Why Solve for a Break-Even Fund Yield?
A sinking-fund loan and a directly amortized loan can have the same principal and term but different payment structures. Under a sinking-fund arrangement, the borrower pays interest on the loan and separately deposits money into a fund that will repay the principal at the end. Under direct amortization, each level payment covers both interest and principal repayment. Finding the break-even fund yield identifies when the two arrangements require the same periodic outlay.
The key is to compare complete periodic outlays, not just the fund deposits. The sinking-fund outlay includes loan interest as well as the deposit. The model below assumes that the loan interest rate and the fund yield are effective rates per payment period, and that all fund deposits are made at the ends of the periods. This treatment is aligned with the SOA Exam FM (Financial Mathematics) syllabus.
Set Up the Two Payment Structures
Let \(i\) be the effective loan rate per period, and let \(j\) be the effective fund yield per period. Assume \(L>0\), \(n\) is a positive integer, \(i\geq 0\), and \(j>-1\). The loan interest is paid each period, so the principal remains outstanding until it is repaid from the fund at the end of period \(n\).
The directly amortized loan has level end-of-period payments. Its payment \(R\) is found by equating the present value of those payments to the loan principal. A sinking fund with equal end-of-period deposits has accumulated value \(D s_{\overline{n}|}\) at time \(n\), where \(D\) is the deposit and the accumulation factor uses the fund rate \(j\).
The sinking-fund outlay is level because the loan interest \(Li\) and the fund deposit \(D\) are each level. Break-even therefore means
This equation can be solved for \(j\). It is important not to set the deposit alone equal to the direct loan payment: the sinking-fund borrower also pays loan interest each period.
Worked Example: Comparing Outlays for a Three-Period Loan
A borrower takes a loan of $24,000 at an effective annual loan rate of 5%, with three annual payments. Find the annual fund yield that makes the sinking-fund outlay equal to the level annual payment on a directly amortized loan. Assume end-of-year payments and deposits.
Method. Calculate the direct payment, then compare it with the sum of annual loan interest and the fund deposit. The periods are annual, so both rates are annual effective rates.
The direct payment is
Thus the direct annual payment is $8,813.01 to the nearest cent. The loan interest due each year under the sinking-fund arrangement is
The fund must therefore receive a level deposit of
The required fund accumulation factor is \(L/D\), which, using the unrounded values, is \(24{,}000/7{,}613.00555\ldots=3.1525\). With three end-of-year deposits, the accumulation factor at rate \(j\) is \(1+(1+j)+(1+j)^2\). At \(j=0.05\), this is \(1+1.05+1.1025=3.1525\). The break-even fund yield is therefore 5% per year.
Check. At 5%, the fund deposits accumulate to \(7{,}613.00555\ldots(3.1525)=24{,}000\). Each year's sinking-fund outlay is \(1{,}200+7{,}613.00555\ldots=8{,}813.00555\ldots\), matching the direct payment before rounding. The common mistake avoided here is comparing the direct payment with the fund deposit while omitting the $1,200 loan-interest payment.
Why the Break-Even Rate Equals the Loan Rate
The equation can be simplified without first calculating any payments. Divide the break-even equation by \(L\), which is positive, and rearrange:
For \(i>0\), the right-hand side is exactly the reciprocal of the loan-rate accumulation factor \(s_{\overline{n}|i}\). To see why, use \(a_{\overline{n}|i}=(1-v^n)/i\), where \(v=1/(1+i)\):
The last equality follows from \(s_{\overline{n}|i}=((1+i)^n-1)/i\). Thus the break-even equation requires \(s_{\overline{n}|j}=s_{\overline{n}|i}\). For more than one period, this determines a unique rate.
For \(i>0\), the derivation above proves that \(j=i\) satisfies the equation. For \(i=0\), the direct payment is \(L/n\), there is no loan-interest payment, and the break-even deposit must be \(L/n\). At \(j=0\), \(s_{\overline{n}|0}=n\), so this deposit accumulates to \(L\); for \(n>1\), the accumulated-value factor is strictly increasing in \(j>-1\), making \(j=0=i\) the unique solution.
For \(n>1\), the strict increase follows from the sum representation \(s_{\overline{n}|j}=\sum_{k=0}^{n-1}(1+j)^k\). As \(j\) increases over \(j>-1\), each term with \(k\geq 1\) increases, and at least one such term is present. Hence two different admissible fund rates cannot produce the same accumulation factor. When \(n=1\), however, \(s_{\overline{1}|j}=1\) for every \(j\): the sole deposit is made at the end of the only period and earns no interest before redemption.
Worked Example: Solving for the Yield From the Required Deposit
A $20,000 loan has an effective annual loan rate of 4% and a term of two years. Payments and fund deposits are annual and made at year-end. Find the fund yield for which the sinking-fund outlay equals the direct annual payment.
Method. First calculate the direct payment. The difference between that payment and the loan interest is the required annual fund deposit; use that deposit to find the necessary fund accumulation factor.
The direct annual payment is
Annual loan interest is \(20{,}000(0.04)=\$800\). The annual deposit required for break-even is therefore
For two annual end-of-year deposits, the accumulation factor is \(s_{\overline{2}|j}=1+(1+j)=2+j\). The required factor is
Consequently, \(2+j=2.04\), so \(j=0.04\), or 4% per year. The calculation has retained unrounded values until the rate is determined.
