Irregular Cash-Flow Dates Within A Policy Year

Why Cash-Flow Dates Matter

A policy year may contain receipts and outgo at different times: a premium can be received early in the year, an expense paid later, and a benefit settled near year-end. Adding those amounts directly ignores the interest earned, or forgone, between each payment date and the end of the year. To calculate an annual profit on a common endpoint basis, first accumulate every dated amount to that endpoint, then combine the results using their signs.

The method here applies to cash flows with known dates within a policy year and a stated interest basis. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. If a contract or question specifies a different fractional-year convention, use that convention rather than assuming one.

Definition: Let time $t=0$ be the start of a policy year and $t=1$ its end. Represent each cash flow by a signed amount $C_j$ paid at time $t_j$, where receipts are positive, outgo is negative, and $0\leq t_j\leq1$. The accumulated annual profit from these cash flows is their combined value at time $1$, after each amount has been accumulated from its payment date to the year-end endpoint.

The Endpoint-Value Formula

Suppose the effective annual interest rate is $i$, and the model uses constant force of interest within the year. The force is $\delta=\ln(1+i)$, so a cash flow accumulates for a fraction $s$ of a year by the factor $e^{\delta s}=(1+i)^s$. A cash flow at time $t_j$ has $1-t_j$ years to accumulate to the endpoint.

\Pi_1=\sum_j C_j(1+i)^{1-t_j}.

Here $\Pi_1$ is the accumulated annual profit at time $1$. If the force of interest is given directly, use $e^{\delta(1-t_j)}$ in place of $(1+i)^{1-t_j}$. A flow at the start of the year accumulates for a full year; a flow at the end accumulates for zero years. Outgo is subtracted because its signed amount is negative.

The fractional-year rule is an assumption, not an automatic consequence of knowing an annual effective rate. In this tutorial, a stated annual effective rate is interpolated using constant force, giving the factor $(1+i)^s$ over $s$ years. If a problem instead specifies simple interest, a nominal rate with a conversion frequency, or another convention, use its specified accumulation function.

Why Accumulating Each Flow Works

The formula follows from the accumulation law and the additivity of value at a common date. Under constant force, an amount $C_j$ invested from $t_j$ to time $1$ grows to $C_je^{\delta(1-t_j)}$. The endpoint value of several amounts is the sum of their individual endpoint values: each amount is carried over the time for which it is invested, and positive and negative amounts retain their signs.

C_j\text{ at }t_j \longrightarrow C_je^{\delta(1-t_j)}\text{ at }1, \qquad \Pi_1=\sum_j C_je^{\delta(1-t_j)} =\sum_j C_j(1+i)^{1-t_j}.

This derivation assumes the stated accumulation law applies throughout the relevant intervals and that the cash flows are valued on a consistent basis. It does not account for uncertainty in whether a future contingent payment occurs; the dated flows in the calculation must already be the flows being valued.

Worked Example: Accumulating a Premium and Two Expenses

A hypothetical policy receives a premium of $1,000 at $t=0.25$. It pays expenses of $300 at $t=0.50$ and $200 at $t=0.75$. The effective annual interest rate is $i=16.985856\%$, with constant force within the year. Calculate the accumulated annual profit at $t=1$. Round only the final amount to the nearest cent.

Method. Use the endpoint-value formula. The annual accumulation factor is $1+i=1.16985856$, and the specified rate gives a quarterly accumulation factor of $1.04$, since $1.04^4=1.16985856$. Thus the premium has three quarters to accumulate, the first expense has two, and the second expense has one.

\begin{aligned} \Pi_1 &=1000(1.04)^3-300(1.04)^2-200(1.04)\\ &=1000(1.124864)-300(1.0816)-200(1.04)\\ &=1124.864-324.48-208\\ &=592.384. \end{aligned}

The accumulated annual profit is $592.38. As a check, the accumulated receipt exceeds the accumulated expenses by $592.384 before rounding. A common mistake is to give the premium four quarters of interest because it is received during the year; it is received at $t=0.25$, so it earns interest for only $1-0.25=0.75$ years, or three quarters.

Worked Example: Combining Receipts and Outgo at Three Dates

A hypothetical contract has receipts of $1,200 at $t=0.25$ and $400 at $t=0.75$, and outgo of $500 at $t=0.50$ and $700 at $t=1$. Use the same effective annual rate, $i=16.985856\%$, and the same constant-force fractional-year convention. Calculate the accumulated annual profit, rounding only the final amount to the nearest cent.

Method. Accumulate each flow separately, using a positive sign for receipts and a negative sign for outgo. The $700 paid at the endpoint has no time to earn interest.

\begin{aligned} \text{Receipts at }t=1 &=1200(1.04)^3+400(1.04)\\ &=1200(1.124864)+400(1.04)\\ &=1349.8368+416\\ &=1765.8368,\\[4pt] \text{Outgo at }t=1 &=500(1.04)^2+700\\ &=500(1.0816)+700\\ &=540.8+700\\ &=1240.8,\\[4pt] \Pi_1&=1765.8368-1240.8=525.0368. \end{aligned}

The accumulated annual profit is $525.04. The endpoint outgo is subtracted at its original amount, not accumulated for a year. The common mistake here is to assign every flow the same accumulation period; each period must instead be measured from that flow’s own payment date to $t=1$.

