Amortization Schedule Calculator

Build a loan schedule row by row with AI-powered step-by-step solutions
Amortization schedule for $30,000 at 5% APR over 60 months
Split the first payment on $30,000 at 5% into interest and principal
Total interest paid on a $250,000 loan at 6% over 30 years
How many payments of $400 clear a $20,000 balance at 6% APR?

What an Amortization Schedule Is

An amortization schedule is a table with one row per payment. Each row records the interest charged that period, the principal repaid, and the balance left over. The payment itself never changes; the split changes every single row.

The level payment comes from

M=PтЛЕr(1+r)n(1+r)nтИТ1M = P \cdot \frac{r(1+r)^n}{(1+r)^n - 1}

with PP the principal, rr the periodic rate (annual rate ├╖\div periods per year) and nn the number of payments. Then, row by row, starting from B0=PB_0 = P:

Ik=BkтИТ1тЛЕr,Pk=MтИТIk,Bk=BkтИТ1тИТPkI_k = B_{k-1} \cdot r, \qquad P_k = M - I_k, \qquad B_k = B_{k-1} - P_k

A correct schedule ends with Bn=0B_n = 0 (to the cent). The whole table follows mechanically from those three lines тАФ everything else, including the totals, is bookkeeping.

This page is a math tool. It shows the arithmetic on the numbers you enter; your actual lender's schedule depends on their rounding, fees and day-count conventions.

Reading and Rearranging the Schedule

Closed forms you can use instead of the table

The remaining balance after kk payments, without building every row:

Bk=P(1+r)kтИТMтЛЕ(1+r)kтИТ1rB_k = P(1+r)^k - M \cdot \frac{(1+r)^k - 1}{r}

Total interest over the whole loan:

total┬аinterest=nMтИТP\text{total interest} = nM - P

And the number of payments needed to clear a balance at a chosen payment MM:

n=тИТlnтБбтАЙтБг(1тИТrPM)lnтБб(1+r)n = \frac{-\ln\!\left(1 - \dfrac{rP}{M}\right)}{\ln(1+r)}

That last one is how you price an extra payment: raise MM, recompute nn, compare nMnM against the original.

Simple interest is a different animal

I=PтЛЕrannualтЛЕtI = P \cdot r_{\text{annual}} \cdot t

Simple interest never touches the balance, so it always overstates the cost of an amortizing loan, where the balance falls every month. Use it only for instruments that genuinely accrue that way.

Why early payments feel wasted

IkI_k is proportional to the balance, which is largest at the start. The crossover тАФ the row where principal first exceeds interest тАФ arrives later the longer the term and the higher the rate.

Common Mistakes to Avoid

  • Annual rate in a monthly row: rr must match the payment frequency. A 5% annual rate is 0.05/12тЙИ0.00416670.05/12 \approx 0.0041667 per month.
  • Charging interest on the original principal every row: interest is on the current balance BkтИТ1B_{k-1}, which is the entire point of amortization.
  • Rounding drift: rounding MM and every row to the cent leaves a small residue. Real lenders adjust the final payment to force the balance to exactly zero тАФ never leave a schedule ending at -\0.07$.
  • Confusing nn with years: a 30-year monthly loan has n=360n = 360, not 30.
  • Assuming an extra payment cuts interest proportionally: it does not. The saving depends on when it is made; the same dollar early is worth far more than late.
  • Treating nMтИТPnM - P as the cost of a loan paid off early: that total assumes every scheduled payment is made. Pay it off in month 40 of 60 and the interest total is much smaller.

