Exercises_12.1.docx - Exercises 12.1 Exercise 12.1 Solution 1 a This is an ordinary simple annuity because Payments are made at the end of each payment

# Exercises_12.1.docx - Exercises 12.1 Exercise 12.1 Solution...

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Exercises 12.1 Exercise 12.1, Solution 1: 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 = 0.04 4 = 0.01 quarterly -n 1- (1+ i) i PV = PMT 10,000 = PMT [ 1 −( 1 + 0.01 ) 4 0.01 ] 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. 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 7537.19 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 0.00 Total \$10,251.24 \$251.24 \$10,000.00 c. Total amount paid to amortize the loan = 2562.81 × 4 = \$10,251.24 Therefore, the total amount paid to amortize the loan is \$10,251.24 d. Total payment of the loan = 2562.81 × 4 = \$10,251.24 Therefore, the cost of financing (total interest paid) in the loan period = 10,251.24 – 10,000 = \$251.24. Exercise 12.1, Solution 3: a. This is an ordinary general annuity because: Payments are made at the end of each payment period (quarterly) Compounding period (monthly) ≠ payment period (quarterly) Last updated: November 19, 2014
c = Number of compounding periods per year Number of payments per year = 12 4 n =? j = 7.55% = 0.0755, m = 12 m j i = 0.0755 12 = 0.006292 monthly i 2 = (1 + i ) c – 1 = (1 + 0.006292) (12/4) – 1 = 0.018994… quarterly. Loan amount =\$75,400 ¿ (1-0.4) = \$45,240 n =− ln ( 1 i 2 × PV PMT ) ln ( 1 + i ) n =− ln ( 1 0.018994 × 45 , 240 2098.25 ) ln ( 1 + 0.018994 ) n = 27.999254 n = 28 payments N I/Y P/Y C/Y PV PMT FV ? 7.55 4 12 45,240 2098.2 0 From the calculator computations shown, we get the N= 27.999254 t = n number of payments t = n m = 28 4 = 7 years Therefore, the amortization period is 7 years. b. Contracting the amortization schedule: Payment Number Amount Paid Interest portion Principal Portion Principal Balance 0 - - - \$45,240.00 1 \$2098.25 \$859.29 \$1238.96 44,001.04 2 2098.25 835.76 1262.49 42,738.54 27 2098.25 77.47 2020.78 2057.62 28 2096.70 39.08 2057.62 0.00 Total \$58,749.45 \$13,509.45 \$45,240.00 c. Total payment of the loan = (2098.25 × 27) + 2096.70= \$58,749.45 Therefore, the cost of financing (total interest paid) in the loan period = 58,749.45– 45,240 = \$13,509.45. Exercise 12.1, Solution 5: This is an ordinary general annuity because: Payments are made at the end of each payment period (monthly) Compounding period (daily) ≠ payment period (monthly) Last updated: November 19, 2014
c = Number of compounding periods per year Number of payments per year = 365 12 n = 12 payments/year × 3 years = 36 monthly payments j = 5.75% = 0.0575, m = 365 m j i = 0.0575 365 = 0.000157534 daily i 2 = (1 + i ) c – 1 = (1 + 0.000157534) (365/12) – 1 = 0.004802786… monthly PV = PMT n 2 2 1 (1 i ) i 24,500 = PMT [ 1 ( 1 + 0.004802786 ) 36 0.004802786 ] Solving for PMT , we get PMT = \$742.7131684… N I/Y P/Y C/Y PV PMT FV 36 5.75 12 365 24,500 ? 0 From the calculator computations shown, we get the PMT = – 742.7131684… After one year he would have made 12 payment. We can find the principal balance on the loan after one year using the retrospective method: BAL Focal Date = FV Original Loan – FV Payment Made Until the Focal Date BAL Focal Date = PV (1 + i ) n – PMT [ ( 1 + i ) n 1 i ] = 24,500 (1 + 0.004802786) 12 – 742.71 12 (1 ) 1 0.004802786 0.004802786 = 25,949.92161… - 9151.796193 = 16,798.12542…. = \$16,798.13 Therefore, the principal balance on the loan after one year is \$16,798.13 Exercise 12.1, Solution 7: This is an ordinary simple annuity because:

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