Deriving a Revised Schedule From Unequal Payments

Why an Unequal Revised Schedule Needs Its Own Calculation

When installments change during a loan, the balance cannot generally be read from a level-payment annuity formula. Each payment reduces the amount outstanding at its own date, and the remaining balance then earns interest until the next payment date. A reliable schedule therefore tracks both the timing and the amount of every installment.

The same calculation can answer two common planning questions: what installment makes a stated balance possible after a sequence of payments, or what balance remains after the planned payments have been made? The essential choice is to use either a recurrence date by date or an equivalent value equation at one common date.

Definition: Let $B_k$ be the loan balance immediately after the payment at the end of period $k$, with $B_0$ the balance at the starting date. Let $R_k$ be the payment made at the end of period $k$, and let $i$ be the effective interest rate per payment period. Assuming interest accrues at rate $i$ for each full period and payments are made at period-end, the balance follows $$ B_k = B_{k-1}(1+i)-R_k. $$

This model assumes a constant rate per period, no fees or additional advances, and payments exactly at the stated period ends. If the rate changes, the rate for each period must be used separately. If payments occur between scheduled dates, the interest-accrual period must also be adjusted. The recurrence records the balance after each payment; the amount immediately before that payment is $B_{k-1}(1+i)$.

From the Recurrence to a Value Equation

A recurrence is often the clearest way to construct a schedule. For solving one unknown installment, it can be efficient to accumulate all amounts to the date of the final known balance. This is not a different loan model; it is the same recurrence written at one common date.

Result: After $n$ periods, the balance is $$ B_n = B_0(1+i)^n-\sum_{k=1}^{n}R_k(1+i)^{n-k}. $$ Each payment is accumulated from its payment date to the end of period $n$. In particular, the payment at the end of period $n$ has no further interest accumulation.

To justify the result, start with the recurrence. At period 1, $B_1=B_0(1+i)-R_1$. At period 2, substitute that expression into $B_2=B_1(1+i)-R_2$, giving $B_2=B_0(1+i)^2-R_1(1+i)-R_2$. Each additional recurrence step multiplies the existing balance by $1+i$ and subtracts the new payment. Thus, by repeating the substitution, the initial balance earns $n$ periods of interest, while $R_k$ earns interest for the $n-k$ periods after it is paid. This gives the stated sum.

The formula makes the timing convention visible. A payment at the end of period 1 is multiplied by $(1+i)^{n-1}$, not by $(1+i)^n$, because it was not outstanding during period 1. A payment at the end of period $n$ is subtracted without accumulation.

Set the payment-period unit.
Identify the effective rate for one payment period and the date at which the starting balance is measured.

Write the payment sequence.
List each $R_k$ with its period number. Mark any unknown installment and the date of the target balance.

Propagate or accumulate.
Use $B_k=B_{k-1}(1+i)-R_k$ period by period, or accumulate every balance and payment to one date.

Solve and check.
Substitute the unknown, calculate the resulting balance sequence, and verify the target after the payment on the stated date.

Finding an Unknown Installment

The first example uses an accumulated-value equation to find an installment that produces a specified balance. It also shows why an earlier payment must be accumulated forward to the target date.

Worked Example: Solve for a Middle Payment

A loan balance is $8,000 immediately after a payment date. The effective annual interest rate is 5%. Payments of $1,000, $x$, and $3,000 are made at the ends of years 1, 2, and 3, respectively. The balance immediately after the third payment is to be $5,000. Find $x$, rounded to the nearest cent.

Method. Accumulate the opening balance and the first two payments to the end of year 3, then subtract the year-3 payment. The target balance is measured immediately after that payment.

The opening balance accumulates for three years: $8000(1.05)^3=8000(1.157625)=9261$. The year-1 payment accumulates for two years: $1000(1.05)^2=1102.50$. The year-2 payment accumulates for one year: $1.05x$. The year-3 payment is made on the target date, so it has no accumulation.

The equation is $$ 5000=9261-1102.50-1.05x-3000=5158.50-1.05x. $$ Thus $1.05x=158.50$, and $$ x=\frac{158.50}{1.05}=150.952380\ldots. $$ Rounded to the nearest cent, the installment is $150.95.

Check. Using the recurrence, $B_1=8000(1.05)-1000=7400$. Then $B_2=7400(1.05)-150.952380\ldots=7629.047619\ldots$. Finally, $B_3=7629.047619\ldots(1.05)-3000=5000$. Using the cent-rounded installment gives a balance of $5000.0025$, which rounds to $5,000.00.

Common mistake avoided. Subtracting the year-1 payment as only $1,000 at year 3 would give $5000=9261-1000-1.05x-3000$, and the wrong installment would be about $248.57. That calculation forgets that the first payment reduces the balance for the next two years; its equivalent value at year 3 is $1000(1.05)^2$.

Using the Recurrence to Clear a Balance

When the unknown is the last payment in a planned sequence, the recurrence gives a direct check of the amount due immediately before that payment. The payment is the accrued balance at that date, provided the intention is to leave no balance afterward.

Worked Example: Determine the Payment That Clears the Balance

A balance of $12,000 is outstanding at the start of a loan period. The effective interest rate is 2% per month. Payments of $2,000 and $3,000 are made at the ends of months 1 and 2. Find the payment at the end of month 3 that clears the balance, rounded to the nearest cent.

