Question

the answer is 124.72

5. A loan of 1,000 is to be repaid by annual payments of 100 to commence at the end of the fifth year and to continue thereafter for as long as necessary. The effective rate of discount is 5%. Find the amount of the final payment if it is to be larger than the regular payments.

Answer 1

Loan Amount = $ 1000, Planned Annual Payments = $ 100, Interest Rate = 5 %, First Payment comes in at the end of Year 5,

Let the number of years required to pay off the loan be N years (beginning from end of year 5) and the final bigger payment be $ P

Therefore, 1000 = 100 x (1/0.05) x [1-{1/(1.05)^(N-1)}] x [1/(1.05)^(4)] + P / (1.05)^(N) = PV of Level Payments + PV of Final Payment

The above equation can only be solved by means of trial and error as it contains two variable. Further, as each annual payment of $ 100 is 1/10th of the loan amount, the value of N is definitely greater than 10

Hence, when N = 13

PV of Level Payments = 100 x (1/0.05) x [1-{1/(1.05)^(12)}] x [1/(1.05)^(4)] = $ 729.182

Remaining Loan = 1000 - 729.182 = $ 270.818

270.818 = P / (1.05)^(17)

P = $ 620.72

One needs to now progressively increase the value of N such that the value of P approaches the level payment value of $ 100 and the required value of N is that at which the final payment value is greater than the level payment value for the last time.

Now when N = 15

PV of Level Payments = 100 x (1/0.05) x [1-{1/(1.05)^(14)}] x [1/(1.05)^(4)] = $ 814.364

Remaining Loan = 1000 - 814.364 = $ 185.636

185.636 = P / (1.05)^(19)

P = $ 469.094

When N = 19

PV of Level Payments = 100 x (1/0.05) x [1-{1/(1.05)^(18)}] x [1/(1.05)^(4)] = $ 961.705

Remaining Loan = 1000 - 961.705 = $ 38.2948

38.2948 = P / (1.05)^(23)

P = $ 117.623