Finding the Number of Payments from Present Value

Introduction

When a level-payment stream has a known present value, payment amount, and effective interest rate, its term can be the unknown. The main step is to isolate the exponential term in the annuity factor and then use logarithms. The resulting answer still needs an interpretation: an equation may produce a noninteger number of periods even though an actual annuity has a whole number of payments.

This tutorial treats payments made at regular annual intervals, with the effective annual rate applying to each interval. The same method applies to another payment interval when the effective rate for that interval is known. The focus is solving the present-value equation, not converting rates between different periods.

Definition: An annuity-immediate with $n$ payments of amount $R$ makes payments at the ends of periods $1,2,\ldots,n$. An annuity-due with $n$ payments of amount $R$ makes payments at the beginnings of periods $0,1,\ldots,n-1$. Here $R>0$, and $n$ is a positive integer for an actual payment stream.

Solving for the Term of an Annuity-Immediate

Let $i>0$ be the effective rate per payment period and $v=1/(1+i)$. The present value of an annuity-immediate is $R a_{\overline{n}|}$, where $a_{\overline{n}|}=(1-v^n)/i$. If the given present value is $P$, then

P=R a_{\overline{n}|} =\frac{R}{i}\left(1-(1+i)^{-n}\right).

Isolate the power, then take logarithms:

(1+i)^{-n}=1-\frac{iP}{R}, \qquad n=-\frac{\ln\left(1-\frac{iP}{R}\right)}{\ln(1+i)}.

This logarithm requires $1-iP/R>0$. For a positive finite term, the full condition is $0<P<R/i$. The upper bound $R/i$ is the present value of a perpetuity-immediate; any finite positive annuity-immediate has a smaller present value.

Formula: For $i>0$, a positive finite real-valued term implied by an annuity-immediate present value $P$ is $$ n=-\frac{\ln\left(1-\frac{iP}{R}\right)}{\ln(1+i)}, \qquad 0<P<\frac{R}{i}. $$ For an actual annuity with payments at period ends, this solution must also be a positive integer.

Why the Logarithm Formula Works

The present value is a finite geometric sum. Each payment is discounted by one additional period, so the first few terms are $Rv, Rv^2,\ldots,Rv^n$. The geometric-sum formula gives $P=R(v+v^2+\cdots+v^n)=R(1-v^n)/i$. Rearranging this equation leaves a power of $1+i$, which is why logarithms recover the term.

Write the present-value equation.
Use $P=R(1-(1+i)^{-n})/i$ for an annuity-immediate.

Isolate the exponential term.
Rearranging gives $(1+i)^{-n}=1-iP/R$. Check that the right-hand side is positive.

Take logarithms and interpret.
Taking natural logarithms gives $-n\ln(1+i)=\ln(1-iP/R)$. Solve for $n$, then check whether it is a whole number if the question asks for an actual payment count.

Solving for the Term of an Annuity-Due

An annuity-due payment occurs one period earlier than the corresponding annuity-immediate payment. Its present value is $R\ddot{a}_{\overline{n}|}=R(1+i)a_{\overline{n}|}$. Therefore, if $P$ is the present value of an annuity-due, the equation and its solution are

P=\frac{R(1+i)}{i}\left(1-(1+i)^{-n}\right), \qquad n=-\frac{\ln\left(1-\frac{iP}{R(1+i)}\right)}{\ln(1+i)}.

For a positive finite term, $0<P<R(1+i)/i$. This upper limit is the present value of a perpetuity-due. The factor $1+i$ in the denominator inside the logarithm is essential: it reflects the earlier timing of every due payment.

Key takeaway: First identify whether the payments are immediate or due. For an immediate annuity, use $iP/R$ inside the logarithm; for a due annuity, use $iP/[R(1+i)]$. Both formulas assume $i>0$ and a rate measured per payment period.

Worked Examples

Worked Example: Recovering an Integer Term for an Annuity-Immediate

A hypothetical contract pays $R=\$100$ at each year-end. Its present value is $P=\$195.20$, using an effective annual rate of $i=25\%$. Find the number of payments, if it is a whole number.

Method. This is an annuity-immediate because payments are at year-end. Use the immediate-annuity term formula. The common mistake to avoid is using the annuity-due formula merely because the payment amount is level.

First check the finite-term bound: $R/i=\$100/0.25=\$400$. Since $0<\$195.20<\$400$, a positive finite real-valued solution is possible.

Substitute the data:

n=-\frac{\ln\left(1-\frac{0.25(195.20)}{100}\right)}{\ln(1.25)} =-\frac{\ln(1-0.488)}{\ln(1.25)} =-\frac{\ln(0.512)}{\ln(1.25)}.

Since $0.512=0.8^3$ and $0.8=1/1.25$, the numerator is $\ln(0.8^3)=3\ln(0.8)=-3\ln(1.25)$. Thus $n=3$.

Check. The three-payment present value is $100(1/1.25+1/1.25^2+1/1.25^3)=100(0.8+0.64+0.512)=\$195.20$. The value matches the data, so there are 3 payments.

Worked Example: Using the Due Formula for Payments at the Start of Each Year

A hypothetical fund pays $\$100$ at the start of each year for a fixed number of years. The present value is $\$244$, calculated at an effective annual rate of $25\%$. Find the number of payments, if it is a whole number.

