Comparing Annual and Monthly Payment Schedules

Why Payment Frequency Changes a Loan Schedule

Two loans can have the same principal, term, and effective annual interest rate yet use different payment frequencies. Their schedules will not have the same number or size of payments: a monthly schedule repays part of the principal throughout each year, whereas an annual schedule makes its payment only at the year-end. To compare them fairly, use an equivalent rate for each payment period and be explicit about when payments occur.

Unless a problem says otherwise, this tutorial assumes level payments in arrears: annual payments are made at each year-end, and monthly payments at each month-end. Interest accumulates at a constant effective rate, with fractional-year accumulation consistent with compound interest. These assumptions make the effective monthly rate equivalent to the stated effective annual rate.

Definition: If $i$ is the effective annual rate, the equivalent effective monthly rate $j$ satisfies $$ 1+j=(1+i)^{1/12}, \qquad j=(1+i)^{1/12}-1. $$ Thus $ (1+j)^{12}=1+i $. The rates are equivalent because they produce the same accumulation over one year. The formula applies for $i>-1$; for ordinary positive-interest loan models, $i\geq 0$.

Let $L$ be the original loan principal and $n$ the term in years. For a positive annual effective rate, the annual payment $R$ and monthly payment $C$ are determined from the present value of the payments:

L=R\,a_{\overline{n}|\,i} \qquad\text{and}\qquad L=C\,a_{\overline{12n}|\,j}.

Here $a_{\overline{n}|\,i}$ is the present value of $n$ unit payments at year-ends, and $a_{\overline{12n}|\,j}$ is the present value of $12n$ unit payments at month-ends. These formulas describe level-payment loans with payments in arrears. If payments are due at the beginnings of periods, the annuity-due value is needed instead.

How the Annual and Monthly Payments Relate

The annual and monthly payment amounts are not directly comparable as single installments: one annual payment covers a full year, while one monthly payment covers one month. The useful comparison is between $R$ and the total $12C$ of monthly installments during a year.

Group the monthly payments into blocks of 12. Within each block, the payments occur at months 1 through 12 after the anniversary at the start of that year. The present value at that anniversary of one block is $C\,a_{\overline{12}|\,j}$. Define $v=1/(1+i)$. At time 0, the values of the $n$ blocks form a geometric series with ratio $v$. The annual schedule's $n$ payments have present value $R\,a_{\overline{n}|\,i}$, where $a_{\overline{n}|\,i}=v(1+v+\cdots+v^{n-1})$. Equating the present values of the two schedules gives the following relationship.

Theorem: For equivalent rates and end-of-period payments, the level payments on otherwise identical annual and monthly schedules satisfy $$ C\,a_{\overline{12}|\,j}=R\,v. $$ For $i>0$, this can also be written as $R/C=(1+i)a_{\overline{12}|\,j}$. If $i=0$, then $j=0$, $a_{\overline{12}|\,0}=12$, and the relationship becomes $R=12C$. The zero-rate case is included without dividing by either rate.

For $i>0$, $12C<R$. To see why, defer every year's 12 monthly payments to that year's end while keeping their total $12C$ unchanged. Deferring each payment reduces its present value, so the present value of the actual monthly schedule is greater than the present value of these deferred annual totals. Therefore $L>12C\,a_{\overline{n}|\,i}$. Since $L=R\,a_{\overline{n}|\,i}$, it follows that $R>12C$. At zero interest, deferral has no effect and $R=12C$.

Balances at Annual Anniversaries

The different payment timing does not mean that the schedules must have different balances at every convenient comparison date. Immediately after the $k$th annual payment, the annual schedule has $n-k$ annual payments remaining. Immediately after month $12k$'s payment, the monthly schedule has $12(n-k)$ monthly payments remaining. Their prospective balances are

B_k^{\text{annual}}=R\,a_{\overline{n-k}|\,i}, \qquad B_k^{\text{monthly}}=C\,a_{\overline{12(n-k)}|\,j}.

