Outstanding Balances on Annuity-Due Loans

Why Payment Timing Changes the Balance Formula

An annuity-due loan has installments at the beginning of each payment period. That timing affects both the original loan value and the outstanding balance. In particular, immediately after an installment is made, the next installment is one full period away. The installments still due are therefore valued as an annuity-immediate at that balance date, even though the original loan was an annuity-due.

The key is to locate the balance date before choosing a formula. A balance immediately after the kth payment excludes that payment and includes only payments that remain. A balance immediately before the kth payment includes the payment due at that date. Confusing those two values shifts the result by one installment.

Definition: Suppose a loan is issued at time 0 and repaid by $n$ equal installments of $R$, paid at times $0,1,\ldots,n-1$. The installments form an annuity-due. The outstanding balance immediately after the kth installment is the amount of principal still owed at time $k-1$, after the payment at that time has been made.

Unless stated otherwise, assume a fixed effective interest rate $i$ per payment period, level installments, and no fees or other cash flows. The formulas below apply at payment dates, immediately after the installment on that date. If the rate changes, the cash flows must instead be valued using the applicable rates for each period; if the balance date is between payment dates, additional timing steps are needed.

The Original Loan and the Balance After a Payment

Let $v=1/(1+i)$. The present value at time 0 of installments $R$ at times $0$ through $n-1$ is $R\ddot{a}_{\overline{n}|}$. Thus the original principal $L$ is $R\ddot{a}_{\overline{n}|}$. After the kth installment, the remaining payments occur at times $k,k+1,\ldots,n-1$. At the balance date $k-1$, these are $n-k$ payments, with the first due one period later.

Prospective balance: Immediately after the kth installment, the balance is the value at that date of the installments still due: $$ B_k=R a_{\overline{n-k}|}, \qquad k=1,2,\ldots,n. $$ Here $a_{\overline{0}|}=0$, so the balance after the final installment is zero. The notation $a_{\overline{m}|}=v+v^2+\cdots+v^m$ describes $m$ payments, the first one period after the valuation date.

The original loan uses an annuity-due value because its first installment is paid at time 0. After a payment at time $k-1$, the next remaining installment is at time $k$, not at time $k-1$. The remaining stream is consequently an annuity-immediate at that date. This timing distinction is the central feature of annuity-due balance calculations.

Worked Example: Balance After the Second Annuity-Due Payment

A loan is repaid by four annual installments of $1,000, paid at the beginning of each year. The effective annual interest rate is 10%. Find the original loan amount and the balance immediately after the second installment.

Method. Value all four installments at time 0 for the original principal. Then value only the installments still due at the balance date. This avoids treating the remaining payments as if one were due immediately.

The discount factor is $v=1/1.10=10/11$. The original loan amount is:

L=1{,}000\ddot{a}_{\overline{4}|} =1{,}000(1+v+v^2+v^3) =1{,}000\left(1+\frac{10}{11}+\frac{100}{121}+\frac{1{,}000}{1{,}331}\right)

The four terms are $1$, $0.9090909$, $0.8264463$, and $0.7513148$. Their sum is $3.4868520$, so the original loan amount is $3,486.85, rounded to the nearest cent.

The second installment is paid at time 1. Immediately after that payment, two installments remain, at times 2 and 3. Their value at time 1 is:

B_2=1{,}000a_{\overline{2}|} =1{,}000(v+v^2) =1{,}000\left(\frac{10}{11}+\frac{100}{121}\right) =1{,}000\left(\frac{210}{121}\right) =\$1{,}735.54

Check. The balance is less than the original principal, and it is the value of exactly the two installments still due. The common mistake is to use $\ddot{a}_{\overline{2}|}$, which would incorrectly place one of those installments at the balance date rather than one period later.

