.08 × 1.07 and hence i2 = 0.07498837. Now (1 + i3)3 = 1.08×1.07×1.06 and hence i3 = 0.06996885. Finally, (1+i4)4 = 1.08×1.07×1.06×1.05 and hence i4 = 0.06494131. (b) The price of the bond is P = 5 [ 1 1 + i1 + 1 (1 + i2)2 + 1 (1 + i3)3 + 1 (1 + i4)4 ] + 100 (1 + i4)4 = 94.67515038 So we need the gross redemption yield of Time 0 1 2 3 4 Cash flow −P 5 5 5 105 So we need to find i with P = 5a4 ,i + 100/(1 + i)4. To get a first approximation, use the usual method of spreading the capital gain over the coupons; this gives 5 + (100− P )/4 = 6.33 which suggests i lies between 6% and 7%. If i = 0.06, then 5a4 ,i + 100/(1 + i)4 − P = 1.8598. If i = 0.07, then 5a4 ,i + 100/(1 + i)4 − P = −1.44960. Using linear interpolation gives i = 0.06 + 0.01× 1.8598/(1.8598 + 1.4496) = 0.06562. or 6.56%. (c) Let K denote the forward price. Hence K (1 + i2)2 = 400− 4 1 + i1 and hence K = 1.08× 1.07× (400− 4/1.08) = 457.96 Page 208 Answers 7.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed 12. Suppose K denotes the non-arbitrage forward price. Then Ke−0.03/2 = 10− 0.5e−0.03/12. Hence K = 9.64484. So the forward price of £9.70 is too high. Hence a risk-free profit can be made as follows. Time 0: borrow £10 and buy one share—the net cash flow is zero. Also enter into a forward contract to sell the share in 6 months’ time for £9.70. Time 1 month: receive dividend of £0.50 and invest for 5 months. Time 6 months: dividend has grown to £0.5e0.03×5/12; also owe £10e0.03/2 to the bank; receive £9.70 for the share. So the return is £(9.70 + 0.5e0.03×5/12 − 10e0.03/2) = £(9.70−K) = £0.0551586. 13. (i) See page 131. (ii) The given information is summarised in the following table: t = 0 t = 1, market up t = 1, market down cost of security A 20 25 15 cost of security B 15 20 12 Now 25/20 = 15/12. Whatever the state at time t = 1, the prices of the shares are in the ratio 5 : 4. Hence for no arbitrage, the prices of the shares at time t = 0 must be in the ratio 5 : 4. Hence if the price of share A at time 0 is 20, then the price of share B must be 16. At 15, the price of B is too low: consider selling three A and buying four B at time 0. This leads to a profit at time 1 whether the market rises or falls. 14. (i) Let K1 denote price of original forward contract of 12 years. Then K1 1.0312 = 95− 5 1.037 − 6 1.039 Let K2 denote price of forward contract arranged now for 7 years. Then K2 1.037 = 145− 5 1.032 − 6 1.034 Hence value of contract is K2 −K1 in 7 years’ time. In units of money now it is K2 −K1 1.037 = 145− 95× 1.035 = 34.86896 (ii) For this case: K1 1.0312 = 95 1.0212 and K2 1.037 = 145 1.027 Hence value of contract is K2 −K1 in 7 years’ time. In units of money now it is K2 −K1 1.037 = 145 1.027 − 95× 1.03 5 1.0212 = 145× 1.025 − 95× 1.035 1.0212 = 39.39365 15. We have 900 = 50 1.051/2 + 50 1.06 + K 1.06 This leads to K = 900× 1.06− 50× 1.06/1.051/2 − 50 = 852.2773 or 852p. 16. We have, in pence, 1000 = 50 1.051/12 + 50 1.057/12 + K 1.0511/12 Hence K = 1.054/12(1000× 1.057/12 − 50× 1.056/12 − 50) = 942.84489 or 9.43. 17. (i) Compare (A): buy asset at price B and (B): buy forward at price K and deposit Ke−δT in risk free asset. At time T , both lead to same value—ownership of the asset. Hence, by no arbitrage, Ke−δT = B. (ii) Let K denote the price of the forward contract. Then 200− 10 1.02 = K 1.022 Hence K = 200× 1.022 − 10× 1.02 = 197.88 or £197.88. 