Examples

Step 1: r=0.05/12тЙИ0.0041667r = 0.05/12 \approx 0.0041667, n=60n = 60, (1+r)60тЙИ1.2833587(1+r)^{60} \approx 1.2833587
Step 2: MтЙИ30000├Ч0.0041667├Ч1.2833587/0.2833587тЙИ566.14M \approx 30000 \times 0.0041667 \times 1.2833587 / 0.2833587 \approx 566.14
Step 3: Row 1: I1=30000├Ч0.0041667=125.00I_1 = 30000 \times 0.0041667 = 125.00, P1=566.14тИТ125.00=441.14P_1 = 566.14 - 125.00 = 441.14, B1=29,558.86B_1 = 29{,}558.86
Step 4: Row 2: I2=29558.86├Ч0.0041667тЙИ123.16I_2 = 29558.86 \times 0.0041667 \approx 123.16, P2тЙИ442.98P_2 \approx 442.98, B2тЙИ29,115.88B_2 \approx 29{,}115.88
Step 5: Row 3: I3тЙИ121.32I_3 \approx 121.32, P3тЙИ444.82P_3 \approx 444.82, B3тЙИ28,671.06B_3 \approx 28{,}671.06
Answer: M \approx \566.14;thefirstthreebalancesare; the first three balances are $29{,}558.86,, $29{,}115.88andand$28{,}671.06$

Step 1: Total interest: nMтИТP=60├Ч566.14тИТ30000=33,968.40тИТ30,000nM - P = 60 \times 566.14 - 30000 = 33{,}968.40 - 30{,}000
Step 2: =3,968.40= 3{,}968.40
Step 3: Balance after 12: B12=P(1+r)12тИТM(1+r)12тИТ1rB_{12} = P(1+r)^{12} - M\dfrac{(1+r)^{12} - 1}{r}
Step 4: (1+r)12тЙИ1.0511619(1+r)^{12} \approx 1.0511619, so P(1+r)12тЙИ31,534.86P(1+r)^{12} \approx 31{,}534.86
Step 5: MтЛЕ0.05116190.0041667тЙИ566.14├Ч12.278855тЙИ6,951.55M \cdot \dfrac{0.0511619}{0.0041667} \approx 566.14 \times 12.278855 \approx 6{,}951.55
Step 6: B12тЙИ31,534.86тИТ6,951.55тЙИ24,583.31B_{12} \approx 31{,}534.86 - 6{,}951.55 \approx 24{,}583.31
Answer: About \3{,}968.40ofinterestintotal;roughlyof interest in total; roughly$24{,}583.31$ still owed after one year

Step 1: Scheduled payment: r=0.005r = 0.005, (1.005)60тЙИ1.3488502(1.005)^{60} \approx 1.3488502, MтЙИ386.66M \approx 386.66
Step 2: Scheduled total interest: 60├Ч386.66тИТ20000тЙИ3,199.3660 \times 386.66 - 20000 \approx 3{,}199.36
Step 3: At M=400M = 400: n=тИТlnтБб(1тИТ0.005├Ч20000/400)/lnтБб(1.005)n = -\ln(1 - 0.005 \times 20000/400)/\ln(1.005)
Step 4: =тИТlnтБб(0.75)/lnтБб(1.005)тЙИ0.2876821/0.0049875тЙИ57.68= -\ln(0.75)/\ln(1.005) \approx 0.2876821 / 0.0049875 \approx 57.68
Step 5: So 58 payments; total paid тЙИ400├Ч57.68тЙИ23,072.05\approx 400 \times 57.68 \approx 23{,}072.05, interest тЙИ3,072.05\approx 3{,}072.05
Answer: About 57.7 payments instead of 60 тАФ roughly two months shorter and about \127$ less interest

Frequently Asked Questions

Interest each period is the current balance times the periodic rate, and the balance is at its largest right at the start. As principal is repaid the interest charge shrinks, so a steadily larger slice of the same fixed payment goes to principal.

Use B_k = P(1+r)^k тИТ M┬╖((1+r)^k тИТ 1)/r, where k is the number of payments already made. It gives the same figure as row k of the schedule, up to rounding to the cent.

Simple interest is I = P┬╖r┬╖t, computed on the original principal and never reduced. An amortizing loan charges interest on the outstanding balance, which falls with every payment, so the total interest is substantially less than the simple-interest figure over the same term.

Yes. If it does not, either the rate and payment frequency are mismatched or rounding has drifted. Lenders resolve the last few cents by adjusting the final payment, so the closing balance is exactly zero.

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