Accumulated Value Under Constant Force

Introduction

A continuous payment stream does not arrive as one lump sum at the end of its payment period. Each small payment arrives at its own time and accumulates for the remaining time until the chosen terminal date. The length of the payment interval and the length of the accumulation interval therefore play different roles.

This tutorial calculates accumulated values when the payment intensity is constant and the force of interest is constant. The method is to identify when payments are made, determine how long each payment accumulates, and add the resulting values. The topic is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: A constant continuous payment intensity $c$ means that $c$ dollars are paid per unit of time throughout the payment interval. If time is measured in years, $c$ has units of dollars per year. Under a constant force of interest $\delta$, a payment made at time $t$ accumulates to time $T$ by the factor $e^{\delta(T-t)}$, provided $T\ge t$.

Payment Interval and Accumulation Interval

Suppose payments occur continuously from time $a$ through time $b$, and the value is required at terminal time $T$, where $T\ge b$. The payment interval is $[a,b]$; its length is $b-a$. The accumulation interval for a payment made at time $t$ is $[t,T]$; its length is $T-t$, which varies across the payment stream.

With a constant force of interest, a payment made at time $t$ accumulates to time $T$ by $e^{\delta(T-t)}$. Thus, a small payment of $c\,dt$ made near time $t$ has terminal value $c e^{\delta(T-t)}dt$. Adding these values over the payment interval gives the accumulated value.

Accumulated value: For constant intensity $c$, constant force $\delta$, payments on $[a,b]$, and terminal time $T\ge b$, the accumulated value is $$ AV_T=\int_a^b c e^{\delta(T-t)}\,dt $$ For $\delta\ne 0$, this evaluates to $$ AV_T=\frac{c}{\delta}\left(e^{\delta(T-a)}-e^{\delta(T-b)}\right). $$ If $\delta=0$, there is no accumulation from interest, so $AV_T=c(b-a)$.

The exponents use the time from each payment date to the terminal date. In particular, $T-a$ and $T-b$ are not the payment-interval endpoints themselves: they measure accumulation from those endpoints to $T$. The formula assumes that $c$ and $\delta$ are constant over the relevant intervals, time is measured in the same units used for $\delta$, and there are no withdrawals or other cash flows included in this stream. If the force changes with time, this constant-force formula does not apply.

Why the Formula Holds

The derivation follows directly from accumulating each small payment separately. The constant-force model gives the accumulation factor $e^{\delta(T-t)}$ for one unit of value invested at time $t$. A continuous payment intensity $c$ supplies $c\,dt$ over a short interval of length $dt$.

Value a small payment at its own date.
A payment of $c\,dt$ at time $t$ has terminal value $c e^{\delta(T-t)}dt$.

Add over all payment dates.
Integrate those terminal values from the start $a$ to the end $b$ of the payment interval.

Evaluate the integral.
For $\delta\ne0$, an antiderivative of $c e^{\delta(T-t)}$ with respect to $t$ is $-\frac{c}{\delta}e^{\delta(T-t)}$. Evaluating at $b$ and $a$ yields $\frac{c}{\delta}(e^{\delta(T-a)}-e^{\delta(T-b)})$.

This derivation also explains why a stream that ends before $T$ still earns interest after its final payment: each payment continues to accumulate until $T$. The terminal date is not automatically the end of the payment interval.

Worked Applications

Worked Example: Payments End Before the Terminal Date

A fund receives a continuous payment intensity of $400 per year from time 0 through time 3. The force of interest is constant at $\delta=0.04$ per year. Find the accumulated value at time 5.

Method. Use the payment interval $[0,3]$ and terminal time $T=5$. Each payment accumulates from its own date to time 5; the last payment occurs at time 3 and still accumulates for two years.

AV_5=\frac{400}{0.04}\left(e^{0.04(5-0)}-e^{0.04(5-3)}\right) =10{,}000(e^{0.20}-e^{0.08})

Using $e^{0.20}=1.221402758$ and $e^{0.08}=1.083287068$, the difference is $0.138115690$. Therefore $AV_5=10{,}000(0.138115690)=\$1{,}381.15690$, or $1,381.16 to the nearest cent.

Check. The total payments are $400(3)=\$1{,}200$. Since the force is positive and every payment is made no later than time 3, the accumulated value at time 5 should exceed $1,200; $1,381.16 does.

Worked Example: A Stream That Starts After Time 0

A project receives $500 per year continuously from time 1 through time 4. The force of interest is constant at $\delta=0.02$ per year. Calculate the accumulated value at time 5.

Method. Here $a=1$, $b=4$, and $T=5$. The first payment accumulates for four years, while the last payment accumulates for one year. Use those accumulation periods in the endpoint exponents.

