Solutions to Extra Problems 4

# Solutions to Extra Problems 4 - ACTSC 231 Extra Problem Set...

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Unformatted text preview: ACTSC 231 Extra Problem Set 4 Solutions 1 1.05 12 1 1.05 5 5  700 1. 05 5  7025. 951 1. a. B 7  500a 5|.05  700v a 12|.05  500 .05 .05 I 8 . 05B 7 . 05 7025. 951  351. 29755 P 8  500 351. 29755  148. 702 45 1 1.05 10 b.B 14  700a 10|.05  700 .05  5405. 2144 I 15 . 05B 14 . 05 5405. 2144  270. 26072 P 15  700 270. 26072  429. 739 28 2. . 1 a. B 0  3700a 25|i , where 1  i 12  1  .08 2 Í i  1. 04 6 1 . 0065582 2 B 0  3700a 25|.65582%  85, 058. 66 b. Amount to refinance  O.P.  penalty  85,058.66  750  85,808.66 Amortize over 25 1 months R  85,808.66 , where 1  j 12  1  .07 Í j  1. 035 6 1 . 005750039 a 25|i 2 R  a85,808.66  3, 694. 80  3, 700  refinance! 25|.5750039% 3. R  6,500 , where i  13.5%  6. 75% Í R  a6,500  1078. 02 a 8 |i 2 8 |6.75% Principal in the 5 th payment  change in O.P. between time 4 and 5. B 4 B 5  6, 500 1. 0675 4 1078. 02 s 4 |6.75% 6, 500 1. 0675 5 1078. 02 s 5 |6.75%  3672. 17 2842. 02  830. 15 OR B 4 B 5  1078. 02 a 4 |6.75% 1078. 02 a 3 |6.75%  3672. 18 2842. 04  830. 14 Interest in the 5 th payment 1078.02 - 830.15  247.87 (247.88) Principal in the 6 th payment B 5 B 6  6, 500 1. 0675 5 1078. 02 s 5 |6.75% 6, 500 1. 0675 6 1078. 02 s 6 |6.75%  2842. 02 1955. 84  886. 18 OR B 5 B 6  1078. 02 a 3 |6.75% 1078. 02 a 2 |6.75%  2842. 02 1955. 85  886. 19 Interest in the 6 th payment 1078.02 - 886.18  191.84 (191.83) NOTE: Principal in the k th payment is: B k 1 B k  R a n k1| R a n k|  R 1i nk 1 4. 13 th payment has 600 principal  200 interest Interest due at the end of the 13 th month is B 12 i, where 1  i 12  1  .10 2  i  1. 05 1/6 1 . 81648% 2 B 12  200/ . 0081648  24, 495. 26 But B 12  A 1  i 12 800 s 12|i solve for A A  B 12  800 s 12|i 1  i 12  24, 495. 26  800 s 12|.81648% 1. 0081648 5. B 12  780a 24|i i  14%  1. 16% 12  780a 24|1.16%  16, 245. 64 Amount needed to refinance, 16, 245. 64  3 780  18, 585. 64 1 i  1. 04 6 1 . 65582% refinance rate 1  i 12  1  .08 2 2 18,585.64 don’t refinance R  a 24|.65582%  839. 48  780 12  31, 327. 29 6. a. j  i 12  1  .07 1 . 575003948% & i  2 200000 R  a 240|.575003948%  1538. 62 12 2/12 1 .07 2 2 1  7. 1225% b. Bal. after 5 years: B 60  200000 1. 00575003948 60 1538. 62s 60|.575003948% 1000s 5 |7.1225%  166, 485. 18 Principal paid during the 1st 5 years: 200, 000 166, 485. 18  33, 514. 82 Int. paid during the 1st 5 years: 1538. 62 60  1000 5 33, 514. 82  63, 802. 38 c. The bal. after n years is: 200000 1. 00575003948 12n 1538. 62s 12n|.575003948% 1000s n |7.1225% We want to find n such that the bal. after n years is zero. I.e. solve: 12n n 200000 1. 00575003948 12n 1538. 62 1.00575003948 1 1000 1.071225 1  0 Note .00575003948 .071225 1. 00575003948 12  1. 071225 1.071225 n 1 1.071225 n 1 0 Í 200000 1. 071225 n 1538. 62 .00575003948 1000 .071225 n n 267584. 2489 1. 071225  267584. 2489 14040. 01404 1. 071225 n Í 200000 1. 071225 3.450251836 Í 81624. 2629 1. 071225 n  281624. 2629 Í 1. 071225 n  3. 450251836 Í n  lnln 1.071225 B 216  200000 1. 00575003948 216 1538. 62s 216|.575003948% 1000s 18|7.1225%  1. 17 At the end of 18 years we don’t need a full extra payment of \$1000, a payment of 998.83 is sufficient. ...
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