Why Payment Dates Matter
A scheduled loan installment is made on a specified due date. An extra principal payment, by contrast, may be made between due dates. Even if the amount is known, its effect depends on when it is paid: an earlier payment reduces the balance for longer and therefore prevents more interest from accruing. To value several such payments correctly, place every cash flow on a common date using the interest earned or charged over the actual time interval.
This tutorial assumes a fixed effective annual interest rate and compound accumulation between transactions. When a transaction occurs at a fraction of a year, the examples use the fractional-year accumulation factor \((1+i)^t\). A loan contract may specify a different convention, such as daily interest or a particular treatment of payments received between due dates; in that case, use the contract’s convention instead.
Why the Dated Value Equation Works
The equation follows by moving each cash flow forward to time \(T\). The original balance grows for \(T\) years to \(B_0(1+i)^T\). A payment \(C_k\) made at time \(t_k\) is no longer owed after it is paid, so its reduction to the balance at time \(T\) is the amount that payment would have accumulated to by \(T\), namely \(C_k(1+i)^{T-t_k}\). Subtracting each accumulated reduction from the accumulated advance gives the balance.
The same reasoning can be applied interval by interval. Between two transaction dates, the balance first accumulates at the applicable rate; at a payment date, subtract the payment immediately after that accumulation. Repeating this process gives the same result as moving all cash flows to one common date. A negative result would mean the payments exceed the amount required to discharge the loan under this model; it should not be interpreted as an ordinary positive outstanding balance.
For a balance question, use the date at which the balance is requested, and be clear whether it is immediately before or after any payment on that date.
Include the original balance, scheduled installments, and each extra principal payment. Keep the actual time of each payment.
Accumulate earlier amounts forward or discount later amounts backward using the stated interest convention.
Find the unknown balance or payment, then substitute the result into the dated equation to verify it.
Finding the Balance After Multiple Extra Payments
For a balance immediately after an installment at time \(T\), include that installment among the payments at or before \(T\). Payments made earlier must be accumulated for different lengths of time. In particular, an extra payment at \(t=0.5\) is accumulated for \(T-0.5\) years, not for the same period as an installment at \(t=1\).
Worked Example: Balance After Two Unscheduled Payments
A loan balance of $4,000 is outstanding at time 0. Interest is 21% effective annually. The borrower pays installments of $1,000 at times 1 and 2, plus extra principal payments of $250 at time 0.5 and $150 at time 1.5. Find the balance immediately after the time-2 installment.
Method. Accumulate every amount to time 2. Because \(1.21=(1.1)^2\), a half-year accumulation factor is \(1.1\), and a one-and-a-half-year factor is \(1.1^3=1.331\).
The original balance accumulated to time 2 is $$ 4000(1.21)^2=4000(1.4641)=5856.40. $$ The time-1 installment accumulates for one year: $$ 1000(1.21)=1210.00. $$ The time-0.5 extra payment accumulates for 1.5 years: $$ 250(1.21)^{1.5}=250(1.331)=332.75. $$ The time-1.5 extra payment accumulates for half a year: $$ 150(1.21)^{0.5}=150(1.1)=165.00. $$ The time-2 installment is made on the valuation date, so it is subtracted without accumulation. Therefore, $$ B_2=5856.40-1210.00-332.75-165.00-1000.00=3148.65. $$ The balance immediately after the time-2 installment is $3,148.65.
Check. Accumulating to time 2 gives an accumulated advance of $5,856.40 and accumulated payments of \(1210+332.75+165+1000=2707.75\). Their difference is \(5856.40-2707.75=3148.65\), as required. The common mistake avoided here is treating an extra payment as if it occurred at time 1 or time 2.
Solving for an Unknown Extra Payment
A dated value equation is also useful when an extra payment must be chosen to produce a specified balance. The unknown payment should be accumulated from its own date to the valuation date, just like every known payment. It is not generally equal to the difference between the target balance and a balance calculated on a nearby scheduled date.
Worked Example: Extra Payment Required for a Target Balance
A $2,500 loan accrues interest at 21% effective annually. Scheduled installments of $700 are paid at times 1 and 2. An extra principal payment of $100 is paid at time 0.5, and an unknown extra payment \(X\) is paid at time 1.5. Find \(X\) so that the balance immediately after the time-2 installment is $1,000.
Method. Accumulate all payments and the initial balance to time 2. The initial balance accumulates to $$ 2500(1.21)^2=2500(1.4641)=3660.25. $$ The time-1 installment accumulates one year, the time-0.5 payment accumulates 1.5 years, and the time-1.5 payment accumulates half a year: $$ 700(1.21)=847.00,\qquad 100(1.21)^{1.5}=100(1.331)=133.10,\qquad X(1.21)^{0.5}=1.1X. $$ The time-2 installment is $700. The target-balance equation is therefore $$ 1000=3660.25-847.00-133.10-1.1X-700. $$ Combining known amounts, $$ 1000=1980.15-1.1X, $$ so $$ 1.1X=980.15,\qquad X=\frac{980.15}{1.1}=891.04545\ldots. $$ To the nearest cent, the required extra payment is $891.05.
