Prospective Balances for Unequal Installments

Why Unequal Installments Need Individual Valuations

A loan may have installments that change over time: a borrower might make smaller payments at first, followed by larger payments, or repay part of the debt with a final balloon payment. If the remaining installments are known, the outstanding balance can be found by valuing those future payments at the balance date. Because their amounts differ, each installment must be discounted separately.

The essential timing question is: how many payment periods remain between the balance date and each installment? A payment due one period after the balance date is discounted once; a payment due three periods afterward is discounted three times. The installments already paid are not part of this calculation.

Definition: The prospective balance at a balance date is the present value, on that date, of all scheduled loan payments still due. For a loan with unequal installments, the balance is calculated by discounting each remaining payment separately.

Formula and Assumptions

Suppose a loan has installments at the ends of regular periods $1,2,\ldots,n$. Let $R_t$ be the installment due at the end of period $t$, and let $i$ be the effective interest rate per payment period. The balance is measured immediately after the installment at the end of period $k$ has been paid. Assume that $i$ is constant, the installments are known, and there are no additional fees, payments, or changes to the loan terms.

B_k=\sum_{t=k+1}^{n} R_t v^{\,t-k} =\sum_{t=k+1}^{n}\frac{R_t}{(1+i)^{t-k}}, \qquad v=\frac{1}{1+i}.

The exponent $t-k$ is the number of periods from the balance date to the payment date. The sum starts at $k+1$, not at $k$, because the installment at time $k$ has already been paid. If a payment is made at the end of every period, the formula uses the rate per period: for monthly installments, for example, $i$ must be the effective monthly rate.

This formula applies to regular payment intervals with a constant rate per period. It does not, by itself, cover payments on irregular dates or rates that change over time. It also cannot determine a balance if future payments are not known or specified.

Why the Prospective Formula Holds

At the balance date, the outstanding amount must be sufficient to meet the remaining installments under the loan’s interest basis. One way to justify the formula is to compare the accumulated value of the balance with the accumulated value of every remaining payment at a common future date.

Choose a common date.
Use the date of the final installment, time $n$. A balance $B_k$ at time $k$ accumulates for $n-k$ periods, giving $B_k(1+i)^{n-k}$.

Accumulate each remaining installment.
An installment $R_t$ paid at time $t$ accumulates from time $t$ to time $n$, giving $R_t(1+i)^{n-t}$. Add these accumulated amounts for $t=k+1,\ldots,n$.

Equate values and solve.
The balance and the remaining installments represent the same obligation, so their accumulated values at time $n$ are equal. Dividing by $(1+i)^{n-k}$ discounts each installment back to time $k$.

B_k(1+i)^{n-k} =\sum_{t=k+1}^{n}R_t(1+i)^{n-t} $$ $$ B_k =\sum_{t=k+1}^{n}R_t(1+i)^{k-t} =\sum_{t=k+1}^{n}\frac{R_t}{(1+i)^{t-k}}.

This derivation also explains why the balance is a sum of individual present values. When the installments are equal, the terms can be grouped using an annuity factor. When they are unequal, each amount and its timing still need to be included; the equal-payment shortcut cannot be applied to the whole sequence.

Worked Applications

Worked Example: Three Different Remaining Installments

A hypothetical loan charges an effective annual interest rate of 10%. Immediately after the borrower pays the first installment, three payments remain: $110 at the end of year 2, $242 at the end of year 3, and $399 at the end of year 4. Find the outstanding balance at the stated date.

Method: prospective valuation. The balance date is immediately after year 1, so the three payments are 1, 2, and 3 years away. With $i=0.10$, discount each payment by the matching power of $1.10$:

B_1 =\frac{110}{1.10}+\frac{242}{1.10^2}+\frac{399}{1.10^3} $$ $$ B_1=100+200+300=\$600.

Check. One year later, immediately after the $110 installment is paid, the remaining payments have a value of $242/1.10+399/1.10^2=220+330=\$550$. Accumulating the original balance for a year and subtracting the $110 payment also gives $600(1.10)-110=\$550$.

Common mistake avoided: using three years of discounting for every installment. The payment at the end of year 2 is only one period after the balance date, not two.

Worked Example: Locate the Balance After the Second Payment

A hypothetical loan has an effective annual interest rate of 10% and four installments: $300 at the end of year 1, $220 at the end of year 2, $363 at the end of year 3, and $484 at the end of year 4. Find the outstanding balance immediately after the second installment.

