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=Pr(1+r)n(1+r)n1M = 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=Bk1r,Pk=MIk,Bk=Bk1PkI_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 — 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)kM(1+r)k1rB_k = P(1+r)^k - M \cdot \frac{(1+r)^k - 1}{r}

Total interest over the whole loan:

total interest=nMP\text{total interest} = nM - P

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

n=ln ⁣(1rPM)ln(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=PrannualtI = 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 — the row where principal first exceeds interest — 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 0.05/120.00416670.05/12 \approx 0.0041667 per month.
  • Charging interest on the original principal every row: interest is on the current balance Bk1B_{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 — 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 nMPnM - 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.

示例题目

Step 1: r=0.05/120.0041667r = 0.05/12 \approx 0.0041667, n=60n = 60, (1+r)601.2833587(1+r)^{60} \approx 1.2833587
Step 2: M30000×0.0041667×1.2833587/0.2833587566.14M \approx 30000 \times 0.0041667 \times 1.2833587 / 0.2833587 \approx 566.14
Step 3: Row 1: I1=30000×0.0041667=125.00I_1 = 30000 \times 0.0041667 = 125.00, P1=566.14125.00=441.14P_1 = 566.14 - 125.00 = 441.14, B1=29,558.86B_1 = 29{,}558.86
Step 4: Row 2: I2=29558.86×0.0041667123.16I_2 = 29558.86 \times 0.0041667 \approx 123.16, P2442.98P_2 \approx 442.98, B229,115.88B_2 \approx 29{,}115.88
Step 5: Row 3: I3121.32I_3 \approx 121.32, P3444.82P_3 \approx 444.82, B328,671.06B_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: nMP=60×566.1430000=33,968.4030,000nM - 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)12M(1+r)121rB_{12} = P(1+r)^{12} - M\dfrac{(1+r)^{12} - 1}{r}
Step 4: (1+r)121.0511619(1+r)^{12} \approx 1.0511619, so P(1+r)1231,534.86P(1+r)^{12} \approx 31{,}534.86
Step 5: M0.05116190.0041667566.14×12.2788556,951.55M \cdot \dfrac{0.0511619}{0.0041667} \approx 566.14 \times 12.278855 \approx 6{,}951.55
Step 6: B1231,534.866,951.5524,583.31B_{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)601.3488502(1.005)^{60} \approx 1.3488502, M386.66M \approx 386.66
Step 2: Scheduled total interest: 60×386.66200003,199.3660 \times 386.66 - 20000 \approx 3{,}199.36
Step 3: At M=400M = 400: n=ln(10.005×20000/400)/ln(1.005)n = -\ln(1 - 0.005 \times 20000/400)/\ln(1.005)
Step 4: =ln(0.75)/ln(1.005)0.2876821/0.004987557.68= -\ln(0.75)/\ln(1.005) \approx 0.2876821 / 0.0049875 \approx 57.68
Step 5: So 58 payments; total paid 400×57.6823,072.05\approx 400 \times 57.68 \approx 23{,}072.05, interest 3,072.05\approx 3{,}072.05
Answer: About 57.7 payments instead of 60 — roughly two months shorter and about \127$ less interest

常见问题

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 − M·((1+r)^k − 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.

相关求解器

免费试用 AI-Math

任何数学问题都能获得分步解答。拍照上传或输入问题即可。

开始解题