Loans with Varying Payments

Nothing about the amortization method requires payments to be equal. When payments go up, down, or stop being level altogether, the same recursion still runs: interest is \(i\) times the balance you carried in, and whatever's left of the payment chips away at principal.

By the end you'll be able to build an amortization schedule for a stream of unequal payments, spot negative amortization when a payment doesn't cover interest, and confirm that the outstanding balance always equals the present value (PV) of whatever payments remain.

Predict: if you move payment 1 up by some amount, does it cut total interest more than moving payment 2 up by that same amount instead? Adjust a payment below and watch the total.

A 5000 loan at \(i = 10\%\) is repaid by two payments you control, then a third payment that automatically clears whatever's left. Drag \(R_1\) or \(R_2\) and watch the balance line and the interest / principal split of each bar update below.

autoClearing payment \(R_3 = B_2(1+i)\) 3245.00

Total interest paid over the loan: 1245.00

  • t = 1: \(I_1=\)500.00, \(P_1=\)500.00, \(B_1=\)4500.00
  • t = 2: \(I_2=\)450.00, \(P_2=\)1550.00, \(B_2=\)2950.00
  • t = 3: \(I_3=\)295.00, \(P_3=\)2950.00, \(B_3=\)0.00

Prospective check at t = 1: PV of remaining payments \(R_2 v + R_3 v^2\) = 4500.00 = \(B_1\) ✓ — the balance is always the present value of what's left, level or not.

Balance path (top) and each payment's interest / principal split (bottom)
Balance \(B_t\) Interest \(I_t\) Principal \(P_t\)

When a loan's payments \(R_1, R_2, \ldots, R_n\) aren't all equal, the balance is still the present value of the remaining stream, and each period still splits into interest on the prior balance plus a principal repayment — only the payment amounts themselves change from period to period.

Formal

With payments \(R_1, \ldots, R_n\) at rate \(i\) and \(v = 1/(1+i)\), the loan amount is \(L = R_1 v + R_2 v^2 + \cdots + R_n v^n\). The balance right after payment \(t\) can be found two ways, and they always agree:

Prospective: \(B_t = R_{t+1} v + R_{t+2} v^2 + \cdots + R_n v^{n-t}\) — discount each remaining payment by its distance from \(t\), not its original time index.

Retrospective: \(B_t = L(1+i)^t - \left(R_1(1+i)^{t-1} + \cdots + R_t\right)\).

The per-period recursion is unchanged from the level case: \(I_t = i \cdot B_{t-1}\), \(P_t = R_t - I_t\), \(B_t = B_{t-1} - P_t\). What's gone is the geometric shortcut \(P_{t+1}=P_t(1+i)\) — that only holds when \(R\) is constant, so with varying payments you carry the balance forward one period at a time. If a payment is smaller than the interest due, \(P_t\) is negative and the balance grows — that's negative amortization.

Applied

This is exactly what happens whenever a borrower's income is irregular — a commission-based salesperson, a farmer with seasonal cash flow, or anyone who throws an extra lump sum at a loan in a good month. The lender doesn't need a different formula; the same schedule-building recursion just gets fed a different payment each period, and a final "clearing" payment settles whatever's left.

Worked example

A 5000 loan at \(i = 10\%\) is repaid by \(R_1 = 1000\), \(R_2 = 2000\), then a final clearing payment. \(B_0 = 5000\). \(I_1 = 500.00\), \(P_1 = 1000-500 = 500.00\), \(B_1 = 4500.00\). \(I_2 = 450.00\), \(P_2 = 2000-450 = 1550.00\), \(B_2 = 2950.00\). Clearing payment \(= B_2(1+i) = 2950 \times 1.1 = \) 3245.00.

Your turn

Same setup, different numbers: an 8000 loan at \(i = 8\%\), repaid by \(R_1 = 2000\), \(R_2 = 3000\), then a final clearing payment. \(B_0 = 8000\). \(I_1 = 640.00\), \(P_1 = 2000-640 = 1360.00\), \(B_1 = 6640.00\). \(I_2 = 8\% \times 6640 = 531.20\), \(P_2 = 3000-531.20 = 2468.80\), \(B_2 = 6640-2468.80 = 4171.20\). Now finish it: the clearing payment \(= B_2(1+i) = 4171.20 \times 1.08 = \) ____

Reveal the answer

\(4171.20 \times 1.08 = \) 4504.90. Try setting the interactive above so \(B_2\) lands near 4171 and check the auto clearing payment against this value.

More info — why the balance is always a present value

Think of the loan as an IOU for the whole payment stream. At any point in time, the fair value of that IOU — what someone would pay to take over the remaining debt — is the present value of the payments still owed, whatever those payments happen to be. That's true whether the stream is level, increasing, decreasing, or completely irregular; "amortization schedule" is just the name for tracking that present value period by period as interest accrues and payments chip it down. See the Finan study guide in Dive deeper for the full derivation with unequal payments.

Check your understanding

Question 1 of 4

A 6000 loan at i = 9% is repaid by two payments: \(R_1 = 1500\), then a final clearing payment \(R_2\). Find \(R_2\).

Question 2 of 4

On a varying-payment loan, one period's payment \(R_t\) turns out to be smaller than the interest due for that period, \(I_t = i \cdot B_{t-1}\). What happens to the principal-repaid portion and the balance?

Question 3 of 4

For a level-payment loan, principal repaid grows geometrically by a factor of \((1+i)\) each period. Does that shortcut also apply once the payments start varying from period to period?

Question 4 of 4

A loan is repaid by four payments \(R_1, \ldots, R_4\). Which expression correctly gives the outstanding balance \(B_2\), right after the 2nd payment, by the prospective method?

Recap

  • The outstanding balance after any payment is the present value of whatever payments remain — true for level payments and varying payments alike.
  • Every period: \(I_t = i \cdot B_{t-1}\), \(P_t = R_t - I_t\), \(B_t = B_{t-1} - P_t\) — the same recursion as the level case, just fed a different \(R_t\) each period.
  • The geometric shortcut \(P_{t+1} = P_t(1+i)\) only holds for level payments; with varying payments you must carry the balance forward one period at a time.
  • If \(R_t < I_t\), principal repaid is negative and the balance grows — negative amortization.

Dive deeper

Sources

  • Loans with Non-Level (Varying) Payments