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Exam FM · Loan Amortization Schedules · Tutorial 234 of 998

Finding the Components of the kth Payment

Use the balance immediately before a selected payment to separate its interest and principal components and determine the post-payment balance.

Intermediate 10 min read

What You'll Learn

  • Define the balance immediately before and after payment \(k\).
  • Find the pre-payment balance from the value of the remaining level payments.
  • Calculate the interest and principal portions of a selected payment.
  • Verify the post-payment balance using both a recurrence and remaining payments.
  • Avoid confusing payments remaining before a payment with payments remaining after it.

Why the Payment Number Matters

A level-payment loan has the same scheduled payment each period, but the amount of that payment applied to interest and the amount applied to principal depend on the balance at that point in the schedule. To analyze payment \(k\), first find the loan balance immediately before it. The interest for the period is calculated on that balance; the rest of the payment reduces principal.

This approach is useful when a problem asks about one particular installment rather than a full amortization table. It also gives a direct way to check the balance after the payment: that balance must equal the value, at that time, of the payments still due.

Definition: Consider a fully amortizing loan with \(n\) equal payments of amount \(R\), made at the end of each period, and a constant effective interest rate \(i\) per payment period. Let \(B_j\) be the outstanding balance immediately after payment \(j\), with \(B_0\) the original loan principal. For payment number \(k\), \(B_{k-1}\) is the balance immediately before that payment, and \(B_k\) is the balance immediately after it. Assume \(1\leq k\leq n\), no fees or extra payments, and no rounding within the calculations.

Finding the Balance Before Payment \(k\)

The balance immediately before payment \(k\) is the value at that time of all payments from \(k\) through \(n\). There are \(n-k+1\) payments in that set: the payment being made now and every payment after it. Because those payments are level and occur at the end of each period, their value is an annuity-immediate.

$$ B_{k-1}=R a_{\overline{n-k+1}|} $$

Here \(a_{\overline{m}|}\) is the present-value annuity factor for \(m\) end-of-period payments at the applicable rate \(i\). The balance is measured immediately before payment \(k\), so the first payment included in its value is due one period later.

Key result: For payment \(k\), interest is calculated on the balance before the payment. The principal portion is the part of the payment left after interest is paid. The balance afterward is the balance beforehand less that principal portion.
$$ \begin{aligned} \text{Interest portion of payment }k,\quad I_k &= iB_{k-1},\\ \text{Principal portion of payment }k,\quad C_k &= R-I_k,\\ \text{Balance immediately after payment }k,\quad B_k &= B_{k-1}-C_k. \end{aligned} $$

Why the Component Formulas Hold

During the period from immediately after payment \(k-1\) to immediately before payment \(k\), the balance \(B_{k-1}\) accrues interest at the effective rate \(i\). The interest charged for that period is therefore \(iB_{k-1}\). The payment \(R\) first covers this interest; the amount remaining, \(R-iB_{k-1}\), reduces the loan principal.

Subtracting that principal reduction from the pre-payment balance gives the post-payment balance. Equivalently, the balance grows for one period and then the full payment is subtracted:

$$ B_k=B_{k-1}-(R-iB_{k-1})=(1+i)B_{k-1}-R. $$

There is also a useful prospective check. Immediately after payment \(k\), the remaining payments are \(k+1\) through \(n\), a total of \(n-k\) payments. Their value at that time is:

$$ B_k=R a_{\overline{n-k}|}. $$

If \(k=n\), no payments remain and \(a_{\overline{0}|}=0\), so the formula gives a zero balance after the final payment. This assumes the scheduled payments fully amortize the loan.

1
Count payments remaining before payment \(k\).
There are \(n-k+1\), including payment \(k\) itself.
2
Calculate the balance before payment \(k\).
Use \(B_{k-1}=R a_{\overline{n-k+1}|}\), or an otherwise known balance at that time.
3
Split the payment.
Calculate interest as \(iB_{k-1}\), then subtract it from \(R\) to get principal repaid.
4
Find and check the balance afterward.
Subtract principal repaid from \(B_{k-1}\), then compare with \(R a_{\overline{n-k}|}\).

