Chapter002

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

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The Theory of Interest - Solutions Manual 14 Chapter 2 1. The quarterly interest rate is () 4 .06 .015 44 i j === and all time periods are measured in quarters. Using the end of the third year as the comparison date 12 42 8 3000 1 2000 5000 j Xv v ++ = + ( ) ( ) ( ) 2000 .94218 5000 .65910 3000 1.19562 $1593.00. X =+− = 2. The monthly interest rate is 12 .18 .015. 12 12 i j == = Using the end of the third month as the comparison date ( ) 32 1000 1 200 1 300 1 1000 1.04568 200 1.03023 300 1.015 $535.13. Xj j j =+ + + =− = 3. We have 51 0 5 200 500 400.94 vv v += 10 5 5 5 .40188 .40188 or 1 2.4883. vi = = Now using time 10 t = as the comparison date ()() 10 5 2 1001 1201 100 2.4883 120 2.4883 $917.76. Pi i = 4. The quarterly discount rate is 1/41 and the quarterly discount factor is 11 / 4 1 4 0 / 4 1 −= . The three deposits accumulate for 24, 16, and 8 quarters, respectively. Thus, ( ) 24 16 8 35 40 40 40 28 100 1.025 1.025 1.025 41 41 41 A −−− ⎡⎤ ⎛⎞ + ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ . However, 1 40 1.025 41 = so that ( ) 25 19 13 28 100 1.025 1.025 1.025 $483.11 A + = .
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The Theory of Interest - Solutions Manual Chapter 2 15 5. ( a ) At time 10 t = , we have ( ) ( ) () 100 1 10 100 1 5 with .05 200 1500 .05 $275. Xi i i =+ ++ = = ( b ) At time 15 t = , we have ( ) ( ) ( ) 1 5 100 1 15 100 1 10 with .05 1.25 200 2500 .05 325 i i i X += + + + = = and 325 $260 1.25 X == . 6. The given equation of value is 1000 1.06 2000 1.04 nn = 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 25 3000 2000 5000 5000 3000 2000 1 5000 1 1 1 and 3000 2000 .94 5000 .94 1 .94 n n vvv dd d + +=+ +− =− + + since .06 d = . Simplifying, we have ( ) 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 vv which is a quadratic in n v . Solving
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The Theory of Interest - Solutions Manual Chapter 2 16 () ( ) 2 10 11 4 15 22 .618034 rejecting the negative root. nn n vv v +− = −± − −+ == = We are given .08 i = , so that 1.08 1/.61803 1.618034 ln1.618034 and 6.25 years.
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This note was uploaded on 11/18/2010 for the course MATH 320 taught by Professor Dr.k during the Spring '10 term at Nevada.

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

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