Method: identify the unpaid installments, then discount them to year 2. The $220 payment at year 2 has just been made, so it is excluded. The $363 payment is one year away, and the $484 payment is two years away:

B_2=\frac{363}{1.10}+\frac{484}{1.10^2} =330+400=\$730.

Check using the balance one period earlier. Immediately after the first installment, the remaining installments have a value of $\frac{220}{1.10}+\frac{363}{1.10^2}+\frac{484}{1.10^3} =200+300+363.63636\ldots$. That is $863.63636\ldots$. Accumulating this amount for one year and subtracting the year-2 installment gives $863.63636\ldots(1.10)-220=950-220=\$730$.

Common mistake avoided: including the second installment in the balance just after it has been paid. The balance date’s position relative to that payment determines whether it is included.

Worked Example: A Balance With a Non-Level Payment Pattern

A hypothetical loan has an effective annual rate of 6%. Immediately after the year-2 payment, the remaining installments are $500 at the end of year 3, $300 at the end of year 4, and $700 at the end of year 5. Find the balance to the nearest cent.

Method: discount each remaining installment for its own number of periods. From the balance date, the payments are 1, 2, and 3 periods away:

B_2=\frac{500}{1.06}+\frac{300}{1.06^2}+\frac{700}{1.06^3} $$ $$ B_2=471.6981132\ldots+266.9989317\ldots+587.7334981\ldots =1326.4305429\ldots

Rounding only the final result to two decimal places, the balance is $1,326.43.

Check. Each positive future payment has a positive present value, and their values add to an amount greater than the first payment’s present value of approximately $471.70. The total, approximately $1,326.43, is therefore consistent with all three remaining obligations.

Common mistake avoided: discounting the $700 payment by two periods because it is the third payment in the list. Its date is year 5, three periods after the year-2 balance date, so its discount factor is $1.06^3$.

Common Errors and Practical Checks

Unequal-installment problems are usually manageable once the timeline is explicit. A short timeline, with the balance date and every unpaid payment marked, prevents most errors. Before calculating, identify the rate per payment period and count the number of periods from the balance date to each remaining payment.

Starting the sum with the wrong payment. Immediately after payment $k$, start with payment $k+1$. Including payment $k$ counts an amount that has already been paid.
Using the payment number as the exponent. Use $t-k$, not $t$. The exponent measures the gap between the payment date and the balance date.
Replacing unequal payments with an average. Discounting an average installment as though every payment were equal generally gives the wrong result, because the payments occur at different times.
Mixing rate and time units. The rate must match the payment period. If installments are monthly, do not use an annual effective rate directly as though it were a monthly rate.
Rounding too early. Keep intermediate present values unrounded when possible, then round the total to the requested precision.

A useful check is to move the balance forward by one period: accumulate the earlier balance at the period rate, then subtract the payment made at the end of that period. The result should agree with the prospective value of the payments still unpaid at the new balance date. This check uses the same payment timeline and can expose a misplaced exponent or an installment included twice.

Key takeaway: At a balance date immediately after payment $k$, the prospective balance is the sum of the present values of installments $k+1$ through $n$. For unequal installments, discount each payment separately using the number of periods from the balance date to its payment date.

Check Your Understanding

For each problem, identify the unpaid payments and their periods from the balance date before calculating.

A loan earns an effective annual rate of 5%. Immediately after the year-2 installment, the remaining payments are $315 at year 3 and $441 at year 4. Find the balance. (Answer: $700)
A loan earns an effective annual rate of 8%. Immediately after the year-1 installment, the remaining payments are $216 at year 2 and $349.92 at year 3. Find the balance. (Answer: $500)
A loan earns an effective annual rate of 4%. Immediately after the year-3 installment, the remaining payments are $520 at year 4 and $676 at year 5. Find the balance. (Answer: $1,125)
A loan earns an effective annual rate of 10%. Immediately after the year-2 installment, the remaining payments are $165 at year 3, $363 at year 4, and $532.40 at year 5. Find the balance. (Answer: $850)
A loan earns an effective annual rate of 5%. Immediately after the year-1 installment, its balance is $600. The only remaining payments are $210 at year 2 and an unknown amount $X$ at year 3. Find $X$. (Answer: $441)

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