Chapter005

# Chapter005 - The Theory of Interest Solutions Manual...

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The Theory of Interest - Solutions Manual 46 Chapter 5 1. The quarterly interest rate is .06/ 4 .015 j = = . The end of the second year is the end of the eighth quarter. There are a total of 20 installment payments, so 20 .015 1000 R a = and using the prospective method ( ) 12 .015 8 12 .015 20 .015 1000 1000 10.90751 \$635.32. 17.16864 p a BR a a == = = 2. Use the retrospective method to bypass having to determine the final irregular payment. We then have () ( ) 5 5 5.12 10,000 1.12 2000 10,000 1.76234 2000 6.35283 \$4918 to the nearest dollar. r Bs =− = 3. The quarterly interest rate is .10/ 2 .025 j = = . Applying the retrospective method we have 4 4 4 1 r j BL j R s =+− and solving for L 4 4 4 12,000 1500 4.15252 1.10381 1 \$16,514 to the nearest dollar. r j s L j + + + = 4. The installment payment is 12 20,000 R a = and the fourth loan balance prospectively is ( ) ( ) 82 4 12 3 8 12 20,000 20,000 1 20,000 1 2 \$17,143 to the nearest dollar. 11 2 p v Ba av −− = = 5. We have 5 15 20 20,000 and . P RB R a a The revised loan balance at time 7 t = is 2 75 1, p BB i =+ since no payments are made for two years. The revised installment payment thus becomes ( ) 2 71 5 13 20 13 1 20,000 . a i B R aa a +

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The Theory of Interest - Solutions Manual Chapter 5 47 6. The installment payment is 25 1 n L R aa == . Using the original payment schedule 20 5 20 25 p a BR a a and using the revised payment schedule 5 15 5 p a K a =+ . Equating the two and solving for K we have 20 15 20 15 52 5 2 5 2 5 5 1 . a a K a a a ⎛⎞ =− = ⎜⎟ ⎝⎠ 7. We have 15 .065 150,000 150,000 15,952.92 9.4026689 R a = and ( ) ( ) 5 10 .065 15,952.92 7.1888302 114,682.83 p a . The revised fifth loan balance becomes 5 114,682.83 80,000 194,682.83 B = and the revised term of the loan is 15 5 7 17. n ′ = −+= Thus, the revised installment payment is 17 .075 194,682.83 194,682.83 \$20,636 to the nearest dollar. 9.4339598 R a === 8. The quarterly interest rate is .12/ 4 .03. j = = Directly from formula (5.5), we have () 15 20 6 1 6. 0 3 1000 1000 1.03 \$641.86. Pv −+ = 9. The installment payment is 20 10,000 R a = and applying formula (5.4) we have ( )( ) 10 20 11 1 11 20 20 10 10 10 10 10,000 10,000 .1 1 1 1 1000 1 1000 . 1 11 v Iv av v v vv = +
The Theory of Interest - Solutions Manual Chapter 5 48 10. The quarterly interest rate is .10/ 4 .025 j = = . The total number of payments is 54 2 0 n =×= . Using the fact that the principal repaid column in Table 5.1 is a geometric progression, we have the answer ()()()()() () 13 14 15 16 17 18 13 100 1 1 1 1 1 100 100 22.38635 15.14044 \$724.59. iiiii ss ⎡⎤ ++ + ⎣⎦ =− = = 11. ( a ) We have 6 4 6 10 p i ij Ba v a =+ so that ( ) 6 54 i Ii Bi a v a =⋅ = + . ( b ) After 10 years, the loan becomes a standard loan at one interest rate. Thus applying formula (5.5) 20 15 1 6 15 . j j P vv −+ = = 12. After the seventh payment we have 7 13 p = . If the principal 20 8 1 13 8 P == in the next line of the amortization schedule is also paid at time 7 t = ; then, in essence, the next line in the amortization schedule drops out and we save 13 1 v in interest over the life of the loan. The loan is exactly prepaid one year early at time 19 t = . 13. ( a ) The amount of principal repaid in the first 5 payments is 5 2 3 5 05 5 10 5 4 9 10 10 11 1 . 4 . p a Lv BBL B L a L L L L aav ⎛⎞ −− −= = = = = = ⎜⎟ ⎝⎠ ( b ) The answer is ( ) 5 5 3 1.

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Chapter005 - The Theory of Interest Solutions Manual...

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