How Rate and Term Affect the Schedule

Why Rate and Term Matter

A fixed-payment loan schedule depends on more than the amount borrowed. The periodic interest rate determines how quickly interest accumulates between payments, while the number of payments determines how long the balance remains outstanding. Changing either assumption can change the payment, the split between interest and principal, and the total interest paid.

For a meaningful comparison, hold the other assumptions constant. To isolate the effect of the rate, compare loans with the same principal and number of payments. To isolate the effect of the term, compare loans with the same principal and periodic rate. A comparison that changes both at once does not show which change caused the difference.

Definition: For a loan with principal $L$, periodic effective interest rate $i$, and $n$ equal payments made at the end of each period, the level payment $R$ is the same amount paid $n$ times. Assume $i \geq 0$, the payments and interest periods coincide, and the final payment repays the loan exactly. When $i>0$, $R=L/a_{\overline{n}|}$, where $a_{\overline{n}|}=(1-v^n)/i$ and $v=1/(1+i)$. When $i=0$, $R=L/n$.

The rate in these comparisons is the rate per payment period. For example, a monthly schedule requires a monthly rate; an annual effective rate should not be inserted as if it were a monthly rate. Converting rates to the payment period is essential before comparing schedules.

Why the Payment Changes

A level-payment loan is valued by equating the loan principal at time 0 to the present value of its payments. The formula follows directly from that equation of value.

L=Rv+Rv^2+\cdots+Rv^n=R a_{\overline{n}|} \qquad\Longrightarrow\qquad R=\frac{L}{a_{\overline{n}|}}.

This is also the key to understanding the direction of the changes. If the rate rises while $n$ stays fixed, each discount factor $v^k$ falls, so $a_{\overline{n}|}$ falls. To have the same present value $L$, the payment $R$ must rise. If the term grows while $i$ stays fixed, additional positive discounted payments are included, so $a_{\overline{n}|}$ grows and the required payment falls.

Result: For a positive loan principal and positive periodic rate, holding the term fixed, a higher rate means a higher level payment and higher total interest. Holding the rate fixed, a longer term means a lower level payment but higher total interest. The interest portion of the first payment is $iL$, and its principal portion is $R-iL$.

Changing the Rate: Payment and Principal Split

When the rate changes with the principal and term fixed, the first-period interest changes from $iL$. Since the payment also changes, its principal portion is not found by comparing payments alone: subtract the interest charge from the payment in each schedule.

Worked Example: Comparing Two Rates for a Two-Payment Loan

A hypothetical lender advances $L=\$1{,}000$, to be repaid with two equal end-of-period payments. Compare periodic effective rates of 5% and 10%. Find each payment, the first payment's interest and principal portions, and total interest.

Method. Use $R=L/a_{\overline{2}|}$, then calculate first-period interest as $iL$, first-period principal as $R-iL$, and total interest as $2R-L$. The common mistake to avoid is comparing the payments without separating out the different interest charges.

At 5%. The annuity factor is $a_{\overline{2}|}=1/1.05+1/1.05^2=820/441\approx1.859410$. Thus $R=1{,}000/(820/441)=441{,}000/820\approx\$537.804878$. First-period interest is $0.05(1{,}000)=\$50$, so first-period principal is $537.804878-50=\$487.804878$. Total interest is $2(537.804878)-1{,}000=\$75.609756$.

At 10%. The annuity factor is $a_{\overline{2}|}=1/1.10+1/1.10^2=210/121\approx1.735537$. Thus $R=1{,}000/(210/121)=121{,}000/210\approx\$576.190476$. First-period interest is $0.10(1{,}000)=\$100$, so first-period principal is $576.190476-100=\$476.190476$. Total interest is $2(576.190476)-1{,}000=\$152.380952$.

Check and conclusion. Both schedules repay the original \$1,000, as total payments less total interest equal the principal. Rounded to the nearest cent only at the end, the higher-rate loan has a payment of \$576.19 instead of \$537.80 and total interest of \$152.38 instead of \$75.61. Its first payment repays less principal (\$476.19 instead of \$487.80), even though its payment is larger.

For a fixed term of at least two periods, the first principal portion can also be written $R-iL=L i/((1+i)^n-1)$. This makes clear that the higher first-period interest charge is not simply added on top of an unchanged principal portion: the principal portion adjusts too. For example, when $n=2$, the 10% schedule above has a larger payment but a smaller first principal portion than the 5% schedule.

Later principal portions need not follow the same direction as the first one. A higher rate changes the pattern of repayment across the whole schedule, not just the first payment. Therefore, state which payment or which cumulative amount is being compared rather than saying broadly that a higher rate means “more principal” or “less principal” in every period.

Changing the Term: Smaller Payments, More Interest

With rate and principal fixed, extending the term spreads repayment across more payments. The payment falls, and the first-period interest remains $iL$, because the initial balance and rate are unchanged. Therefore, the first payment's principal portion falls along with the payment.

Total interest over the full loan is total payments less the original principal:

I_{\text{total}}=nR-L.

Although the payment is lower for a longer term, it is made more times. For positive rates, the longer schedule has more total interest. One way to verify this is to write $x=\ln(1+i)>0$. For a positive integer term $n$, $nR/L=ni/(1-e^{-nx})$. The corresponding expression is increasing with $n$: for positive $y$, $e^y>1+y$, so $1-e^{-y}-ye^{-y}>0$. Taking $y=nx$ shows the derivative of $n/(1-e^{-nx})$ is positive. Thus $nR-L$ increases with the term. This result assumes a positive rate and exact level payments; at zero interest, total interest is zero for every term.

