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Unformatted text preview: The Theory of Interest  Solutions Manual 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 ( 29 12 .015 8 12 .015 20 .015 1000 1000 10.90751 $635.32. 17.16864 p a B Ra a = = = = 2. Use the retrospective method to bypass having to determine the final irregular payment. We then have ( 29 ( 29 ( 29 ( 29 ( 29 5 5 5 .12 10,000 1.12 2000 10,000 1.76234 2000 6.35283 $4918 to the nearest dollar. r B s = = = 3. The quarterly interest rate is .10/ 2 .025 j = = . Applying the retrospective method we have ( 29 4 4 4 1 r j B L j Rs = + and solving for L ( 29 ( 29 4 4 4 12,000 1500 4.15252 1.10381 1 $16,514 to the nearest dollar. r j B Rs L j + + = = + = 4. The installment payment is 12 20,000 R a = and the fourth loan balance prospectively is ( 29 ( 29 8 2 4 12 3 8 12 20,000 20,000 1 20,000 1 2 $17,143 to the nearest dollar. 1 1 2 p v B a a v = = = = 5. We have 5 15 20 20,000 and . P R B Ra a = = The revised loan balance at time 7 t = is ( 29 2 7 5 1 , p B B i ′ = + since no payments are made for two years. The revised installment payment thus becomes ( 29 2 7 15 13 20 13 1 20,000 . a i B R a a a + ′ = = 46 The Theory of Interest  Solutions Manual Chapter 5 6. The installment payment is 25 1 n L R a a = = . Using the original payment schedule 20 5 20 25 p a B Ra a = = and using the revised payment schedule 5 15 5 p B Ra Ka = + . Equating the two and solving for K we have 20 15 20 15 5 25 25 25 5 1 . a a a a K a a a a a = = 7. We have 15 .065 150,000 150,000 15,952.92 9.4026689 R a = = = and ( 29 ( 29 5 10 .065 15,952.92 7.1888302 114,682.83 p B Ra = = . 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 ( 29 15 20 6 1 6 .03 1000 1000 1.03 $641.86. P v + = = = 9. The installment payment is 20 10,000 R a = and applying formula (5.4) we have ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 10 20 11 1 11 20 20 10 10 10 10 10,000 10,000 .1 1 1 1 1000 1 1000 . 1 1 1 v I v a v v v v v + = = = = + + 47 The Theory of Interest  Solutions Manual Chapter 5 10. The quarterly interest rate is .10/ 4 .025 j = = . The total number of payments is 5 4 20 n = × = . Using the fact that the principal repaid column in Table 5.1 is a geometric progression, we have the answer ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 13 14 15 16 17 18 13 100 1 1 1 1 1 100 100 22.38635 15.14044 $724.59....
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This document was uploaded on 12/10/2011.
 Fall '09

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