AV_5=\frac{500}{0.02}\left(e^{0.02(5-1)}-e^{0.02(5-4)}\right) =25{,}000(e^{0.08}-e^{0.02})

Using $e^{0.08}=1.083287068$ and $e^{0.02}=1.020201340$, the difference is $0.063085728$. Thus $AV_5=25{,}000(0.063085728)=\$1{,}577.14320$, or $1,577.14 to the nearest cent.

Check. The payment interval lasts $4-1=3$ years, so total payments are $500(3)=\$1{,}500$. A positive force makes the accumulated value exceed total payments, consistent with the result.

Worked Example: Two Payment Segments and an Unknown Intensity

A fund receives $600 per year continuously from time 0 through time 2, then receives an unknown constant intensity $k$ dollars per year from time 2 through time 3. The force of interest is constant at $\delta=0.02$ per year. The accumulated value at time 4 is $2,500. Find $k$.

Method. Accumulate each payment segment to time 4 separately, then add the values. The first segment has two years of payments, but those payments accumulate for between two and four years. The second segment has one year of payments, accumulating for between one and two years.

AV_{\text{first}}=\frac{600}{0.02}\left(e^{0.02(4-0)}-e^{0.02(4-2)}\right) =30{,}000(e^{0.08}-e^{0.04})

Using $e^{0.08}=1.083287068$ and $e^{0.04}=1.040810774$, the difference is $0.042476294$. Hence $AV_{\text{first}}=30{,}000(0.042476294)=\$1{,}274.28882$.

AV_{\text{second}}=\frac{k}{0.02}\left(e^{0.02(4-2)}-e^{0.02(4-3)}\right) =k\frac{e^{0.04}-e^{0.02}}{0.02} \approx 1.030471709k

The two accumulated values must total $2,500. Therefore,

$$ 1{,}274.28882+1.030471709k=2{,}500 $$

Subtracting gives $1.030471709k=1{,}225.71118$, so $k=1{,}225.71118/1.030471709\approx \$1{,}189.47$ per year, rounded to the nearest cent.

Check. At this intensity, the second segment accumulates to approximately $1.030471709(1{,}189.47)=\$1{,}225.72$. Adding the first segment's $1,274.29 gives $2,500.01; the one-cent discrepancy is due to rounding $k$ to cents. Using the unrounded value of $k$ gives $2,500.00.

Common mistake avoided. Multiplying $k$ by one year gives the second segment's payments, not their value at time 4. The second segment's payments accumulate after each payment date, so its terminal value includes the factor $1.030471709$.

Common Timing Errors

A useful way to avoid timing errors is to mark the payment start $a$, payment end $b$, and terminal time $T$ separately before substituting into the formula. The following checks help distinguish the intervals.

Using the payment length as every accumulation time. The stream may last $b-a$ years, but a payment at time $t$ accumulates for $T-t$ years.
Stopping accumulation when payments stop. If $T>b$, payments made before $b$ continue to earn interest through $T$.
Using the wrong endpoints in the exponent. The correct exponents are $\delta(T-a)$ and $\delta(T-b)$, not simply $\delta a$ and $\delta b$.
Mixing time units. If $\delta$ is per year, the times in the exponent must be in years. A force quoted per month must instead be paired with time in months.
Ignoring the zero-force case. The expression with division by $\delta$ is not defined at $\delta=0$; in that case the accumulated value is just total payments $c(b-a)$.

For $\delta>0$, the accumulated value cannot be below the total payments $c(b-a)$. This is a quick reasonableness check, though it does not replace checking the payment and accumulation intervals individually. If a problem has several consecutive payment intensities, calculate the terminal value of each segment using its own start and end dates, then add them.

Key takeaway: For a constant intensity on $[a,b]$, accumulate each payment to the specified terminal time $T$. The payment interval determines which payments are included; $T-t$ determines how long the payment made at time $t$ accumulates.

Check Your Understanding

For each problem, identify the payment interval and terminal time before calculating. Round dollar answers to the nearest cent.

A continuous intensity of $300 per year is paid from time 0 through time 2. The constant force is $0.05$ per year. Find the accumulated value at time 3. (Answer: $663.38)
A continuous intensity of $800 per year is paid from time 1 through time 3. The constant force is $0.04$ per year. Find the accumulated value at time 5. (Answer: $1,804.48)
A fund receives $500 per year continuously from time 0 through time 2, followed by $700 per year from time 2 through time 4. The constant force is $0.03$ per year. Find the accumulated value at time 4. (Answer: $2,537.19)
A continuous intensity $c$ dollars per year is paid from time 1 through time 4. Under a constant force of $0.02$ per year, its accumulated value at time 6 is $4,000. Find $c$, rounded to the nearest cent. (Answer: $1,243.02 per year)
A continuous intensity of $250 per year is paid from time 2 through time 5. The force is $0.03$ per year, and the value is needed at time 7. Find the accumulated value. (Answer: $833.31)

What You'll Learn