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Chapter007

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

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The Theory of Interest - Solutions Manual 73 Chapter 7 1. The maintenance expense at time 6 t = is () 60 3000 1.06 4255.56 = . The projected annual return at time 6 t = is 61 30,000 .96 24,461.18 = . Thus, 6 24,461.18 4255.56 \$20,206 to the nearest dollar. R =− = 2. ( a ) ( ) 23 7000 4000 1000 5500 . ii i vv v Pi + + Thus, ( ) ()() .09 1000 7 4 .91743 .91743 5.5 .91743 75.05. P ⎡⎤ + + = ⎣⎦ ( b ) ( ) .10 1000 7 4 .90909 .90909 5.5 .90909 57.85. P + + = 3. Net cash flows are: Time NCF 0 3000 1 2000 1000 = 1000 2 4000 The IRR is found by setting ( ) 0 = , i.e. ( ) 2 2 3000 1000 4000 0 43 4 3 1 0 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, 5 10 5 600,000 300,000 2,000,000 1.12 \$544,037. Xaa = 5. Project P: ( ) 2 4000 2000 4000 . + + Project Q: ( ) 2 2000 4000 . vX v =+ Now equating the two expressions, we have ( ) ( ) ( ) 2 2 4000 2000 6000 0 4000 2000 1.1 6000 1.1 0 Xv v X +− = = and 2200 7260 4000 \$5460. X =+−=

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The Theory of Interest - Solutions Manual Chapter 7 74 6. ( a ) This Exercise is best solved by using the NPV functionality on a financial calculator. After entering all the NCF’s and setting I 15%, = we compute ( ) NPV .15 \$498,666. P == ( b ) We use the same NCF’s as in part ( a ) and compute IRR 13.72% = . 7. ( a ) The formula for ( ) 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 () 2 2 100 1 208 1 108.15 0 1 2.08 1 1.0815 rr +− + + = + + which can be factored as ( ) [ ] ( ) [ ] 11 . 0 5 . 0 3 . Thus, .05 r = and .03, so that .02. ij −= 9. Using one equation of value at time 2, t = we have 2 2 1000 1.2 1.2 0 1000 1.4 1.4 0 AB ++ = = 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: ( ) ( ) 5 5.04 600 1.04 600 5.416323 1.216653 3953.87 s Fund C: ( )( ) 5.05 600 600 5.525631 3315.38 s . A+B+C = 10,000 3953.87 3315.38 \$17,269 ++= to the nearest dollar.
The Theory of Interest - Solutions Manual Chapter 7 75 ( b ) We then have the equation of value () 10 10,000 1 17,269 i += s o t h a t 1 10 1.7269 1 .0562, or 5.62%. i =− = 11. If the deposit is D , then the reinvested interest is .08 , .16 , .24 , , .80 DDD D . We must adapt formula (7.7) for an annuity-due rather than an annuity-immediate. Thus, we have the equation of value ( ) 10 .04 10 .08 1000 DD I s so that 10 .04 10 .04 11.04 .08 .04 1000 1000 1000 . 10 10 2 10 12 D ss s == = +− ±± 12. The lender will receive a total accumulated value of 20 .05 1000 33,065.95 s = at the end of 20 years in exchange for the original loan of 10,000. Thus, we have the equation of value applying formula (7.9) 20 10,000 1 33,065.95 i and 1 20 3.306595 1 .0616, or 6.16%. i = 13. From formula (7.7) the total accumulated value in five years will be 5.03 5 5 1000 40 5412.18 .03 s . The purchase price P to yield 4% over these five years is 5 5412.18 1.04 \$4448 to the nearest dollar. P 14. Applying formula (7.10) we have 24 24 .035 110 1 5 100 283.3326 is + = so that 1 24 24 1 2.57575 and 2.57575 1 .04021. ii ′′ = = The answer is ( ) 2 2 .04021 .0804, or 8.04%. i

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The Theory of Interest - Solutions Manual Chapter 7 76 15. The yield rate is an annual effective rate, while the bond coupons are semiannual.
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Chapter007 - The Theory of Interest Solutions Manual...

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