Retrospective Balances After the kth Payment

What a Retrospective Balance Measures

A loan balance immediately after an installment is the amount still owed at that date, under the loan’s stated interest and payment terms. The retrospective method determines it by carrying the original principal and all payments made so far forward to the balance date. The method is especially direct when the past payments are known.

The timing convention matters. In this tutorial, the loan is advanced at time 0, interest accrues at a constant effective rate per payment period, and each installment is made at the end of a period. A balance “after the kth payment” is measured at time $k$, immediately after that payment has been applied. Thus the kth payment earns no interest before the balance is measured; the first payment earns interest for $k-1$ periods, the second for $k-2$ periods, and so on.

Definition: Let $L$ be the original loan principal, $i$ the effective interest rate per payment period, and $R_j$ the payment made at the end of period $j$. Under the timing assumptions above, the retrospective balance $B_k$ immediately after payment $k$ is $$ B_k=L(1+i)^k-\sum_{j=1}^{k}R_j(1+i)^{k-j}. $$ The formula assumes a constant rate per period and that the listed payments are applied to the loan at the stated period ends. It does not directly describe a balance between payment dates or under changing rates.

Why the Formula Works

At time $k$, the original principal would have grown to $L(1+i)^k$ if no payments had been made. Each payment reduces the amount carried forward: a payment $R_j$ made at time $j$ is accumulated for the remaining $k-j$ periods, so its time-$k$ value is $R_j(1+i)^{k-j}$. Subtracting the accumulated value of all payments from the accumulated original principal gives the balance.

The same result follows from the period-by-period balance recurrence. Just before payment $j$, the balance from the previous period has earned interest; then the payment is subtracted. In symbols, $B_0=L$ and $B_j=B_{j-1}(1+i)-R_j$. Repeated substitution gives the retrospective formula:

\begin{aligned} B_1&=L(1+i)-R_1,\\ B_2&=L(1+i)^2-R_1(1+i)-R_2,\\ B_k&=L(1+i)^k-R_1(1+i)^{k-1}-R_2(1+i)^{k-2}-\cdots-R_k. \end{aligned}

The exponent on a payment is determined by how long that payment remains invested between its payment date and the balance date. This is why a payment at time $k$ has exponent zero, and therefore appears as $R_k(1+i)^0=R_k$.

A Reliable Calculation Process

Fix the balance date.
State that the balance is immediately after payment $k$, at time $k$.

Use the rate per payment period.
If a nominal rate is given, convert it to an effective rate for one payment period before accumulating cash flows.

Accumulate the principal.
Carry $L$ forward $k$ periods to obtain $L(1+i)^k$.

Accumulate each payment.
Carry payment $R_j$ forward $k-j$ periods, then subtract the sum from the accumulated principal.

Check the timing and result.
The latest payment is subtracted without interest accumulation. A positive result is debt remaining under the model.

Worked Applications

Worked Example: Two Annual Payments

A loan of $5,000 is advanced at time 0. The effective annual interest rate is 4%, and the borrower makes annual payments of $1,800 at the ends of years 1 and 2. Find the balance immediately after the second payment. Use the retrospective method.

Method. Accumulate the original principal and each payment to the end of year 2. The per-period rate is $i=0.04$, and $k=2$.

The principal accumulates for two years:

5{,}000(1.04)^2=5{,}000(1.0816)=5{,}408.

The year 1 payment accumulates for one year, while the year 2 payment is already at the balance date:

$$ 1{,}800(1.04)+1{,}800=1{,}872+1{,}800=3{,}672. $$

Therefore,

B_2=5{,}408-3{,}672=\boxed{\$1{,}736.00}.

Check. The recurrence gives $B_1=5{,}000(1.04)-1{,}800=3{,}400$, then $B_2=3{,}400(1.04)-1{,}800=3{,}536-1{,}800=\$1{,}736$. The two calculations agree. A common mistake is to accumulate the second payment for a full year; that would use the wrong balance date.

Worked Example: A Level Payment after Three Periods

A $10,000 loan has an effective annual rate of 5%. Annual payments of $2,300 are made at the ends of years 1 through 5. Find the balance immediately after the third payment. Assume the first three payments have been made as scheduled.

Method. Only the first three payments enter the balance after payment 3. Accumulate the principal for three years and the payments for three, two, and one fewer periods, respectively.

The accumulated principal is

10{,}000(1.05)^3=10{,}000(1.157625)=11{,}576.25.

The three payment values at time 3 are

\begin{aligned} R_1(1.05)^2&=2{,}300(1.1025)=2{,}535.75,\\ R_2(1.05)&=2{,}300(1.05)=2{,}415.00,\\ R_3&=2{,}300. \end{aligned}

Their total accumulated value is $2{,}535.75+2{,}415.00+2{,}300.00=\$7{,}250.75$. Hence

B_3=11{,}576.25-7{,}250.75=\boxed{\$4{,}325.50}.

