DRV-001
The annuity payment, derived from scratch
The setup
A loan of principal P carries a periodic interest rate i and is repaid in n equal instalments of E, each paid at the end of a period. Interest accrues on the reducing balance. The formula answers exactly one question: what fixed payment, repeated n times, drives the balance to precisely zero?
The derivation
Step 1: Track the balance month by month
Each month the balance grows by one month of interest, then falls by the payment.
B1 = P(1+i) - EStep 2: Do it again
The second month applies the same operation to the first month's closing balance.
B2 = B1(1+i) - E = P(1+i)^2 - E(1+i) - EStep 3: Generalise
The pattern is now visible. After k payments the principal has compounded k times, and each instalment has compounded for however many periods have passed since it was paid.
B_k = P(1+i)^k - E * [ (1+i)^(k-1) + (1+i)^(k-2) + ... + 1 ]Step 4: Collapse the geometric series
That bracket is a geometric series with first term 1 and ratio (1+i). Summing it gives the outstanding-balance formula, which is useful in its own right. It is how you compute a payoff amount mid-loan.
B_k = P(1+i)^k - E * ((1+i)^k - 1) / iStep 5: Impose the condition that defines the EMI
The loan is repaid exactly when the balance after the final payment is zero. That single condition is what pins E down.
P(1+i)^n = E * ((1+i)^n - 1) / iStep 6: Solve for E
Rearranging gives the formula that appears on every lending site in the world, and now with a provenance.
E = P * i(1+i)^n / ((1+i)^n - 1)E = P * i * (1+i)^n / ((1+i)^n - 1)Checks that prove it is right
A formula you cannot test is a formula you have to trust. These take seconds, and they are what separate a derivation from a formula restatement.
n = 1 → E = P(1+i)
One payment repays the principal plus one period of interest. Borrow ₹1,00,000 at 1% monthly, repay in a single instalment, and you owe ₹1,01,000.
i → 0 → E → P/n
Expanding (1+i)^n ≈ 1 + ni makes the interest terms cancel. A zero-interest loan is repaid in equal principal slices. Note for implementers: the closed form divides by zero here, so this branch must be guarded explicitly.
n → ∞ → E → Pi
A perpetual loan services interest only and never touches the principal. This is also why very long tenures barely reduce the instalment while enormously increasing total interest.
Worked example
₹50,00,000 at 8.5% for 20 years.
The monthly rate is 0.085 ÷ 12 = 0.00708333 and n is 240. Substituting gives an EMI of ₹43,391. Total paid over the term is ₹1,04,13,879, of which ₹54,13,879 is interest, 108% of the amount borrowed. Month one splits ₹35,417 interest against ₹7,974 principal, because interest in any period is the rate times the previous balance and the balance starts at its maximum.
Computed by the amortization engine and cross-checked against the closed-form annuity formula in an independent implementation.
What it assumes
- The rate is fixed for the whole term.
- Interest compounds once per payment period, on the reducing balance.
- Every instalment is paid in full and on time.
- No fees, insurance or statutory charges are included.
Where the formula stops being valid
Stating limits plainly is more useful than pretending there are none, and it is the item competitors most consistently omit.
- Floating rates. Most Indian home loans are repo-linked and reset periodically. On a reset most lenders hold the EMI constant and adjust the tenure instead, so a rate rise can silently extend a loan by years without changing the monthly outgo.
- Broken periods. A loan disbursed mid-month may carry pre-EMI interest for the partial period, which this formula does not model.
- Part-disbursement. Under-construction property disburses in tranches and interest accrues only on what has been released, so early instalments differ from this schedule.
- Rounding. Lenders round the instalment to the currency unit and true up the final payment against the residual balance, which is why a real schedule shows an odd last instalment.
Calculators built on this
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