Chapter007

# Chapter007 - The Theory of Interest - Solutions Manual...

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Unformatted text preview: The Theory of Interest - Solutions Manual Chapter 7 1. The maintenance expense at time 6 t = is ( 29 6 0 3000 1.06 4255.56- = . The projected annual return at time 6 t = is ( 29 6 1 30,000 .96 24,461.18- = . Thus, 6 24,461.18 4255.56 \$20,206 to the nearest dollar. R =- = 2. ( a ) ( 29 2 3 7000 4000 1000 5500 . i i i v v v P i = - +- + Thus, ( 29 ( 29 ( 29 ( 29 2 3 .09 1000 7 4 .91743 .91743 5.5 .91743 75.05. P =- +- + = ( b ) ( 29 ( 29 ( 29 ( 29 2 3 .10 1000 7 4 .90909 .90909 5.5 .90909 57.85. P =- +- + = - 3. Net cash flows are: Time NCF 3000 1 2000 1000 = 1000 2 4000 The IRR is found by setting ( 29 P i = , i.e. ( 29 ( 29 2 2 3000 1000 4000 4 3 4 3 1 v v v v v v- + + = +- =- + = so that 3 , 4 v = rejecting the root 1. v = - Finally, 4 1 , 3 i + = and 1 , 3 i = so 3. n = 4. The equation of value equating the present values of cash inflows and cash outflows is 5 10 5 2,000,000 600,000 300,000 at 12% Xv a a i + =- = . Therefore, ( 29 5 10 5 600,000 300,000 2,000,000 1.12 \$544,037. X a a =-- = 5. Project P: ( 29 2 4000 2000 4000 . v v P i = - + + Project Q: ( 29 2 2000 4000 . v Xv P i = +- Now equating the two expressions, we have ( 29 ( 29 ( 29 ( 29 2 2 4000 2000 6000 0 4000 2000 1.1 6000 1.1 X v v X +-- = +-- = and 2200 7260 4000 \$5460. X = +- = 73 The Theory of Interest - Solutions Manual Chapter 7 6. ( a ) This Exercise is best solved by using the NPV functionality on a financial calculator. After entering all the NCFs and setting I 15%, = we compute ( 29 NPV .15 \$498,666. P = = - ( b ) We use the same NCFs as in part ( a ) and compute IRR 13.72% = . 7. ( a ) The formula for ( 29 P i in Exercise 2 has 3 sign changes, so the maximum number of positive roots is 3. ( b ) Yes. ( c ) There are no sign changes in the outstanding balances, i.e. 7000 to 3000 to 4000 at 0. i = Taking into account interest in the range of 9% to 10 % would not be significant enough to cause any sign changes. 8. The equation of value at time 2 t = is ( 29 ( 29 ( 29 ( 29 2 2 100 1 208 1 108.15 0 1 2.08 1 1.0815 r r r r +- + + = +- + + which can be factored as ( 29 [ ] ( 29 [ ] 1 1.05 1 1.03 . r r +- +- Thus, .05 r = and .03, so that .02. i j- = 9. Using one equation of value at time 2, t = we have ( 29 ( 29 ( 29 ( 29 2 2 1000 1.2 1.2 1000 1.4 1.4 A B A B + + = + + = or 1.2 1440 1.4 1960. A B A B + = - + = - Solving two equations in two unknowns gives 2600 A = - and 1680. B = 10. ( a ) Adapting formula (7.6) we have: Fund A: 10,000 Fund B: ( 29 ( 29 ( 29 ( 29 5 5 .04 600 1.04 600 5.416323 1.216653 3953.87 s = = Fund C: ( 29 ( 29 5 .05 600 600 5.525631 3315.38 s = = . A+B+C = 10,000 3953.87 3315.38 \$17,269 + + = to the nearest dollar....
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Chapter007 - The Theory of Interest - Solutions Manual...

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