Continuous Payments Versus Equal-Total Installments

Introduction

A payment stream and a schedule of installments can have the same total nominal amount and still have different financial values. The reason is timing: money paid earlier has more time to earn interest, or is discounted for less time. This tutorial isolates that effect by comparing a continuous stream with installments paid at each year-end under a constant force of interest.

The comparison uses one-year periods. In each year, the continuous stream pays a constant amount per year, while the installment schedule pays one lump sum at the end of that year. The total nominal payment during each year is the same in both cases. We compare values at time zero and at the end of the final year.

Definition: Let $R$ be a constant payment intensity in dollars per year, and let $C=R$ be the end-of-year installment. Over each one-year period, the continuous stream pays $R$ dollars in total, and the installment schedule pays $C$ dollars. Under a constant force of interest $\delta$, the accumulation factor over $t$ years is $e^{\delta t}$, and the present-value factor is $e^{-\delta t}$. Unless stated otherwise, assume $\delta>0$, payments are certain, and there are no fees, taxes, or other adjustments.

Present Values: Earlier Payments Are Worth More

Suppose there are $n$ years of payments. The continuous stream pays at intensity $R$ from time zero through time $n$. The installment schedule pays $C=R$ at times $1,2,\ldots,n$. A payment made at time $t$ has present value equal to its amount multiplied by $e^{-\delta t}$.

\begin{aligned} PV_{\text{continuous}} &=\int_0^n R e^{-\delta t}\,dt =R\frac{1-e^{-\delta n}}{\delta},\\ PV_{\text{installments}} &=\sum_{k=1}^{n} C e^{-\delta k} =C\frac{1-e^{-\delta n}}{e^\delta-1}. \end{aligned}

The installment expression is also $C a_{\overline{n}|}$ when the annual effective rate is $i=e^\delta-1$. This link is useful, but the two schedules must not be confused: $a_{\overline{n}|}$ values payments at year-ends, whereas the integral values payments arriving throughout each year.

Theorem: If $\delta>0$ and $C=R>0$, then $PV_{\text{continuous}}>PV_{\text{installments}}$. For each year, the continuous payments arrive before that year’s end-of-year installment, so each portion is discounted for less time.

The following derivation makes the comparison precise. Over one year, the continuous payments have present value $R\int_0^1 e^{-\delta t}\,dt$, while the year-end installment has present value $R e^{-\delta}$. Their ratio is $(e^\delta-1)/\delta$, which exceeds one for positive $\delta$.

Write each one-year value.
The continuous stream’s present value during the year is $R(1-e^{-\delta})/\delta$. The installment’s present value is $R e^{-\delta}$.

Form their ratio.
Dividing the continuous value by the installment value gives $\frac{R(1-e^{-\delta})/\delta}{R e^{-\delta}}=\frac{e^\delta-1}{\delta}$.

Establish the inequality.
For $\delta>0$, $e^\delta-1=\int_0^\delta e^x\,dx>\int_0^\delta 1\,dx=\delta$, because $e^x>1$ for $x>0$. Thus the ratio exceeds one. Repeating this comparison across years preserves the inequality because each year’s payments are discounted using the same force.

For $n$ years, the same result follows directly from the formulas: $1-e^{-\delta n}>0$ and $(e^\delta-1)/\delta>1$. The present-value difference can also be written as

PV_{\text{continuous}}-PV_{\text{installments}} = R(1-e^{-\delta n}) \left(\frac{1}{\delta}-\frac{1}{e^\delta-1}\right)>0.

Worked Example: Present Value of Three Years of Equal-Total Payments

A hypothetical contract pays continuously at $R=\$1{,}200$ per year for three years, or alternatively pays $\$1{,}200$ at each of the three year-ends. The constant force is $\delta=\ln(1.10)=0.0953102$ per year. Find and compare the present values. Round dollar amounts to the nearest cent.

Method. Use the continuous-payment integral for the stream and the year-end annuity formula for the installments. This avoids treating the continuous payments as though they were all made at year-end.

First, $e^{-\delta}=1/1.10$, so $e^{-3\delta}=1/(1.10)^3=1/1.331=0.7513148$. The continuous present value is

PV_{\text{continuous}} =1{,}200\frac{1-0.7513148}{0.0953102} =1{,}200\frac{0.2486852}{0.0953102} =\$3{,}131.06.

The annual effective rate for the installments is $i=e^\delta-1=0.10$. Therefore,

PV_{\text{installments}} =1{,}200\frac{1-0.7513148}{0.10} =1{,}200(2.486852) =\$2{,}984.22.

The continuous stream has the greater present value, by $\$3{,}131.06-\$2{,}984.22=\$146.84$. As a check, each alternative pays a total nominal amount of $3(\$1{,}200)=\$3{,}600$; the higher present value for the continuous stream is consistent with its payments arriving earlier.

Accumulated Values: Earlier Payments Earn More

The same timing comparison applies if the values are measured at time $n$. A payment at time $t$ accumulates for $n-t$ years, so its value at time $n$ is multiplied by $e^{\delta(n-t)}$. Integrating the accumulated continuous payments and accumulating the installments gives

\begin{aligned} AV_{\text{continuous}} &=\int_0^n R e^{\delta(n-t)}\,dt =R\frac{e^{\delta n}-1}{\delta},\\ AV_{\text{installments}} &=\sum_{k=1}^{n} C e^{\delta(n-k)} =C\frac{e^{\delta n}-1}{e^\delta-1}. \end{aligned}

With $C=R$ and $\delta>0$, the continuous accumulated value is greater. Indeed, the ratio of the two accumulated values is again $(e^\delta-1)/\delta>1$. Each continuous payment is made before the corresponding year-end installment and therefore earns interest for longer by the time both schedules are valued at year $n$.