Check. At 4%, the two deposits accumulate by the factor \(1+1.04=2.04\), giving \(9{,}803.9215686\ldots(2.04)=\$20{,}000\). The outlay is \(\$9{,}803.9215686\ldots+\$800=\$10{,}603.9215686\ldots\), equal to the direct payment.
The One-Period Exception
The condition \(n>1\) is essential to the uniqueness result. With one period, the direct loan payment is \(L(1+i)\). The sinking-fund arrangement requires the borrower to pay \(Li\) in interest and deposit \(L\) at the end of the period. The deposit earns no interest before it is needed, regardless of the fund rate. Therefore the total outlay is \(Li+L=L(1+i)\) for every \(j>-1\).
Worked Example: A One-Year Loan Has No Unique Break-Even Yield
A hypothetical borrower takes a $12,500 loan at an effective annual loan rate of 8%, due in one year. The borrower pays interest annually and makes one end-of-year sinking-fund deposit. Determine the break-even fund yields and verify the outlay.
Method. Compare the one-period direct payment with the one-period sinking-fund outlay. A single end-of-period deposit accumulates by a factor of 1, whatever the fund yield.
The direct payment is
The sinking-fund loan requires interest of \(12{,}500(0.08)=\$1{,}000\) and a deposit of \(\$12{,}500\). Its total outlay is
The deposit is made at the end of the only period, so its accumulated value at that same date is \(\$12{,}500\) for any \(j>-1\). Thus every admissible fund yield is break-even; there is no unique yield to solve for. For example, a fund yield of 0% still gives the same outlay. The common mistake avoided is applying the uniqueness conclusion for \(n>1\) to a one-period loan.
Common Exam Trap: Equating the Deposit to the Direct Payment
A frequent error is to set \(D=R\). That omits loan interest from the sinking-fund outlay and typically produces an incorrect fund yield. The correct comparison is \(Li+D=R\).
Use the loan principal, loan rate, and number of periods to find \(R=L/a_{\overline{n}|i}\).
The break-even fund deposit is \(D=R-Li\), not \(R\).
Set \(D s_{\overline{n}|j}=L\) and solve for \(j\), checking whether \(n=1\).
Worked Example: Correcting an Omitted-Interest Calculation
A $10,000 loan has an effective annual loan rate of 5% and a two-year term. An analyst finds the direct payment and then incorrectly treats that entire amount as the required fund deposit. Show the resulting error and find the correct break-even yield.
Method. Calculate the direct payment, subtract the annual loan interest to obtain the correct fund deposit, and then solve the two-period accumulation equation.
The direct annual payment is
Annual loan interest is \(\$10{,}000(0.05)=\$500\), so the required fund deposit is
For two end-of-year deposits, \(s_{\overline{2}|j}=2+j\). The correct target accumulation factor is
The analyst’s incorrect calculation sets \(D=R\), giving \(10{,}000/R=1.8594104\ldots\), and would then imply \(2+j=1.8594104\ldots\), or \(j\approx-0.1406\). This is not the break-even yield: it funds the principal using deposits that are too large because it has failed to include the separate $500 interest outlay. At the correct yield of 5%, the deposit accumulates to \(\$10{,}000\), and the total outlay is \(\$4{,}878.0487805\ldots+\$500=R\).
Interpretation and Scope
Under the stated assumptions and for \(n>1\), a fund yield below the loan rate produces a smaller accumulated value per dollar deposited. More must therefore be deposited to build the required principal, increasing the total outlay above the directly amortized payment. A yield above the loan rate has the opposite effect. The break-even point is the loan rate itself.
This conclusion depends on matching the period basis and timing: the loan rate and fund yield are effective rates per the same period, loan interest is paid each period, deposits are level and made at period-end, and the full principal is funded at the term’s end. If the loan interest frequency and deposit frequency differ, or the fund yield changes over time, this equation is not directly applicable; the cash flows must be expressed on a compatible timeline under the relevant assumptions.
For exam problems, identify the payment period before using any accumulation factor, distinguish interest from principal funding, and check whether the term contains more than one period. Those checks determine both the correct equation and whether a unique break-even yield exists.
Check Your Understanding
Use the stated period rates and end-of-period timing. Answers are given at the end of each question for checking.
- A $16,000 loan has a 3.5% effective annual loan rate and five annual periods. What is the break-even annual fund yield? (Answer: 3.5% per year)
- A $20,000 loan has a 4% effective annual loan rate and a two-year term. At the break-even yield, find the annual fund deposit and total annual outlay. (Answer: Deposit \(20{,}000/2.04=\$9{,}803.92\); total outlay \(\$10{,}603.92\))
- A $7,500 loan at a 7% effective annual rate is due after one year. Find the break-even fund yields and the total one-year outlay. (Answer: Every \(j>-1\) is break-even; the outlay is \(\$8{,}025.00\))
- A $25,000 loan has a 5% effective annual loan rate and a three-year term. The annual sinking-fund deposit is exactly \(25{,}000/3.1525\), and deposits are made at each year-end. Find the fund yield that makes the total outlay equal the direct annual payment. (Answer: 5% per year)
- A $12,000 loan has a 6% effective annual loan rate and a two-year term. A borrower proposes an annual fund deposit of \(\$12{,}000/2.06\). Determine whether this deposit gives break-even outlay and calculate that outlay. (Answer: Yes; the deposit is \(\$5{,}825.24\), and the total annual outlay is \(\$6{,}545.24\))