Retrospective Calculation and Why the Methods Agree

A retrospective calculation starts with the original loan, accumulates it to the balance date, and subtracts the accumulated value of payments already made. For a balance immediately after the kth payment, the original loan has accumulated for $k-1$ periods. The $k$ payments have accumulated to that date for $k-1,k-2,\ldots,0$ periods. Since $s_{\overline{k}|}=1+(1+i)+\cdots+(1+i)^{k-1}$, the retrospective formula is:

Retrospective balance: Under the fixed-rate and level-payment assumptions, immediately after the kth installment, $$ B_k=L(1+i)^{k-1}-R s_{\overline{k}|}. $$ The accumulated value subtracted includes the kth payment, which was made at the balance date and therefore accumulates for zero periods.

The prospective and retrospective approaches agree because both value the same cash flows at the same date. The following calculation shows the cancellation explicitly. Substitute $L=R\ddot{a}_{\overline{n}|}$ into the retrospective expression. Accumulating the original annuity-due terms to time $k-1$ produces terms with exponents $k-1,k-2,\ldots,0,-1,\ldots,-(n-k)$. The first $k$ terms sum to $s_{\overline{k}|}$, exactly the accumulated value of the payments already made. Subtracting those terms leaves the discounted values of the $n-k$ remaining installments:

\begin{aligned} B_k &=R\ddot{a}_{\overline{n}|}(1+i)^{k-1}-R s_{\overline{k}|}\\ &=R\left(\sum_{j=0}^{n-1}(1+i)^{k-1-j} -\sum_{r=0}^{k-1}(1+i)^r\right)\\ &=R\left((1+i)^{-1}+(1+i)^{-2}+\cdots+(1+i)^{-(n-k)}\right)\\ &=R a_{\overline{n-k}|}. \end{aligned}

This derivation also explains the timing rule: after subtracting all installments already paid, the remaining terms begin with a discount of $v$, not with $1$. A term of $1$ would represent a payment due immediately at the valuation date.

Worked Example: Comparing the Balance Before and After a Payment

A loan is repaid by five equal installments of $2,000 at times 0, 1, 2, 3, and 4. The effective interest rate is 25% per period. Find the balance immediately after the second installment, then find the balance immediately before and immediately after the third installment.

Method. Use the prospective method just after each payment. To move the balance from just after the second payment to just before the third, accumulate it for one period. The third installment is then subtracted. This makes the payment-date distinction explicit.

Since $i=0.25$, $v=0.8$. The original loan is:

L=2{,}000\ddot{a}_{\overline{5}|} =2{,}000(1+0.8+0.64+0.512+0.4096) =2{,}000(3.3616) =\$6{,}723.20

Immediately after the second payment, three installments remain, at times 2, 3, and 4. The balance at time 1 is:

B_2=2{,}000a_{\overline{3}|} =2{,}000(0.8+0.64+0.512) =2{,}000(1.952) =\$3{,}904.00

Immediately before the third payment at time 2, this balance has accumulated one period:

3{,}904(1.25)=\$4{,}880.00

The third installment of $2,000 is then paid. Immediately after it, two installments remain:

4{,}880-2{,}000=\$2{,}880.00, \qquad 2{,}000a_{\overline{2}|}=2{,}000(0.8+0.64)=\$2{,}880.00

Check. Both methods give $2,880 immediately after the third payment. The common mistake is to report $4,880 as the post-payment balance; that is the balance immediately before the third installment.

A Step-by-Step Procedure

A reliable calculation begins with payment dates, not with a memorized annuity symbol. Use this sequence whenever the question asks for an annuity-due loan balance at a payment date.

Set the period and rate.
Identify one payment period and use the effective interest rate for that period. Convert a quoted rate if it is not already expressed per payment period.

Mark the payment dates.
For an annuity-due with $n$ installments, place them at times $0$ through $n-1$. Mark the balance date as immediately before or after the specified payment.

Count the installments still due.
Immediately after payment $k$, there are $n-k$ installments left. Immediately before payment $k$, that payment is still due, so there are $n-k+1$ installments including it.

