Why Equal Rates Match Direct Amortization

Why the Equal-Rate Result Matters

A sinking-fund loan separates the borrower’s periodic outlay into two parts: interest paid to the lender and a deposit into a separate fund intended to accumulate to the loan principal. A directly amortized loan combines interest and principal repayment in one level payment. When the fund earns the same effective rate charged on the loan, and the cash flows occur at matching period-ends, these two payment structures have the same level periodic outlay.

The result is an equality of scheduled amounts, not an assertion that the two arrangements have identical cash flows or risks in every respect. In the sinking-fund arrangement, deposits accumulate in a fund while the original loan principal remains outstanding until redemption. In direct amortization, each payment reduces the loan balance.

Definition: Suppose a loan of principal $L$ is repaid after $n$ equal periods. The loan charges an effective rate $i$ per period. Under a sinking-fund arrangement, the borrower pays loan interest $iL$ at the end of each period and deposits a level amount $R$ into a separate fund at the end of each period. The fund also earns effective rate $i$ per period. The total periodic outlay is $iL+R$.

For the equal-outlay result, assume the loan principal is $L$ at the start and remains unchanged until redemption at time $n$; loan interest is paid in full at each period-end; the fund starts with no balance; and the level deposits are made at times $1,2,\ldots,n$. There are no fees, taxes, withdrawals, or other cash flows. The loan and fund rates must be the same effective rate for each period, and the period lengths must match.

The Result and Its Derivation

Under the assumptions above, the sinking-fund deposits accumulate to $R s_{\overline{n}|}$ at time $n$, where $s_{\overline{n}|}$ is the accumulated value of an annuity-immediate of 1 per period at rate $i$. To have exactly $L$ available for redemption, the deposit is $R=L/s_{\overline{n}|}$. Meanwhile, the level payment $P$ on a directly amortized loan satisfies $P a_{\overline{n}|}=L$, where $a_{\overline{n}|}$ is the present value of an annuity-immediate of 1 per period at rate $i$.

Theorem: If the loan and sinking fund both earn or charge the same effective rate $i$ per period, and loan-interest payments and fund deposits are both made at the end of each of the $n$ periods, then $$ iL+\frac{L}{s_{\overline{n}|}}=\frac{L}{a_{\overline{n}|}}. $$ Thus, the total sinking-fund outlay each period equals the level payment on a directly amortized loan.

The identity follows from the standard formulas for these two annuities. For $i>0$, let $v=1/(1+i)$. Then

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

Taking reciprocals gives

\frac{1}{a_{\overline{n}|}} =\frac{i(1+i)^n}{(1+i)^n-1} =i+\frac{i}{(1+i)^n-1} =i+\frac{1}{s_{\overline{n}|}}.

The last equality holds because $1/s_{\overline{n}|}=i/((1+i)^n-1)$. Multiplying the identity by $L$ proves the theorem: the sinking-fund outlay $iL+L/s_{\overline{n}|}$ equals the direct payment $L/a_{\overline{n}|}$. If $i=0$, the same conclusion follows by taking limits: the deposit and direct payment are both $L/n$, and there is no loan interest.

The derivation also explains why matching rates matter. The relationship between $a_{\overline{n}|}$ and $s_{\overline{n}|}$ uses the same $i$ in both formulas. If either the fund yield or the timing changes, this identity does not establish equality of the outlays.

Worked Applications

Worked Example: Comparing Two-Year Payments

A borrower takes a loan of $2,100 for two years at an effective annual rate of 10%. Interest is paid at each year-end. The borrower also makes equal year-end deposits into a fund earning 10% effective annually, with the fund used to redeem the principal at the end of year 2. Find the deposit, the total annual outlay, and verify the direct level payment.

Method. First accumulate the two end-of-year deposits to find the required sinking-fund deposit. Then compare the total outlay with the present-value equation for direct amortization.