Check. Applying the recurrence gives $B_1=10{,}000(1.05)-2{,}300=8{,}200$, $B_2=8{,}200(1.05)-2{,}300=6{,}310$, and $B_3=6{,}310(1.05)-2{,}300=6{,}625.50-2{,}300=\$4{,}325.50$. The fifth-year payment is irrelevant to the balance at time 3; including future installments is a timing error.

Worked Example: Monthly Payments and a Nominal Rate

A borrower receives $20,000 and repays it with monthly installments of $2,000. The nominal annual interest rate is 12% convertible monthly. Find the balance immediately after the fourth monthly payment. Assume the nominal rate and payment schedule remain unchanged.

Method. Convert the nominal rate to the effective monthly rate. Since the nominal rate is convertible monthly,

i_{\text{month}}=\frac{0.12}{12}=0.01.

The original principal accumulates for four months:

20{,}000(1.01)^4=20{,}000(1.04060401)=20{,}812.0802.

The first payment accumulates for three months, the second for two, the third for one, and the fourth for zero:

\begin{aligned} 2{,}000(1.01)^3&=2{,}060.602,\\ 2{,}000(1.01)^2&=2{,}040.200,\\ 2{,}000(1.01)&=2{,}020.000,\\ 2{,}000&=2{,}000.000. \end{aligned}

The accumulated payments total $2{,}060.602+2{,}040.200+2{,}020.000+2{,}000.000=\$8{,}120.802$. Thus

B_4=20{,}812.0802-8{,}120.802=\boxed{\$12{,}691.28}.

The final answer is rounded to the nearest cent. Check. The fourth payment is subtracted at its payment-date amount, not multiplied by $1.01$. Using 12% as the monthly rate would also be incorrect; the contractual nominal rate is divided by 12 to get the monthly rate.

Using the Formula with Unequal Payments

The retrospective calculation does not require equal installments. If the payment amounts vary, use the actual payment $R_j$ at each date and its own accumulation period. For example, the payment at time 1 is carried forward $k-1$ periods, while the payment at time $k$ is subtracted at face value. A level-payment shortcut is only appropriate when the payments included in the calculation are in fact level.

The accumulation also requires a consistent rate and time unit. If the payment frequency is monthly, use the effective monthly rate. If payments are annual, use the effective annual rate. A rate stated as nominal convertible $m$-thly converts to a per-conversion-period rate $i^{(m)}/m$; this rate is appropriate only when payments occur at that conversion frequency under the stated model.

Common Mistakes and Interpretation

Using the wrong endpoint. “After payment $k$” means the payment at time $k$ has already been subtracted. It is not accumulated for another period.
Giving every payment the same exponent. The exponent is $k-j$ for payment $j$, not $k$. Earlier payments have more time to accumulate.
Mixing time units. A monthly payment schedule requires a monthly rate and a number of months, not an annual rate applied once per payment.
Including payments after time $k$. Future payments are not past cash flows at the retrospective valuation date.
Assuming the payment amount determines the balance by itself. The principal, rate, payment timing, and all payments through the valuation date are needed.

A computed negative balance means that, under the assumed cash-flow model, accumulated payments exceed the accumulated principal at the valuation date. It should not automatically be interpreted as a negative contractual debt: an actual loan may instead be paid off, subject to its contract and any applicable settlement terms. The formula calculates the mathematical balance implied by the cash flows supplied.

Key takeaway: For a balance immediately after payment $k$, carry the original principal forward $k$ periods, carry each payment $R_j$ forward $k-j$ periods, and subtract the accumulated payments from the accumulated principal. The latest payment has exponent zero.

Check Your Understanding

For each problem, use the retrospective method and treat the balance date as immediately after the specified payment. Round money to the nearest cent.

A $6,000 loan accrues interest at an effective annual rate of 3%. The borrower pays $1,500 at the end of each year. Find the balance just after the second payment. (Answer: $3,320.40)
A $12,000 loan has a 4% effective annual rate and annual payments of $3,000. Find the balance immediately after payment 3. (Answer: $4,133.57)
A $8,000 loan charges a nominal annual rate of 9% convertible quarterly. Quarterly payments are $1,100. Find the balance immediately after the third payment. (Answer: $5,177.43)
A $15,000 loan accrues interest at an effective annual rate of 6%. Payments at the ends of years 1, 2, and 3 are $2,000, $3,000, and $4,000, respectively. Find the balance after the third payment. (Answer: $8,438.04)
A $9,000 loan has an effective annual rate of 8% and annual installments of $2,500. Find the balance immediately after the third payment. (Answer: $3,221.41)

What You'll Learn