Solving for an Unknown Payment Date

Why an Unknown Payment Date Can Be Solved

An uneven cash-flow stream may have known amounts and known dates, yet still leave one practical question unanswered: when must a particular payment occur so that the stream reaches a specified value? This is a time-value-of-money problem. The payment date changes how long the payment earns interest or is discounted, even though its amount stays fixed.

The method is to move every known cash flow to one common date, calculate the value still needed from the unknown payment, and solve the resulting exponential equation with logarithms. This tutorial uses a constant effective annual interest rate and fractional-year compounding. These assumptions matter: the equation changes if rates vary by period or a different compounding convention is specified.

Definition: Let a cash-flow stream contain signed amounts $C_j$ at times $t_j$, measured in years from time 0. Let a nonzero payment $P$ occur at an unknown time $t$, and let the stream have target value $V$ at time $T$. Under a constant effective annual rate $i>0$, the unknown date is found by valuing all amounts at the same date. A positive amount is an inflow to the account; a negative amount is an outflow.

The Date-Solving Equation

For a target value at time $T$, a cash flow at time $s$ is accumulated by the factor $(1+i)^{T-s}$. Thus, if the known flows are $C_1,\ldots,C_n$, the target equation is

\sum_{j=1}^{n} C_j(1+i)^{T-t_j}+P(1+i)^{T-t}=V.

First calculate the accumulated value of the known flows. Define the residual amount required from the unknown payment at time $T$ as

R=V-\sum_{j=1}^{n} C_j(1+i)^{T-t_j}.

Then $P(1+i)^{T-t}=R$. When $P$ and $R$ have the same sign, divide by $P$, take natural logarithms, and solve:

t=T-\frac{\ln(R/P)}{\ln(1+i)}.

The ratio $R/P$ must be positive for this real-valued logarithm to exist. The resulting date must also meet any restrictions in the question, such as $0\le t\le T$. A date outside the allowed interval is a mathematical solution to the equation, but not a feasible payment date for that scenario.

Why the Logarithm Gives the Date

The unknown payment’s value at time $T$ is its amount multiplied by the number of annual accumulation factors between $t$ and $T$. Fractional-year compounding is part of the model here, so the number of factors may be non-integer.

Isolate the unknown payment’s target-date value.
Subtract the accumulated known cash flows from the target. This leaves $R=P(1+i)^{T-t}$.

Divide by the payment amount.
Because $P\ne 0$, $R/P=(1+i)^{T-t}$. The ratio must be positive.

Take logarithms and isolate time.
$\ln(R/P)=(T-t)\ln(1+i)$. Dividing by $\ln(1+i)$ and rearranging gives $t=T-\ln(R/P)/\ln(1+i)$.

This also explains the direction of the result when $i>0$. For a positive payment, moving its date later reduces its value at $T$, because fewer periods of interest remain. The logarithm translates the required accumulation factor into the number of years.

Worked Applications

Worked Example: Finding a Deposit Date From a Future Target

A fund earns an effective annual rate of 5%, with fractional-year accumulation at the same rate. It receives $3,000 at year 1 and $2,000 at year 3. A further deposit of $5,000 must be made at an unknown date $t$. The fund is to have $11,932.22 at year 6. Find $t$, rounded to two decimal places, and check the result.

Method. Accumulate the known deposits to year 6, subtract their value from the target, and solve for the unknown deposit’s accumulation period.

The year-1 deposit earns interest for 5 years: $1.05^5=1.2762815625$, so its year-6 value is $3{,}000(1.2762815625)=\$3{,}828.8446875$. The year-3 deposit earns interest for 3 years: $1.05^3=1.157625$, so its year-6 value is $2{,}000(1.157625)=\$2{,}315.25$.

The known deposits therefore have a combined year-6 value of $\$3{,}828.8446875+\$2{,}315.25=\$6{,}144.0946875$. The unknown deposit must supply $\$11{,}932.22-\$6{,}144.0946875=\$5{,}788.1253125$ at year 6. Thus,

5{,}000(1.05)^{6-t}=5{,}788.1253125, \qquad 6-t=\frac{\ln(5{,}788.1253125/5{,}000)}{\ln(1.05)}.

The ratio inside the logarithm is $1.1576250625$, giving $6-t\approx3.000001$ years and $t\approx2.999999$. Rounded to two decimal places, the deposit date is year 3.00.

Check. At year 3, the $5,000 deposit grows for 3 years to $5{,}000(1.157625)=\$5{,}788.125$. Adding the known year-6 value gives $\$11{,}932.2196875$, which rounds to the stated target of $11,932.22.

Common mistake avoided: using $t$, rather than $6-t$, as the number of years the deposit earns interest. The deposit earns interest from its payment date to the target date.

Worked Example: Finding a Receipt Date From a Present-Value Target

At an effective annual rate of 10%, a set of receipts must have present value $5,000 at time 0. Known receipts are $1,100 at year 1 and $1,210 at year 2. A further receipt of $3,993 will occur at an unknown date $t\ge0$. Find its date.

Method. Value all receipts at time 0. A cash flow at time $s$ has present value equal to its amount divided by $(1+i)^s$.

The known receipts have present values $\$1{,}100/1.10=\$1{,}000$ and $\$1{,}210/1.10^2=\$1{,}210/1.21=\$1{,}000$. Together they contribute $2,000 of the $5,000 target, leaving $3,000 of present value for the unknown receipt. Therefore,

\frac{3{,}993}{(1.10)^t}=3{,}000, \qquad (1.10)^t=\frac{3{,}993}{3{,}000}=1.331.

