Beginning-of-Period Installments and Annuity-Due Loans

Why Payment Timing Changes a Loan

Two loans can have the same principal, rate, and number of installments but different payment amounts because the installments occur at different times. An installment paid at the beginning of a period is made one period earlier than an otherwise comparable installment paid at the end. Since earlier payments are discounted for fewer periods, a loan with beginning-of-period installments can be repaid with smaller level installments.

The standard model here assumes a fixed effective rate $i$ per payment period, a loan principal advanced at time 0, and equal installments made exactly at times $0,1,\ldots,n-1$. There are no fees, skipped payments, or additional payments, and the final installment fully repays the loan. If the quoted rate is not already expressed per payment period, first convert it to the effective rate for that interval.

Definition: An annuity-due loan has installments at the beginning of each payment period. For $n$ equal installments of amount $R$, the payment dates are $0,1,\ldots,n-1$. Its present-value factor is $\ddot{a}_{\overline{n}|}$. An ordinary-payment loan instead has installments at the end of each period, at times $1,2,\ldots,n$, with present-value factor $a_{\overline{n}|}$.

Present Value and the Installment Formula

Let $L$ be the amount advanced at time 0. Use $v=1/(1+i)$, where $i$ is the effective rate per payment period. A payment at time 0 is not discounted; a payment at time 1 is discounted once; and a payment at time $n-1$ is discounted $n-1$ times. Therefore, the present value of the installments is

L=R\ddot{a}_{\overline{n}|} =R(1+v+v^2+\cdots+v^{n-1}).

The terms in parentheses form a finite geometric series. Multiplying the series by $v$ and subtracting gives $(1-v)(1+v+\cdots+v^{n-1})=1-v^n$. Thus

\ddot{a}_{\overline{n}|} =\frac{1-v^n}{1-v} =\frac{1-v^n}{d} =(1+i)a_{\overline{n}|}, \qquad d=\frac{i}{1+i}.

The identity $\ddot{a}_{\overline{n}|}=(1+i)a_{\overline{n}|}$ also follows by shifting every ordinary-annuity payment one period earlier: each payment’s present value is multiplied by $1+i$. Solving the loan equation for the level installment gives

R=\frac{L}{\ddot{a}_{\overline{n}|}}.

For $i=0$, the geometric-series expression involving $d$ is not used; the present value is $nR$, so $R=L/n$. The annuity-due formula applies only when the installments really are at the beginnings of consecutive equal-length periods. It does not account for fees, irregular dates, or a different amount at the final payment.

Worked Example: Three Installments Due at the Beginning

A borrower receives a loan repaid by three installments of $3,600 at times 0, 1, and 2. The effective interest rate is 20% per period. Find the original loan amount and the balance immediately before the second installment. Assume there are no fees and the third installment fully repays the loan.

Method. Value each installment at time 0, then track the balance from just after the first installment to just before the second. This avoids treating the time-0 installment as though it were paid at time 1.

The discount factor is $v=1/1.20=5/6$. Thus the original loan amount is the present value of the three due installments:

L=3600(1+v+v^2) =3600\left(1+\frac{5}{6}+\frac{25}{36}\right) =3600\left(\frac{91}{36}\right) =\$9{,}100.

Immediately after the first installment at time 0, the balance is $9{,}100-3{,}600=\$5{,}500$. One period of interest accrues before the second installment, so the balance immediately before it is

5{,}500(1.20)=\$6{,}600.

As a check, the second installment reduces that balance to $6{,}600-3{,}600=\$3{,}000$. One more period of interest gives $3{,}000(1.20)=\$3{,}600$, exactly the final installment. The loan is fully repaid.

Common mistake avoided: $5,500 is the balance just after the first payment, not the balance just before the second. The intervening period’s interest must be included.

Balances Around a Due Installment

A balance must always be tied to a specific time and payment position. At time 0, the original balance immediately before the first installment is $L$. Immediately after that installment, the balance is $L-R$, because no interest period has yet elapsed. Before the next installment, interest accrues on that reduced balance. If $B$ is the balance just after a payment, then the balance just before the next payment is $B(1+i)$, and paying $R$ leaves $B(1+i)-R$.

\text{balance before next payment}=B(1+i), \qquad \text{balance after next payment}=B(1+i)-R.

The same balance can be found prospectively: at any date, it is the value at that date of the installments still unpaid. For example, immediately after a payment at time $t$, the remaining due installments, if any, fall at times $t+1,t+2,\ldots$. Their value at time $t$ is an ordinary-annuity value, because the next payment is one period away. This timing distinction is why a balance after a due installment should not automatically be calculated using an annuity-due factor.

Locate the balance date.
Decide whether the requested balance is before or after the installment. At a payment date, these balances differ by exactly $R$.

Apply interest only over elapsed periods.
A balance just after a payment grows by $1+i$ before the next payment. The time-0 payment has no preceding loan period in this model.

Check prospectively when useful.
Value the installments still unpaid at the stated date. Their first payment date determines whether the appropriate factor is an annuity-due or ordinary-annuity factor.

