Retrospective Balances Under Changing Rates

Why Changing Rates Require Period-by-Period Accumulation

When a loan’s effective rate changes over time, an outstanding balance cannot generally be found by applying one rate to the original principal and all payments. Each amount earns interest only during the periods after it occurs, and those periods may have different rates. The retrospective method handles this by accumulating the original loan and each past payment separately to the balance date.

This tutorial considers a loan issued at time 0, with interest charged once per period and payments made at period ends. The period rates and payments may vary. All examples are hypothetical, and assume that the stated payments are made in full on their scheduled dates, with no fees or other cash flows.

Definition: The retrospective balance at a specified date is the accumulated value at that date of the original loan proceeds, less the accumulated value at that date of all payments made before or at that date. It measures the remaining debt by carrying past cash flows forward under the rates that actually apply in each period.

The Retrospective Formula with Changing Rates

Let $L$ be the loan amount at time 0. Let $i_j$ be the effective rate for period $j$, from time $j-1$ to time $j$, and let $P_t$ be the payment at the end of period $t$. The balance $B_n$ is measured immediately after the payment at time $n$. Rates and payments can differ from one period to the next.

Retrospective balance formula: Under these assumptions, accumulate each cash flow from its own date to time $n$:

B_n = L\prod_{j=1}^{n}(1+i_j) - \sum_{t=1}^{n} P_t\prod_{j=t+1}^{n}(1+i_j).

The loan proceeds earn interest in every period through $n$. A payment $P_t$, however, earns interest only after time $t$, so its accumulation factor includes rates $i_{t+1},\ldots,i_n$, not $i_t$. For the payment made at time $n$, the product has no factors and is defined to equal 1. Thus, the payment at the balance date is subtracted without further accumulation.

Why the Formula Holds

The loan balance can also be built one period at a time. Immediately after payment $t$, the previous balance has earned the period’s interest and the payment is then subtracted:

B_0=L, \qquad B_t=B_{t-1}(1+i_t)-P_t.

Expanding this recurrence shows why each cash flow receives exactly the accumulation factor in the retrospective formula. For example, after two periods,

\begin{aligned} B_1 &= L(1+i_1)-P_1,\\ B_2 &= \bigl[L(1+i_1)-P_1\bigr](1+i_2)-P_2\\ &=L(1+i_1)(1+i_2)-P_1(1+i_2)-P_2. \end{aligned}

The original loan accumulates through both periods; the first payment accumulates only through period 2; and the second payment is already at the balance date. Repeating this expansion through period $n$ gives the stated formula: each past amount is multiplied by the rates in the periods after its date. This also gives a useful check: computing $B_n$ recursively period by period must agree with the retrospective calculation.

Set the balance date.
Identify the period $n$ and determine whether the balance is immediately before or after the payment at that date. Here it is immediately after.

Accumulate the original loan.
Multiply $L$ by $1+i_j$ for every period from 1 through $n$.

Accumulate each payment separately.
For a payment at time $t$, multiply $P_t$ only by the factors for periods $t+1$ through $n$.

Subtract and check.
Subtract the accumulated payments from the accumulated loan, then verify the result with the recurrence if practical.

Worked Applications

Worked Example: Six Half-Year Periods with Two Rates

A hypothetical loan of $20,000 is repaid over six half-year periods. The effective rate per half-year is 1.5% in periods 1 through 3 and 2% in periods 4 through 6. Payments of $3,000, $3,000, $3,500, $3,500, $4,000, and $4,000 are made at the respective period ends. Find the balance immediately after the sixth payment.

Method. Use the retrospective formula, with each payment accumulated only through later periods. The six period rates are $i_1=i_2=i_3=0.015$ and $i_4=i_5=i_6=0.02$.

The original loan accumulates through all six periods:

20{,}000(1.015)^3(1.02)^3 =20{,}000(1.109682256977) =22{,}193.64513954.

Now accumulate each payment to the end of period 6. The first payment has two remaining periods at 1.5% and three at 2%. The second has one at 1.5% and three at 2%. The third has three at 2%, and so on:

\begin{aligned} P_1 &: 3{,}000(1.015)^2(1.02)^3=3{,}279.8490354,\\ P_2 &: 3{,}000(1.015)(1.02)^3=3{,}231.37836,\\ P_3 &: 3{,}500(1.02)^3=3{,}714.228,\\ P_4 &: 3{,}500(1.02)^2=3{,}641.40,\\ P_5 &: 4{,}000(1.02)=4{,}080,\\ P_6 &: 4{,}000. \end{aligned}

Their accumulated total is $3{,}279.8490354+3{,}231.37836+3{,}714.228+3{,}641.40+4{,}080+4{,}000=21{,}946.8553954$. Therefore,

B_6=22{,}193.64513954-21{,}946.8553954 =246.78974414.

Rounded to the nearest cent, the balance is $246.79. As a check, the recurrence gives $B_1=17{,}300$, $B_2=14{,}559.50$, $B_3=11{,}277.8925$, $B_4=8{,}003.45035$, $B_5=4{,}163.519357$, and $B_6=246.78974414$, agreeing with the retrospective result. The method avoids treating the two rate regimes as one rate.

