Prospective Balances Under Changing Rates

Why Changing Rates Matter

A prospective loan balance is the value, at a chosen date, of the payments still owed. If the effective rate differs from one future period to another, each payment must be discounted using the rates that apply between the balance date and that payment date. Applying one rate to every payment can materially misstate the balance, even when the payment schedule itself is unchanged.

The method below uses a discrete timeline with one effective rate assigned to each period. It focuses on a balance immediately after a scheduled payment, so that payment is no longer part of the remaining cash flows. This timing convention makes it clear which payments and which rates belong in the calculation.

Definition: Let $C_j$ be the payment at time $j$, and let $i_r$ be the effective rate for the period from time $r-1$ to time $r$. The prospective balance $B_k$, immediately after the payment at time $k$, is the value at time $k$ of all payments after time $k$, discounted through the rates applicable to their periods. Assume payments occur at integer times, each $i_r>-1$, and the rates and remaining payments are known.

The Changing-Rate Prospective Formula

A payment at time $j$ is separated from the balance date $k$ by the periods $k+1,k+2,\ldots,j$. Its discount factor back to time $k$ is the reciprocal of the product of the accumulation factors for those periods. Thus, the balance is

B_k=\sum_{j=k+1}^{n}\frac{C_j}{\displaystyle\prod_{r=k+1}^{j}(1+i_r)}.

The rate $i_r$ is attached to a particular interval, not to a payment. For example, the discount factor for a payment at time $k+3$ includes three rates: $i_{k+1}$, $i_{k+2}$, and $i_{k+3}$. If a rate changes after the first remaining period, the first rate applies only to the first interval; the later rate applies to later intervals.

Timing convention: In this tutorial, $B_k$ is measured immediately after the payment at time $k$. The payment at time $k$ is excluded, and the first rate used is $i_{k+1}$, for the interval from $k$ to $k+1$.

Why the Formula Holds

The formula follows from the ordinary accumulation rule for effective interest, applied one period at a time. A dollar at time $k$ accumulates to $1+i_{k+1}$ dollars at time $k+1$, then to $(1+i_{k+1})(1+i_{k+2})$ dollars at time $k+2$, and so on. Consequently, one dollar due at time $j$ has value at time $k$ equal to the reciprocal of that accumulated amount.

\text{Accumulation from }k\text{ to }j =\prod_{r=k+1}^{j}(1+i_r), \qquad \text{discount factor from }j\text{ to }k =\frac{1}{\displaystyle\prod_{r=k+1}^{j}(1+i_r)}.

Each remaining payment is a separate cash flow. Its value at time $k$ is its amount multiplied by the discount factor from its own due date to $k$. Adding those values gives $B_k$. This also explains why a single average rate is generally not a substitute: the relevant discount factor is a product of the period-specific factors, not a discount factor built from an arbitrary average.

Fix the balance date.
State whether the balance is immediately before or after a payment. Here it is immediately after payment $k$.

List future payments.
Include only payments after the balance date, with each payment’s due time and amount.

Trace the rate path for each payment.
Multiply the accumulation factors for every period between the balance date and that payment date.

Discount and add.
Divide each payment by its cumulative accumulation factor, then sum the present values.

Worked Applications

Worked Example: Unequal Payments Under Three Rates

A loan has just received its payment at time 1. The remaining payments are $660 at time 2, $990 at time 3, and $1,650 at time 4. The effective rates for periods 1 to 2, 2 to 3, and 3 to 4 are 10%, 20%, and 25%, respectively. Find the prospective balance at time 1.

Method. Discount each remaining payment to time 1 through every rate interval on its path. The payment at time 1 has already been made, so it is not included.

The time-2 payment crosses only the first period. Its value at time 1 is

\frac{660}{1.10}=600.

The time-3 payment crosses the first and second periods. The cumulative accumulation factor is $1.10(1.20)=1.32$, giving

\frac{990}{1.10(1.20)}=\frac{990}{1.32}=750.

The time-4 payment crosses all three periods. The cumulative accumulation factor is $1.10(1.20)(1.25)=1.65$, so its value at time 1 is

\frac{1{,}650}{1.10(1.20)(1.25)} =\frac{1{,}650}{1.65}=1{,}000.

Add the three present values: $600+750+1{,}000=2{,}350$. The prospective balance immediately after the time-1 payment is $2,350.00.

Check. Accumulating the three values from time 1 to their respective due dates gives $600(1.10)=660$, $750(1.10)(1.20)=990$, and $1{,}000(1.10)(1.20)(1.25)=1{,}650$, exactly the specified payments. The common mistake avoided is using just one of the three rates for every remaining payment.

Worked Example: Level Payments with a Rate Reset

Immediately after a scheduled payment, a borrower owes three remaining annual payments of $1,000 each, due one, two, and three years later. The effective rates for those three successive years are 5%, 10%, and 20%. Calculate the prospective balance, rounding to the nearest cent.