Worked Example: Find the Installment and Compare Payment Timing

A loan of $9,760 is to be repaid in three equal installments at an effective rate of 25% per period. Find the installment if the payments are due at the beginning of each period, and compare it with the installment for three end-of-period payments. Assume both schedules fully repay the loan and have no fees.

Method. Use the present value of the specified payment dates. At 25%, $v=1/1.25=0.8$. For beginning-of-period payments,

\ddot{a}_{\overline{3}|} =1+0.8+0.8^2 =1+0.8+0.64 =2.44.

The due installment is therefore $R=9{,}760/2.44=\$4{,}000$. Its present value checks directly: $4{,}000+3{,}200+2{,}560=\$9{,}760$.

For end-of-period payments, the dates are 1, 2, and 3, so the factor is

a_{\overline{3}|} =0.8+0.64+0.512 =1.952.

The ordinary-payment installment is $9{,}760/1.952=\$5{,}000$. Its present value is $5{,}000(0.8+0.64+0.512)=5{,}000(1.952)=\$9{,}760$. The ordinary installments are larger because each is paid one period later, and each dollar of installment has a lower present value.

Common mistake avoided: Using $a_{\overline{3}|}$ for the due schedule would incorrectly produce $5,000. The first due payment occurs at time 0, so its discount factor is 1, not 0.8.

Worked Example: Balance After Two Due Payments

A $2,952 loan is repaid by four installments of $1,000 at times 0, 1, 2, and 3. The effective rate is 25% per period. Find the balance immediately after the second installment, which is paid at time 1, and verify it by a prospective valuation.

Method. First use the recurrence to move the balance through the two payments. Then value the remaining installments at time 1.

At time 0, the first installment reduces the balance from $2,952 to $2{,}952-1{,}000=\$1{,}952$. Interest over the first period is $1{,}952(0.25)=\$488$, so the balance immediately before the second installment at time 1 is $1{,}952+488=\$2{,}440$. After that installment, it is $2{,}440-1{,}000=\$1{,}440$.

For the prospective check, the two unpaid installments occur at times 2 and 3. Discounting them to time 1 gives

1{,}000v+1{,}000v^2 =1{,}000(0.8)+1{,}000(0.64) =\$1{,}440.

The recurrence and prospective methods agree. As a further schedule check, the balance grows to $1{,}440(1.25)=\$1{,}800$ just before the third payment, falls to $800 after it, and grows to $800(1.25)=\$1{,}000$ just before the final payment. The final $1,000 installment clears the loan.

Common mistake avoided: The balance after the second payment is $1,440, while the balance immediately before it is $2,440. Both are valid balances, but they refer to different sides of the payment.

Practical Checks and Common Traps

Beginning-of-period payment schedules are particularly sensitive to date labels. A dependable calculation starts with a timeline, even if it contains only the dates $0$ through $n-1$. Then check the result using a second method when practical: compare the loan amount with the present value of payments, or compare a recurrence-based balance with the present value of payments still unpaid.

Using an ordinary-annuity factor by habit. An ordinary annuity has its first payment at time 1; an annuity-due loan has its first payment at time 0.
Leaving out the first payment at origination. The time-0 installment contributes $R$ to present value, without discounting.
Ignoring interest between installments. A balance after one payment is not the balance before the next; multiply by $1+i$ between those dates.
Reporting a balance without its timing. State whether it is immediately before or immediately after the scheduled payment.
Using an annual rate as a period rate without conversion. The rate in these formulas is the effective rate for one payment interval.

These timing and valuation checks are aligned with the SOA VEE Accounting and Finance syllabus. They also apply when reviewing a hypothetical loan offer: compare payment amounts only after confirming both the rate per payment period and the installment dates.

Key takeaway: For $n$ equal installments $R$ made at times $0$ through $n-1$, the loan amount is $L=R\ddot{a}_{\overline{n}|}$. Track balances by date: a payment at the beginning reduces the balance first, and interest accrues on the remaining balance until the next payment.

Check Your Understanding

For each problem, identify the payment timing before selecting an annuity factor. Unless stated otherwise, assume a fixed effective rate per period, no fees, and full repayment by the final stated installment.

A loan is repaid by four beginning-of-period installments of $500 at an effective rate of 10% per period. Find the original principal, rounded to the nearest cent. (Answer: $1,743.43)
A $7,200 loan is repaid by three equal installments at times 0, 1, and 2. The effective rate is 20% per period. Find the installment, rounded to the nearest cent. (Answer: $2,848.35)
Four installments of $1,500 are due at times 0, 1, 2, and 3 on a loan charging 8% effective per period. Find the balance immediately before the third installment at time 2, rounded to the nearest cent. (Answer: $2,888.89)
A $10,000 loan has three equal beginning-of-period installments and an effective rate of 10% per period. Find the installment, rounded to the nearest cent. (Answer: $3,655.59)
A $5,000 loan is repaid in two equal payments at an effective rate of 5% per period. Calculate the installment for (a) beginning-of-period payments and (b) end-of-period payments, each rounded to the nearest cent. Which schedule has the larger installment? (Answer: Beginning-of-period: $2,560.98; end-of-period: $2,689.02; the end-of-period schedule has the larger installment.)

What You'll Learn