Question

Ms. Cressida bought a car for $48,000 exactly three years ago. After making an up-front equity payment of $5,000, she borrowed the rest of the car value from her bank in the form of a five-year loan. She negotiated a loan rate of 2.5% APR with semi-annual compounding. She makes loan payments of an equal dollar amount every two weeks (i.e., biweekly), and her first loan payment was due two weeks after she signed the loan contract.

1. What is Cressida’s biweekly loan payment?                                       (2 marks)
2. What is Cressida’s current loan balance?                                           (2 marks)
3. What is the total amount of interest that Cressida would have paid to the bank after five years of loan payments?            (2 marks)
4. Show the amortization schedule (table) for the first five payments and the last five payments in the amortization table provided below. Round your answers in the table to two decimal places.              (2 marks)

Answer 1

Number of payments a year is assumed at 26.

2.5% APR compounded semi annually is equal to 2.485691% compounded biweekly as follows:

B 1 Given Given 2.50% Semi annual 2 Yearly Nominal Rate (R) 3 Compounding frequency 4 Times compounded a year (t) 5 EAR=((1+R/t)^t)-1 ((1+C2/C4)^C4)-1 0.0251563 6 7 Frequency 8 Times compounded a year (t) 9 Weekly rate 10 Yearly rate (Weekly compounding) 11 ((1+C5)^1/08)-1 C9*C8 Biweekly 26 0.000956 0.0248569 2.485691%

Loan amount= Cost - Upfront payment= $48,000 - $5,000 = $43,000.

At the above rate,

(a) Biweekly loan payment= $ 351.91 Details as follows:

7 8 A B C 1 Biweekly payments 2 Assuming payments at the end of each period 3 EMI is calculated using the formula EMI=[P*r*(1+r)^n]/[(1+r)^n)-1 4 Where P= Principal (loan amount), 5 n= Number of installments. ie if N= number of years, n=N*26 6 r=Rate of interest per month in decimals ie. If R= Yearly rate, r=R/(100*26) All amounts in $ Given that 9 Principal (loan amount) (P)= 43,000 No. of years (N)= 10 Rate of interest (% p.a)=R= 2.485691 No. of months (n)= 130 11 12 Calculations: 13 Formula Cell reference Value 14 Interest rate per month in decimals (r )= R/(100*12) D10/(100*26) 0.000956 15 No. of months (n)= N*12 G9*26 130 16 (1+r)= 1+G14 1.000956 17 (1+r)^n= G164G10 1.1322708 18 [(1+r)^n]-1= G17-1 0.1322708 19 P*r*(1+r)^n= D9*G14*G17 46.547102 20 Biweekly instalments [P*r*(1+r)^n]/[(1+r)^n]-1 G19/618 351.90754 21 Rounded to $ 351.91 22

(b) Balance after 3 years= $17,843.46

Relevant portion of amortization schedule as follows:

Shedule of amortization Period Beginning Biweekly Biweekly interest Principal End Biweek balance interest Payments component Component balance rate =Col (2)*(3) Col (4)-(5) =Col (2)-(6) (2) (3) (4) (5) (6) (7) 43000 0.000956 351.91 41.1095124 310.80 4 2689.20 2 42689.20 0.000956 351.91 40.81237855 311.10 42378.11 77 78 79 18512.20 0.000956 18177.99 0.000956 17843.46 0.000956 351.91 17.69830987 351.91 17.37879409 351.91 17.05897285 334.21 334.53 334.85 18177.99 17843.46 17508.61

(c) Total amount to be paid over 5 years= $351.91*130 installments = $45,747.98

Loan amount= $43,000

Therefore, total amount of interest that would have been paid over 5 years

= $45,747.98-$43,000= $2,747.98

(d) Amortization schedule for the first and last five payments is given below:

(5) (6) (7) Shedule of amortization Period Beginning Biweekly Biweekly Interest Principal End Biweek balance interest Payments component Component balance rate =Col (2)*(3) = Col (4)-(5) = Col (2)-(6) (1) (2) (3) (4) 1 43000 0.000956 351.91 41.11 310.80 42689.20 2 42689.20 0.000956 351.91 40.81 311.10 42378.11 3 42378.11 0.000956 351.91 40.51 311.39 42066.71 4 42066.71 0.000956 351.91 40.22 311.69 41755.02 5 41755.02 0.000956 351.91 39.92 311.99 41443.04 126 127 128 129 130 1754.50 0.000956 1404.27 0.000956 1053.71 0.000956 702.81 0.000956 351.57 0.000956 351.91 351.91 351.91 351.91 351.91 1.68 1.34 1.01 350.23 350.57 350.90 351.24 351.57 1404.27 1053.71 702.81 351.57 0.00 0.67 0.34