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Unformatted text preview: Investment Science Chapter 3 Dr. James A. Tzitzouris <[email protected]> 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 ∞ X 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 That is, A is directly proportional to P ∞ . 3.3 (a) To find the life expectancy, we multiply each age of death by its probability. Thus the life expectancy is L = 90 × . 07 + 91 × . 08 + ··· + 101 × . 04 = 95 . 13 years. (b) To find the present value of an annuity that ends at age 95.13, we calculate the values for ages 95 and 96. From the standard formula P = A r 1 1 (1 + r ) n , with n = 5 and n = 6, we find that P 95 = $39 , 927 and P 96 = $46 , 228. Then, taking P = . 87 × P 95 +0 . 13 × P 96 (the average of the two values, weighted by fraction of the interval), we have...
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This note was uploaded on 04/02/2010 for the course EE 204 taught by Professor Won during the Spring '10 term at 카이스트, 한국과학기술원.
 Spring '10
 Won
 Electromagnet

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