Solving for Constant Force from a Continuous-Annuity Value

Introduction

A continuous-annuity problem may give the payment pattern, term, and present value while leaving the force of interest unknown. The task is then to find the force that makes the discounted value of the entire payment stream equal to the stated amount. The equation is usually an integral equation in the unknown force; even when the integral can be evaluated, solving the resulting equation may require numerical methods.

The central discipline is to keep the payment intensity, time unit, and valuation date consistent. Once these are specified, the value equation can be built directly from the definition of present value under a constant force. This method is aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Definition: Suppose a payment stream has nonnegative payment intensity $c(t)$ dollars per year over $0\le t\le n$, and the constant force of interest is $\delta$ per year. Its present value at time 0 is $$ V(\delta)=\int_0^n c(t)e^{-\delta t}\,dt. $$ The unknown-force problem is to find a value of $\delta$ for which $V(\delta)$ equals the stated present value.

Build the Value Equation

Under a constant force $\delta$, one dollar payable at time $t$ has present value $e^{-\delta t}$. During a short interval of length $dt$, the payment intensity $c(t)$ produces approximately $c(t)\,dt$ dollars of payment. Discounting that amount to time 0 and adding across the term gives the integral in the definition. The payment intensity is a rate of payment, not an amount paid at each instant.

For a level payment intensity $C$ over $n$ years, the integral can be evaluated:

V(\delta)=\int_0^n Ce^{-\delta t}\,dt =\begin{cases} C\dfrac{1-e^{-\delta n}}{\delta},&\delta\ne 0,\\[6pt] Cn,&\delta=0. \end{cases}

When the intensity varies with time, first integrate the specified function. For example, a linear intensity $c(t)=A+Bt$ gives

V(\delta)=A\frac{1-e^{-\delta n}}{\delta} +B\frac{1-(1+\delta n)e^{-\delta n}}{\delta^2}, \qquad \delta\ne 0.

The value equation is $V(\delta)=V_0$, where $V_0$ is the stated present value. The resulting equation may not have a useful algebraic solution. In that case, a numerical root is an appropriate answer, provided the method, precision, and check are clear.

Existence, Uniqueness, and Numerical Methods

Assume that the payment intensity is nonnegative and that some positive amount is paid at times strictly greater than 0. Differentiating the value with respect to the constant force gives

V'(\delta)=-\int_0^n t\,c(t)e^{-\delta t}\,dt<0.

The inequality holds because the integrand $t\,c(t)e^{-\delta t}$ is nonnegative and is positive over some part of the payment term. Thus, increasing a nonnegative force strictly decreases the present value. There can be at most one nonnegative force that matches a given value. For a finite term, $V(0)=\int_0^n c(t)\,dt$, while $V(\delta)$ approaches 0 as $\delta$ increases without bound, provided the stream has no lump sum at time 0. Therefore a target strictly between 0 and $V(0)$ has one nonnegative solution under these assumptions.

A reliable numerical approach is to find two forces that give values on opposite sides of the target, then refine the bracket using bisection. Newton’s method can converge more quickly when a suitable initial estimate is available. For the equation $F(\delta)=V(\delta)-V_0=0$, its update is

\delta_{\text{new}}=\delta_{\text{old}}- \frac{V(\delta_{\text{old}})-V_0}{V'(\delta_{\text{old}})}.

The derivative here is useful for computing a root; the important practical check is to substitute the estimated force into the original value equation. If the stated present value is rounded, the inferred force is also approximate and should be reported to a precision supported by the data.

Describe the stream.
Write its payment intensity and term, and use consistent time units for the intensity and force.

Form the present value.
Discount each payment at time $t$ by $e^{-\delta t}$ and integrate over the payment period.

Set the value equal to the target.
Use the stated present value and solve $V(\delta)-V_0=0$.

Find and verify a root.
Bracket the solution or use a numerical iteration, then substitute the result into the original value equation.

Worked Examples

Worked Example: Inferring Force from a Level Continuous Payment

A hypothetical fund pays continuously at an intensity of $1,200 per year for 4 years. Its stated present value is $4,107.76. Find the nonnegative constant force of interest, to four decimal places.

Method. Use the level-intensity present-value formula. The equation is

1200\frac{1-e^{-4\delta}}{\delta}=4107.76.

There is no convenient algebraic isolation of $\delta$, so use a numerical root. At $\delta=0.08$,

V(0.08)=1200\frac{1-e^{-0.32}}{0.08} =1200\frac{1-0.7261490371}{0.08} =4107.7644435.

This is $0.0044435 above the stated value. The derivative for this level stream is

V'(\delta)=-1200\frac{1-(1+4\delta)e^{-4\delta}}{\delta^2}, \qquad V'(0.08)\approx -7778.11.

One Newton update is therefore

\delta_{\text{new}} =0.08-\frac{4107.7644435-4107.76}{-7778.11} \approx 0.08000057.

Check. Substitution of $0.08000057$ gives a present value of approximately $4,107.76. The force to four decimal places is $\delta=0.0800$ per year. The stated present value is rounded to cents, so the extra digits in the intermediate root do not imply extra precision in the final answer. This calculation also avoids the common mistake of treating the cent-rounded target as if it were the unrounded value at exactly $0.08$.

