Schedules With Payments in Advance

Why Payment Timing Changes a Loan Schedule

A level-payment loan schedule depends not only on the payment amount and interest rate, but also on when each payment is made. If payments are made in advance, the first payment occurs when the loan is issued. It reduces the principal immediately, before any interest has accrued. Each later payment is made after the remaining balance has accumulated interest for one period.

This timing changes both the payment formula and the schedule. For the same number and amount of payments at a positive interest rate, payments in advance have a greater present value than payments at the ends of the periods. Therefore, a given loan can be repaid with smaller payments in advance than with payments in arrears.

Definition: A level-payment loan with $n$ payments in advance has equal payments $R$ at times $0,1,\ldots,n-1$. Assume a loan amount $L$, no fees or other cash flows, and an effective interest rate $i$ per payment period. The first payment is made at time 0, when the loan is issued; each subsequent payment is made at the beginning of the next period.

Valuing the Payments and Building the Schedule

Let $v=1/(1+i)$. At the loan issue date, the first payment is worth $R$, the payment at time 1 is worth $Rv$, and the payment at time $n-1$ is worth $Rv^{n-1}$. Equating the loan amount to the present value of all scheduled payments gives the payment in advance annuity formula.

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

Here $\ddot{a}_{\overline{n}|}$ denotes the present value of $n$ unit payments in advance at effective rate $i$ per period. For $i>0$, it is also equal to $(1-v^n)/d$, where $d=i/(1+i)$. When $i=0$, its value is $n$, so the payment is $L/n$.

To describe the schedule, let $B_k$ be the loan balance immediately after the $k$th payment. Thus the first payment occurs at time 0 and leaves $B_1=L-R$. If $k<n$, the balance $B_k$ earns interest for one period before payment $k+1$. The next payment then reduces that amount by $R$.

B_1=L-R, \qquad B_{k+1}=(1+i)B_k-R \quad (1\leq k<n).

This recurrence makes the timing explicit. The interest accruing between payments $k$ and $k+1$ is $iB_k$, not $i$ times the balance before payment $k$. The interest portion of the first payment is zero because that payment occurs at the loan issue date. For each later payment, the interest portion is $iB_{k-1}$, and the principal portion is $R-iB_{k-1}$.

Key result: With payments in advance, the first payment is made immediately and has no accrued interest portion. Between every pair of subsequent payments, the post-payment balance accumulates one period of interest; the next payment first covers that accrued interest and then reduces principal.

Why the Recurrence and Payment Formula Agree

The present-value formula and the schedule recurrence are two views of the same cash flows. One values the promised payments at the issue date; the other carries the balance forward one period at a time.

Start at issue.
The lender advances $L$, and the borrower immediately pays $R$. The balance after payment 1 is $L-R$.

Accumulate to the next payment date.
Over one period, the balance becomes $(L-R)(1+i)$. Payment 2 then reduces it by $R$, leaving $B_2=(L-R)(1+i)-R$.

Continue through the term.
Applying the same step after each payment gives $B_{k+1}=(1+i)B_k-R$. The payment formula ensures that the final payment leaves no balance.

The final statement can also be checked directly. Immediately after payment $k$, the remaining payments occur one period or more later. Their value at that date is $R a_{\overline{n-k}|}$, because there are $n-k$ payments and the first remaining one is due one period later. Thus, for $k<n$, the balance is $B_k=R a_{\overline{n-k}|}$, and after payment $n$ it is zero.

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

This is why using an annuity-immediate present value for payments that start at time 0 gives the wrong loan payment: it discounts the first payment by one period even though that payment is due immediately.

Worked Example: A Three-Payment Schedule at 10%

A hypothetical loan of $\$331$ is repaid by three equal annual payments in advance. The effective annual interest rate is $10\%$. Find the payment and construct the balance and interest schedule.

Method. First value the three payments at time 0, then apply the balance recurrence. This keeps the immediate first payment separate from interest that accrues later.

At $i=0.10$, $v=1/1.10=10/11$. Therefore $\ddot{a}_{\overline{3}|}=1+10/11+100/121=331/121$. The level payment is

R=\frac{331}{331/121}=\$121.

Payment 1 is made at time 0. It has no accrued interest, so its entire $\$121$ reduces principal: $B_1=331-121=\$210$.

During year 1, interest is $0.10(210)=\$21$. The balance just before payment 2 is $210+21=\$231$. After payment 2, $B_2=231-121=\$110$.

During year 2, interest is $0.10(110)=\$11$. The balance just before payment 3 is $110+11=\$121$, so the final payment of $\$121$ repays it in full.

The payment portions are therefore: payment 1, $\$0$ interest and $\$121$ principal; payment 2, $\$21$ interest and $\$100$ principal; payment 3, $\$11$ interest and $\$110$ principal. As a check, principal repaid totals $121+100+110=\$331$, equal to the original loan, and the ending balance is zero.

Worked Example: Locating the First Interest Charge

A hypothetical loan of $\$244$ is repaid by three equal annual payments in advance at an effective annual rate of $25\%$. Find the payment, identify the interest portions, and calculate the balance immediately after payment 2.

Method. The present value of three payments in advance is $R(1+v+v^2)$. Then the balance after each payment is found by accumulating interest between payment dates and subtracting the next payment.

