Question

Beaver & Crampson took out a loan for specialized manufacturing equipment requiring 7 large loan payments in the future. The company will make the first loan payment of $19034 at the end of year 10, with each subsequent payment decreasing by 9%.

The company would like to save for these large loan payments by making 9 equal deposits at the end of each year (beginning in year 1). What amount must the company deposit each year in order to make the future loan payments? Assume an interest rate of 6% compounded annually.

Note: The first loan payment will be made at the end of year 10. The company will make a total of 7 loan payments. The first deposit will be made at the end of year 1. The company will make a total of 9 deposits.

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Answer 1

i = 6% = 0.06

Loan payment from 10th yr = 19034

Total 7 payment dereasing by 9% each yr

Present value of geometric series = C * [(1+g)^n / (1+i)^n - 1] / (g-i)

Here C = 19034, g = -0.09, n = 7, i = 0.06

The present value of loan payment at EOY 9 = 19034 * [(1-0.09)^7 / (1+0.06)^7 - 1] / (-0.09-0.06)

= 19034 * [(0.91)^7 / (1.06)^7 - 1] / (-0.15)

= 19034 * 4.375496

= 83283.19

Let deposit be A, then

Future value of deposits at EOY 9 = A *(F/A,6%,9) = A* 11.491316

Future value of deposits at EOY 9 = present value of loan payment at EOY 9

A* 11.491316 = 83283.19

A = 7247.49