Worked Example: Unequal Payments and Three Different Rates

A hypothetical $12,000 loan has effective rates of 5%, 4%, and 3% in its first, second, and third periods. Payments are $4,000, $4,500, and $2,500, respectively, at the period ends. Find the balance immediately after the third payment.

Method. Accumulate all cash flows to time 3. The loan accumulates through all three periods: $12{,}000(1.05)(1.04)(1.03)=12{,}000(1.12476)=13{,}497.12$.

The first payment earns interest in periods 2 and 3, the second earns interest only in period 3, and the last is already at time 3:

\begin{aligned} P_1 &: 4{,}000(1.04)(1.03)=4{,}284.80,\\ P_2 &: 4{,}500(1.03)=4{,}635,\\ P_3 &: 2{,}500. \end{aligned}

The accumulated payments total $4{,}284.80+4{,}635+2{,}500=11{,}419.80$, so $B_3=13{,}497.12-11{,}419.80=\mathbf{\$2{,}077.32}$. To check, the recurrence gives $B_1=12{,}000(1.05)-4{,}000=8{,}600$, $B_2=8{,}600(1.04)-4{,}500=4{,}444$, and $B_3=4{,}444(1.03)-2{,}500=2{,}077.32$. The balance is positive, consistent with the payments not yet having repaid the full debt.

Worked Example: Verifying a Balance with the Recurrence

A hypothetical $10,000 loan has period rates of 2%, 3%, 1%, and 4%. Payments of $1,000, $2,000, $1,500, and $5,000 are made at the ends of periods 1 through 4. Find the balance immediately after period 4.

Method. Accumulate to time 4. The original loan becomes $10{,}000(1.02)(1.03)(1.01)(1.04)=10{,}000(1.10355024)=11{,}035.5024$. The payments accumulate as follows:

\begin{aligned} P_1 &: 1{,}000(1.03)(1.01)(1.04)=1{,}081.912,\\ P_2 &: 2{,}000(1.01)(1.04)=2{,}100.80,\\ P_3 &: 1{,}500(1.04)=1{,}560,\\ P_4 &: 5{,}000. \end{aligned}

The accumulated payments total $1{,}081.912+2{,}100.80+1{,}560+5{,}000=9{,}742.712$. Thus, $B_4=11{,}035.5024-9{,}742.712=1{,}292.7904$, or $1,292.79 to the nearest cent.

For the check, the recursive balances are $B_1=10{,}000(1.02)-1{,}000=9{,}200$, $B_2=9{,}200(1.03)-2{,}000=7{,}476$, $B_3=7{,}476(1.01)-1{,}500=6{,}050.76$, and $B_4=6{,}050.76(1.04)-5{,}000=1{,}292.7904$. The two methods agree. In particular, the period-4 payment is subtracted in full because the balance is measured immediately after it.

Common Exam Trap: Subtracting Payments Without Accumulating Them

A payment made earlier than the balance date is not worth just its original amount at that later date: it would have earned interest under the applicable intervening rates. In the $12,000 example, subtracting the nominal total of payments from the accumulated loan gives $13{,}497.12-(4{,}000+4{,}500+2{,}500)=2{,}497.12$. That is incorrect because the first two payments were not accumulated to time 3. Their correct accumulated values are $4,284.80 and $4,635, producing the correct balance of $2,077.32.

Another error is to apply the rate for the balance period to every earlier payment. The rate $i_j$ applies only to money outstanding during period $j$; a payment made at time $t$ is already subtracted at that time and earns interest only in later periods. A reliable exam approach is to mark each cash flow’s date on a timeline, then count its remaining accumulation periods.

Key takeaway: With changing effective rates, carry each cash flow forward using only the rates for the periods after its date. The retrospective balance is the accumulated original loan minus the accumulated past payments. Payments need not be level, but the formula assumes the stated dates, rates, and cash flows are applied consistently.

Check Your Understanding

For each hypothetical loan, find the requested balance or payment. Assume payments occur at period ends, rates apply for the stated periods, and balances are measured immediately after any payment on that date.

A $5,000 loan has period rates of 4% and 5%, with payments of $1,000 and $2,000. Find the balance after the second payment. (Answer: $2,410.00)
An $8,000 loan has period rates of 3%, 2%, and 4%, with payments of $1,500, $2,000, and $2,500. Find the balance after the third payment. (Answer: $2,569.79)
A $15,000 loan has period rates of 2%, 2%, 3%, and 1%, with payments of $3,000, $4,000, $2,000, and $5,000. Find the balance after the fourth payment. (Answer: $1,870.40)
A $9,000 loan has period rates of 5%, 1%, and 4%. Payments of $2,000 and $3,000 are made at the ends of the first two periods. Find the balance immediately after the second payment. (Answer: $4,524.50)
A $20,000 loan has period rates of 1%, 2%, and 3%. Payments of $5,000 and $6,000 are made at the ends of periods 1 and 2. What payment at the end of period 3 will reduce the balance to zero immediately after that payment? (Answer: $9,789.12)

What You'll Learn