Why the Two Balance Formulas Agree

Why Compare the Two Balance Methods?

A level-payment loan can be described in two useful ways. One method starts with the original principal and updates the balance after each payment by adding interest and subtracting the payment. The other method values the payments that remain, discounting them to the balance date. These methods look different, but under the standard level-payment loan assumptions they give exactly the same amount.

That agreement is more than a convenient check. It confirms that the balance carried forward by the schedule is consistent with the value, at that date, of the borrower’s remaining scheduled payments. The main source of apparent disagreement is usually not the algebra; it is a mismatch in timing, such as comparing a balance just after a payment with a value just before it.

Definition: Assume a loan is repaid by $n$ equal payments of $R$, each made at the end of a payment period. Let $i$ be the effective interest rate per payment period, and let $B_k$ be the balance immediately after the $k$th payment, for $k=0,1,\ldots,n$. Thus $B_0$ is the original principal, and $B_n=0$ for a fully amortizing loan. Unless stated otherwise, assume the payments and balances are calculated without intermediate rounding.

The rate must match the payment period: for example, if payments are monthly, $i$ in the formulas is the effective monthly rate. The formulas below assume level payments in arrears, meaning each payment is made at the end of its period. They do not directly describe payments made in advance or a loan with irregular payments.

The Two Balance Formulas

The schedule recurrence follows from one period of loan activity. During a period, the balance earns interest at rate $i$; the payment $R$ is then subtracted. Consequently, the balance after the next payment is

B_k=(1+i)B_{k-1}-R,\qquad k=1,2,\ldots,n.

Iterating this recurrence gives the balance after $k$ payments in terms of the original principal:

B_k=B_0(1+i)^k-Rs_{\overline{k}|}, \qquad s_{\overline{k}|}=\frac{(1+i)^k-1}{i}.

The other method values the payments still to come at time $k$. There are $n-k$ payments remaining, with the first due one period after time $k$. Their value at time $k$ is therefore

B_k=R a_{\overline{n-k}|}, \qquad a_{\overline{m}|}=\frac{1-v^m}{i}, \qquad v=\frac{1}{1+i}.

This second expression is a present value: it discounts each remaining end-of-period payment back to the date immediately after payment $k$. In particular, it does not include a payment already made at time $k$.

Why the Formulas Agree

For a fully amortizing loan, the original principal is the present value of all $n$ scheduled payments, so $B_0=R a_{\overline{n}|}$. Substituting this into the iterated recurrence expresses the schedule balance using only $R$, $i$, $n$, and $k$. The following algebra reduces it to the present value of the remaining payments.

Write the recurrence-based balance.
Use the original loan value $B_0=R a_{\overline{n}|}$ in $B_k=B_0(1+i)^k-Rs_{\overline{k}|}$.

Substitute the annuity expressions.
Use $a_{\overline{n}|}=(1-v^n)/i$, $s_{\overline{k}|}=((1+i)^k-1)/i$, and $v^n=(1+i)^{-n}$.

Simplify the result.
The terms involving $(1+i)^k$ cancel, leaving the annuity value for the $n-k$ payments that remain.

\begin{aligned} B_k &=R a_{\overline{n}|}(1+i)^k-Rs_{\overline{k}|}\\ &=\frac{R}{i}\left[(1+i)^k\left(1-(1+i)^{-n}\right) -\left((1+i)^k-1\right)\right]\\ &=\frac{R}{i}\left[1-(1+i)^{k-n}\right]\\ &=R a_{\overline{n-k}|}. \end{aligned}

The last equality follows because $(1+i)^{k-n}=(1+i)^{-(n-k)}=v^{n-k}$. Thus the recurrence balance and the present value of the remaining payments are not competing approximations: under the stated assumptions, they are algebraically identical. At the endpoints, the result also makes sense: when $k=0$, the balance is the value of all $n$ payments; when $k=n$, no payments remain and the balance is zero.

Theorem: For a fully amortizing loan with equal end-of-period payments, a constant effective rate $i>0$ per payment period, and no intermediate rounding, the balance immediately after payment $k$ satisfies $B_k=B_0(1+i)^k-Rs_{\overline{k}|}=R a_{\overline{n-k}|}$.

Worked Examples

Worked Example: Checking a Balance After Two Payments

A hypothetical equipment loan is repaid by four annual payments of $100 at the end of each year. The effective annual interest rate is 25%, and the loan is fully amortizing. Find the balance immediately after the second payment using both methods.

Method. First find the original principal as the present value of all four payments, then apply the schedule recurrence twice. Finally, value the two remaining payments at the same date.

The discount factor is $v=1/1.25=0.8$. The original principal is

B_0=100a_{\overline{4}|} =100(0.8+0.8^2+0.8^3+0.8^4) =100(0.8+0.64+0.512+0.4096) =\$236.16.

After the first payment,

B_1=1.25(236.16)-100=295.20-100=\$195.20.

After the second payment,

B_2=1.25(195.20)-100=244.00-100=\$144.00.

There are two payments remaining, due one and two years after this balance date. Their present value is

100a_{\overline{2}|} =100(0.8+0.64) =\$144.00.