1. Find the deposit. The first $1 deposit earns interest for one year and the second earns none before redemption. Thus,

s_{\overline{2}|}=1+1.10=2.10, \qquad R=\frac{2100}{2.10}=\$1{,}000.

2. Find the sinking-fund outlay. Annual loan interest is $0.10(2100)=\$210$. Adding the fund deposit gives

iL+R=210+1000=\$1{,}210.

3. Verify using direct amortization. The annuity-immediate present value factor is

a_{\overline{2}|} =\frac{1}{1.10}+\frac{1}{1.10^2} =0.9090909\ldots+0.8264463\ldots =1.7355372\ldots.

Therefore, the direct level payment is $2100/a_{\overline{2}|}=\$1{,}210$. As a schedule check, the first direct payment contains $210 interest and $1,000 principal, leaving $1,100 outstanding. The second period’s interest is $110, so the second payment contains $1,100 principal. The sinking-fund balance is $1000(1.10)+1000=\$2{,}100$ at year 2. Both checks agree. The common mistake avoided here is treating the fund deposit as if it were the whole loan payment; the loan interest must be added to it.

Worked Example: Checking the Equality Against a Full Amortization Schedule

A loan of $6,100 is due after three years. It charges 25% effective annual interest, payable at each year-end. Equal year-end deposits are made to a separate fund earning 25% effective annually. Find the annual fund deposit and outlay, then verify the outlay against the three payments on a directly amortized loan.

1. Accumulate the deposits. The deposits earn interest for three, two, and one years, respectively. Their accumulated value per dollar of deposit is

s_{\overline{3}|}=1+1.25+1.25^2=1+1.25+1.5625=3.8125.

The required deposit is $R=6100/3.8125=\$1{,}600$, since $3.8125(1600)=6100$. Annual loan interest is $0.25(6100)=\$1{,}525$, so the total outlay is

iL+R=1525+1600=\$3{,}125.

2. Find the direct payment. The present value factor is

a_{\overline{3}|} =\frac{1}{1.25}+\frac{1}{1.25^2}+\frac{1}{1.25^3} =0.8+0.64+0.512 =1.952.

The direct level payment is $6100/1.952=\$3{,}125$. To check with an amortization schedule: the first payment has $1,525 interest and $1,600 principal, leaving $4,500. The second has $0.25(4500)=\$1{,}125$ interest and $2,000 principal, leaving $2,500. The third has $625 interest and $2,500 principal. Each payment is $3,125, and the balance is fully repaid. The schedule confirms that the equal-rate result is consistent with the changing interest and principal portions of direct payments.

Worked Example: Finding the Loan Principal From the Fund Deposit

A borrower makes deposits of $1,200 at each year-end for three years into a fund earning 20% effective annually. The loan charges the same effective annual rate, and loan interest is paid at each year-end. The fund is intended to redeem the loan principal after the third deposit. Find the principal, the annual total outlay, and the corresponding direct payment.

1. Find the principal accumulated in the fund. The accumulated value factor is

s_{\overline{3}|}=1+1.20+1.20^2=1+1.20+1.44=3.64.

Therefore the principal is $L=1200(3.64)=\$4{,}368$. This uses the deposits’ accumulated value at year 3, not their sum of $3,600.

2. Find the sinking-fund outlay. Annual loan interest is $0.20(4368)=\$873.60$, so the total annual outlay is

873.60+1200=\$2{,}073.60.

3. Verify the direct payment. The present-value factor is

a_{\overline{3}|} =\frac{1}{1.20}+\frac{1}{1.20^2}+\frac{1}{1.20^3} =\frac{5}{6}+\frac{25}{36}+\frac{125}{216} =\frac{455}{216}.

Thus the direct payment is $4368/(455/216)=4368(216)/455=\$2{,}073.60$, since $4368/455=9.6$ and $9.6(216)=2073.6$. A schedule check gives direct-loan interest and principal portions of $873.60 and $1,200 in year 1; $633.60 and $1,440 in year 2; and $345.60 and $1,728 in year 3. The principal portions sum to $4,368, and each payment is $2,073.60.

