The Sinking-Fund Method
Instead of shrinking the balance a little each period like amortization, the sinking fund method keeps the whole loan outstanding and pays it off in one lump sum at the end, funded by a side account building up in parallel.
By the end you'll be able to compute the periodic interest, the sinking-fund deposit, and the total cost of this method — and compare it directly against the level amortization payment for the same loan.
Predict: if the sinking fund earns less than the loan rate (j < i), is the borrower's total cost higher or lower than plain amortization? Drag j and check.
A $10,000, 5-year loan. The debt owed to the lender (dashed line) never moves — every period the borrower pays interest on the full principal. Meanwhile the sinking fund (bars) grows from nothing up to that same $10,000 by the last period, when it's handed over to clear the loan in one shot. Drag i and j to see how the two rates drive the total yearly cost.
Loan interest: $800.00/yr + Sinking-fund deposit: $1,773.96/yr = Total $2,573.96/yr — amortization equivalent: $2,504.56/yr ( +$69.40/yr, since j < i)
Two accounts, two rates: the lender charges \(i\) on the full loan every period while the borrower separately saves toward the payoff at rate \(j\) — usually a lower rate, which is exactly why this method usually costs more than amortization.
Picture two accounts running side by side for the whole loan term. One is the debt — it just sits there at the full amount, and every period you hand the lender a check for interest on it. The other is your sinking fund — you feed it a deposit every period, and it grows on its own, aiming to hit the loan amount exactly when the term ends. On the last day, you sweep the fund out and use it to pay off the debt in one shot.
For a loan of \(L\) at rate \(i\) over \(n\) periods, with sinking-fund rate \(j\): the deposit \(D\) solves \(D \cdot s_{n@j} = L\), so \(D = L/s_{n@j}\) where \(s_{n@j} = ((1+j)^n-1)/j\) is the accumulated-value annuity factor. The total periodic cost is \(L\cdot i + D = L\cdot(i + 1/s_{n@j})\). When \(j = i\), the identity \(1/s_n + i = 1/a_n\) makes this exactly equal to the level amortization payment \(L/a_{n@i}\) — the two methods tie. When \(j < i\) the fund grows too slowly and the sinking-fund method costs strictly more; when \(j > i\) it would cost less.
This is the same mechanism corporations use with bond sinking funds: rather than amortizing a bond's face value, the issuer pays coupon interest each period and separately funds a sinking fund that matures to the face value at redemption. The exam version is the same math on a personal loan, and it always tests whether you can keep the loan interest (on the untouched \(L\)) and the fund deposit (on the separately-growing \(D\)) from getting mixed up.
Borrow \(L = 10{,}000\) for \(n=5\) years. Loan rate \(i=8\%\); sinking fund earns \(j=6\%\). Interest to the lender each year: \(10000 \times 0.08 = 800.00\). \(s_{5@6\%} = (1.06^5-1)/0.06 = 5.63709\). Deposit: \(D = 10000/5.63709 = 1773.96\). Total annual outlay: \(800 + 1773.96 = \) 2573.96. Compare to the amortization payment \(R = 10000/a_{5@8\%} = 2504.56\): the sinking-fund method costs $69.40 more per year, precisely because \(j = 6\% < i = 8\%\).
Same idea, different numbers: \(L = 20{,}000\), \(n=4\) years, loan rate \(i=10\%\), sinking fund rate \(j=7\%\). Interest to the lender each year: \(20000 \times 0.10 = 2000.00\). \(s_{4@7\%} = (1.07^4-1)/0.07 = 4.43994\). Deposit: \(D = 20000/4.43994 = 4504.56\). Now finish it: total annual outlay = \(2000 + 4504.56 = \) ____
Reveal the answer
Total = \(2000 + 4504.56 = \) 6504.56. Compare to the amortization payment \(R = 20000/a_{4@10\%} = 20000/3.16987 = 6309.41\): the sinking-fund method costs $195.15 more per year, again because \(j = 7\% < i = 10\%\). Set \(i=10\%\) and \(j=7\%\) on the sliders above (as close as the steps allow) and check the shape of the gap.
More info — the identity behind the j = i tie
\(1/a_n\) and \(1/s_n\) always differ by exactly \(i\): \(1/a_n - 1/s_n = i\), so \(1/a_n = i + 1/s_n\). Multiply both sides by \(L\) and the left side is the level amortization payment while the right side is the sinking-fund method's total outlay — when both methods use the same rate \(i = j\), they're forced to be the same number. This is a favorite exam check: plug in \(j=i\) and confirm the two totals match before trusting either calculation. See the principal/interest split for how the amortization side builds that payment period by period.
Check your understanding
A sinking fund must accumulate to $15,000 in 6 years, earning \(j = 5\%\). What annual deposit \(D\) is required?
If the sinking-fund rate exactly equals the loan rate (\(j = i\)), how does the sinking-fund method's total periodic cost compare to the level amortization payment for the same loan?
Amortization and the sinking-fund method both repay the same loan. What happens to the amount owed to the lender before the final period, under the sinking-fund method?
A borrower assumes that because they're paying interest every period — just like an amortized loan — their debt to the lender must be shrinking too. Why is this wrong under the sinking-fund method?
Recap
- The sinking-fund method pays interest \(L\cdot i\) on the full, untouched principal every period, while separately depositing \(D\) into a fund earning \(j\) that accumulates to \(L\) by term end.
- \(D = L / s_{n@j}\), where \(s_{n@j} = ((1+j)^n-1)/j\).
- Total periodic cost: \(L\cdot i + D = L\cdot(i + 1/s_{n@j})\).
- When \(j = i\) this exactly equals the level amortization payment \(L/a_{n@i}\); when \(j < i\) (the usual case) it costs strictly more.
Dive deeper
- Sinking fund — Wikipedia Explain setting aside money over time to retire debt principal.
- Sinking Fund — Corporate Finance Institute Illustrate regular deposits accumulating to repay principal.
Sources
- The Sinking-Fund Method of Loan Repayment