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...

This preview shows page 1 - 4 out of 11 pages.

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
Image of page 1
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
Image of page 2
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:
Image of page 3
Image of page 4

You've reached the end of your free preview.

Want to read all 11 pages?

  • Fall '16
  • xyz

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes