Question

A loan of amount L is to be repaid by n(n > 1) payments, starting one period after the loan is made, with interest at rate i per period. Two repayment schemes are considered: (i) level payments for the lifetime of the loan; (ii) each payment consists of principal repaid of L n plus interest on the previous outstanding balance.

(a) Find the total interest repaid under each scheme and show algebraically that the interest paid under scheme (i) is larger than that paid under scheme (ii).

(b) Show that for each t = 1, 2, . . . , n − 1, OBt is larger under scheme (i) than under scheme (ii).

(c) Verify algebraically that L is the present value at the time of the loan, at rate of interest i per payment period, of all payments made under scheme (ii).

Answer 1

Let L (Loan amount) be $1,000,000 and i (rate of interest) be 10% per annum and N (no. of periods) be 5 years

Now, there are two ways in which the repayment of loan can be done:

(i) Level payments (Equal) for the lifetime of the loan, or

(ii) Each payment which consists of equal principal payments and interest on the outstanding balance

Case (i) - Level payments

The amount of equal payments, is given by A = L / sum of annuity factors

Where, A = annual level payments, L = Loan amount

Now sum of annuity factors can be calculated by present value of 10% interest for 5 years = 3.79079

Therefore, A = $1,000,000/3.79079 = $263,797 per annum

The total payments will be = 263,797 * 5 = $1,318,985

Thus interest paid = $1,318,985 - $1,000,000 = $3,18,985

Case (ii) - Equal principal payments and interest on the outstanding balance

Equal annual principal payments in 5 years can be made by = $1,000,000/5 = $200,000

Therefore, total interest will be on the outstanding balances, so at the end of year 1 payment will be = 200,000 + 1,000,000 * 10% = 300,000

at the end of year 2 = 200,000 + 800,000 * 10% = 280,000

at the end of year 3 = 2,00,000 + 600,000 * 10% = 260,000

at the end of year 4 = 2,00,000 + 400,000 * 10% = 240,000

at the end of year 5 = 2,00,000 + 200,000 * 10% = 220,000

Thus, total payment = $1,300,000

Interest paid = $1,300,000 - $1,00,000 = $3,00,000

(A) Therefore, higher interest of $18,985 is paid in Case (i) as compared to Case (ii).

(B) Following is the schedule of payments for both the cases

Case (i) Payment Schedule

Year	Outstanding at the beginning (A)	Payment amount (B)	Interest amount (C)	Principal Amount (D)	Outstanding at the end (E = A-D)
1	1,000,000	263,797	100,000	163,797	836,203
2	836,203	263,797	83,620	180,177	656,026
3	656,026	263,797	65,603	198,194	457,832
4	457,832	263,797	45,783	218,014	239818
5	239818	263,797	23979	239818	0

Case (iI) Payment Schedule

Year	Outstanding at the beginning (A)	Payment amount (B)	Interest amount (C)	Principal Amount (D)	Outstanding at the end (E = A-D)
1	1,000,000	300,000	100,000	200,000	800,000
2	800,000	280,000	80,000	200,000	600,000
3	600,000	260,000	60,000	200,000	400,000
4	400,000	240,000	40,000	200,000	200,000
5	200,000	220,000	20,000	200,000	0

Thus, we can see that the outstanding amount in Case (i) is always higher than Case (ii)

(C) We know that in level payments, the sum of the present value of lease amounts paid is equal the loan value as presented in the table below

Year	Payment Amount	Interest rate factor @ 10%	Present Value
1	263,797	0.9091	239,818
2	263,797	0.8264	218,002
3	263,797	0.7513	198,190
4	263,797	0.6830	180,173
5	263,797	0.6209	163,792
Total	1,318,958		999,974 or ~1000,000