Worked Example: Accumulated Value at the End of Four Years

A hypothetical fund receives either a continuous stream at $R=\$600$ per year or $\$600$ at each year-end for four years. The constant force is $\delta=\ln(1.10)$ per year. Find both accumulated values at time 4 and their difference. Round to the nearest cent.

Method. Accumulate each payment to time 4. A year-end payment at time $k$ earns interest for $4-k$ years; a continuous payment at time $t$ earns interest for $4-t$ years.

Since $e^{4\delta}=1.10^4=1.4641$, the continuous accumulated value is

AV_{\text{continuous}} =600\frac{1.4641-1}{0.0953102} =600\frac{0.4641}{0.0953102} =\$2{,}921.62.

The installment schedule accumulates to

\begin{aligned} AV_{\text{installments}} &=600(1.10^3+1.10^2+1.10+1)\\ &=600(1.331+1.210+1.100+1.000)\\ &=600(4.641) =\$2{,}784.60. \end{aligned}

The continuous schedule is greater by $\$2{,}921.62-\$2{,}784.60=\$137.02$. The result is consistent with the timing: the continuous payments are not all postponed to the year-end.

Finding an Equivalent Year-End Installment

Sometimes the question is not to compare equal nominal totals, but to find the year-end installment that has the same present value as a continuous stream. Let the stream have intensity $R$, and let the equivalent installment be $C^*$ at each year-end. Equating the two present values over $n$ years gives

R\frac{1-e^{-\delta n}}{\delta} = C^*\frac{1-e^{-\delta n}}{e^\delta-1} \quad\Longrightarrow\quad C^*=R\frac{e^\delta-1}{\delta}.

For $\delta>0$, this equivalent year-end installment exceeds $R$. That does not contradict the equal-total comparison: the larger installment compensates for arriving later. The equivalence factor does not depend on $n$, provided both alternatives cover the same positive number of complete one-year periods under the same constant force.

Worked Example: Matching Present Values With Larger Installments

A hypothetical continuous payment stream has intensity $R=\$2{,}000$ per year for three years. The force is $\delta=\ln(1.10)=0.0953102$ per year. Find the constant year-end installment $C^*$ with the same present value. Then verify the equality. Round to the nearest cent.

Method. Equate the continuous and installment present-value formulas. The common factor $1-e^{-3\delta}$ cancels, leaving $C^*=R(e^\delta-1)/\delta$.

C^* =2{,}000\frac{1.10-1}{0.0953102} =2{,}000\frac{0.10}{0.0953102} =\$2{,}098.41.

To check, $e^{-3\delta}=0.7513148$. The continuous present value is

2{,}000\frac{1-0.7513148}{0.0953102} =\$5{,}218.44.

The installment annuity factor is $(1-0.7513148)/0.10=2.486852$. Using the unrounded equivalent installment, $2{,}000(0.10/0.0953102)=\$2{,}098.4117$, gives

\$2{,}098.4117(2.486852)=\$5{,}218.44.

The present values agree. The common mistake avoided here is to set $C^*=R$: equal annual totals do not imply equal present values when payment dates differ.

Common Mistake and Scope of the Comparison

A frequent error is to use the end-of-year annuity factor $a_{\overline{n}|}$ directly for a continuous stream simply because the total paid in each year is the same. That factor assumes payments occur at the year-ends. The integral is needed for payments spread continuously through the year.

For example, in the three-year case above, the incorrect calculation would be

PV_{\text{wrong}} =\$1{,}200\frac{1-(1.10)^{-3}}{0.10} =\$2{,}984.22.

This is the correct value for the three year-end installments, but not for the continuous stream. The continuous value is $\$3{,}131.06$, as found by integrating $1{,}200e^{-\delta t}$ from time zero to time three. The correction is to match each formula to its payment timing before calculating.

The strict comparison assumes positive interest. If $\delta=0$, all payment dates have the same value and the two schedules are equal when their nominal totals match. If $\delta<0$, the timing comparison reverses: later payments have greater present value, and earlier payments have lower accumulated value. The formulas remain valid when $\delta\ne0$, but the inequality must reflect the sign of the force.

Key takeaway: Under a positive constant force, a continuous stream and year-end installments with equal total payments do not have equal value. The continuous stream has the higher present and accumulated values because its payments occur earlier within each year.

Check Your Understanding

For each problem, compare like-for-like values under the stated constant force. Round dollar answers to the nearest cent.

A hypothetical one-year stream pays continuously at $\$100$ per year. The alternative pays $\$100$ at the year-end, and $\delta=\ln 2$. Find the present value of each alternative. (Answer: continuous $\$72.13$; installment $\$50.00$.)
A hypothetical two-year stream pays continuously at $\$500$ per year, or alternatively $\$500$ at each year-end. If $\delta=\ln(1.10)$, find both accumulated values at time 2 and the difference. (Answer: continuous $\$1{,}101.67$; installments $\$1{,}050.00$; difference $\$51.67$.)
For one year, a continuous stream pays $\$1{,}200$ in total at constant intensity, while the alternative pays $\$1{,}200$ at year-end. If $\delta=0.05$, find the difference in present values. (Answer: $\$29.02$, with the continuous stream higher.)
A continuous stream has intensity $\$3{,}000$ per year. Under $\delta=\ln(1.06)$, find the constant year-end installment with the same present value over any positive whole number of years. (Answer: $\$3{,}089.13$.)

What You'll Learn