Solving for the Number of Annuity Payments

Why Solve for the Number of Payments?

An annuity problem often gives the payment amount and interest rate, then asks how many payments are needed to reach a target value. The answer may be a whole number, but the algebra can also produce a fractional term. That fractional result is useful as an equivalent duration; it does not by itself mean that a contract makes a partial payment.

The formulas below apply to a level ordinary annuity: equal payments are made at the end of each period, the interest rate is constant, and the stated value is measured at the beginning or end of the payment stream as indicated. Let $R>0$ be the payment each period, $i$ the effective interest rate per payment period, and $n$ the number of payments. If payments are annual, $i$ must be the effective annual rate. A quoted rate for another frequency must first be expressed per payment period.

Definition: An ordinary annuity with $n$ payments of $R$ has payments at times $1,2,\ldots,n$. Its present value is measured at time 0, before the first payment, and its accumulated value is measured at time $n$, just after the final payment. For a positive integer $n$ and $i\ne 0$, these values are $R a_{\overline{n}|}$ and $R s_{\overline{n}|}$, respectively.

Solving from Present Value

For $i\ne 0$, the present value of the ordinary annuity is $PV=R(1-(1+i)^{-n})/i$. Dividing by $R$, rearranging, and taking logarithms gives a direct expression for $n$. The rate must satisfy $i>-1$, so that $1+i$ is positive and logarithms are defined.

\begin{aligned} PV&=R\frac{1-(1+i)^{-n}}{i}\\ 1-\frac{iPV}{R}&=(1+i)^{-n}\\ n&=-\frac{\ln\left(1-\frac{iPV}{R}\right)}{\ln(1+i)}. \end{aligned}

The logarithm step is valid only when $1-iPV/R>0$. With a positive rate, a finite positive payment count requires $0<PV<R/i$. The upper bound is the limiting present value as the number of payments grows without bound; no finite number of positive payments reaches that limit. With a negative rate greater than $-1$, each later payment has a larger present value than the previous one, and a positive finite target can be reached for a suitable term. If $i=0$, the logarithmic formula is not used: $PV=nR$, so $n=PV/R$.

Worked Example: Finding a Fractional Term from Present Value

A hypothetical fund makes annual payments of $500 at each year-end and earns 10% effective interest annually. Its present value is $600. Find the equivalent term to three decimals. Assume the payments and valuation date satisfy the ordinary-annuity timing described above.

Method. Use the present-value equation and solve for $n$ by logarithms. The target is feasible: $0<$600<$500/0.10=$5,000$.

Substitute $PV=600$, $R=500$, and $i=0.10$:

\begin{aligned} n &=-\frac{\ln\left(1-\frac{0.10(600)}{500}\right)}{\ln(1.10)}\\ &=-\frac{\ln(0.88)}{\ln(1.10)}\\ &=\frac{0.127833\ldots}{0.095310\ldots}\\ &=1.341235\ldots \end{aligned}

Rounded to three decimals, the equivalent term is 1.341 periods. A check using the rearranged equation gives $(1.10)^{-1.341235\ldots}=0.88$, so $PV=500(1-0.88)/0.10=$600$. This fractional term is an equivalent-value result, not a literal schedule of 1.341 annual payments. The method avoids rounding the term before checking the value.

Solving from Accumulated Value

The accumulated value at time $n$ includes the final payment at that date. For a positive rate, earlier payments earn interest up to time $n$, while the final payment earns none. The equation can be rearranged in the same way, but the logarithm expression differs from the present-value formula.

\begin{aligned} FV&=R\frac{(1+i)^n-1}{i}\\ 1+\frac{iFV}{R}&=(1+i)^n\\ n&=\frac{\ln\left(1+\frac{iFV}{R}\right)}{\ln(1+i)}. \end{aligned}

For $i=0$, $FV=nR$, so $n=FV/R$. For a positive rate, every positive target accumulated value gives a positive equivalent term. If $-1<i<0$, the accumulated value of a positive ordinary annuity has a finite upper limit $R/(-i)$; a finite positive term requires $0<FV<R/(-i)$. These conditions follow from the geometric-series formulas and prevent taking a logarithm of a nonpositive quantity.

Worked Example: Finding the Term and Whole-Payment Threshold from Present Value

A hypothetical fund pays $400 at each year-end and earns 6% effective interest annually. Its present value is $1,500. Find the equivalent term to three decimals, then find how many whole payments are required for the present value to reach at least $1,500. Use the same time-0 valuation date for each comparison.

Method. First solve the present-value equation for the equivalent term. Since $0<$1,500<$400/0.06=$6,666.67$, the target is feasible for a finite term.

\begin{aligned} n &=-\frac{\ln\left(1-\frac{0.06(1,500)}{400}\right)}{\ln(1.06)}\\ &=-\frac{\ln(0.775)}{\ln(1.06)}\\ &=\frac{0.254892\ldots}{0.058269\ldots}\\ &=4.37441\ldots \end{aligned}

The equivalent term is 4.374 periods, rounded to three decimals. To decide the whole-payment requirement, compare the present values for four and five payments:

\begin{aligned} PV_4 &=400\frac{1-(1.06)^{-4}}{0.06} =400(3.4651056\ldots) =$1,386.04,\\ PV_5 &=400\frac{1-(1.06)^{-5}}{0.06} =400(4.2123638\ldots) =$1,684.95. \end{aligned}

Four payments fall short of $1,500, while five exceed it. Therefore, five whole payments are required to reach at least the target. The present value increases with each additional positive payment, so checking the adjacent integer values establishes the threshold. The method avoids interpreting 4.374 as a permitted fractional payment count.

