A Diagnostic Workshop for Irregular Loan Cash Flows

Why Irregular Loan Balances Need a Diagnostic Method

For a level-payment loan, a familiar annuity formula may give an outstanding balance directly. Unequal payments, irregular payment dates, and changing rates often make that shortcut unavailable. The core principle does not change: at a chosen balance date, the outstanding debt is the value of the remaining obligation, or equivalently the original debt accumulated to that date less the value of payments already made.

The practical challenge is choosing a calculation that matches the information given. A prospective method is often most direct when the remaining payments are known. A retrospective method is often most direct when the loan and payments already made are known. A period-by-period recurrence is usually easiest when rates change or the payment schedule is irregular. These methods should agree when they use the same timing convention and accumulation assumptions.

Definition: Let $B_t^+$ be the outstanding balance immediately after any payment made at time $t$, and let $B_t^-$ be the balance immediately before a payment due at time $t$. If a payment $C_t$ is made at that time, then $B_t^-=B_t^++C_t$. Payments $C_j$ are positive amounts paid by the borrower.

Let $A(s,t)$ denote the accumulation factor from time $s$ to a later time $t$, with $A(s,s)=1$. The factor must follow the interest-rate model in the problem. With a constant effective annual rate $i$, and integer-year dates, $A(s,t)=(1+i)^{t-s}$. With changing rates, multiply the accumulation factors for the successive rate intervals. For fractional dates, use the fractional-period convention specified in the problem; if none is given, do not silently assume a convention.

Three Routes to the Same Balance

Retrospective balance: Immediately after all payments at or before $t$ have been made, accumulate the original loan and subtract the accumulated value of those payments.

B_t^+ = L A(0,t)-\sum_{t_j\le t} C_j A(t_j,t)

Here $L$ is the original loan at time 0. A payment made at $t$ is included with accumulation factor $A(t,t)=1$, so it is subtracted in full. If calculating immediately before a payment due at $t$, omit that payment from the retrospective sum; it has not yet been made.

Prospective balance: Immediately after all payments at time $t$, the balance is the value at $t$ of payments due strictly after $t$.

B_t^+ = \sum_{t_j>t}\frac{C_j}{A(t,t_j)}

For a balance immediately before a payment due at $t$, include that payment in full as well as discounting the later payments:

B_t^- = \sum_{t_j\ge t}\frac{C_j}{A(t,t_j)}

These distinctions matter even when the date is the same. A payment due at the balance date belongs in the before-payment value, but not in the after-payment value.

Period-by-period recurrence: If the balance immediately after one cash flow at time $s$ is $B_s^+$, then immediately before the next payment at $t$, the balance is $B_s^+A(s,t)$. Subtract a payment made at $t$ to obtain $B_t^+$.

Why the Methods Agree

The cash-flow identity is the reason retrospective and prospective calculations match. Suppose the loan is fully repaid by the last scheduled payment date $T$. Accumulate the initial loan and every borrower payment to $T$. The loan value at $T$ equals the accumulated value of those payments, with each payment occurring at its stated date:

L A(0,T)=\sum_j C_j A(t_j,T)

To obtain the balance just after time $t$, separate the payments made by $t$ from those still due. Divide the identity by $A(t,T)$, or, equivalently, value each cash flow at $t$ using the same accumulation model. The past payments and original loan produce the retrospective difference; the future payments produce the prospective value. Thus the two expressions represent the same net obligation at $t$. This argument requires a consistent interest model and a cash-flow schedule that satisfies the loan equation; it does not make an arbitrary proposed final payment sufficient to repay the loan.

Fix the balance convention.
Write whether the requested value is immediately before or after any payment at the balance date.

Map the cash flows and rates.
List dates in order, mark payments, and identify the accumulation factor for each interval.

Choose the direct route.
Use prospective value when remaining payments are known, retrospective value when past cash flows are known, or a recurrence when each interval has a different rate or timing.

Check independently.
Use a second method or the accumulated cash-flow identity, and verify that the balance changes by exactly the payment made at a payment date.

Worked Examples: Choosing and Checking a Method

Worked Example: Unequal Payments on Irregular Dates

A hypothetical loan of $20,000 accrues interest at a constant force of interest $\delta=\ln(1.21)$ per year. Thus the annual effective rate is 21%, and accumulation over $t-s$ years is $A(s,t)=1.21^{t-s}$. The borrower pays $7,000 at time 1 and $8,000 at time 2.5. Find the balance immediately after the payment at time 2.5 and the payment required at time 4 to repay the loan. Use the recurrence, then verify prospectively.

Method. The recurrence is direct because the dates are irregular and the balance is required just after a specified payment. The force assumption gives unambiguous accumulation over the 1.5-year intervals: $1.21^{1.5}=1.21\sqrt{1.21}=1.21(1.1)=1.331$.

From time 0 to time 1, accumulate the loan and subtract the first payment:

B_1^+=20{,}000(1.21)-7{,}000 =24{,}200-7{,}000 =17{,}200.

Accumulate this balance 1.5 years to time 2.5, then subtract the payment due there:

B_{2.5}^+=17{,}200(1.331)-8{,}000 =22{,}893.20-8{,}000 =14{,}893.20.

There are another 1.5 years from time 2.5 to time 4, with the same accumulation factor. The required final payment is therefore:

C_4=14{,}893.20(1.331) =14{,}893.20+4{,}929.6492 =19{,}822.8492.