18. Note: this is a poorly phrased question! (i) See page 131. (ii) Let t = 0 denote the time four years ago and let K0 denote the price of the initial forward contract at that time. The price of the security was £7.20 at t = 0 with dividends of £1.20 at t = 5, 6, 7, 8 and 9. So we are comparing the following 2 cash flows: Time 0 1 2 3 4 5 6 7 8 9 c1 −7.20 0 0 0 0 1.20 1.20 1.20 1.20 1.20 c2 −K0 Hence K0 = 7.20− 1.2a5 ,0.025/1.0254. Let K1 denote the price of a forward contract at t = 4. Then K1 = 10.45− 1.2a5 ,0.025. Hence the value of the original contract at t = 4 is K1 − 1.0254K0 = 10.45− 7.20× 1.0254 = 2.502547 or £2.50. Appendix 2 Sep 27, 2016(9:51) Answers 7.3 Page 209 (iii) Assume 1 unit of the security grows to 1.03t units of the security over the interval (0, t). Then the price of the original forward contract at t = 0 is 7.2/1.039 which has value 7.2 × 1.0254/1.039 at t = 4. The price of a new forward contract at t = 4 is 10.45/1.035. Hence the value of the original forward contract at t = 4 is 10.45/1.035 − 7.2× 1.0254/1.039 = 2.923201 or £2.92. 19. Let ν = 1/1.042. Date 1/9/12 1/12/12 1/3/13 1/6/13 1/7/13 Cash flow: −10 1 1 1 K So 10 = ν1/4 + ν1/2 + ν3/4 +Kν10/12. Hence K = 7.595645867212 or £7.5956. (ii) The forward price is the price quoted today (1/9/12) for buying 1 share on 1/7/13. Whether or not the share price increases or decreases between 1/9/12 and 1/7/13 does not affect the forward price and indeed this information is unknown when the cost of the forward contract is determined. Owning a forward contract on 1/9/12 is like owning the share on 1/9/12 except that no dividends before 1/7/13 are received and there is no need to pay the seller of the share until 1/7/13. 20. (i)(a) A futures contract is a legally binding contract to buy or sell a specified quantity of an asset at a specified price at a specified time in the future. An option gives the right but not the obligation to buy or sell a specified quantity of an asset at a specified price at a specified time in the future. (b) A call option gives the owner the right but not the obligation to buy a specified asset for a specified price at some specified time in the future. A put option gives the owner the right but not the obligation to sell a specified asset for a specified price at some specified time in the future. (ii) We are comparing the following 2 cash flows: Time 1/4/2013 30/9/2013 31/3/2014 c1 −10.50 1.10 1.10 c2 −K Assuming no arbitrage, we must have NPV(c1) = NPV(c2). Hence −10.50 + 1.1 1.0225 + 1.1 1.0252 = −K 1.0252 Hence K = 8.801306 or £8.80. 21. Let K denote the forward price in pence. Then 180 = 10 1.041/2 + K 1.043/4 Hence K = 1.041/4 [ 180× 1.041/2 − 10] = 175.274906066 or 175.275p. 22. (i) See page 131. (ii) See section 2.6 on page 136. Equating time 0 values gives Ke−0.09×0.75 = 6e−0.035×0.75 and hence K = 6e(0.09−0.035)×0.75 = 6.2526756 or £6.25268. (iii) Suppose that at time 0, the investor borrows the amount £6e−0.035×0.75 for 9 months and uses this to buy the stock; he reinvests the dividends in purchasing further units of the stock. Hence at time 0, the investor has e−0.035×0.75 units of the stock which grow to 1 unit after 9 months. Suppose further that at time 0, he enters into a forward contract to sell one unit of the stock at time 9 months for the amount £6.30. After 9 months, the investor will owe £6e−0.035×0.75 × e0.09×0.75 = £6e(0.09−0.035)×0.75 = £6.25268. Hence he makes a profit of £6.30− £6.25268. 23. Let δ = 0.05, the force of interest. The NPV at 1 October 2015 of the dividend at 1 November 2015 is 0.5e−δ/12 and the NPV of the dividend at 1 May 2016 is 0.5e−7δ/12. Finally, the NPV at 1 October 2015 of the amount K, the price of the forward contract received at 1 July 2016 is Ke−9δ/12. Hence 10 = 0.5e−δ/12 + 0.5e−7δ/12 + Ke−9δ/12 giving K = 9.360988337958 or £9.36099. (ii) The investor has 2 choices: buy the share now, which costs £10 or agree now to a forward contract to buy the share in 9 months’ time. The price of this forward contract is fixed now and so cannot depend on the price of the share in 9 months’ time which is unknown at this time. 欢迎咨询51作业君