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M12_MCDO8122_01_ISM_C12

# M12_MCDO8122_01_ISM_C12 - Chapter 12 Financial Engineering...

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Chapter 12 Financial Engineering and Security Design n Question 12.1 Let 06 . R e . = The present value of the dividends is 1 2 3 4 5 (1 50) 2 (2 50) 3 8 1317 R R R R R - - - - - + . + + . + = . . The note originally sells for 100 8 1317 91 868. - . = . With the 50 cent permanent increase, the present value of dividends rises by 1 2 3 4 5 2.0957 2 R R R R R - - - - - + + + + = to 10.2274, leading the note value to fall to100 10 2274 89 773. - . = . n Question 12.2 For this problem, let 1 1 03 . B . = 1. The prepaid forward price is 015 3 0 1200 1147 20. T S e e δ - -. × = = . 2. We have to solve the coupon, c , that solves 6 1 52 8 1147 20 1200 9 7467 5 4172 i i c B c = . + . = = = . . . 3. The prepaid forward price for 1 share at time t is 015 1200 ; P t t F e -. = for each semi-annual share, we can write the relevant prepaid forward price as 1200 D i where 015 2 . D e -. / = With this formulation we have a similar analysis for the fractional shares, c * : 6 * * 1 52 8 1200 1147 20 1200 007528 shares. 1200 5 845 i i c D c = . + . = = = . × . To interpret, we will receive 007528 . units of the index every six months. This has a current value of 1200 ( 007528) 9 0336. . = . We could quote c * in dollars (\$9.0336) instead of units.

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146 McDonald • Fundamentals of Derivatives Markets n Question 12.3 1. This is the two-year prepaid forward price: 015(2) 0 1200 1164 5. T S e e δ - -. = = . 2. As in Equation (12.5), 015(2) 0 8 1 1200(1 ) 4 762 7 4475 i P T t i S F e c P -. = - - = = = . . . 3. As in the Problem 12.2c, letting 015 4 , D e -. / = 8 8 * 8 * 8 1 1 1 1200 1200 1200 003757 shares, i i i i D c D D c D = = - + = = = . which is currently worth ( 29 003757 \$1200 \$4 5084. . = . n Question 12.4 The relevant two-year interest rate is ln(1 8763) 2 6 6023%. /. / = . 1. The embedded option is worth 247.91. The prepaid forward is worth 015(2) 1200 1164 53. e -. = . The bond price is worth the sum 1164 53 247 91 1412 44 . + . = . . 2. λ must solve 1164 53 247 91 1200 35 47 247 91 1431. λ λ . + . = = . / . = . n Question 12.5 As in the previous question, we use 6 6023%. r = . 1. The embedded option is worth 247.91. The bond price is worth 1200 ( 8763) 247 91 1299 47 . + . = . . 2. λ must solve 1200 ( 8763) 247 88 1200 59884. λ λ . + . = = . n Question 12.6 We continue to use 6.6023% as the relevant two-year interest rate. 1. The out-of-the-money option (i.e., K = 1500) is worth 141.56, making the bond have a value of 1164 53 247 91 141 56 1270 89. . + . - . = . 2. We must solve 1164 53 (247 91 141 56) 1200 λ . + . - . = for a solution of 3335. λ = . 3. If 1, λ = we have to adjust the strike (from Part (1), we know we have to lower K from 1500) to make the out-of-the-money option worth ( ) 1164 53 247 91 1200 212 44 C K = . + . - = . 1284 50. K = .
Chapter 12 Financial Engineering and Security Design 147 n Question 12.7 Let 06 5 5 5 5 7189 B P e -. × .

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