Payment Intensities and Continuous Cash-Flow Timelines

Introduction

A continuous cash-flow description gives a payment rate, not a single payment amount. For example, “\$600 per month continuously” describes cash arriving throughout each month. To value that stream, the rate must be matched to a time variable and the cash arriving at each time must be valued at the correct date.

The central tool is an integral over the payment timeline. The integral adds up the values of small amounts paid at different times. Before writing one, identify the payment intensity, the interval when payments occur, the valuation date, and the accumulation function. This tutorial develops that setup without relying on a special closed-form annuity factor.

Definition: A payment intensity $r(t)$ is the rate at which cash is paid at time $t$, measured in currency units per unit of time. Over a short interval of length $\Delta t$ near $t$, the amount paid is approximately $r(t)\Delta t$. If $r(t)$ is constant on an interval, the amount paid over that interval is the rate multiplied by the interval length.

Read the Timeline Before Valuing the Stream

Use one time unit consistently throughout a calculation. If $t$ is measured in years, an intensity used in an integral must be in dollars per year, and the integration limits must be in years. A payment rate quoted per month can be converted to dollars per year by multiplying it by 12. Equivalently, the time variable and limits may be measured in months, provided the accumulation function is expressed using that same unit.

A continuous stream over $0\leq t\leq T$ pays throughout that interval. The amount paid during a small interval of width $dt$ is $r(t)\,dt$. A payment stated to occur at a single instant is instead a discrete cash flow; it is not represented by a finite payment intensity over an interval. An endpoint does not add a separate lump sum to a continuous stream unless the problem explicitly specifies one.

Definition: Let $A(t)$ be the accumulation function, where one dollar invested at time 0 has value $A(t)$ at time $t$, with $A(0)=1$ and $A(t)>0$. The value at time 0 of a stream with intensity $r(t)$ over $0\leq t\leq T$ is $$ PV_0=\int_0^T \frac{r(t)}{A(t)}\,dt. $$ The accumulated value at time $T$ of the same stream is $$ AV_T=\int_0^T r(t)\frac{A(T)}{A(t)}\,dt. $$ These formulas assume a deterministic payment intensity and a specified accumulation function over the relevant dates. The integrals must exist and be finite. If the stream is discontinuous but piecewise integrable, integrate over its pieces and add the results.

The factor $1/A(t)$ discounts a payment at time $t$ to time 0. The factor $A(T)/A(t)$ carries that payment from time $t$ forward to time $T$. In both expressions, $r(t)\,dt$ is the amount paid at that time, while the remaining factor changes its valuation date.

Why the Integral Represents the Stream

The integral can be justified by approximating the continuous stream with many small payments. Divide $[0,T]$ into intervals, choose a representative time $t_j$ in each interval, and let $\Delta t_j$ be its length. The approximate payment in that interval is $r(t_j)\Delta t_j$. Its value at time 0 is approximately $r(t_j)\Delta t_j/A(t_j)$; its value at $T$ is approximately $r(t_j)\Delta t_j A(T)/A(t_j)$.

Partition the payment interval.
Split the timeline into short pieces with widths $\Delta t_j$, using the same time unit as the rate and accumulation function.

Estimate each small payment.
On a piece near $t_j$, the cash paid is approximately $r(t_j)\Delta t_j$.

Value each small payment at the target date.
Multiply by $1/A(t_j)$ for time 0, or by $A(T)/A(t_j)$ for time $T$.

Take the limit.
As the pieces become arbitrarily short, the sums become the corresponding integrals.

\sum_j \frac{r(t_j)\Delta t_j}{A(t_j)} \longrightarrow \int_0^T \frac{r(t)}{A(t)}\,dt, \qquad \sum_j r(t_j)\Delta t_j\frac{A(T)}{A(t_j)} \longrightarrow \int_0^T r(t)\frac{A(T)}{A(t)}\,dt.

This also explains why the rate is multiplied by $dt$, rather than used as though it were a lump-sum payment. The intensity alone has units of dollars per year; multiplying by a time increment in years gives dollars. After discounting or accumulation, the integrand still has units of dollars per year, and integration over years gives dollars.