These are equal at each annual anniversary. For $m=n-k$, the monthly remaining payments can be grouped into $m$ blocks of 12. Valued at the anniversary, their present value is $C\,a_{\overline{12}|\,j}(1+v+\cdots+v^{m-1})$. The annual remaining payments have value $R\,v(1+v+\cdots+v^{m-1})$. The theorem above makes these expressions equal. This argument also works when $i=0$: the block value is $12C=R$, and both schedules have the same anniversary balance.

Between anniversaries, however, the schedules generally differ. The monthly borrower makes payments during the year, while the annual borrower makes none until year-end. A balance comparison must therefore specify its valuation date and whether it is just before or just after a payment.

Set the payment dates.
Identify whether installments are at period-end or period-beginning, and count the payments over the term.

Convert the rate.
For monthly payments, use $j=(1+i)^{1/12}-1$, not $i/12$ unless the problem explicitly gives a nominal rate convertible monthly.

Calculate each payment.
Use $R=L/a_{\overline{n}|\,i}$ for annual payments or $C=L/a_{\overline{12n}|\,j}$ for monthly payments in arrears.

Compare on a common date.
Use each schedule's remaining payments to value its balance, with the same valuation date and payment status.

Worked Example: Comparing One-Year Payments and Midyear Balances

A hypothetical borrower takes a loan of $10,000 for one year at an effective annual rate of 12.682503%. Compare a single year-end payment with 12 equal month-end payments. Find each payment, the balance after six months in each schedule, and the total interest paid. Report dollar amounts to the nearest cent.

Method. Convert the effective annual rate to its equivalent monthly rate, then value each schedule as an annuity-immediate. The key trap is dividing the annual effective rate by 12.

The annual accumulation factor is $1.12682503$, which is $1.01^{12}$. Thus $j=0.01$, or 1% per month. The annual payment is

R=10{,}000(1.12682503)=\$11{,}268.25.

For the monthly schedule,

a_{\overline{12}|\,0.01} =\frac{1-(1.01)^{-12}}{0.01} =11.255078, \qquad C=\frac{10{,}000}{11.255078} =\$888.49.

After six months, the annual schedule has made no payment. Its balance has accumulated for half a year, so the balance is $10{,}000(1.01)^6=\$10{,}615.20$. The monthly schedule has made six payments. Since $(1.01)^6=1.06152015$, the accumulated value of those payments is $888.4878[(1.01)^6-1]/0.01$, or about $5,465.99. Therefore,

B_6^{\text{monthly}} =10{,}000(1.06152015)-888.4878(6.152015) =\$5{,}149.21.

The monthly schedule's 12 payments total $12(\$888.4878)=\$10{,}661.85$, so its total interest is $661.85. The annual schedule's one payment totals $11,268.25, so its interest is $1,268.25. At year-end, each schedule has paid off the loan. The check is consistent: monthly payments repay principal earlier and produce less total interest, even though the schedules have the same initial principal and equivalent annual accumulation.

Worked Example: Checking the Balance at an Anniversary

A hypothetical $10,000 loan is repaid over two years at a monthly effective rate of 1%, with either annual year-end payments or monthly month-end payments. Compare the payments and the balance immediately after the first year's payment in each schedule. Round dollar amounts to the nearest cent.

Method. Convert the monthly rate to its equivalent effective annual rate, value both payment streams, then compare their prospective balances at the one-year anniversary.

The annual effective rate is $$ i=(1.01)^{12}-1=0.12682503. $$ The needed annuity values are $a_{\overline{12}|\,0.01}=11.255078$ and $$ a_{\overline{24}|\,0.01} =11.255078\bigl(1+(1.01)^{-12}\bigr) =21.243388. $$ The annual payment and monthly payment are therefore $$ R=\frac{10{,}000}{a_{\overline{2}|\,0.12682503}} =\frac{10{,}000}{1.675015} =\$5{,}970.09, \qquad C=\frac{10{,}000}{21.243388} =\$470.73. $$

At the anniversary, the annual schedule has one annual payment remaining, so its balance is $$ B_1^{\text{annual}}=R\,a_{\overline{1}|\,i} =R(1+i)^{-1} =\$5{,}298.16. $$ The monthly schedule has 12 monthly payments remaining, so its balance is $$ B_1^{\text{monthly}}=C\,a_{\overline{12}|\,0.01} =\$5{,}298.16. $$ The balances agree, as predicted. This check depends on comparing both schedules immediately after their payments at the anniversary; comparing at another date need not give equal balances.

