Investment Science Solutions

# Investment Science Solutions - Investment Science Chapter 3...

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Investment Science Chapter 3 Dr. James A. Tzitzouris 3.1 Use A = rP 1 - 1 (1+ r ) n with r = 7 / 12 = 0 . 58%, P = \$25 , 000, and n = 7 × 12 = 84, to obtain A = \$377 . 32. 3.2 Observe that since the net present value of X is P , the cash flow stream arrived at by cycling X is equivalent to one obtained by receiving payment of P every n + 1 periods (since k = 0 , . . . , n ). Let d = 1 / (1 + r ). Then P = P k =0 ( d n +1 ) k . Solving explicitly for the geometric series, we have that P = P 1 - d n +1 . Denoting the annual worth by A , we must have A = rP 1 - d n , so that solving for P as a function of P and substituting the result into the equation for A , we arrive at A = r 1 - d n +1 1 - d n P . 1

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Investment Science Chapter 4 Solutions to Suggested Problems Dr. James A. Tzitzouris 4.1 (One forward rate) f 1 , 2 = (1 + s 2 ) 2 (1 + s 1 ) - 1 = 1 . 069 2 1 . 063 - 1 = 7 . 5% 4.2 (Spot Update) Use f 1 ,k = (1 + s k ) k 1 + s 1 1 / ( k - 1) - 1 . Hence, for example, f 1 ,k = (1 . 061) 6 1 . 05 1 / 5 - 1 = 6 . 32% . All values are f 1 , 2 f 1 , 3 f 1 , 4 f 1 , 5 f 1 , 6 5 . 60 5 . 90 6 . 07 6 . 25 6 . 32 1
4.3 (Construction of a zero) Use a combination of the two bonds: let x be the number of 9% bonds, and y teh number of 7% bonds. Select x and y to satisfy 9 x + 7 y = 0 , x + y = 1 . The first equation makes the net coupon zero. The second makes the face value equal to 100. These equations give x = - 3 . 5, and y = 4 . 5, respectively. The price is P = - 3 . 5 × 101 . 00 + 4 . 5 × 93 . 20 = 65 . 90. 4.5 (Instantaneous rates) (a) e s ( t 2 ) t 2 = e s ( t 1 ) t 1 e f t 1 ,t 2 ( t 2 - t 1 ) = f t 1 ,t 2 = s ( t 2 ) t 2 - s ( t 1 ) t 1 t 2 - t 1 (b) r ( t ) = lim t t 1 s ( t ) t - s ( t 1 ) t 1 t - t 1 = d [ s ( t ) t ] dt = s ( t ) + s ( t ) (c) We have d (ln x ( t )) = r ( t ) dt, = s ( t ) dt + s ( t ) dt, = d [ s ( t ) t ] . Hence, ln x ( t ) = ln x (0) + s ( t ) t, and finally that x ( t ) = x (0) e s ( t ) t . This is in agreement with the invariance property of expectation dynamics. Investing continuously give the same result as investing in a bond that matures at time t . 4.6 (Discount conversion) 2

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The discount factors are found by successive multiplication. For example, d 0 , 2 = d 0 , 1 d 1 , 2 = 0 . 950 × 0 . 940 = 0 . 893 . The complete set is 0 . 950, 0 . 893, 0 . 770, 0 . 707, 0 . 646. 4.7 (Bond taxes) Let t be the tax rate, x i be the number of bond i purchased, c i be the coupon of bond i , p i be the price of bond i . To create a zero coupon bond, we require, first, that the after tax coupons match. Hence x 1 (1 - t ) c 1 + x 2 (1 - t ) c 2 = 0 , which reduces to x 1 c 1 + x 2 c 2 = 0 . Next, we require that the after tax final cash flows match. Hence p 0 = x 1 p 1 + x 2 p 2 . Using this last relation in the equationfor final cash flow, we find x 1 + x 2 = 1 . Combining these equations, we find that p 0 = c 2 p 1 - c 1 p 2 c 2 - c 1 . After plugging in the given values, we find that p 0 = 37 . 64 . 4.8 (Real zeros) We assume that with coupon bonds there is a capital gains tax at maturity. We replicate the zero- coupon bond’s after-tax cash flows using bonds 1 and 2. Let x i be the amount of bond i required 3
(a) d 0 , 1 d 0 , 2 d 0 , 3 d 0 , 4 d 0 , 5 d 0 , 6 0 . 9524 0 . 9018 0 . 8492 0 . 7981 0 . 7472 0 . 7010 = NPV = 9 . 497 (b) Year 0 1 2 34 5 6 Cash Flow - 40 10 10 10 10 10 10 Discount 0 . 9524 0 . 9469 0 . 9416 0 . 9399 0 . 9362 9381 PV( n ) 9 . 497 51 . 970 44 . 324 36 . 453 28 . 144 19 . 381 10 . 000 4.12 (Pure duration) P ( λ ) = n k =0 x k (1 + s k /m ) - k = n k =0 x k (1 + s 0 k /m ) e λ/m - k , dP ( λ ) = n k =0 x k - k m (1 + s 0 k /m ) e λ/m - k - 1 (1 + s 0 k /m ) e λ/m , = n k =0 x k - k m (1 + s k /m ) - k , - 1 P ( λ ) dP ( λ ) = n k =0 x k ( k m ) (1 + s k /m ) - k n k =0 x

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