Check. Both methods give $144.00. The schedule also reaches zero after the remaining payments: $B_3=1.25(144)-100=80$, and $B_4=1.25(80)-100=0$. A common mistake is to value four payments at the date after the second payment; only two payments remain at that date.

Worked Example: Keeping the Balance Date Straight

A hypothetical loan has two annual end-of-year payments of $110. The effective annual interest rate is 10%, and the loan is fully amortizing. Find the original principal, the balance immediately after the first payment, and the amount due immediately before the second payment.

Method. Value both payments at time 0 for the principal. Then calculate the balance just after payment 1; separately accumulate that balance for one period to find the amount immediately before payment 2.

The discount factor is $v=1/1.10$. Hence

B_0=110a_{\overline{2}|} =110\left(\frac{1}{1.10}+\frac{1}{1.10^2}\right) =110(0.9090909+0.8264463) =\$190.91

The exact value before rounding is $23100/121$, or approximately $190.9091. Applying the recurrence after the first payment gives

B_1=1.10\left(\frac{23100}{121}\right)-110 =210-110 =\$100.00.

The present value at time 1 of the one remaining $110 payment is

110v=\frac{110}{1.10}=\$100.00.

Immediately before the second payment, one period has passed since the balance date, so the amount has accumulated to $1.10(100)=\$110.00$. This equals the payment due at that instant. Check. The post-payment balance is $100.00, while the pre-payment amount is $110.00. Treating these as the same balance is the timing error this example is designed to prevent.

Worked Example: Verifying Both Remaining-Payment Values

A hypothetical three-year loan is repaid by three annual payments of $54, made at each year-end. The effective annual interest rate is 20%. Find the balances immediately after payments 1 and 2, and verify each using the remaining-payment method.

Method. Calculate the original principal from the three-payment present value, apply the recurrence in sequence, and compare each result with the present value of the payments then remaining.

Here $v=1/1.20=5/6$. The original principal is

B_0=54a_{\overline{3}|} =54\left(\frac{5}{6}+\frac{25}{36}+\frac{125}{216}\right) =54\left(\frac{455}{216}\right) =\$113.75.

After the first payment, the recurrence gives

B_1=1.20(113.75)-54=136.50-54=\$82.50.

Two payments remain, so their value at time 1 is

54a_{\overline{2}|} =54\left(\frac{5}{6}+\frac{25}{36}\right) =54\left(\frac{55}{36}\right) =\$82.50.

After the second payment,

B_2=1.20(82.50)-54=99.00-54=\$45.00.

The one remaining payment has present value $54/1.20=\$45.00$ at time 2. Check. The balance after the final payment is $1.20(45)-54=0$, as required. The rate, payment timing, and balance dates are the same in both methods; changing any of those while comparing the results would invalidate the check.

Common Traps and Practical Checks

The agreement is useful for checking a calculation, but only if both methods refer to the same contract and the same date. In a hand calculation, carry sufficient precision through the steps and round money only at the final reporting stage. If a real payment schedule rounds each payment or interest charge to cents, it may use a small final-payment adjustment; the exact unrounded formulas need not reproduce that adjusted final amount.

Using the wrong number of remaining payments. Immediately after payment $k$, there are $n-k$ payments left, so use $a_{\overline{n-k}|}$, not $a_{\overline{n}|}$.
Discounting from the wrong date. The first remaining payment is one full payment period after the balance date. The remaining-payment value is not an annuity-due value.
Mixing before-payment and after-payment balances. A balance immediately before a scheduled payment includes the amount due at that instant; the balance immediately after that payment does not.
Using a rate for a different period. The $i$ in the recurrence and annuity value must be the effective rate for one payment period.
Rounding too early. Rounding each intermediate balance can create a small discrepancy between the methods. Use unrounded values for the comparison, then round the final dollar amount to cents.

Key takeaway: At a date immediately after payment $k$, the recurrence balance and the value of the remaining level payments agree because both equal $R a_{\overline{n-k}|}$. Align the payment count, rate period, and balance date before comparing calculations.

Check Your Understanding

For each problem, identify the balance date carefully and use the stated effective rate per payment period.

A fully amortizing loan has three annual end-of-year payments of $50, with an effective annual rate of 10%. Find the original principal and the balance immediately after the first payment. (Answer: original principal $124.34; balance $86.78.)
A fully amortizing loan has four annual payments of $80, with an effective annual rate of 25%. Find the balance immediately after the third payment using the value of the remaining payments. (Answer: $64.00.)
A loan has four semiannual end-of-period payments of $300. The effective rate is 5% per half-year. Find the balance immediately after the second payment using the remaining-payment method. (Answer: $557.82.)
A fully amortizing loan has four annual payments of $400, with an effective annual rate of 10%. Calculate the balance immediately after payment 2 using the recurrence formula, starting from the original principal, and check it by valuing the remaining payments. (Answer: $694.21.)
A fully amortizing loan has two annual payments of $110 and an effective annual rate of 10%. The balance immediately after payment 1 is $100. What is the scheduled amount due immediately before payment 2? (Answer: $110.00.)

What You'll Learn