Method. Payments occur at times $0,1,2,\ldots$, so this is an annuity-due. Apply the due formula. A common error is to treat the first payment as occurring at year-end and use the immediate formula.

The due-annuity upper bound is $R(1+i)/i=100(1.25)/0.25=\$500$, and $\$244$ is between zero and that bound. Now substitute:

n=-\frac{\ln\left(1-\frac{0.25(244)}{100(1.25)}\right)}{\ln(1.25)} =-\frac{\ln(1-0.488)}{\ln(1.25)} =-\frac{\ln(0.512)}{\ln(1.25)} =3.

Check. The payment times are $0,1,2$, so their present value is $100+100(0.8)+100(0.8^2)=100(1+0.8+0.64)=\$244$. Therefore the stream has 3 payments.

Worked Example: A Noninteger Solution Does Not Give a Whole Payment Count

An annuity-immediate pays $\$100$ at each year-end. The effective annual rate is $25\%$, and the stated present value is $\$180$. Find the implied term and determine whether a whole number of payments fits the data.

Method. Use the annuity-immediate logarithm formula, then compare the result with nearby integer terms. The common mistake to avoid is rounding a noninteger result to the nearest payment count and claiming that the rounded count has the stated present value.

The finite-term bound is $R/i=\$400$, so the data allow a positive finite real-valued solution. The calculation is

n=-\frac{\ln\left(1-\frac{0.25(180)}{100}\right)}{\ln(1.25)} =-\frac{\ln(0.55)}{\ln(1.25)} \approx 2.67916.

Rounded to three decimal places, the implied real-valued term is 2.679 periods. This is not a whole number of payments. Checking the adjacent integer terms confirms the distinction: two payments have present value $100(0.8+0.64)=\$144.00$, while three have present value $100(0.8+0.64+0.512)=\$195.20$. The stated $\$180$ lies between those values, so no whole-number payment count has present value $\$180$.

The real-valued solution is useful for diagnosing the equation, but it does not describe a stream that makes 2.679 actual annual payments.

Worked Example: The Zero-Interest Case

A hypothetical annuity-immediate pays $\$420$ at each year-end. Its present value is $\$2{,}940$, and the effective annual rate is $0\%$. Find the number of payments.

Method. When $i=0$, do not substitute into the logarithm formula, which divides by $i$ in its derivation. With no discounting, each payment contributes its full amount to present value.

Thus $P=nR$, so $n=P/R=\$2{,}940/\$420=7$.

Check. Seven undiscounted payments total $7(\$420)=\$2{,}940$. The answer is 7 payments.

Common Mistakes and Interpretation

The equation can give a useful mathematical answer without establishing that the original cash-flow description is possible. A term for a payment count is ordinarily an integer; if the logarithm gives a noninteger value, check the neighboring integer present values rather than rounding the count. Similarly, a present value at or above the relevant perpetuity limit cannot come from any positive finite term.

Using the wrong timing. An annuity-immediate begins at the end of period 1; an annuity-due begins at time 0. Identify the first payment before selecting the formula.
Dropping the minus sign. The exponent is $-n$, so solving gives a negative logarithm ratio. With $i>0$ and a valid positive present value, the final $n$ is positive.
Using an incompatible rate period. The formula uses the effective rate for one payment interval. Do not insert an annual rate unchanged if the payment periods are not annual.
Accepting an invalid logarithm argument. The quantity inside the logarithm must be positive. For a positive finite term, the present value must also be positive and below the applicable perpetuity bound.
Rounding to manufacture a payment count. A noninteger result is not an exact whole-number term. Verify any proposed integer by substituting it into the original present-value equation.

This procedure is aligned with the SOA Exam FM (Financial Mathematics) syllabus: set up the correct annuity present-value equation, solve it carefully, and interpret the result in the context of the stated cash flows.

Check Your Understanding

For each problem, identify the payment timing, solve for the implied term, and check whether the data describe a whole number of payments.

An annuity-immediate pays $\$400$ each year. At an effective annual rate of $10\%$, its present value is $\$1{,}000$. Find the implied term to three decimal places and state whether an integer payment count fits. (Answer: $3.018$ periods; no whole-number count fits, since the values for 3 and 4 payments are approximately $\$994.74$ and $\$1{,}267.95$.)
An annuity-due pays $\$250$ at the start of each year. Its present value is approximately $\$776.62$ at an effective annual rate of $20\%$. Find the number of payments, assuming it is an integer. (Answer: 4 payments.)
An annuity-immediate pays $\$600$ annually at an effective annual rate of $8\%$. Can its present value be $\$8{,}000$? Justify your conclusion using the appropriate upper bound. (Answer: No; the perpetuity-immediate limit is $\$600/0.08=\$7{,}500$, below $\$8{,}000$.)
An annuity-due pays $\$300$ at the start of each year. Its present value is $\$540$ at an effective annual rate of $25\%$. Solve for the term and verify the payment count from its cash flows. (Answer: 2 payments; $300+300/1.25=\$540$.)
An annuity-immediate pays $\$420$ at each year-end and has present value $\$2{,}940$ at a $0\%$ effective annual rate. Explain why logarithms are not appropriate and find the number of payments. (Answer: 7 payments; at zero interest, $P=nR$, so $n=2{,}940/420=7$.)

What You'll Learn