Since $1.331=1.10^3$, it follows that $t=3$. The receipt must occur at year 3.00.

Check. Its present value is $\$3{,}993/1.331=\$3{,}000$. Adding the two known present values gives $\$1{,}000+\$1{,}000+\$3{,}000=\$5{,}000$, as required.

Common mistake avoided: discounting a receipt at year $t$ by multiplying by $(1.10)^t$. Present value uses division by that factor; multiplication would move the value forward instead.

Worked Example: Solving With a Withdrawal and Checking the Date Range

An account earns an effective annual rate of 8%, with fractional-year accumulation. It receives $2,500 at year 1 and pays out $1,000 at year 4. A positive deposit of $4,000 is to be made at an unknown date $t$. The account must have $6,600 at year 6. Find the deposit date, rounded to two decimal places, and verify that it is between years 0 and 6.

Method. Treat the withdrawal as a negative cash flow, accumulate all known flows to year 6, and solve for the time remaining after the deposit.

The deposit of $2,500 accumulates for $6-1=5$ years: $1.08^5=1.4693280768$, so its value at year 6 is $2{,}500(1.4693280768)=\$3{,}673.320192$. The withdrawal is a negative $1,000 cash flow at year 4. Its value at year 6 is $-1{,}000(1.08^2)=-1{,}000(1.1664)=-\$1{,}166.40$.

The known flows have combined value $\$3{,}673.320192-\$1{,}166.40=\$2{,}506.920192$. The unknown deposit must therefore accumulate to $\$6{,}600-\$2{,}506.920192=\$4{,}093.079808$. Hence,

4{,}000(1.08)^{6-t}=4{,}093.079808, \qquad 6-t=\frac{\ln(4{,}093.079808/4{,}000)}{\ln(1.08)}.

The ratio is $1.023269952$. Thus $6-t\approx0.2989$ years and $t\approx5.7011$ years. To two decimal places, the deposit date is year 5.70, which is within the permitted interval from year 0 to year 6.

Check. The deposit grows for about 0.2989 years, producing approximately $4{,}000(1.023269952)=\$4{,}093.079808$ at year 6. Adding the known-flow value of $2,506.920192 gives $6,600.

Common mistake avoided: adding the withdrawal’s future value as a positive amount. The account pays out the $1,000, so its signed contribution to the target is negative.

Feasibility and Limits of the Method

A calculated date should never be accepted without checking its domain and the cash-flow model. For $i>0$, a positive payment $P$ at a date between 0 and $T$ has target-date value between $P$ and $P(1+i)^T$. Therefore, if the residual $R$ is positive, a date in that interval requires

P\le R\le P(1+i)^T.

If $R<P$, the equation places the payment after $T$; if $R>P(1+i)^T$, it places the payment before time 0. The same checks can be made using the solved date. When $P$ and $R$ are both negative, their ratio is positive and the algebra still works, but interpret the result as the date of an outflow, not an inflow.

If $P$ and $R$ have opposite signs, no real date solves this single-payment equation under a positive accumulation factor: multiplying $P$ by $(1+i)^{T-t}$ cannot change its sign. If $i=0$, the payment’s value is independent of its date, so this logarithmic method does not identify a unique date. If interest rates vary by year or the problem specifies another accumulation model, replace $(1+i)^{T-t}$ with the appropriate accumulation factor before solving; the single logarithm formula need not apply.

Why Date Errors Matter

Solving for a payment date is useful whenever a fixed payment must be scheduled to meet a funding target, settlement amount, or account balance. The key is not the logarithm alone; it is the setup. A correct answer depends on a consistent timeline, a common valuation date, correct signs for receipts and withdrawals, and the correct number of years between each cash flow and that valuation date. This method is aligned with the SOA VEE Accounting and Finance syllabus.

Use one valuation date. Do not add amounts stated at different dates without first moving them to a common date.
Use elapsed time. At target date $T$, a payment at $t$ accumulates for $T-t$ years.
Keep signs consistent. Receipts and deposits are positive; withdrawals and payments out are negative when measuring an account’s value.
Check the result in the original equation. The date should reproduce the target, within any stated rounding precision.

Key takeaway: Move known cash flows to the target date, isolate the unknown payment’s required value, and solve $P(1+i)^{T-t}=R$ with logarithms. Then verify the sign conditions, date restrictions, and target value.

Check Your Understanding

For each problem, use the stated constant effective annual rate and round dates to two decimal places where needed.

An account earns 5% effective annually. A $2,000 deposit is made at year 1, and a $3,000 deposit is made at an unknown date $t$. The target is $5,903.89 at year 5. Find $t$ and check that it is within years 0 and 5. (Answer: $t=2.00$ years.)
At a 4% effective annual rate, known flows are a $2,500 deposit at year 1 and a $1,000 withdrawal at year 2. A $5,000 deposit occurs at an unknown date $t$. The account target is $7,000 at year 4. Find $t$. (Answer: $t\approx2.66$ years.)
A stream must have present value $7,500 at time 0, using a 10% effective annual rate. It includes receipts of $2,200 at year 1 and $2,420 at year 2, plus a $5,124.35 receipt at an unknown date $t\ge0$. Find $t$. (Answer: $t=4.00$ years.)
At a 6% effective annual rate, an account receives $4,000 at year 2 and pays out $1,200 at year 5. A positive deposit of $6,000 is planned at date $t$. The target is $15,000 at year 7. Determine the mathematical date and decide whether it is feasible if the deposit must occur between years 0 and 7. (Answer: $t\approx-3.40$ years, so no feasible date exists in the permitted interval.)

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