Loan Balances Across Rate-Change Dates

Why a Rate-Change Date Needs Careful Timing

A loan balance evolves as interest accrues and payments reduce the debt. When the interest rate changes, the balance must be carried through each rate segment using that segment’s rate. The calculation also depends on whether a payment occurs on the change date: if it does, interest accrues to the date, the payment is made, and the new rate applies after the date.

This tutorial assumes a loan with deterministic interest accumulation, no fees, and payments made at specified times. Unless a problem states otherwise, payments remain at the originally specified amounts; a rate change does not itself change the payment. Determining a new payment after a reset is a separate calculation. The methods here are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: The outstanding balance immediately after a payment is the debt remaining once that payment has been applied. The balance immediately before a payment includes the interest accumulated to that payment date but does not yet subtract the payment. At a rate-change date that is also a payment date, use the old rate to accumulate up to the date, subtract the payment, and use the new rate only for accumulation after that date.

Carrying a Balance Through Rate Segments

Let $B(s)$ be the balance immediately after any payment at time $s$. Suppose the balance is subject to annual effective rate $i_1$ for a duration $t_1-s$, then to rate $i_2$ for a duration $t-t_1$. Under the standard compound-interest model, the accumulation factor across the two segments is the product of their factors. If no payment occurs between $s$ and $t$, the balance at $t$ is $B(s)(1+i_1)^{t_1-s}(1+i_2)^{t-t_1}$.

If a payment $R$ is made at an intermediate date $u$, first accumulate the balance to $u$, subtract $R$, and then accumulate the remainder. This order follows directly from what the balance represents: the payment reduces the debt on its payment date, not before interest has accrued to that date.

B(t)=\left[B(s)(1+i_1)^{u-s}-R\right](1+i_2)^{t-u}

For fractional durations, this article uses compound accumulation $(1+i)^r$ over a duration of $r$ years. That convention must be supported by the problem’s interest model. If a problem specifies a different method for fractional periods, use the stated method instead.

Theorem: Starting with a balance immediately after time $s$, the balance immediately after time $t$ equals the starting balance accumulated to $t$, less each payment made after $s$ and by $t$, with each payment accumulated from its payment date to $t$ using the rates that apply over that interval: $$ B(t)=B(s)A(s,t)-\sum_j R_j A(t_j,t). $$ Here $A(a,b)$ is the product of the applicable accumulation factors from $a$ to $b$, $R_j$ is a payment at time $t_j$, and a payment at $t$ has $A(t,t)=1$. This form assumes the balance at $s$ is already after any payment made at $s$.

The formula can be justified by treating the initial debt and every payment as separate cash flows. The initial debt grows forward to $t$. Each payment reduces the debt, so it is subtracted after being grown from its own date to $t$. Equivalently, the balance can be calculated one interval at a time by applying interest and then subtracting payments. The two procedures agree because each earlier payment receives precisely the interest accumulation it would have avoided by reducing the balance on its payment date.

Mark the dates.
Identify payment dates, the rate-change date, and the date at which the balance is requested.

Split the timeline.
Use a separate accumulation factor for each period with a constant rate. Do not carry one rate across a reset.

Apply each payment at its date.
Accumulate to the payment date first, then subtract the payment. A payment on a reset date is subtracted before any post-reset accumulation.

Check at a common date.
Accumulate the initial balance and each earlier payment to the requested date. The resulting balance should match the interval-by-interval calculation.

Worked Applications

Worked Example: Annual Payments Before and After a Reset

A borrower owes $10,000 at time 0. The annual effective rate is 5% for the first two years and 8% thereafter. The borrower pays $2,000 at the end of each year. Find the balance immediately after the payment at time 4. Assume the $2,000 payments continue unchanged after the reset. Use the interval-by-interval method and report the final balance to the nearest cent.

Method. Calculate the balance after each annual payment, switching rates only for interest periods after time 2.

At time 1, interest is calculated at 5%: $$ 10{,}000(1.05)=10{,}500,\qquad 10{,}500-2{,}000=8{,}500. $$ The balance immediately after the first payment is $8,500.

At time 2, the 5% rate still applies to the second year: $$ 8{,}500(1.05)=8{,}925,\qquad 8{,}925-2{,}000=6{,}925. $$ The time-2 payment is made at the reset date. The balance after that payment is $6,925; the 8% rate applies to the next year, not the year just completed.

At time 3, use 8%: $$ 6{,}925(1.08)=7{,}479,\qquad 7{,}479-2{,}000=5{,}479. $$ At time 4, use 8% again: $$ 5{,}479(1.08)=5{,}917.32,\qquad 5{,}917.32-2{,}000=3{,}917.32. $$ The balance immediately after the time-4 payment is $3,917.32.

Check. Accumulating the original principal to time 4 gives $10{,}000(1.05)^2(1.08)^2=12{,}859.56$. The four payments accumulate to $2{,}000[(1.05)(1.08)^2+(1.08)^2+1.08+1]=8{,}942.24$. Their difference is $12{,}859.56-8{,}942.24=3{,}917.32$, as calculated.

Common mistake avoided: Applying 8% to the period ending at time 2 would incorrectly treat the new rate as though it began before the reset date.

Worked Example: A Reset Between Payment Dates

A loan balance is $20,000 at time 0. The annual effective rate is 4% through time 1.5 and 10% from time 1.5 onward. A $6,000 payment is made at time 1, and another $6,000 payment is due at time 2. Find the balance immediately after the time-2 payment. Use compound fractional-year accumulation and report the final balance to the nearest cent.

Method. Apply the 4% rate from time 0 to 1, subtract the first payment, apply the old rate for the next half-year, and then apply the new rate for the half-year from time 1.5 to 2.

