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Chapter002

# Chapter002 - The Theory of Interest Solutions Manual...

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The Theory of Interest - Solutions Manual Chapter 2 1. The quarterly interest rate is ( 29 4 .06 .015 4 4 i j = = = and all time periods are measured in quarters. Using the end of the third year as the comparison date ( 29 12 4 28 3000 1 2000 5000 j X v v + + = + ( 29 ( 29 ( 29 2000 .94218 5000 .65910 3000 1.19562 \$1593.00. X = + - = 2. The monthly interest rate is ( 29 12 .18 .015. 12 12 i j = = = Using the end of the third month as the comparison date ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 3 2 1000 1 200 1 300 1 1000 1.04568 200 1.03023 300 1.015 \$535.13. X j j j = + - + - + = - - = 3. We have 5 10 5 200 500 400.94 v v v + = ( 29 10 5 5 5 .40188 .40188 or 1 2.4883. v v v i = = + = Now using time 10 t = as the comparison date ( 29 ( 29 ( 29 ( 29 10 5 2 100 1 120 1 100 2.4883 120 2.4883 \$917.76. P i i = + + + = + = 4. The quarterly discount rate is 1/41 and the quarterly discount factor is 1 1/ 41 40/ 41 - = . The three deposits accumulate for 24, 16, and 8 quarters, respectively. Thus, ( 29 ( 29 ( 29 ( 29 24 16 8 3 5 40 40 40 28 100 1.025 1.025 1.025 41 41 41 A - - - = + + . However, 1 40 1.025 41 - = so that ( 29 ( 29 ( 29 ( 29 25 19 13 28 100 1.025 1.025 1.025 \$483.11 A = + + = . 14

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The Theory of Interest - Solutions Manual Chapter 2 5. ( a ) At time 10 t = , we have ( 29 ( 29 ( 29 100 1 10 100 1 5 with .05 200 1500 .05 \$275. X i i i = + + + = = + = ( b ) At time 15 t = , we have ( 29 ( 29 ( 29 ( 29 ( 29 1 5 100 1 15 100 1 10 with .05 1.25 200 2500 .05 325 X i i i i X + = + + + = = + = and 325 \$260 1.25 X = = . 6. The given equation of value is ( 29 ( 29 1000 1.06 2000 1.04 n n = so that [ ] 1.06 2 1.04 ln1.06 ln1.04 ln 2 n n = - = and .693147 36.4 years .058269 .039221 n = = - . 7. The given equation of value is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 5 2 5 2 5 3000 2000 5000 5000 3000 2000 1 5000 1 1 1 and 3000 2000 .94 5000 .94 1 .94 n n n n v v v d d d + + = + + - = - + - + = + since .06 d = . Simplifying, we have ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 4767.20 8669.52 .94 4767.20 .94 .54988 8669.52 ln .94 ln .54988 ln .54988 and 9.66 years. ln .94 n n n n = = = = = = 8. The given equation of value is 2 100 100 100 n n v v = + which is a quadratic in n v . Solving 15
The Theory of Interest - Solutions Manual Chapter 2 ( 29 ( 29 ( 29 2 1 0 1 1 4 1 1 1 5 2 2 .618034 rejecting the negative root.

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