Chapter010

# Chapter010 - The Theory of Interest Solutions Manual...

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The Theory of Interest - Solutions Manual Chapter 10 1. ( a ) We have ( 29 ( 29 ( 29 ( 29 ( 29 1 2 3 4 5 1000 1.095 1.0925 1.0875 1.08 1.07 \$3976.61. - - - - - + + + + = ( b ) The present value is greater than in Example 10.1 (1), since the lower spot rates apply over longer periods while the higher spot rates apply over shorter periods. 2. We have ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 3 4 1 2 3 4 1 2 3 4 2 3 4 1000 1 1.05 1 1.05 1 1.05 1 1.05 1 1.05 1.05 1.05 1.05 1000 1 \$4786.78. 1.09 1.081 1.0729 1.06561 s s s s - - - - + + + + + + + + = + + + + = 3. Since k s is differentiable over 0 4, k .002 .001 0 at 2 k d s k k dk = - = = which is a relative maximum or minimum. Computing values for 0,1,2,3,4 k = we obtain 0 1 2 3 4 .09 .0915 .092 .0915 .09. s s s s s = = = = = ( a ) Normal. ( b ) Inverted. 4. 117 Payment at Spot rate Accumulated value 0 t = .095 ( 29 5 1.095 1.57424 = 1 t = .0925 − .0025 ( 29 4 1.09 1.41158 = 2 t = .0875 − .0050 ( 29 3 1.0825 1.26848 = 3 t = .0800 − .0075 ( 29 2 1.0725 1.15026 = 4 t = .0700 − .0100 ( 29 1 1.06 1.06000 = 6.4646

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Chapter 10 5. Adapting Section 9.4 to fit this situation we have ( 29 ( 29 ( 29 4 2 2 10,000 1.0925 \$11,946.50. 1.05 1.04 = 6. ( a ) ( 29 2 2 930.49 1019.31 \$841.67. B A P P - = - = ( b ) ( 29 2 2 1000.00 1000.00 \$1000.00. B A C C - = - = ( c ) We have 2 .09 s = and ( 29 2 841.67 1.09 1000.00 confirming the statement. = 7. The price of the 6% bond per 100 is ( 29 6 .06 6 .12 6 100 1.12 75.33. P a - = + = The price of the 10% bond per 100 is ( 29 6 .10 6 .08 10 100 1.08 109.25. P a - = + = We can adapt the technique used above in Exercise 6. If we buy 10/6 of the 6% bonds, the coupons will exactly match those of the 10% bond. The cost will be ( 29 ( 29 10 10 75.33 125.55 and will mature for 100 . 6 6 = Thus, we have ( 29 ( 29 ( 29 6 6 4 125.55 109.25 1 100 6 s - + = and solving 6 .2645, s = or 26.45%. 8.
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Chapter010 - The Theory of Interest Solutions Manual...

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