The balance immediately after the time-1 payment is $$ 20{,}000(1.04)-6{,}000=20{,}800-6{,}000=14{,}800. $$ From time 1 to time 1.5, the balance accumulates for half a year at 4%: $$ 14{,}800(1.04)^{0.5}=14{,}800(1.0198039027)=15{,}093.0978. $$ No payment is made on the reset date, so $15,093.0978 is also the balance immediately after the rate changes at time 1.5.

From time 1.5 to time 2, the balance accumulates for half a year at 10%: $$ 15{,}093.0978(1.10)^{0.5} =15{,}093.0978(1.0488088482) =15{,}829.7745. $$ This is the balance immediately before the time-2 payment. Subtract that payment: $$ 15{,}829.7745-6{,}000=9{,}829.7745. $$ The balance immediately after the time-2 payment is $9,829.77.

Check. The same balance can be written from the time-1 balance as $14{,}800(1.04)^{0.5}(1.10)^{0.5}-6{,}000=9{,}829.7745$. The factors correspond to the two separate half-year segments, and the payment at time 2 has no further accumulation to time 2.

Common mistake avoided: Using 10% for the entire half-year from time 1 to 1.5 would move the reset backward. The stated rates apply on opposite sides of time 1.5.

Worked Example: A Payment Made on the Reset Date

A loan of $15,000 is repaid with annual payments of $5,000 at times 1, 2, and 3. The annual effective rate is 4% through time 2; the rate becomes 9% immediately after time 2. Find the balance immediately after the time-3 payment. Report the final balance to the nearest cent.

Method. Accumulate each year separately, treating the time-2 payment as occurring after interest at 4% for the second year and before interest at 9% for the third year.

At time 1: $$ 15{,}000(1.04)=15{,}600,\qquad 15{,}600-5{,}000=10{,}600. $$ At time 2, the old rate applies up to the reset date: $$ 10{,}600(1.04)=11{,}024,\qquad 11{,}024-5{,}000=6{,}024. $$ Thus, $6,024 remains immediately after the payment at time 2. Accumulate that balance for one year at 9%, then subtract the time-3 payment: $$ 6{,}024(1.09)=6{,}566.16,\qquad 6{,}566.16-5{,}000=1{,}566.16. $$ The balance immediately after the time-3 payment is $1,566.16.

Check. The original loan accumulated to time 3 is $15{,}000(1.04)^2(1.09)=17{,}684.16$. The time-1 payment accumulates by $1.04(1.09)$, the time-2 payment by $1.09$, and the time-3 payment by 1. Therefore, the accumulated payments total $5{,}000[1.04(1.09)+1.09+1]=16{,}118$. The difference, $17{,}684.16-16{,}118=1{,}566.16$, checks the result.

Common mistake avoided: Charging 9% on the balance from time 1 to time 2 would give $10{,}600(1.09)-5{,}000=6{,}554$ after the time-2 payment, rather than the correct $6,024. The 9% rate begins after the time-2 payment.

Common Timing Traps

A rate-change problem is primarily a timeline problem. Before calculating, state which rate applies to each interval and whether the requested balance is before or after any payment on the valuation date. A compact ledger often makes the order unambiguous.

Situation	Correct operation	Frequent error
Payment date occurs before a reset	Accumulate to the date at the old rate, then subtract the payment.	Apply the new rate to an interval that ends at the reset date.
Payment date occurs after a reset	Use the new rate for the post-reset interval, then subtract the payment.	Continue using the old rate after the reset.
Payment occurs on the reset date	Accumulate to the date using the rate that applied before it; subtract the payment; then use the new rate.	Accumulate the boundary payment or balance for time after its date.
Reset occurs between payments	Split the interest period at the reset and use each rate for its own duration.	Apply one rate to the whole payment period.

A second useful check is to calculate the balance from a common focal date. At that date, the initial principal has accumulated forward and each earlier payment has also accumulated forward from its payment date. Subtracting accumulated payments from the accumulated principal must give the same balance as the step-by-step ledger. If the answers differ, check the rate segments, payment timing, and whether the requested balance is before or after the final payment.

Key takeaway: At a reset date, interest up to the date belongs to the rate that applied before the reset. A payment on that date is subtracted at the date itself. Only the remaining balance accumulates at the new rate after the reset.

Check Your Understanding

For each problem, identify the balance just before or after the stated payment as requested. Assume compound fractional-year accumulation where needed, and use the stated rates on their specified intervals.

A $12,000 loan earns 5% for the first year and 7% for the second year. The borrower pays $3,000 at each year-end. Find the balance immediately after the time-2 payment. (Answer: $7,272.00)
A $9,000 loan earns 4% through time 2, then 8% afterward. Payments of $2,500 are made at times 1 and 2. Find the balance immediately after the time-2 payment. (Answer: $4,634.40)
A balance of $11,600 immediately after the time-1 payment earns 4% until time 1.5. No payment occurs at the reset date. Find the balance at time 1.5, using compound fractional-year accumulation. (Answer: $11,829.73)
A $10,000 loan earns 6% for the first two years and 3% for the third year. Payments of $2,000 are made at times 1 and 2, and a $3,000 payment is made at time 3. Find the balance immediately after the time-3 payment. (Answer: $4,329.48)
A $15,000 loan has payments of $5,000 at times 1 and 2. The rate is 4% through time 2 and becomes 9% after time 2. What is the error in the balance immediately after the time-2 payment if someone incorrectly uses 9% for the second year instead of 4%? (Answer: The incorrect balance is $6,554.00, which is $530.00 too high; the correct balance is $6,024.00.)

What You'll Learn