Worked Example: Present Value of a One-Year Stream

A hypothetical service contract pays continuously at an intensity of \$1,200 per year from time 0 through time 1. Its accumulation function is $A(t)=e^{0.08t}$, with $t$ measured in years. Find the present value at time 0, to the nearest cent.

Method. Use the present-value integral, with the payment intensity and time measured in dollars per year and years, respectively. This avoids treating the annual rate as a single payment at time 1.

The intensity is $r(t)=1{,}200$, the interval is $0\leq t\leq1$, and the discount factor for a payment at $t$ is $1/A(t)=e^{-0.08t}$. Therefore,

PV_0=\int_0^1 1{,}200e^{-0.08t}\,dt =1{,}200\left[\frac{-e^{-0.08t}}{0.08}\right]_0^1 =1{,}200\frac{1-e^{-0.08}}{0.08}.

Using $e^{-0.08}\approx0.9231163464$, the factor is $(1-0.9231163464)/0.08\approx0.9610456700$. Thus $PV_0\approx1{,}200(0.9610456700)=\$1{,}153.2548$, or \$1,153.25 to the nearest cent.

Check. The undiscounted total is \$1,200. Since each payment occurs at time 0 or later and the accumulation function increases, the present value must be less than \$1,200. The result satisfies that check.

Convert Units and Split Changing Streams

The intensity can vary over time. If the rate changes at a specified date, define it piecewise and use the appropriate rate on each interval. The accumulation function continues to value each payment according to its actual payment time; a change in the payment rate does not reset the timeline.

A useful timeline record includes the valuation date, each interval’s start and end, the intensity on that interval, and the units. For accumulated value, the target date $T$ must be at or after the payments being accumulated. For present value, each payment is brought back to the specified valuation date. If that date is not time 0, use the accumulation ratio between the payment date and the valuation date rather than silently using the time-0 formula.

Worked Example: Monthly Rate on a Year-Based Timeline

A hypothetical equipment lease pays continuously at \$2,400 per month for the first six months. Time is measured in years, the accumulation function is $A(t)=e^{0.08t}$, and the present value is required at time 0. Find it to the nearest cent.

Method. Convert the monthly intensity to an annual intensity because $t$ and the accumulation function use years. Six months is $0.5$ years, and \$2,400 per month is $12(2{,}400)=\$28{,}800$ per year. Thus $r(t)=28{,}800$ for $0\leq t\leq0.5$.

PV_0=\int_0^{0.5}28{,}800e^{-0.08t}\,dt =28{,}800\frac{1-e^{-0.08(0.5)}}{0.08} =28{,}800\frac{1-e^{-0.04}}{0.08}.

Using $e^{-0.04}\approx0.9607894392$, the factor is $(1-0.9607894392)/0.08\approx0.4901320100$. Therefore, $PV_0\approx28{,}800(0.4901320100)=\$14{,}115.8019$, or \$14,115.80 to the nearest cent.

Check. The undiscounted amount paid is \$2,400 per month for six months, or \$14,400. The present value is below \$14,400, as it must be under the stated increasing accumulation function.

Worked Example: Accumulated Value of a Piecewise Stream

A hypothetical project receives continuous funding at \$2,400 per year from time 0 to time 1, then at \$1,200 per year from time 1 to time 2. The accumulation function is $A(t)=e^{0.05t}$, with time in years. Find the accumulated value at time 2, to the nearest cent.

Method. Split the stream at time 1. A payment at time $t$ accumulates to time 2 by the factor $A(2)/A(t)=e^{0.05(2-t)}$. Apply the first intensity only over $[0,1]$ and the second only over $[1,2]$.

AV_2= \int_0^1 2{,}400e^{0.05(2-t)}\,dt +\int_1^2 1{,}200e^{0.05(2-t)}\,dt.