Worked Calculations

Worked Example: The Tenth Payment on a Monthly Loan

A loan is repaid by 12 end-of-month payments of $500 at an effective monthly interest rate of 3%. Find the interest portion, principal portion, and balance immediately after payment 10. Keep full precision during the calculation and round the final dollar amounts to the nearest cent.

Method. Find the balance immediately before payment 10 from the value of payments 10, 11, and 12. This is an annuity of three payments, not two.

$$ \begin{aligned} B_9&=500a_{\overline{3}|}\\ &=500\left(\frac{1}{1.03}+\frac{1}{1.03^2}+\frac{1}{1.03^3}\right)\\ &=500(0.9708737864+0.9425959091+0.9151416593)\\ &=500(2.8286113548)\\ &=1414.3056774. \end{aligned} $$

The interest for the tenth month is 3% of that pre-payment balance:

$$ I_{10}=0.03(1414.3056774)=42.4291703. $$

The remainder of the $500 payment reduces principal:

$$ C_{10}=500-42.4291703=457.5708297. $$

Subtract the principal portion from the balance before the payment:

$$ B_{10}=1414.3056774-457.5708297=956.7348477. $$

Check. After payment 10, payments 11 and 12 remain, so their value is \(500a_{\overline{2}|}\). Since \(a_{\overline{2}|}=0.9708737864+0.9425959091=1.9134696955\), the remaining-payment value is \(500(1.9134696955)=956.7348478\), agreeing up to the last displayed decimal. Rounded to cents, the interest portion is $42.43, the principal portion is $457.57, and the post-payment balance is $956.73. The common mistake this method avoids is using the post-payment balance as the base for the interest charge.

Worked Example: A Payment Near the Middle of the Schedule

A fully amortizing loan has six end-of-year payments of $1,000 and an effective annual interest rate of 5%. The original principal is set equal to the present value of those six payments. Find the interest portion, principal portion, and post-payment balance for payment 4. Round final dollar amounts to the nearest cent.

Method. Payment 4 is preceded by three payments, so the balance immediately before it is the value of payments 4, 5, and 6. The original principal, for context, is \(1000a_{\overline{6}|}\).

$$ \begin{aligned} L&=1000a_{\overline{6}|}\\ &=1000(0.9523809524+0.9070294785+0.8638375985\\ &\qquad\quad+0.8227024748+0.7835261665+0.7462153966)\\ &=1000(5.0756920673)\\ &=5075.6920673. \end{aligned} $$

The balance just before payment 4 is the value of the three remaining payments, including payment 4:

$$ \begin{aligned} B_3&=1000a_{\overline{3}|}\\ &=1000(0.9523809524+0.9070294785+0.8638375985)\\ &=2723.2480294. \end{aligned} $$

The interest portion is \(5\%\) of this balance, and principal repaid is the difference between the payment and that interest:

$$ \begin{aligned} I_4&=0.05(2723.2480294)=136.1624015,\\ C_4&=1000-136.1624015=863.8375985. \end{aligned} $$

Therefore the post-payment balance is:

$$ B_4=2723.2480294-863.8375985=1859.4104309. $$

Check. Payments 5 and 6 remain, and \(1000a_{\overline{2}|}=1000(0.9523809524+0.9070294785)=1859.4104309\). The check agrees. Rounded to cents, the interest portion is $136.16, the principal portion is $863.84, and the post-payment balance is $1,859.41. The key is to count payment 4 among the payments valued in \(B_3\), but not among the payments valued in \(B_4\).

Worked Example: Verifying a Selected Payment Two Ways

A hypothetical loan is repaid by eight end-of-period payments of $1,200 at an effective rate of 4% per period. Find the components and post-payment balance for payment 6. Round final dollar amounts to the nearest cent.