Here $v=1/1.25=0.8$, so $\ddot{a}_{\overline{3}|}=1+0.8+0.64=2.44$. Thus $R=244/2.44=\$100$.

Payment 1 occurs at time 0. Its interest portion is $\$0$, and the balance immediately after it is $244-100=\$144$. During the first year, this balance accrues $0.25(144)=\$36$ interest, giving a balance of $144+36=\$180$ just before payment 2. After payment 2, the balance is $180-100=\$80$.

The interest portion of payment 2 is $\$36$, so its principal portion is $100-36=\$64$. The balance after payment 2 is also consistent with the schedule update $B_2=1.25(144)-100=\$80$. In the next year, the balance earns $0.25(80)=\$20$, and the final $\$100$ payment repays the resulting $\$100$ balance.

Check. The interest portions are $0,36,20$, and the principal portions are $100,64,80$. Principal totals $\$244$, and the balance after the final payment is zero. The key timing point is that the $\$36$ is added to the balance before payment 2; it is not subtracted from the balance.

Worked Example: Completing a Four-Payment Schedule

A hypothetical loan of $\$671$ is repaid by four equal annual payments in advance at an effective annual rate of $20\%$. Find the payment, the interest and principal portions of each payment, and verify the ending balance.

Method. Calculate the payment using the advance annuity value, then apply the recurrence to the post-payment balances.

At $i=0.20$, $v=5/6$. The annuity value is $\ddot{a}_{\overline{4}|}=1+5/6+25/36+125/216=671/216$. Hence $R=671/(671/216)=\$216$.

Payment 1 is immediate, so its interest portion is $\$0$ and $B_1=671-216=\$455$. In the first year, interest is $0.20(455)=\$91$; the balance before payment 2 is $\$546$. Payment 2 includes $\$91$ interest and $\$125$ principal, leaving $B_2=546-216=\$330$.

In the second year, interest is $0.20(330)=\$66$. Payment 3 includes $\$66$ interest and $\$150$ principal, leaving a balance of $330+66-216=\$180$. In the third year, interest is $0.20(180)=\$36$. Payment 4 includes $\$36$ interest and $\$180$ principal, leaving zero.

The principal portions are $216,125,150,180$, which total $\$671$. The interest portions are $0,91,66,36$, and the final balance is zero. These checks confirm that the schedule both repays the original principal and accounts for interest over exactly three periods.

Common Mistakes and Schedule Checks

A frequent error is to treat an advance payment as if it were made at the end of the first period. That incorrectly assigns interest to the first payment and uses the wrong present value. Another error is to subtract accrued interest from the balance. Interest is a charge added to the amount owed between payment dates; it must be added before the next payment is subtracted.

Common trap: For the $\$244$ loan above, a mistaken calculation might write $244-100-(144)(0.25)-100=\$8$. This subtracts the interest charge from the balance. The correct sequence after payment 1 is $144$, then $144(1.25)=180$ before payment 2, then $180-100=\$80$ after payment 2. Always apply $B_{k+1}=(1+i)B_k-R$, so interest increases the outstanding balance before a later payment reduces it.

When auditing a schedule, check the timing and the arithmetic in this order:

The first payment is at time 0 and has no preceding interest period.
Each balance between payments is multiplied by $1+i$.
Each payment is then subtracted from the balance immediately before it.
Each later payment's interest portion equals the interest accrued since the previous payment.
The final payment should reduce the balance to zero, subject only to stated rounding conventions.

These checks are useful whenever a problem asks for a balance “after a payment.” That phrase refers to the balance after subtracting the payment at that date, not the balance before the payment or after a further period of interest. A clear timeline prevents these distinct values from being confused.

Key takeaway: For $n$ level payments in advance, use $L=R\ddot{a}_{\overline{n}|}$. The initial payment reduces principal immediately; interest first accrues on the balance that remains after that payment. Thereafter, multiply the post-payment balance by $1+i$, then subtract the next payment.

Check Your Understanding

For each problem, use the stated effective rate per payment period. Payments described as “in advance” occur at times $0,1,\ldots,n-1$.

A hypothetical loan is repaid by three annual payments in advance of $\$40$ each. The effective annual rate is $25\%$. Find the loan amount. (Answer: $\$97.60$)
A $\$91$ loan is repaid by three equal annual payments in advance at $20\%$ effective annual interest. Find the payment, the interest accrued before payment 2, and the balance after payment 2. (Answer: payment $\$36$; interest before payment 2 is $\$11$; balance after payment 2 is $\$30$)
A $\$331$ loan is repaid by three equal annual payments in advance at $10\%$ effective annual interest. Find the payment, the interest portion of payment 2, and the balance after payment 2. (Answer: payment $\$121$; payment 2 interest $\$21$; balance after payment 2 is $\$110$)
A $\$671$ loan is repaid by four equal annual payments in advance at $20\%$ effective annual interest. Find the payment and the principal portion of payment 3. (Answer: payment $\$216$; payment 3 principal portion $\$150$)
A hypothetical $\$244$ loan is repaid by three equal annual payments in advance at $25\%$ effective annual interest. Find the balance immediately after payment 2, showing the balance just before that payment as an intermediate step. (Answer: balance before payment 2 is $\$180$; balance after payment 2 is $\$80$)

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