For the first integral, an antiderivative of $e^{0.05(2-t)}$ is $-20e^{0.05(2-t)}$. Hence its value is $2{,}400(e^{0.10}-e^{0.05})/0.05$. For the second, the same antiderivative gives $1{,}200(e^{0.05}-1)/0.05$. Together,

AV_2= 2{,}400\frac{e^{0.10}-e^{0.05}}{0.05} +1{,}200\frac{e^{0.05}-1}{0.05}.

Using $e^{0.10}\approx1.1051701860$ and $e^{0.05}\approx1.0512710964$, the first contribution is $2{,}400(1.0779817920)\approx\$2{,}587.16$. The second is $1{,}200(1.0254219280)\approx\$1{,}230.51$. Their sum is approximately \$3,817.66, to the nearest cent.

Check. The total undiscounted funding is \$2,400+\$1,200=\$3,600. Every payment before time 2 accumulates under an increasing accumulation function, so the value at time 2 should exceed \$3,600. It does.

Common Mistakes and Exam Checks

A unit mismatch can change an answer by a factor of 12, 52, or another conversion factor even when the integral is otherwise set up correctly. Another frequent error is to use $A(t)$ rather than $1/A(t)$ for present value, or to use $A(T)$ rather than $A(T)/A(t)$ for accumulated value. Each payment starts accumulating at its own payment time, not at time 0.

For example, in the monthly-rate example, a common incorrect setup is $\int_0^{0.5}2{,}400e^{-0.08t}\,dt$. It treats \$2,400 per month as though it were \$2,400 per year. That wrong integral is one-twelfth of the correct value, approximately $\$1{,}176.32$, rather than $\$14{,}115.80$. The correction is to convert the intensity to \$28,800 per year when integrating over years. Alternatively, one could use months as the time unit, but then the accumulation function would also have to be expressed in months.

Check the interval: a six-month stream on a year timeline has length $0.5$, not 6.
Check the rate: the units of $r(t)\,dt$ must be currency.
Check the valuation date: use $1/A(t)$ for time-0 value and $A(T)/A(t)$ for value at time $T$.
Check the rate changes: write a separate integral for each interval on which the intensity has a different expression.
Check the scale: compare a present value with the undiscounted total, or an accumulated value with that total, when the accumulation function increases.

Key takeaway: A continuous stream is valued by integrating its intensity times the factor that moves each small payment from its payment date to the chosen valuation date. Keep the time unit, intensity unit, integration limits, and accumulation function consistent.

Check Your Understanding

For each problem, identify the time unit and use it consistently in the value integral.

A hypothetical grant pays continuously at \$900 per year from time 0 to time 1. The accumulation function is $A(t)=e^{0.06t}$, with time in years. Find the present value at time 0, to the nearest cent. (Answer: $900(1-e^{-0.06})/0.06=\$873.53$.)
A hypothetical studio pays \$350 per week continuously for 20 weeks. Time is measured in years, and $A(t)=e^{0.04t}$. Set up the present-value integral at time 0. (Answer: $\int_0^{20/52}18{,}200e^{-0.04t}\,dt=18{,}200(1-e^{-0.04(20/52)})/0.04$.)
A hypothetical fund receives \$1,000 per year continuously from time 0 to time 1, then \$2,000 per year from time 1 to time 1.5. If $A(t)=e^{0.08t}$, set up its accumulated value at time 1.5. (Answer: $\int_0^1 1{,}000e^{0.08(1.5-t)}\,dt+\int_1^{1.5}2{,}000e^{0.08(1.5-t)}\,dt$.)
A hypothetical maintenance plan pays \$500 per month continuously from month 3 through month 9. Time is measured in months, and the accumulation function is $A(m)=1.006^m$. Set up the present value at month 0. (Answer: $\int_3^9 500(1.006)^{-m}\,dm$.)
A hypothetical scholarship stream pays \$1,500 per year continuously from time 0 to time 0.5 and \$800 per year from time 0.5 to time 2. With $A(t)=1.03^t$, write the accumulated-value integral at time 2. (Answer: $\int_0^{0.5}1{,}500(1.03)^{2-t}\,dt+\int_{0.5}^{2}800(1.03)^{2-t}\,dt$.)

What You'll Learn