Check. Using the unrounded solution, the balance is \(1980.15-1.1(891.04545\ldots)=1000\). The common mistake avoided is subtracting \(X\) without accumulating it for the half-year between time 1.5 and time 2.
Recasting Installments After Several Extra Payments
Once the balance has been found at a date, it can be used as the starting amount for a revised installment calculation. Suppose the balance \(B_T\) is measured immediately after all payments at time \(T\), and the borrower will make \(n\) equal installments \(R\) at the ends of the next \(n\) years. The value at time \(T\) of those installments must equal \(B_T\): $$ B_T=R\left(\frac{1}{1+i}+\frac{1}{(1+i)^2}+\cdots+\frac{1}{(1+i)^n}\right)=R a_{\overline{n}|}. $$ This equation assumes the new installments are level, annual, and paid at the ends of periods, with the same effective annual rate continuing to apply. If the actual dates or rate differ, the value equation must be changed accordingly.
Worked Example: Revised Installment After Multiple Extra Payments
An $8,000 loan accrues interest at 21% effective annually. The borrower pays $2,000 at each of times 1 and 2, an extra principal payment of $400 at time 0.5, and another extra principal payment of $250 at time 1.5. Immediately after the time-2 payment, the loan is recast into two equal annual installments at times 3 and 4. Find the revised installment, to the nearest cent.
Step 1: Find the balance at time 2. The original balance accumulates to \(8000(1.21)^2=8000(1.4641)=11712.80\). The time-1 installment accumulates to \(2000(1.21)=2420.00\). The extra payments accumulate to $$ 400(1.21)^{1.5}=400(1.331)=532.40,\qquad 250(1.21)^{0.5}=250(1.1)=275.00. $$ The time-2 installment is $2,000. Thus, $$ B_2=11712.80-2420.00-532.40-275.00-2000.00=6485.40. $$
Step 2: Value the revised installments at time 2. Each revised installment is \(R\), paid at times 3 and 4. Its value at time 2 is \(R/1.21+R/1.4641\). Therefore, $$ 6485.40=R\left(\frac{1}{1.21}+\frac{1}{1.4641}\right). $$ Since \(1/1.21+1/1.4641=(1.21+1)/1.4641=2.21/1.4641\), $$ R=\frac{6485.40(1.4641)}{2.21} =\frac{9495.27414}{2.21} =4296.5041\ldots. $$ Rounded to the nearest cent, the revised annual installment is $4,296.50.
Check. The time-2 value of two installments of the unrounded amount is \(4296.5041\ldots(1/1.21+1/1.4641)=6485.40\). The common mistake avoided is treating the two extra payments as though they were made on times 1 and 2, which would give them the wrong accumulation periods.
Common Timing Traps
When a loan has both regular installments and irregular extra principal payments, the difficult part is usually not the algebra. It is the timeline. A useful calculation records each payment date before choosing whether to accumulate or discount it.
- Using the wrong exponent. A payment at time \(t\) accumulated to time \(T\) uses exponent \(T-t\), not \(T\). For example, a payment at time 1.5 accumulated to time 2 earns interest for 0.5 years.
- Mixing before-payment and after-payment balances. A balance immediately before an installment includes that installment as still unpaid; a balance immediately after it subtracts the installment. State which balance is requested.
- Using a regular-date recurrence across an irregular interval. If the balance is carried from time 1 to time 1.5, the accumulation period is half a year under the stated fractional-year convention, not one full year.
- Ignoring the contract’s interest convention. Fractional powers are appropriate here only because the model specifies compound accumulation at a fixed effective annual rate with fractional-year periods. A different contractual convention can produce a different result.
These dated equations are useful whenever a loan has payments outside its regular schedule. They separate the timing question from the arithmetic: first place every cash flow on a timeline, then value each amount on the same date, and only then solve for the unknown. This approach is aligned with the SOA Exam FM (Financial Mathematics) syllabus.
Check Your Understanding
For each problem, use the actual date of every payment. Unless stated otherwise, assume a fixed effective annual rate and fractional-year accumulation under compound interest.
- A $5,000 balance accrues interest at 21% effective annually. An extra principal payment of $200 is made at time 0.5. Find the balance immediately before any payment at time 1. (Answer: $5,830.00)
- A $6,000 loan accrues interest at 21% effective annually. Installments of $1,500 are paid at times 1 and 2, and extra payments of $250 and $100 are made at times 0.5 and 1.5, respectively. Find the balance immediately after the time-2 installment. (Answer: $5,026.85)
- A $5,000 loan accrues interest at 21% effective annually. Installments of $1,200 are paid at times 1 and 2. An extra payment of $300 is made at time 0.5, and an unknown extra payment \(X\) is made at time 1.5. Find \(X\) if the balance immediately after the time-2 installment is $4,000. (Answer: $244.73)
- A loan balance immediately after all payments at time 2 is $3,148.65. At 21% effective annually, it will be repaid by two equal annual installments at times 3 and 4. Find the installment to the nearest cent. (Answer: $2,085.94)