Method. Immediately before payment 6, the payments still due are 6, 7, and 8. Thus the pre-payment balance is the present value of three payments:

$$ \begin{aligned} B_5&=1200a_{\overline{3}|}\\ &=1200\left(\frac{1}{1.04}+\frac{1}{1.04^2}+\frac{1}{1.04^3}\right)\\ &=1200(0.9615384615+0.9245562130+0.8889963587)\\ &=1200(2.7750910332)\\ &=3330.1092398. \end{aligned} $$

Interest accrues on \(B_5\) for one period. The rest of the payment reduces principal:

$$ \begin{aligned} I_6&=0.04(3330.1092398)=133.2043696,\\ C_6&=1200-133.2043696=1066.7956304. \end{aligned} $$

Subtracting principal repaid gives the balance after payment 6:

$$ B_6=3330.1092398-1066.7956304=2263.3136094. $$

Check. There are two payments left, and \(1200a_{\overline{2}|}=1200(0.9615384615+0.9245562130)=2263.3136094\). The two calculations agree. To the nearest cent, the interest portion is $133.20, the principal portion is $1,066.80, and the post-payment balance is $2,263.31. This check catches arithmetic errors and off-by-one errors in counting the remaining payments.

Common Exam Trap: Using the Wrong Balance

Interest for payment \(k\) is based on the balance before payment \(k\), not the balance after it. Also, the balance before the payment includes the value of payment \(k\) itself. These are distinct counting decisions.

For the 12-payment example, an incorrect approach would use the two payments after payment 10 to get \(B_{10}=500a_{\overline{2}|}\), then calculate interest as \(0.03B_{10}\). That gives an interest charge of about \(0.03(956.73)=$28.70\), which is not the interest in payment 10: $956.73 is the balance after payment 10. The correction is to use \(B_9=500a_{\overline{3}|}=$1,414.31\) as the interest base. Then \(I_{10}=0.03B_9=$42.43\). The two-payment value is a valid balance, but it belongs after payment 10.

Scope and Practical Checks

These formulas apply to level payments made at the end of each period, with a constant effective rate per payment period and a loan that is fully amortized by the scheduled payments. The rate \(i\) must match the payment period: for monthly payments, \(i\) is the effective monthly rate; for quarterly payments, it is the effective quarterly rate. If payments vary, occur in advance, or the rate changes, the balance and payment components must be valued using the actual cash-flow timing and applicable rates rather than this level-payment formula.

A reliable answer should pass three checks. First, interest plus principal must equal the scheduled payment: \(I_k+C_k=R\). Second, the post-payment balance must equal the pre-payment balance less principal repaid: \(B_k=B_{k-1}-C_k\). Third, for a fully amortizing level-payment loan, the post-payment balance must equal the value of the remaining payments, \(R a_{\overline{n-k}|}\). Together, these checks help identify a misplaced payment, an incorrect interest period, or a subtraction error.

Key takeaway: For payment \(k\), start with \(B_{k-1}=R a_{\overline{n-k+1}|}\). Charge interest \(iB_{k-1}\), apply the rest of \(R\) to principal, and verify that the balance left is \(R a_{\overline{n-k}|}\).

Check Your Understanding

For each level-payment loan, find the interest portion, principal portion, and balance immediately after the stated payment. Assume end-of-period payments, a constant effective rate per payment period, no fees or extra payments, and no intermediate rounding. Round dollar answers to the nearest cent.

  1. A five-payment loan has payments of $900 per year at an effective annual rate of 4%. Find the components and post-payment balance for payment 3. (Answer: interest $99.90; principal $800.10; balance $1,697.49.)
  2. A loan has seven quarterly payments of $650 at an effective quarterly rate of 2%. Find the components and balance after payment 7. (Answer: interest $12.75; principal $637.25; balance $0.00.)
  3. A loan has nine monthly payments of $300 at an effective monthly rate of 0.5%. Find the components and balance after payment 5. (Answer: interest $7.39; principal $292.61; balance $1,185.15.)
  4. A loan has six quarterly payments of $1,500 at an effective quarterly rate of 3%. Find the components and balance after payment 2. (Answer: interest $206.09; principal $1,293.91; balance $5,575.65.)
  5. A loan is repaid by ten semiannual payments of $800 at an effective rate of 2.5% per half-year. Find the components and balance after payment 8. (Answer: interest $57.12; principal $742.88; balance $1,541.94.)

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