AI Math
Apri app

Bank Loan Calculator

Payment, balance, total interest and early-payoff time, worked step by step
Monthly payment on $18,000 at 7% over 5 years
Balance remaining after 24 payments on that loan
How long to clear $12,000 at 9% paying $350 a month?
Total interest saved by paying $350 instead of the scheduled payment

The Loan Payment Formula

A bank loan repaid in nn equal instalments carries the level payment

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

  • PP — the amount borrowed
  • rr — the interest rate per payment period: an annual rate divided by the payments per year
  • nn — the total number of payments, not the number of years
  • MM — the payment that reduces the balance to exactly zero at payment nn

The formula comes from setting the present value of all nn payments equal to the amount borrowed, so it is the annuity present-value relation solved for MM. Once you have it, the total paid is nMnM and the total interest is

nM−PnM - P

This is arithmetic on the figures you enter. A lender's actual quote also reflects fees, day-count conventions and rounding, all of which vary by institution and jurisdiction.

Balance, Payoff Time and Paying Extra

The balance after kk payments, without building the whole schedule:

Bk=P(1+r)k−M⋅(1+r)k−1rB_k = P(1+r)^k - M \cdot \frac{(1+r)^k - 1}{r}

That figure is what remains owing on the principal - a lender's payoff quote may add accrued interest to the date of settlement.

To find how many payments a chosen amount MM needs, invert the payment formula:

n=−ln⁡ ⁣(1−rPM)ln⁡(1+r)n = \frac{-\ln\!\left(1 - \dfrac{rP}{M}\right)}{\ln(1+r)}

A fractional nn means the final payment is smaller than the rest. This is how you price an early payoff: raise MM, recompute nn, and compare nMnM with the original total.

The expression requires M>rPM > rP - a payment at or below one period's interest never touches the principal, and the logarithm is undefined.

Common Mistakes to Avoid

  • Leaving the annual rate in rr: with monthly payments, 7% a year is r=0.07/12≈0.0058333r = 0.07/12 \approx 0.0058333.
  • Setting nn to years: a 5-year monthly loan has n=60n = 60.
  • Charging interest on the original principal: interest each period is on the current balance, which falls with every payment.
  • Assuming a 10% larger payment saves 10% of the interest: the saving is non-linear and depends on when the extra is paid.
  • Reading nM−PnM - P as the cost of a loan settled early: that total assumes every scheduled payment is made.
  • Ignoring rounding: round MM up to the cent, or the final balance lands slightly above zero.

Examples

Step 1: r=0.07/12≈0.0058333r = 0.07/12 \approx 0.0058333, n=60n = 60
Step 2: (1.0058333)60≈1.4176253(1.0058333)^{60} \approx 1.4176253
Step 3: Numerator: 0.0058333×1.4176253≈0.00826950.0058333 \times 1.4176253 \approx 0.0082695; denominator: 1.4176253−1=0.41762531.4176253 - 1 = 0.4176253
Step 4: M≈18000×0.0082695/0.4176253≈356.42M \approx 18000 \times 0.0082695 / 0.4176253 \approx 356.42
Step 5: Total paid: 60×356.42=21,385.2060 \times 356.42 = 21{,}385.20; interest =21,385.20−18,000=3,385.20= 21{,}385.20 - 18{,}000 = 3{,}385.20
Answer: M \approx \356.42amonth,a month,\approx $3{,}385.20$ of interest in total

Step 1: (1.0058333)24≈1.1498060(1.0058333)^{24} \approx 1.1498060
Step 2: P(1+r)24≈18,000×1.1498060≈20,696.51P(1+r)^{24} \approx 18{,}000 \times 1.1498060 \approx 20{,}696.51
Step 3: Annuity factor: (1.1498060−1)/0.0058333≈25.681032(1.1498060 - 1)/0.0058333 \approx 25.681032
Step 4: M×25.681032≈356.42×25.681032≈9,153.23M \times 25.681032 \approx 356.42 \times 25.681032 \approx 9{,}153.23
Step 5: B24≈20,696.51−9,153.23≈11,543.28B_{24} \approx 20{,}696.51 - 9{,}153.23 \approx 11{,}543.28
Answer: About \11{,}543.28$ still owed - only 36% of the principal repaid in the first 40% of the term

Step 1: Scheduled: r=0.0075r = 0.0075, (1.0075)60≈1.5656810(1.0075)^{60} \approx 1.5656810, so M≈249.10M \approx 249.10 and interest =60×249.10−12,000=2,946.00= 60 \times 249.10 - 12{,}000 = 2{,}946.00
Step 2: At M=350M = 350: rP/M=0.0075×12000/350≈0.2571429rP/M = 0.0075 \times 12000/350 \approx 0.2571429
Step 3: n=−ln⁡(0.7428571)/ln⁡(1.0075)≈0.2972515/0.0074720≈39.782n = -\ln(0.7428571)/\ln(1.0075) \approx 0.2972515 / 0.0074720 \approx 39.782
Step 4: So 39 full payments plus a smaller 40th; total paid ≈350×39.782≈13,923.70\approx 350 \times 39.782 \approx 13{,}923.70
Step 5: Interest ≈13,923.70−12,000=1,923.70\approx 13{,}923.70 - 12{,}000 = 1{,}923.70, against 2,946.002{,}946.00 scheduled
Answer: Cleared in about 39.8 months instead of 60, saving roughly \1{,}022.30$ of interest

Frequently Asked Questions

M = P · r(1+r)^n / ((1+r)^n − 1), where P is the amount borrowed, r the rate per payment period and n the number of payments. For monthly payments, divide the annual rate by 12 and count n in months.

Use n = −ln(1 − rP/M)/ln(1+r). A fractional result means the last payment is smaller than the rest. The formula needs M greater than rP; a payment at or below one period's interest never reduces the balance.

Recompute n at the higher payment and compare the two totals nM. On $12,000 at 9%, paying $350 instead of the scheduled $249.10 clears the loan in about 39.8 months rather than 60 and cuts interest from about $2,946 to about $1,924.

B_k = P(1+r)^k − M((1+r)^k − 1)/r, where k is the number of payments already made. That is the outstanding principal; a lender's payoff figure may add interest accrued since the last payment, plus any fees their contract specifies.

Try AI-Math for Free

Get step-by-step solutions to any math problem. Upload a photo or type your question.

Start Solving