Annuities, sinking funds and loan instalments
An annuity is a series of equal payments at equal intervals. Its future value plans savings; its present value prices a loan and sets the instalment.
Annuity formulas (R = payment, i = rate per period, n = number of payments)
Ordinary annuity vs annuity due
Future value: a savings plan
A worker saves Rs. 10,000 at the end of every month for 2 years. The account pays 12% a year compounded monthly. How much will she have?
- R = 10,000, i = 0.12 ÷ 12 = 0.01, n = 24.
- 1.01²⁴ = 1.269734649.
- S = 10,000 × (1.269734649 − 1) ÷ 0.01 = 10,000 × 26.9734649 = 269,734.65.
- She paid in 24 × 10,000 = 240,000, so interest earned = 29,734.65.
- If she saved at the start of each month (annuity due): 269,734.65 × 1.01 = 272,432.00.
Rs. 269,734.65 (Rs. 272,432.00 if she pays at the start of each month).
Sinking fund
A company must replace a machine costing Rs. 5,000,000 in 5 years. It will make equal deposits at the end of each year into a fund paying 10% a year compounded yearly. Find the yearly deposit.
- R = S × i ÷ [(1 + i)ⁿ − 1], with S = 5,000,000, i = 0.1, n = 5.
- 1.1⁵ = 1.61051, so (1.1⁵ − 1) = 0.61051.
- R = 5,000,000 × 0.1 ÷ 0.61051 = 500,000 ÷ 0.61051 = 818,987.40.
- Total deposited = 5 × 818,987.40 = 4,094,937.00; the fund earns the rest as interest.
Rs. 818,987.40 a year.
Loan instalment and schedule
A shop owner borrows Rs. 331,000 at 10% a year, repaid in 3 equal instalments at the end of each year. Find the instalment and draw up the repayment schedule.
- R = A × i ÷ [1 − (1 + i)^(−n)] = 331,000 × 0.1 ÷ [1 − 1 ÷ 1.331].
- 1 − 1 ÷ 1.331 = 0.331 ÷ 1.331, so R = 33,100 × 1.331 ÷ 0.331 = 133,100.
- Year 1: interest 331,000 × 0.1 = 33,100; principal repaid 133,100 − 33,100 = 100,000; balance 231,000.
- Year 2: interest 23,100; principal 110,000; balance 121,000.
- Year 3: interest 12,100; principal 121,000; balance 0.
- Total interest = 3 × 133,100 − 331,000 = 68,300 (= 33,100 + 23,100 + 12,100).
Instalment Rs. 133,100 a year; total interest Rs. 68,300.
Repayment (amortisation) schedule for the Rs. 331,000 loan
| Year | Opening (Rs.) | Interest 10% (Rs.) | Instalment (Rs.) | Principal repaid (Rs.) | Closing (Rs.) |
|---|---|---|---|---|---|
| 1 | 331,000 | 33,100 | 133,100 | 100,000 | 231,000 |
| 2 | 231,000 | 23,100 | 133,100 | 110,000 | 121,000 |
| 3 | 121,000 | 12,100 | 133,100 | 121,000 | 0 |
A monthly vehicle loan
A three-wheeler loan of Rs. 1,200,000 is repaid monthly over 3 years at 18% a year compounded monthly. Find the monthly instalment and the total interest.
- i = 0.18 ÷ 12 = 0.015, n = 36.
- 1.015³⁶ = 1.709139538, so (1.015)^(−36) = 0.585089735 and 1 − 0.585089735 = 0.414910265.
- R = 1,200,000 × 0.015 ÷ 0.414910265 = 18,000 ÷ 0.414910265 = 43,382.87.
- Total paid = 36 × 43,382.87 = 1,561,783.32.
- Total interest = 1,561,783.32 − 1,200,000 = 361,783.32.
Rs. 43,382.87 a month; total interest Rs. 361,783.32.
Leasing and finance companies in Sri Lanka quote equal monthly rentals for vehicles: each rental is an annuity payment found from the present value formula. A company setting money aside to replace machinery uses a sinking fund (future value formula).
Short answer: “Define an annuity due and give one example.” Equal payments made at the start of each period, e.g. rent or lease rentals paid in advance.
Structured question: “Find the yearly sinking-fund deposit to reach Rs. 5,000,000 in 5 years at 10% compounded yearly.” R = 500,000 ÷ 0.61051 = Rs. 818,987.40.
Essay: “A Rs. 331,000 loan at 10% is repaid by 3 equal year-end instalments. Find the instalment, draw the schedule and find total interest.” R = Rs. 133,100; balances 231,000, 121,000, 0; interest Rs. 68,300.
Saving towards a target → future value (sinking fund). Borrowing now → present value (loan). Each instalment pays the period's interest first, the rest cuts the balance.