Worked Example: Finding the Term from Accumulated Value

A hypothetical reserve receives $400 at each year-end and earns 6% effective interest annually. Find the equivalent term for an accumulated value of $2,000, rounded to three decimals. Also determine the least whole number of payments whose accumulated value is at least $2,000.

Method. Use the ordinary-annuity accumulated-value formula. The rate is positive, so the accumulated value increases with the number of payments.

\begin{aligned} n &=\frac{\ln\left(1+\frac{0.06(2,000)}{400}\right)}{\ln(1.06)}\\ &=\frac{\ln(1.30)}{\ln(1.06)}\\ &=\frac{0.262364\ldots}{0.058269\ldots}\\ &=4.502646\ldots \end{aligned}

The equivalent term is 4.503 periods, to three decimals. Compare the accumulated values at four and five payments:

\begin{aligned} FV_4 &=400\frac{(1.06)^4-1}{0.06} =400(4.374616) =$1,749.85,\\ FV_5 &=400\frac{(1.06)^5-1}{0.06} =400(5.637093) =$2,254.84. \end{aligned}

Four payments do not reach the target, but five do. Thus the least whole number of payments is five. The check confirms that the equivalent fractional term lies between the two integer terms and that the required whole-payment answer is not four.

Why the Logarithm Formulas Work

The formulas are consequences of geometric series. For present value, the payment at time $k$ is discounted $k$ periods, so the value is the sum of $R(1+i)^{-k}$ for $k=1$ through $n$. For accumulated value, the payment at time $k$ earns interest for $n-k$ periods, so its value at time $n$ is $R(1+i)^{n-k}$. Summing each series produces the standard annuity factors.

Write the cash-flow sum.
At time 0, the present value is $R\sum_{k=1}^{n}(1+i)^{-k}$. At time $n$, the accumulated value is $R\sum_{j=0}^{n-1}(1+i)^j$.

Apply the finite geometric-series identity.
For ratio $q\ne1$, $\sum_{j=0}^{n-1}q^j=(1-q^n)/(1-q)$. Applying this identity to the accumulated-value sum gives $R((1+i)^n-1)/i$; applying it to the discounted payments gives $R(1-(1+i)^{-n})/i$.

Isolate the power of $1+i$.
For present value, rearrangement yields $(1+i)^{-n}=1-iPV/R$. For accumulated value, it yields $(1+i)^n=1+iFV/R$.

Take logarithms and solve.
Taking natural logarithms converts the exponent $n$ into a multiplier. Dividing by $\ln(1+i)$, with the negative sign for present value, gives the two term formulas.

When $i=0$, the common ratio in the geometric series is 1, so the geometric-series identity in its quotient form does not apply. There are simply $n$ payments of $R$ with no interest adjustment, giving $PV=FV=nR$. That is why the zero-rate case must be handled separately rather than substituted into a formula containing division by $i$ or $\ln(1+i)$.

Interpreting the Answer and Avoiding Traps

The annuity formulas are defined for a whole number of payments in a contract. The logarithmic rearrangement extends the value equation to a real-valued equivalent term. If the problem asks for a threshold met by whole payments, use the direction of the question: for a positive payment and rate, take the least integer whose value is at least the target, and verify it by evaluating that integer and the preceding one. Do not simply report a rounded fractional term.

A common error is to drop the negative sign when solving from present value. For the $500 payment, 10% rate, and $600 present value example, that mistake gives

\frac{\ln(0.88)}{\ln(1.10)}=-1.341\ldots\ \text{periods},

which cannot be a positive payment count. The correct rearrangement is $(1+i)^{-n}=1-iPV/R$, so taking logarithms gives $-n\ln(1+i)=\ln(1-iPV/R)$, and therefore $n=-\ln(1-iPV/R)/\ln(1+i)=1.341\ldots$. A second frequent error is to use the accumulated-value logarithm for a present-value target: the expressions $1+iFV/R$ and $1-iPV/R$ arise from different equations and are not interchangeable.

These calculations are useful when comparing a target reserve, loan repayment stream, or other level cash-flow arrangement with a specified rate. The result is aligned with the SOA VEE Accounting and Finance syllabus. Always state the valuation date, payment timing, rate period, feasibility, and whether the requested answer is an equivalent term or a whole-payment requirement.

Check Your Understanding

For each problem, use level end-of-period payments and the stated effective rate per payment period. Treat a fractional result as an equivalent term unless the question asks for whole payments.

A hypothetical account receives $300 at each year-end and earns 5% effective interest annually. Its present value is $800. Find the equivalent term to three decimals. (Answer: 2.933 periods)
A hypothetical reserve receives $250 at each year-end and earns 4% effective interest annually. Find the equivalent term for an accumulated value of $1,000 to three decimals, and give the least whole number of payments that reaches at least that value. (Answer: 3.784 periods; 4 whole payments)
A positive-rate ordinary annuity pays $500 each period at an effective rate of 8%. Is a present value of $7,000 possible for a finite positive number of payments? Explain using the limiting present value. (Answer: No; the limit is $500/0.08=$6,250, and the target exceeds it)
A hypothetical zero-interest fund receives $180 at each year-end. Find the equivalent term for an accumulated value of $990 and the least whole number of payments that reaches at least that amount. (Answer: 5.5 periods; 6 whole payments)
A hypothetical account receives $100 at each year-end and earns $-10\%$ effective interest annually. Find the equivalent term for an accumulated value of $250 to three decimals, then find the least whole number of payments that reaches at least $250. (Answer: 2.730 periods; 3 whole payments)

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