Ch.12_Solution_Manual_Ed.1_v8_

# Ch.12_Solution_Manual_Ed.1_v8_ - Exercises 12.1 Exercise...

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Exercises 12.1 Exercise 12.1, Solution 1: This is an ordinary simple annuity because: Payments are made at the end of each payment period (quarterly) Compounding period (quarterly) = payment period (quarterly) n = 4 payments/year × 5 years = 20 quarterly payments. j = 5.5% = 0.055, m = 4 m j i = = = 0.01375 quarterly -n 1- (1+ i) i PV = PMT 35,560 = PMT Solving for PMT , we get PMT = \$2045.785884… N I/Y P/Y C/Y PV PMT FV 20 5.5 4 4 35,560 ? 0 From the calculator computations shown, we get the PMT = 2045.785884 Therefore, the size of the payments is \$2045.79. Total payment of the loan = 2045.79 × 20 = \$40,915.80 Therefore, the total interest paid in the loan period = 40,915.80 – 35,560 = \$5355.80 Exercise 12.1, Solution 3: a. This is an ordinary simple annuity because: Payments are made at the end of each payment period (quarterly) Compounding period (quarterly) = payment period (quarterly) n = 4 payments/year × 1 years = 4 quarterly payments. j = 4% = 0.04, m = 4 m j i = = = 0.01 quarterly -n 1- (1+ i) i PV = PMT 10,000 = PMT Solving for PMT , we get PMT = \$2562.810939. .. N I/Y P/Y C/Y PV PMT FV 4 4 4 4 10,000 ? 0 From the calculator computations shown, we get the PMT = 2562.810939
Therefore, the size of the payments is \$2562.81. b. Total payment of the loan = 2562.81 × 4 = \$10,251.24 Therefore, the total interest paid in the loan period = 10,251.24 – 10,000 = \$251.24 c. Contracting an amortization schedule: Payment Number Amount Paid Interest portion Principal Portion Principal Balance 0 - - - \$10,000.00 1 \$2562.81 \$100.00 \$2462.81 7527.19 2 2562.81 75.37 2487.44 5049.75 3 2562.81 50.50 2512.31 2537.44 4 2562.81 25.37 2537.44 0 Exercise 12.1, Solution 5: a. This is an ordinary simple annuity because: Payments are made at the end of each payment period (annually) Compounding period (annually) = payment period (annually) n = 1 payment/year × 5 years = 5 annual payments. i = j = 7.54% = 0.0754 The amount of the loan = 750,400(1-0.4) = \$450.240.00 -n 1- (1+ i) i PV = PMT 450,240 = PMT Solving for PMT , we get PMT = \$111,401.7157… N I/Y P/Y C/Y PV PMT FV 5 7.54 1 1 450,240 ? 0 From the calculator computations shown, we get the PMT = 111,401.7157 Therefore, the size of the payments is \$111,401.72. b. Total payment of the loan = 111,401.72 × 5 = \$557,008.60 c. Contracting an amortization schedule: Payment Number Amount Paid Interest portion Principal Portion Principal Balance 0 - - - \$450,240.00 1 \$111,401.72 \$33,948.10 \$77,453.62 372,786.38 2 111,401.72 28,108.09 83,293.63 289,492.75 3 111,401.72 21,827.75 89,573.97 199,918.78 4 111,401.72 15,073.88 96,327.84 103,590.94 5 111,401.70 7810.76 103,590.94 0.00

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Exercise 12.1, Solution 7: This is an ordinary simple annuity because: Payments are made at the end of each payment period (monthly) Compounding period (monthly) = payment period (monthly) n = 12 payments/year × 7 years = 84 monthly payments. j
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## Ch.12_Solution_Manual_Ed.1_v8_ - Exercises 12.1 Exercise...

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