Method. Discount the first payment through one rate, the second through two rates, and the third through all three. A level payment amount does not mean that all payments have the same discount factor.

The first payment has value

\frac{1{,}000}{1.05}=952.380952\ldots

The second payment’s cumulative factor is $1.05(1.10)=1.155$, so its value is

\frac{1{,}000}{1.05(1.10)} =\frac{1{,}000}{1.155} =865.800866\ldots

The third payment’s cumulative factor is $1.05(1.10)(1.20)=1.386$, giving

\frac{1{,}000}{1.05(1.10)(1.20)} =\frac{1{,}000}{1.386} =721.500722\ldots

The unrounded sum is $952.380952\ldots+865.800866\ldots+721.500722\ldots=2{,}539.682540\ldots$. Rounded to the nearest cent, the balance is $2,539.68.

Check. The three discounted values accumulate to $1,000 at years 1, 2, and 3 using the stated rates. A common incorrect calculation discounts every payment at 5%, the rate for only the first remaining period. That would give $1{,}000/1.05+1{,}000/1.05^2+1{,}000/1.05^3=2{,}723.25$ to the nearest cent, which does not use the specified later rates.

Worked Example: Locating the Rate Change Correctly

A loan is valued immediately after the payment at time 1. The remaining payments are $400 at time 2, $500 at time 3, and $600 at time 4. The effective rate is 4% for the period from time 1 to time 2, then 8% for each of the next two periods. Find the balance at time 1, to the nearest cent.

Method. Start the rate path at the balance date. The first payment uses 4%; the later payments use 4% for the first interval and 8% for each subsequent interval.

The payment at time 2 is discounted for one period:

\frac{400}{1.04}=384.615385\ldots

The payment at time 3 crosses one 4% period and one 8% period. Its cumulative factor is $1.04(1.08)=1.1232$, so its value is

\frac{500}{1.04(1.08)} =\frac{500}{1.1232} =445.156695\ldots

The payment at time 4 crosses one 4% period and two 8% periods. Its cumulative factor is $1.04(1.08)(1.08)=1.213056$, giving

\frac{600}{1.04(1.08)(1.08)} =\frac{600}{1.213056} =494.618550\ldots

Adding the unrounded values gives $384.615385\ldots+445.156695\ldots+494.618550\ldots=1{,}324.390630\ldots$. The balance is $1,324.39 to the nearest cent.

Check. The first interval is included in every discount path because every payment occurs after it. The 8% rate is included once for the time-3 payment and twice for the time-4 payment. Using 8% for the first interval as well would misplace the reset and produce a different balance.

Common Traps and Practical Checks

Changing-rate balance questions are primarily questions about timing and rate assignment. Before doing arithmetic, write the periods crossed by each payment. This simple step makes it harder to omit a rate, apply a rate before it begins, or count a period twice.

Using the rate at the balance date for the whole term. The rate for the next interval applies only to that interval unless the problem states it remains in force.
Discounting the first future payment by too many periods. A payment one period after the balance date uses one factor, not a product extending back to the loan’s original issue date.
Including a payment already made. For a balance immediately after payment $k$, begin with $C_{k+1}$, not $C_k$.
Averaging the rates without justification. In general, the cumulative factor is a product of the period-specific factors. It is not obtained by discounting all payments at an arithmetic average rate.
Rounding each present value too early. Keep full precision through the sum and round only the final balance to cents.

A useful independent check is to take each discounted payment value and accumulate it forward to its due date using the corresponding future rates. It should reproduce the original payment. This check catches mismatched periods even when the final sum looks plausible.

Key takeaway: A prospective balance under changing rates is the sum of the remaining payments, each discounted by the product of the effective accumulation factors for exactly the periods between the balance date and that payment’s due date.

Check Your Understanding

For each problem, treat the balance as immediately after the stated payment, include only later payments, and round money to the nearest cent where necessary.

Immediately after a payment, two payments remain: $500 in one year and $600 in two years. The effective rates for the next two periods are 10% and 20%. Find the prospective balance. (Answer: $909.09)
After a payment, three payments remain: $550, $1,320, and $2,475, due in one, two, and three years. The successive effective rates are 10%, 20%, and 25%. Find the balance. (Answer: $3,000.00)
After a payment, payments of $330, $720, and $1,650 are due in one, two, and three years. The successive effective rates are 10%, 20%, and 25%. Find the prospective balance. (Answer: $1,845.45)
A balance is measured immediately after time 0. Payments of $660, $990, and $1,650 are due at times 1, 2, and 3. The effective rates are 10% for the first period, 20% for the second, and 25% for the third. Find the balance and identify how many rate factors discount the time-2 payment. (Answer: $2,350.00; two rate factors)
Immediately after a payment, three payments remain: $416 in one year, $1,102.40 in two years, and $952.4736 in three years. The effective rates for the next three periods are 4%, 6%, and 8%. Find the prospective balance. (Answer: $2,200.00)

What You'll Learn