Rounded to the nearest cent, the final payment is $19,822.85. For a prospective check, its value at time 2.5 is $19{,}822.8492/1.331=14{,}893.20$, exactly the balance found after the time-2.5 payment. The balance convention is important: the $8,000 payment was subtracted at time 2.5, so it must not be subtracted again.

Worked Example: Changing Rates and a Payment on the Balance Date

A hypothetical $25,000 loan earns 4% effective interest in year 1, 6% in year 2, and 5% in year 3. Payments of $5,000 and $8,000 are made at the ends of years 1 and 2, respectively. Find the balance immediately before and immediately after the year-2 payment, then find the payment required at the end of year 3 to clear the loan.

Method. Use a recurrence because the rate changes by year. Each year’s accumulation factor is $1+i$ for that year; do not apply one rate to the whole loan duration.

At the end of year 1, the balance immediately before the $5,000 payment is:

B_1^-=25{,}000(1.04)=26{,}000.

Subtracting the payment gives the year-1 balance immediately after payment:

B_1^+=26{,}000-5{,}000=21{,}000.

Accumulate one year at 6% to obtain the year-2 balance before its payment:

B_2^-=21{,}000(1.06)=22{,}260.

After the $8,000 payment, the balance is:

B_2^+=22{,}260-8{,}000=14{,}260.

The final year accrues 5%, so the payment at time 3 must be $14{,}260(1.05)=14{,}973$. The check is $14{,}973/1.05=14{,}260$, the balance immediately after the year-2 payment. The before-payment balance exceeds the after-payment balance by exactly $8,000, as required.

Worked Example: Testing a Proposed Final Payment

A hypothetical $10,000 loan earns 8% effective interest per year. The borrower pays $3,000 at the end of year 1 and $2,000 at the end of year 2. A proposed final payment of $6,000 is scheduled for the end of year 3. Find the required final payment and the shortfall in the proposal.

Method. The recurrence is simplest: carry the balance forward one year at a time, subtracting each payment only at its actual date. The shortfall is the required payment less the proposed payment.

At time 1, before the first payment, the balance is $10{,}000(1.08)=10{,}800$. After that payment:

B_1^+=10{,}800-3{,}000=7{,}800.

At time 2, before the second payment, the balance is $7{,}800(1.08)=8{,}424$. After the payment:

B_2^+=8{,}424-2{,}000=6{,}424.

Accumulate the remaining balance for one year. The required year-3 payment is:

C_3=6{,}424(1.08)=6{,}937.92.

The proposal is short by $6{,}937.92-6{,}000=$$937.92. Check retrospectively at time 3: the original loan accumulates to $10{,}000(1.08)^3=12{,}597.12$; the year-1 payment accumulates to $3{,}000(1.08)^2=3{,}499.20$; and the year-2 payment accumulates to $2{,}000(1.08)=2{,}160$. Their difference is $12{,}597.12-3{,}499.20-2{,}160=6{,}937.92$, matching the recurrence.

Common Traps and a Practical Decision Rule

The most common error is not algebra; it is valuing a cash flow at the wrong date. A payment at the balance date is included in $B_t^-$ and excluded from $B_t^+$. Another frequent error is to use a single rate when the model specifies different rates over successive intervals. Build each accumulation factor from the rate that applies during that interval.

A typical incorrect calculation for the second example would call $22,260 the balance immediately after the year-2 payment. That value is the balance before the $8,000 payment. The correction is $22{,}260-8{,}000=14{,}260$. If the final payment is calculated from $22,260, it will be too large by $8{,}000(1.05)=\$8{,}400$, because the omitted year-2 payment would itself have accumulated for one year at 5%.

Use this decision rule:

If the remaining payments are listed and the balance is after the current payment, discount only payments strictly later than the balance date.
If the original loan and past payments are listed, accumulate those cash flows to the balance date and subtract.
If dates or rates change from interval to interval, update the balance sequentially and record each before-payment and after-payment value clearly.
If the question asks whether a proposed final payment is adequate, calculate the required payment first and compare amounts at that same payment date.

These methods are aligned with the SOA Exam FM (Financial Mathematics) syllabus. Their usefulness is broader than a particular formula: a well-labeled timeline and a consistent balance convention make irregular loan problems manageable and provide a reliable check against timing and rate errors.

Check Your Understanding

For each problem, identify whether the requested balance is before or after any payment at that date, and use the stated interest assumptions.

A $12,000 loan earns 5% effective interest annually. Payments of $2,500 at the end of year 1 and $3,000 at the end of year 2 have been made. Find the balance immediately after the year-2 payment. (Answer: $7,605.00)
A $9,000 loan earns 4% effective interest annually. A $2,000 payment is made at the end of year 1, and a $2,500 payment is due at the end of year 2. Find the balance immediately before the year-2 payment. (Answer: $7,654.40)
A $15,000 loan earns 3% in year 1 and 7% in year 2. Payments of $4,000 and $5,000 are made at the ends of years 1 and 2. Find the balance immediately after the year-2 payment. (Answer: $7,251.00)
A $10,000 loan uses a half-year accumulation factor of 1.05, equivalent to an annual effective rate of $1.05^2-1=10.25\%$. A $3,000 payment is made at time 0.5, and a $2,000 payment at time 1.5. Find the balance immediately after the time-1.5 payment. (Answer: $6,268.75)
A $20,000 loan earns 6% effective interest annually. Payments of $5,000 and $6,000 are made at the ends of years 1 and 2. Find the required final payment at the end of year 3 and the shortfall if the proposed payment is $10,000. (Answer: Required payment $11,842.32; shortfall $1,842.32)

What You'll Learn