Prospective Balances for Changing Payment Amounts

Why Changing Payments Need a Prospective Balance

A loan’s scheduled installments need not stay level. They may increase after a set date, decrease for a period, or follow a payment sequence specified one installment at a time. At a balance date, the key question is still the same: what amount at that date is equivalent, at the loan’s interest rate, to the payments that remain?

The prospective method answers this by valuing the future installments from the balance date. Each payment is discounted for the number of periods between that date and the payment date. This is especially useful when the amounts change, because it does not require the schedule to fit a level-annuity formula.

Definition: The prospective outstanding balance immediately after the $k$th payment is the value, at time $k$, of all installments due after time $k$, discounted at the loan’s rate. A payment made at time $k$ is not included in this post-payment balance.

Valuing the Remaining Installments

Assume annual installments are made at the ends of years $1,2,\ldots,n$, the effective annual loan rate is a fixed $i$, and there are no fees or other cash flows. Let $C_t$ be the installment due at time $t$. With $v=1/(1+i)$, the prospective balance immediately after the payment at time $k$ is the sum of the present values of the installments still due.

B_k=\sum_{t=k+1}^{n} C_t v^{\,t-k} =\sum_{j=1}^{n-k} C_{k+j}v^j

The two forms index the same payments in different ways. In the first, $t$ is the payment’s time measured from the start of the loan, so the discount exponent is $t-k$, its time from the balance date. In the second, $j$ counts periods forward from the balance date. The installment $C_k$, already paid at time $k$, does not appear.

The balance can be calculated even if the payment schedule has several different levels or no simple pattern at all. If payments are level over a particular future block, an annuity formula may simplify that block; otherwise, discount and add the individual amounts. The fixed-rate assumption matters: if the rate changes over time, each payment must instead be discounted using the applicable sequence of rates.

Why the Formula Holds

At the balance date, a dollar of outstanding debt is equivalent to one dollar invested for the loan’s rate and period. Thus a payment due one period later has value $C_{k+1}v$ at time $k$; a payment due two periods later has value $C_{k+2}v^2$. Applying the same reasoning to each later installment and adding their values gives the prospective balance.

Set the balance date.
Identify the exact time $k$ and whether the balance is required immediately before or after any payment on that date.

List unpaid installments.
For a post-payment balance, start with the first installment due after time $k$; omit installments already paid.

Discount by the remaining time.
For each installment, use an exponent equal to the number of periods from the balance date to its due date.

Add the values and check.
Sum the discounted installments. When useful, carry the balance forward one period, credit interest, and subtract the next payment.

Worked Examples

Worked Example: A Step Increase in the Installments

A loan has annual installments due at the ends of years 1 through 5. The effective annual interest rate is 25%. The installments are $1,000, $1,200, $1,500, $2,500, and $4,000, respectively. Find the outstanding balance immediately after the second installment, and check it by carrying the balance forward from the original loan amount.

Method. Use the prospective value of the payments due after time 2. The discount factor is $v=1/1.25=0.8$. The remaining installments are $1,500 at time 3, $2,500 at time 4, and $4,000 at time 5; their times from the balance date are 1, 2, and 3 periods.

B_2=1500(0.8)+2500(0.8)^2+4000(0.8)^3 =1200+1600+2048 =\$4{,}848

Check using the original loan amount. The amount borrowed is the value at time 0 of all five installments:

\begin{aligned} B_0 &=1000(0.8)+1200(0.8)^2+1500(0.8)^3+2500(0.8)^4+4000(0.8)^5\\ &=800+768+768+1024+1310.72\\ &=\$4{,}670.72. \end{aligned}

After one year, interest accrues on this amount and the first payment is made: $4670.72(1.25)=5838.40$, then $5838.40-1000=4838.40$. Over the next year, $4838.40(1.25)=6048.00$, then $6048.00-1200=4848.00$. This agrees with the prospective balance. The check also illustrates why the balance can rise: the installments already made are smaller than the interest accrued over these periods.

Common mistake avoided: The payment increase does not change the discount rule. Each installment is discounted by its time from year 2, not by the time from the start of the loan.

Worked Example: Payments That Increase Geometrically

Immediately after a loan payment at time 1, three installments remain. They are $1,100 at time 2, $1,210 at time 3, and $1,331 at time 4. The effective annual interest rate is 10%. Find the balance at time 1 and verify the present value of each installment.