Using a Different Fractional-Year Grid

Cash-flow dates need not fall on quarter-year boundaries. The essential steps are to identify each date as a fraction of the policy year and apply the appropriate accumulation factor for the remaining time. The following example uses dates measured in thirds and a rate chosen to give a convenient factor over each third of a year.

Worked Example: Irregular Dates Measured in Thirds

A hypothetical policy receives $900 at $t=\frac13$ and $300 at $t=\frac23$. It pays an expense of $100 at $t=0$ and a benefit of $250 at $t=\frac23$. The effective annual interest rate is $i=33.1\%$, with constant force within the year. Calculate the accumulated annual profit at $t=1$, rounding only the final amount to the nearest cent.

Method. Since $1+i=1.331=(1.1)^3$, the accumulation factor for one third of a year is $1.1$. The receipt at one third accumulates for two thirds of a year, or two such intervals. The flows at two thirds accumulate for one interval, and the start-of-year expense accumulates for three.

\begin{aligned} \text{Receipts at }t=1 &=900(1.1)^2+300(1.1)\\ &=900(1.21)+330\\ &=1089+330\\ &=1419,\\[4pt] \text{Outgo at }t=1 &=100(1.331)+250(1.1)\\ &=133.1+275\\ &=408.1,\\[4pt] \Pi_1&=1419-408.1=1010.9. \end{aligned}

The accumulated annual profit is $1,010.90. The check is that receipts and outgo have both been valued at $t=1$, and their difference is $1,010.90. Applying the annual factor $1.331$ to the receipt at $t=\frac13$ would be an error: that receipt has only two thirds of a year remaining, not a full year.

A Reliable Calculation Process

A short sequence of checks helps prevent timing and sign errors, especially when several types of cash flow appear in the same policy year.

Fix the timeline.
Set the policy-year start to $t=0$ and its endpoint to $t=1$. Translate every stated date into a time measured from the start.

Set the accumulation convention.
Identify the stated rate and any fractional-year rule. Under constant force and effective annual rate $i$, use $(1+i)^s$ for an interval of $s$ years.

Assign signs and remaining times.
Use positive amounts for receipts and negative amounts for outgo. For a flow at $t_j$, the time to the endpoint is $1-t_j$.

Accumulate and combine.
Apply the accumulation factor to each signed amount, add the resulting endpoint values, and round the final monetary result as instructed.

Common Traps and Limits

The most important check is that both the amount and the time period in each term refer to the same flow. A correct accumulation factor attached to the wrong date still gives an incorrect annual profit.

Using elapsed time from the wrong origin. A flow at $t_j$ accumulates for $1-t_j$, not for $t_j$. The latter is the time since the policy-year start.
Prorating the effective annual rate linearly. Under constant force, the factor for $s$ years is $(1+i)^s$, not $1+is$. Use a different rule only when the problem states one.
Giving an endpoint flow a full year of interest. A flow at $t=1$ has accumulation factor $(1+i)^0=1$.
Subtracting outgo twice. Either use signed amounts in the sum, or accumulate receipts and outgo separately and subtract the latter. Do not use negative outgo and then subtract it again.

This method puts dated cash flows on a common endpoint basis; it does not by itself define every component of an insurer’s profit measure. If a stated profit definition includes other items, calculate those according to that definition and include each item only once. A different timing or interest convention changes the accumulated values, so the convention must always be explicit.

Check Your Understanding

For each problem, use constant force within the policy year and $i=16.985856\%$, so that a quarter-year accumulation factor is $1.04$. Round monetary answers to the nearest cent.

A hypothetical policy receives $500 at $t=0.50$ and pays an expense of $100 at $t=0.75$. Find the accumulated annual profit at $t=1$. (Answer: $436.80)
A contract receives $1,000 at $t=0.25$, pays $300 at $t=0.75$, and receives $200 at $t=1$. Find the accumulated annual profit. (Answer: $1,012.86)
A policy pays an expense of $250 at $t=0$, receives $700 at $t=0.50$, and pays a benefit of $200 at $t=0.75$. Find the accumulated annual profit. (Answer: $256.66)
A contract receives $600 and pays an expense of $250, both at $t=0.25$. It also receives $400 at $t=0.50$ and pays $500 at $t=0.75$. Find the accumulated annual profit. (Answer: $306.34)
A policy pays an expense of $400 at $t=0$, receives a premium of $900 at $t=0.25$, and pays an expense of $300 at $t=0.75$. What receipt $B$ at $t=0.50$ makes the accumulated annual profit exactly $500? (Answer: $247.38)

What You'll Learn