Worked Example: Inferring Force from a Linearly Increasing Intensity

A hypothetical continuous payment stream has intensity $c(t)=300+100t$ dollars per year at time $t$, for $0\le t\le2$, where $t$ is measured in years. Its present value is $719.04. Find the nonnegative constant force to four decimal places.

Method. Integrate the level and linearly increasing parts separately. For $\delta\ne0$,

V(\delta)=300\frac{1-e^{-2\delta}}{\delta} +100\frac{1-(1+2\delta)e^{-2\delta}}{\delta^2}.

At $\delta=0.10$, using $e^{-0.2}=0.8187307531$, the two contributions are

300\frac{1-0.8187307531}{0.10} =543.8077407, $$ $$ 100\frac{1-(1.2)(0.8187307531)}{0.10^2} =175.2309631.

Thus $V(0.10)=543.8077407+175.2309631=719.0387038$, which is $0.0012962 below the target. The derivative is obtained by differentiating the defining integral:

V'(\delta)=-\int_0^2 t(300+100t)e^{-\delta t}\,dt, \qquad V'(0.10)\approx-755.39.

A Newton update from $0.10$ gives

\delta_{\text{new}} =0.10-\frac{719.0387038-719.04}{-755.39} \approx0.0999983.

Check. The value at this updated force is approximately $719.04, and the force rounded to four decimal places is $0.1000$ per year. The increasing part of the intensity must remain in the integrand; replacing $300+100t$ with a level payment equal to its average would change the timing and present value.

Worked Example: Inferring Force from a Polynomial Intensity

A hypothetical stream pays at intensity $c(t)=100(1+t^2)$ dollars per year from time 0 through time 3. Its present value is $803.48. Find the nonnegative constant force to four decimal places.

Method. Integrate the constant and quadratic terms. For $\delta\ne0$, with $x=3\delta$,

V(\delta)=100\left[ \frac{1-e^{-x}}{\delta} +\frac{2-e^{-x}(x^2+2x+2)}{\delta^3} \right].

At $\delta=0.20$, $x=0.60$ and $e^{-0.60}=0.5488116361$. The two integrals are

\int_0^3 e^{-0.20t}\,dt =\frac{1-0.5488116361}{0.20} =2.2559418195, $$ $$ \int_0^3 t^2e^{-0.20t}\,dt =\frac{2-0.5488116361(3.56)}{0.20^3} \approx5.7788220.

Consequently, $V(0.20)\approx100(2.2559418195+5.7788220)=803.47638$, which is about $0.00362 below the stated value. Differentiation of the integral gives $V'(0.20)\approx-1564.03$. One Newton update is

\delta_{\text{new}} =0.20-\frac{803.47638-803.48}{-1564.03} \approx0.1999977.

Check. Substitution gives a value of approximately $803.48, and the force to four decimal places is $0.2000$ per year. The polynomial factor $t^2$ matters: discounting only the constant component would omit a substantial part of the payment stream.

Common Traps and Practical Checks

An unknown-force problem is not complete when a numerical solver displays a root. The model must match the cash flows, and the reported force must be supported by the accuracy of the input value. Check the following before finalizing a result:

Use the intensity, not a total payment amount. An intensity in dollars per year is integrated over time. Do not multiply it by the term and then discount that total as if it were payable at one date.
Match time units. If $t$ is in years, $\delta$ must be per year and an intensity must be expressed per year. A mismatch changes the exponent and the value.
Include every part of a varying pattern. For a linear or polynomial intensity, retain all terms inside the integral and evaluate them over the full payment term.
Check the direction of the answer. For nonnegative payments made after time 0, a larger force means a smaller present value. If a computed root moves in the opposite direction from the value residual, recheck the numerical update.
Do not overstate precision. A present value rounded to cents generally does not determine a force to many decimal places. Give the requested rounding and verify the value to the precision of the input.

The monotonicity result provides a useful diagnostic: under its stated assumptions, a target cannot correspond to two different nonnegative forces. If a numerical method produces apparently separate roots, inspect the function evaluation, the permitted force range, and the formulation of the payment stream.

Check Your Understanding

For each problem, write the present-value equation before solving for the force. Amounts and payment intensities are hypothetical.

A stream pays continuously at $250 per year for 3 years and has present value $706.75. Find the nonnegative constant force to four decimal places. (Answer: $\delta\approx0.0400$ per year)
A stream has intensity $100t$ dollars per year for 2 years and present value $175.23. Find the nonnegative constant force to four decimal places. (Answer: $\delta\approx0.1000$ per year)
A stream pays at intensity $200e^{0.02t}$ dollars per year for 5 years and has present value $884.80. Find the nonnegative constant force to four decimal places. (Answer: $\delta\approx0.0700$ per year)
A stream has intensity $50(1+t^2)$ dollars per year over 2 years and present value $193.04. Find the nonnegative constant force to four decimal places. (Answer: $\delta\approx0.1500$ per year)
A stream pays at $100 per year over years 0 to 1 and at $200 per year over years 1 to 2. Its present value is $267.38. Find the nonnegative constant force to four decimal places. (Answer: $\delta\approx0.1000$ per year)

What You'll Learn