Method. Discount each specified installment to time 1. The payment amounts happen to increase by 10% each year, but the prospective calculation still values each one at the balance date.

\begin{aligned} B_1 &=\frac{1100}{1.10}+\frac{1210}{(1.10)^2}+\frac{1331}{(1.10)^3}\\ &=1000+1000+1000\\ &=\$3{,}000. \end{aligned}

The intermediate values follow directly: $1100/1.10=1000$; $(1.10)^2=1.21$, so $1210/1.21=1000$; and $(1.10)^3=1.331$, so $1331/1.331=1000$. Thus the displayed total is exact, not the result of adding rounded approximations.

Check. The time-1 balance grows to $3000(1.10)=3300$ by time 2. After the $1,100 payment, the balance is $2,200. The remaining installments, valued at time 2, are $1210/1.10=1100$ and $1331/(1.10)^2=1100$, totaling $2,200.

Common mistake avoided: Do not treat the three installments as level payments of $1,100. Their amounts differ; each must be discounted as specified. The convenient equal present values here are a consequence of the chosen growth and interest rates.

Worked Example: An Explicit Nonlevel Payment Sequence

A loan’s effective annual interest rate is 25%. Immediately after the payment at time 4, the remaining installments are $900 at time 5, $2,000 at time 6, and $3,500 at time 7. Find the balance at time 4, then check it one year later, immediately after the time-5 installment.

Method. This is an explicitly specified nonlevel sequence, so discount and sum the three amounts using $v=1/1.25=0.8$.

\begin{aligned} B_4 &=900(0.8)+2000(0.8)^2+3500(0.8)^3\\ &=720+1280+1792\\ &=\$3{,}792. \end{aligned}

From time 4 to time 5, the balance accrues interest: $3792(1.25)=4740$. Subtract the $900 payment at time 5 to obtain $4740-900=3840$. Directly valuing the installments still due at time 5 gives $2000(0.8)+3500(0.8)^2=1600+2240=3840$, the same balance.

Common mistake avoided: The installment at time 5 is one period from time 4, not five periods from time 4. The exponent measures the interval between the balance date and the payment date.

Timing, Payment Changes, and Exam Traps

A payment due on the balance date requires careful timing language. A balance immediately after that payment excludes it. A balance immediately before the payment includes its full amount at that date, with no discounting. For example, if the balance immediately after time $k$ is $B_k$ and a payment $C_k$ is due at time $k$, the value immediately before that payment is $B_k+C_k$, assuming no other cash flow occurs at that instant.

Step changes are handled by preserving the actual due amounts. If payments are level at one amount for several periods and then change, one may value each block separately or discount every installment individually. The method is the same in either case. A level-annuity formula applies only to a block whose payment amount and timing meet that formula’s assumptions.

A frequent error is to use the loan’s original time origin for discount exponents after a balance date. Suppose a balance is required at time 2 and a payment is due at time 3. Its value at time 2 is $C_3v$, not $C_3v^3$. The latter discounts to time 0, not time 2. Before adding terms, check that every exponent equals the number of periods remaining from the date at which the balance is requested.

Key takeaway: A prospective balance is the value, at the stated balance date, of the unpaid installments. Keep the actual payment sequence, count each payment’s remaining periods from that date, and follow the wording about whether a payment at the balance date has already been made.

This approach is useful whenever the remaining obligation changes over time, whether the change is a planned step-up, a step-down, or a fully listed sequence. It avoids the false assumption that the remaining installments form a level annuity, while keeping the calculation systematic and easy to verify. These balance calculations are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated annual effective interest rate and identify exactly which installments remain at the balance date.

A loan bears interest at 8% effective annually. Immediately after the payment at time 2, installments of $500 at time 3 and $1,000 at time 4 remain. Find the balance. (Answer: $1,320.30)
The effective annual rate is 20%. Immediately after the payment at time 1, the remaining installments are $720 at time 2, $1,440 at time 3, and $2,160 at time 4. Find the balance. (Answer: $2,850.00)
At an effective annual rate of 5%, the balance is required immediately after the payment at time 3. The only remaining installments are $1,050 at time 4 and $2,205 at time 5. Find the balance. (Answer: $3,000.00)
The effective annual rate is 10%. Find the balance immediately before a $500 payment due at time 3, given that further payments of $1,100 at time 4 and $1,210 at time 5 are due. (Answer: $2,500.00)
At an effective annual rate of 25%, find the balance immediately after the payment at time 2 if the remaining installments are $750 at time 3, $2,500 at time 4, and $1,250 at time 5. (Answer: $2,840.00)

What You'll Learn