Worked Example: Zero Interest and the Payment-Timing Trap

A hypothetical $24,000 loan is repaid over two years with no interest. Compare annual year-end payments with monthly month-end payments. Then explain how the monthly installment changes if the 24 payments are instead due at the beginning of each month. Report dollar amounts to the nearest cent.

Method. At zero interest, divide the principal equally among the scheduled payments. Do not apply formulas that divide by an interest rate, and do not treat an advance-payment schedule as an ordinary annuity.

With annual payments in arrears, each payment is $\$24{,}000/2=\$12{,}000$. With monthly payments in arrears, each payment is $\$24{,}000/24=\$1{,}000$. After one year, both schedules have repaid $12,000 of principal and have a balance of $12,000.

If the 24 monthly payments are due in advance, the first is made immediately. At zero interest, this changes the dates but not the total amount needed to repay the principal: each payment remains $1,000. The first advance payment reduces the balance at time 0, rather than after one month. Thus the payment amount is unchanged in this special zero-interest case, but the balance at intermediate dates differs from the month-end schedule. The check is that 24 payments of $1,000 still total $24,000, with no interest to add.

Common Mistakes and Exam Checks

A frequent error is to convert an effective annual rate by dividing it by 12. In the first worked example, the incorrect calculation would use $0.12682503/12=0.01056875$ as the monthly rate and then calculate $10{,}000/a_{\overline{12}|\,0.01056875}$. That rate does not produce the stated annual accumulation: $(1+0.01056875)^{12}\neq1.12682503$. The correction is to use $j=(1+i)^{1/12}-1$, which gives 1% per month in that example.

Comparing unlike amounts. $R$ is one year's payment, while $C$ is one month's payment. Compare $R$ with $12C$ when considering a year's nominal payments.
Ignoring payment dates. A month-end schedule is an annuity-immediate. A beginning-of-month schedule is an annuity-due and has different payment timing.
Assuming all balances match. The schedules have equal balances at annual anniversaries immediately after their corresponding payments, but not generally at dates between anniversaries.
Dividing by zero at $i=0$. The usual annuity formulas expressed as quotients involving $i$ or $j$ need their zero-rate limits. At zero interest, $a_{\overline{n}|\,0}=n$, $a_{\overline{12n}|\,0}=12n$, and the monthly payment is $R/12$.

A reliable comparison states the effective annual rate, the equivalent period rate, the number and timing of payments, and the valuation date for any balance. These checks keep the payment calculation and the balance comparison on a consistent basis, aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

Use end-of-period payments unless a question specifies otherwise. Give dollar answers to the nearest cent when applicable.

A $36,000 loan has a three-year term and zero interest. Find the annual and monthly payments and both schedules' balance immediately after the second year's payment. (Answer: Annual payment $12,000; monthly payment $1,000; each balance $12,000.)
An effective annual rate is 6%. Find the equivalent effective monthly rate. (Answer: $(1.06)^{1/12}-1$, approximately 0.4868% per month.)
A $5,000 one-year loan has an effective annual rate of 12%. Give the annual year-end payment and the monthly month-end payment in exact annuity form. (Answer: Annual $5{,}000(1.12)=\$5{,}600$; monthly $5{,}000/a_{\overline{12}|\,((1.12)^{1/12}-1)}$.)
For equivalent rates and end-of-period payments, state the relationship between the annual payment $R$, the monthly payment $C$, and the monthly annuity value over 12 months. (Answer: $C\,a_{\overline{12}|\,j}=R/(1+i)$.)
A borrower compares two schedules at a date six months after an annual payment date. Is equality of their annual-anniversary balances enough to conclude the balances are equal on this date? Explain. (Answer: No. Equality is established at the annual anniversaries immediately after the corresponding payments; between anniversaries, the monthly schedule has additional payments and the balances generally differ.)

What You'll Learn