Value the appropriate cash flows.
After a payment, the remaining stream begins one period later and has value $R a_{\overline{n-k}|}$. Before a payment, include the payment due now as well as the later payments.

Check the result.
If useful, accumulate the original principal and subtract the accumulated value of all payments made through the balance date. Confirm that the retrospective and prospective results agree.

Worked Example: Finding the Installment and the Balance

A loan of $1,476 is repaid by four equal installments at times 0, 1, 2, and 3. The effective interest rate is 25% per period. Find the installment and the balance immediately after the third installment.

Method. First equate the loan principal to the present value of the four annuity-due payments. Then value the one remaining payment at the balance date.

Here $v=1/1.25=0.8$, and the present-value factor is:

\ddot{a}_{\overline{4}|}=1+0.8+0.64+0.512=2.952

Therefore the level installment is:

R=\frac{1{,}476}{2.952}=\$500.00

The third installment is paid at time 2. One installment, at time 3, remains. Its value at time 2 is:

B_3=500a_{\overline{1}|}=500(0.8)=\$400.00

Retrospective check. The original loan accumulates for two periods to $1{,}476(1.25)^2=\$2{,}306.25$. The three payments made at times 0, 1, and 2 accumulate to time 2 as follows:

500(1.25)^2+500(1.25)+500 =781.25+625+500 =\$1{,}906.25

Subtracting gives $2{,}306.25-1{,}906.25=\$400.00$, which agrees with the prospective result. A frequent error is to discount the last payment by zero periods; that would value it as $500 at time 2, although it is not due until time 3.

Common Timing Traps

Outstanding-balance questions often test the cash-flow timeline as much as the arithmetic. Keep these distinctions in view:

After payment $k$ is not before payment $k$. Immediately before that installment, it is still unpaid. Immediately after it, it is excluded from the remaining-payment value.
The original annuity-due factor is not automatically the balance factor. The original loan has a payment at time 0, so its value uses $\ddot{a}_{\overline{n}|}$. After a payment, the next installment is one period away, so the remaining value uses $a_{\overline{n-k}|}$.
Count payments, then value them. After the final payment, $n-k=0$ and the balance is zero. A nonzero result usually signals that a payment was counted or timed incorrectly.
Use the rate per payment period. If installments are quarterly, an annual rate cannot be inserted as though it were automatically a quarterly effective rate.

These balances are useful whenever a loan is being reviewed at a scheduled installment date, including a hypothetical comparison of repayment options. For this calculation, however, the model remains the stated fixed-rate loan with scheduled level payments. Any added fee, rate change, skipped installment, or valuation between payment dates changes the cash-flow valuation and must be handled explicitly.

Key takeaway: Immediately after the kth payment on an $n$-payment annuity-due loan, $n-k$ installments remain, and the first is one period away. The balance is $R a_{\overline{n-k}|}$. The retrospective check accumulates the original loan to the balance date and subtracts all $k$ payments already made.

Check Your Understanding

For each problem, identify the balance date carefully. Answers are rounded to the nearest cent where needed.

A loan has five annual installments of $300, paid at the beginning of each year. The effective annual rate is 10%. Find the balance immediately after the third installment. (Answer: $520.66)
A loan is repaid by four installments of $500 at times 0 through 3. The effective rate is 20% per period. Find the balance immediately after the first installment. (Answer: $1,053.24)
A loan has seven annual installments of $900 at the beginning of each year. The effective annual rate is 5%. Find the balance immediately after the fourth installment. (Answer: $2,450.92)
A loan of $976 is repaid by three equal installments at times 0, 1, and 2. The effective rate is 25% per period. Find the installment and the balance immediately after the second installment. (Answer: installment $400.00; balance $320.00)
A loan of $1,200 is repaid by three equal installments at times 0, 1, and 2. The effective rate is 25% per period. Find the installment, then find the balance immediately after the second installment. (Answer: installment $491.80; balance $393.44)

What You'll Learn