Worked Example: Extending a Loan by One Payment Period

A hypothetical \$2,100 loan charges 10% effective interest per payment period. Compare repayment over two periods with repayment over three periods. Find the level payment, first-period principal, and total interest for each term.

Method. Calculate each payment from its own annuity factor, then use $R-iL$ for first-period principal and $nR-L$ for total interest. This avoids the trap of treating a lower payment as evidence that the longer loan costs less overall.

Two periods. The factor is $a_{\overline{2}|}=210/121$, so $R=2{,}100(121/210)=\$1{,}210$. First-period interest is $0.10(2{,}100)=\$210$, giving first-period principal of $1{,}210-210=\$1{,}000$. Total interest is $2(1{,}210)-2{,}100=\$320$.

Three periods. Here $a_{\overline{3}|}=(1-1/1.1^3)/0.10=331/133.1\approx2.486852$. Thus $R=2{,}100/a_{\overline{3}|}=\$844.441088$. First-period interest is still \$210, so first-period principal is $844.441088-210=\$634.441088$. Total interest is $3(844.441088)-2{,}100=\$433.323263$.

Check and conclusion. For each term, total payments minus total interest equal \$2,100. Rounded to the nearest cent at the end, extending the loan reduces the payment from \$1,210.00 to \$844.44, but increases total interest from \$320.00 to \$433.32. The first payment's principal portion also falls, from \$1,000.00 to \$634.44, because the first-period interest is unchanged.

Rate and Term Together

In a real comparison, both the rate and term may differ. The formula still gives a payment for each schedule, but the result combines both effects. A lower payment might reflect a lower rate, a longer term, or both; by itself it does not establish that the loan is less costly. Compare total interest and the payment schedule as well.

Worked Example: Comparing Complete Schedules

Two hypothetical offers finance the same \$3,310 purchase. Offer A charges 10% per payment period and requires two end-of-period payments. Offer B charges 10% per payment period and requires three end-of-period payments. Compare the payments, first-period principal, and total interest.

Method. The principal and rate match, so this comparison isolates the effect of the term. The two-period payment factor is $0.10(1.10)^2/((1.10)^2-1)=121/210$. The three-period factor is $0.10(1.10)^3/((1.10)^3-1)=1331/3310$.

Offer A. The payment is $3{,}310(121/210)=\$1{,}906.047619$. First-period interest is $0.10(3{,}310)=\$331$; first-period principal is $1{,}906.047619-331=\$1{,}575.047619$. Total interest is $2(1{,}906.047619)-3{,}310=\$502.095238$.

Offer B. The payment is $3{,}310(1331/3310)=\$1{,}331$. First-period interest remains \$331, so first-period principal is $1{,}331-331=\$1{,}000$. Total interest is $3(1{,}331)-3{,}310=\$683$.

Check and conclusion. In each case, subtracting total interest from all payments returns the \$3,310 principal. Rounded to the nearest cent, Offer B requires \$575.05 less per payment than Offer A, but costs \$180.90 more in total interest. Its first payment repays \$575.05 less principal. The longer term, not a change in the rate, accounts for these differences.

Common Mistakes and Exam Traps

Using the wrong rate period. The formula needs the effective rate per payment period. Convert the rate before calculating if the payment frequency differs.
Comparing different assumptions as if only one changed. Hold principal, rate, or term fixed as appropriate to isolate the effect being studied.
Calling a lower payment a lower borrowing cost. Calculate total interest as $nR-L$; smaller payments over more periods can mean more interest overall.
Confusing interest with principal in the first payment. The first-period interest is $iL$, and first-period principal is $R-iL$. A larger payment does not guarantee a larger first principal portion if the rate also rises.
Rounding too early. Keep full precision during the schedule calculations, then round reported dollar amounts to the nearest cent.

Key takeaway: For a fixed principal, a higher periodic rate raises the level payment and total interest; a longer term lowers the level payment but raises total interest when the rate is positive. To compare principal repayment, subtract the period's interest charge from its payment.

Check Your Understanding

For each problem, assume end-of-period level payments, an effective rate per payment period, and exact repayment at the end of the stated term. Round dollar answers to the nearest cent.

A hypothetical \$5,000 loan is repaid in two payments at 10% per period. Find the payment, first-payment principal portion, and total interest. (Answer: payment \$2,880.95; first-payment principal \$2,380.95; total interest \$761.90)
A \$3,310 loan at 10% per period is repaid over either two or three periods. How much lower is the three-period payment, and how much higher is its total interest? (Answer: payment is \$575.05 lower; total interest is \$180.90 higher)
A \$2,000 loan is repaid in two payments at 5% per period. Find the payment, first-payment interest, first-payment principal, and total interest. (Answer: payment \$1,075.61; first-payment interest \$100.00; first-payment principal \$975.61; total interest \$151.22)
A \$1,100 loan at 10% per period has either one payment or two level payments. Find each payment and total interest, then state which term has the lower payment and which has lower total interest. (Answer: one period: payment \$1,210.00 and interest \$110.00; two periods: payment \$633.81 and interest \$167.62; the two-period term has the lower payment, while the one-period term has lower total interest)
A borrower compares two schedules for the same principal and term, but the second schedule has a higher periodic rate. What must happen to the level payment and total interest? Can the first-period principal portion be found by comparing the payments alone? (Answer: the payment and total interest increase; no, subtract first-period interest $iL$ from each payment to compare principal portions)

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