Question

Part D and E thank you

You're buying a used car for $8000, but paying $1500 in cash immediately. You'll be borrowing the difference from a local bank. Your first payment to the bank will occur at the end of the 8th month. The last payment will occur at the beginning of the 43rd month. The payments will all be equal in size. The interest rate on the car loan is 0.50% permonth (the equivalent of 600 per year-when annualized as an APR). i.e. Use r-0.50% (per month) in your computations.

Answer 1

(d) Car Price = $ 8000, Cash Payment = $ 1500, Borrowing = $ 6500, Interest Rate = 6 % per annum or 0.5 % per month

First Payment Occurs at the end of the 8th month or at t = 8 and the last payment occurs at the beginning of the 43rd month or end of 42nd month. Total Payment Tenure = (42-7) = 35 months

Growth Rate = - 0.5 % (payment decreases hence negative)

Let the first payment be $ K

Therefore, Present Value of Payments at the end of month 7 = PV(7) = K x [1/(0.005 - (-0.005)) x [1-{(0.995) / (1.005)}^(35)]

PV(7) when discounted at 0.5 % for another 7 months will give the payment values current PV which should equal the original borrowing. Therefore, 6500 = PV(7) / (1.005)^(7)

6500 = K x [1/(0.005 - (-0.005))] x [1-{(0.995) / (1.005)}^(35)] x [1/(1.005)^(7)]

K = $ 227.925

The ninth payment would be the one coming in at the end of month (7+9) = 16.

Therefore, Ninth Payment = 227.925 x (1-0.005)^(8) = $ 218.97

(e) Growth Rate = + 0.5 % and let the first payment be $ K

The present value at the end of month 7 of the repayments can be depicted as: PV(7) = K / (1.005) + [K x (1.005) / (1.005)^(2)] + [K x (1.005)^(2) / (1.005)^(3)] + [K x (1.005)^(3) / (1.005)^(4)] + ................+ [K x (1.005)^(34) / (1.005)^(35)]

PV(7) = K / (1.005) + K / (1.005) + ...............35 times

PV(7) = 35 K / (1.005)

Present Value of PV(7) at current time = 35K / (1.005)^(8) = 6500

6500 = 35K / (1.005)^(8)

K = $ 193.27

Ninth Payment Comes in at the end of Month 16 = 193.27 x (1.005)^(8) = $ 201.142