Method. Apply the balance recurrence to each month. This makes the date at which interest is charged explicit.

After month 1, the balance is $$ B_1=12000(1.02)-2000=12240-2000=10240. $$ After month 2, it is $$ B_2=10240(1.02)-3000=10444.80-3000=7444.80. $$ The amount due immediately before the month-3 payment is $$ 7444.80(1.02)=7593.696. $$ Therefore, the exact clearing payment is $7,593.696, which is $7,593.70 to the nearest cent.

Check. With the exact payment, $B_3=7593.696-7593.696=0$. If the payment is recorded to cents, the $7,593.70 payment exceeds the exact amount due by $0.004, a difference that rounds to $0.00. In practice, the final amount due would be settled according to the lender’s rounding convention.

Common mistake avoided. The month-3 payment is not the month-2 balance. One more month of interest is charged first, so using $7,444.80 would leave a balance of $7593.696-7444.80=148.896$, or $148.90 to the nearest cent.

Constructing a Revised Schedule With an Unknown Payment

A useful schedule records the opening balance for each period, the interest for that period, the payment, and the balance after payment. This is particularly helpful when an unknown installment must fit between other unequal payments.

Worked Example: Find an Installment in a Four-Payment Plan

A loan balance is $10,000 immediately after a payment date. The effective interest rate is 1% per month. Payments of $1,500, $2,300, $x$, and $2,500 are planned at the ends of months 1 through 4. The required balance immediately after the fourth payment is $3,000. Find $x$, rounded to the nearest cent, and check the schedule.

Method. First use the recurrence for the known installments, then express the final balance in terms of $x$.

At the end of month 1, $$ B_1=10000(1.01)-1500=10100-1500=8600. $$ At the end of month 2, $$ B_2=8600(1.01)-2300=8686-2300=6386. $$ Before the month-3 payment, the balance is $6386(1.01)=6449.86$, so immediately after that payment it is $B_3=6449.86-x$. Before the month-4 payment, the balance is $1.01(6449.86-x)=6514.3586-1.01x$. After the $2,500 payment, $$ B_4=4014.3586-1.01x. $$ Set this equal to the required $3,000: $$ 3000=4014.3586-1.01x,\qquad x=\frac{1014.3586}{1.01}=1004.315445\ldots. $$ Therefore, $x$ is $1,004.32 to the nearest cent.

Check. Before the month-3 payment, the balance is $6,449.86. Using the unrounded $x$, the balance after that payment is $6449.86-1004.315445\ldots=5445.544554\ldots$. It grows to $5,500 immediately before the month-4 payment; subtracting $2,500 leaves $3,000. With the cent-rounded payment, the final balance is $4014.3586-1.01(1004.32)=2999.9954$, which rounds to $3,000.00.

Common mistake avoided. The month-3 installment is multiplied by 1.01 once when calculating the month-4 balance, while the month-4 installment is not multiplied at all. Accumulating both payments for the same number of months would place at least one of them at the wrong date.

Interpreting and Checking a Planned Schedule

A calculated installment should be checked both algebraically and against the meaning of the loan arrangement. Under the stated recurrence, $R_k$ is subtracted from the amount due at the end of period $k$. If an installment is greater than that amount, the computed balance becomes negative. A negative result is not an ordinary outstanding loan balance: it indicates an overpayment or credit, unless the contract explicitly permits that treatment.

Likewise, an equation that produces a negative value for an unknown $x$ does not describe a positive installment. It says that reaching the proposed target would require a payment in the opposite direction—an additional advance or other change to the cash flows. Before accepting a result, check that the amount is feasible under the stated arrangement.

A reliable final review asks three questions:

Is the interest rate stated per payment period, with the correct period count?
Is each installment placed at its actual payment date, with interest applied only after that date?
Does the balance after the last listed payment equal the specified target when the recurrence is recalculated?

The accumulated-value equation is compact, but the recurrence is often easier to audit. Both methods must give the same answer when their dates and payment signs are consistent. These techniques are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated effective rate per payment period and round money to the nearest cent.

A balance of $6,000 is outstanding immediately after a payment date. The effective rate is 4% per quarter. Payments of $1,200 and $1,500 are made at the ends of quarters 1 and 2. Find the balance immediately after the second payment. (Answer: $3,741.60)
A loan balance is $10,000 at the start of a monthly period. The effective monthly rate is 0.5%. Payments of $1,000 and $1,500 are made at the ends of months 1 and 2. What payment at the end of month 3 leaves a balance of $7,000 immediately afterward? (Answer: $633.23)
A balance of $5,000 is outstanding at the start of a monthly period. The effective monthly rate is 1%. Payments of $500, $700, and $900 are made at the ends of months 1, 2, and 3. Find the balance immediately after the third payment. (Answer: $3,034.46)
A loan balance is $20,000 immediately after a payment date. The effective quarterly rate is 2%. Payments of $3,000, $x$, and $4,000 are made at the ends of the next three quarters. If the balance immediately after the third payment is $14,000, find $x$. (Answer: $100.94)
A balance of $12,000 is outstanding at the start of a half-year period. The effective rate is 1.5% per half-year. Payments of $2,500, $1,800, and $2,200 are made at the ends of the next three half-years. Find the balance immediately after the third payment. (Answer: $5,945.58)

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