Timing and Rate Conditions

The word “same” in the theorem applies to the effective rate for each matching period, not merely to two rates that happen to have similar labels. A loan rate quoted in one convention and a fund yield quoted in another must be converted to effective rates over the actual payment period before comparing them. For example, if both cash flows are quarterly, the rate used in the formulas must be the effective quarterly rate for both the loan and the fund.

Timing must match as well. The derivation assumes deposits at times $1,2,\ldots,n$, together with interest payments at those same period-ends. A deposit made at the beginning of each period has more time to earn interest than an end-of-period deposit. Its required amount and its dates differ, so the equal-outlay theorem cannot be applied by simply substituting an annuity-due factor while retaining the same payment comparison.

The most reliable approach is to put both arrangements on a timeline before using the formula. Record when interest is due, when deposits occur, when the fund is used, and the effective rate per interval. If any of these differ from the theorem’s assumptions, calculate the relevant cash flows directly rather than assuming the level outlays match.

Confirm the periods.
Use the same period length for loan interest, fund deposits, and the fund’s effective rate.

Confirm the dates.
Check that both the interest payments and fund deposits occur at each period-end and that the final deposit is available at redemption.

Calculate the fund deposit.
With an empty fund and end-of-period deposits, set $R s_{\overline{n}|}=L$.

Add loan interest and verify.
Calculate $iL+R$, then check it against $L/a_{\overline{n}|}$.

Common Exam Trap

A frequent error is to divide the principal evenly by the number of periods and use $L/n$ as the fund deposit. That ignores interest earned by the deposits. For the $2,100 loan in the first example, this incorrect approach gives $2100/2=\$1{,}050$ per deposit. The two deposits would accumulate to $1050(2.10)=\$2{,}205$, which exceeds the $2,100 redemption amount by $105.

The correct deposit is $2100/2.10=\$1{,}000$, and the total outlay is $1,000 plus $210 interest, or $1,210. A second trap is to compare only the fund deposit with the direct payment. The sinking-fund outlay includes both the deposit and the loan interest; omitting either component breaks the comparison.

Key takeaway: With equal effective rates and aligned end-of-period timing, $R=L/s_{\overline{n}|}$ and $iL+R=L/a_{\overline{n}|}$. The result depends on matching both the rate and the timing; it is not a general equivalence for every sinking-fund arrangement.

Check Your Understanding

For each problem, assume no fees or other cash flows. Give dollar answers to the nearest cent.

A $5,000 loan is due after two years at 5% effective annual interest. Interest is paid annually, and equal year-end fund deposits earn 5% effective annually. Find the deposit and total annual outlay. (Answer: deposit $2,439.02; outlay $2,689.02.)
A $7,800 loan is repaid after three years. The loan rate and the year-end sinking-fund yield are both 10% effective annually. Find the fund deposit and total annual outlay. (Answer: deposit $2,356.50; outlay $3,136.50.)
A borrower deposits $900 at each year-end for four years into a fund earning 5% effective annually. The loan rate is also 5%, with interest paid annually. Find the loan principal and the total annual outlay. (Answer: principal $3,879.11; outlay $1,093.96.)
A loan charges 8% nominal annual interest convertible semiannually. Interest is paid each half-year, and six equal half-year-end deposits are made to a fund earning the same nominal rate convertible semiannually. For a principal of $12,000, find the half-yearly deposit and total half-yearly outlay. (Answer: deposit $1,809.14; outlay $2,289.14.)
A $2,200 loan is due in two years at 10% effective annual interest. Equal fund deposits are instead made at times 0 and 1, and the fund earns 10% effective annually. Find each deposit required to accumulate to $2,200 at time 2. Does the equal-rate theorem show that this deposit plus loan interest is the level end-of-year payment on a directly amortized loan? Explain briefly. (Answer: deposit $1,047.62; no—the fund deposits are at the beginnings of periods, not at the matching period-ends assumed by the theorem.)

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