Why the Force Determines Accumulation
When interest is described by a force that changes over time, the accumulated value depends on the force throughout the investment period, not just its value at the beginning or end. The accumulation function records the growth of one unit invested at time zero. Once it is known, it can be used to determine how any initial principal grows over a given interval.
The central operation is integration: the total growth over an interval is determined by the accumulated force across that interval. This tutorial derives that relationship and applies it to constant, piecewise-constant, and linear forces. Its focus is the accumulation of a fund invested at a point in time, rather than the valuation of a stream of payments.
Deriving the Accumulation Function
By definition, the force of interest is the instantaneous rate of growth relative to the fund value. If \(A(t)\) is the value of a fund at time \(t\), then over a short time interval its proportional growth is approximately \(\delta(t)\,dt\). In differential form, this is \(dA(t)/A(t)=\delta(t)\,dt\). Integrating gives the accumulation relationship.
Exponentiating the final expression yields the accumulation function. The same argument applies between any two times \(s\) and \(t\), provided the force is integrable between them.
This result requires the force to describe the fund’s instantaneous proportional growth throughout the interval. The force may vary and may have jumps, as long as its integral exists. The formula does not mean that the force at one selected instant can be multiplied by the whole interval length unless the force is constant over that interval.
Constant Force
If the force is constant, say \(\delta(t)=\delta\), then its integral from zero to \(t\) is \(\delta t\). The accumulation function becomes \(a(t)=e^{\delta t}\). For one year, the corresponding effective annual rate is \(i=e^\delta-1\), since one unit accumulates to \(e^\delta=1+i\).
Worked Example: Accumulation Under a Constant Force
A hypothetical fund receives a deposit of $10,000 at time zero. It earns a constant force of interest of \(0.04\) per year for \(2.5\) years. Find its accumulated value and the equivalent effective annual rate. Use continuous accumulation with no additional deposits or withdrawals.
Method. Integrate the constant force over the investment period, then apply the resulting accumulation factor to the principal. This avoids treating the force as though it were itself an effective annual rate.
The integrated force over \(2.5\) years is $$ \int_0^{2.5}0.04\,du=0.04(2.5)=0.10. $$
Thus the accumulation function at \(2.5\) years is $$ a(2.5)=e^{0.10}=1.10517018599. $$
The accumulated fund is $$ 10{,}000(1.10517018599)=11{,}051.7018599. $$ Rounded to the nearest cent, the accumulated value is $11,051.70. The increase is \(11{,}051.70-10{,}000.00=$1{,}051.70\).
For one year, the effective annual rate \(i\) satisfies \(1+i=e^{0.04}\). Therefore, $$ i=e^{0.04}-1=1.040810774-1=0.040810774. $$ The equivalent effective annual rate is 4.081% to three decimal places. As a check, this rate accumulated for \(2.5\) years gives \((1+i)^{2.5}=e^{0.10}\), the same factor used above.
Piecewise-Constant Forces
A force may be constant over successive periods but change at one or more specified times. The integrated force is then the sum of the force on each period multiplied by that period’s duration. Although the force can jump at a boundary, the accumulated value remains continuous: the fund does not gain or lose value merely because the rate changes.
Identify every time at which the force changes, and measure each interval in years.
For a constant force on an interval, multiply that force by the interval’s length.
The sum is the exponent in the accumulation factor over the full time span.
Multiply the initial principal by the exponential of the sum, and verify that the force-time products use the correct durations.
Worked Example: A Force Change After Two Years
A hypothetical investment of $10,000 earns force \(0.03\) per year for the first \(2\) years and force \(0.06\) per year for the next \(3\) years. Find its value at the end of year \(5\) and the equivalent effective annual rate over the full five-year period. Assume no cash flows occur after the initial investment.
Method. Add the integrated force over the two intervals before exponentiating. This avoids applying the second force to the full five years.
During the first \(2\) years, the integrated force is $$ 0.03(2)=0.06. $$ During the next \(3\) years, it is $$ 0.06(3)=0.18. $$ The total integrated force is \(0.06+0.18=0.24\). Therefore, the five-year accumulation factor and fund value are $$ a(5)=e^{0.24}=1.271249150, \qquad 10{,}000(1.271249150)=12{,}712.49150. $$ The accumulated value is $12,712.49 to the nearest cent.
Let \(i_{\mathrm{eq}}\) be the effective annual rate that produces the same five-year accumulation. It must satisfy $$ (1+i_{\mathrm{eq}})^5=e^{0.24}. $$ Taking fifth roots gives $$ i_{\mathrm{eq}}=e^{0.24/5}-1=e^{0.048}-1=0.049170656. $$ Thus the equivalent effective annual rate is 4.917% to three decimal places. As a check, \((1.049170656)^5\) equals \(e^{0.24}\), apart from rounding.
At the transition time \(t=2\), the accumulation function has value \(e^{0.06}\) whether calculated using the first interval or as the starting value for the second. The force changes at that instant, but the accumulated fund value does not jump.
Linear Forces
A linearly changing force can be written as \(\delta(t)=\alpha+\beta t\), where \(\alpha\) is the force at time zero and \(\beta\) is the rate at which the force changes per year. Integrating from zero to \(t\) gives \(\alpha t+\tfrac{1}{2}\beta t^2\). This expression, rather than the force at the endpoint multiplied by the duration, determines accumulation.
Worked Example: Accumulation Under a Linear Force
A hypothetical fund invests $8,000 at time zero. For the first \(4\) years its force of interest is \(\delta(t)=0.02+0.01t\), with \(t\) measured in years. Find the accumulated value at year \(4\), the force at year \(4\), and the equivalent effective annual rate over the four-year interval. Assume the force follows this formula throughout the period and there are no other cash flows.
Method. Integrate the force as a function of time, then exponentiate. This avoids substituting the endpoint force for the force over the entire period.
The integrated force is $$ \int_0^4(0.02+0.01u)\,du =0.02(4)+0.01\left(\frac{4^2}{2}\right) =0.08+0.08 =0.16. $$ Therefore, $$ a(4)=e^{0.16}=1.173510871, \qquad 8{,}000(1.173510871)=9{,}388.086968. $$ The accumulated value is $9,388.09 to the nearest cent.
The force at year \(4\) is $$ \delta(4)=0.02+0.01(4)=0.06, $$ or \(6\%\) per year at that instant. This is not the average force over the whole interval. The equivalent constant force over four years is \(0.16/4=0.04\) per year, so the equivalent effective annual rate is $$ e^{0.04}-1=0.040810774, $$ or 4.081% to three decimal places. Checking, \(e^{4(0.04)}=e^{0.16}\), which reproduces the accumulated value.
Common Mistake: Using the Endpoint Force for the Whole Term
In the linear-force example, a tempting but incorrect calculation is to take the year-4 force, \(0.06\), as if it applied throughout all four years. That calculation would give $$ 8{,}000e^{0.06(4)}=8{,}000e^{0.24}=10{,}169.99. $$ It overstates the correct value of $9,388.09 by $781.90. The error occurs because the force increased from \(0.02\) to \(0.06\); it was below \(0.06\) during most of the interval. The correct integrated force is \(0.16\), not \(0.24\).
| Force model | Integrated force from zero to \(t\) | Accumulation function |
|---|---|---|
| Constant: \(\delta(u)=\delta\) | \(\delta t\) | \(e^{\delta t}\) |
| Piecewise constant | Sum of force times duration on each interval | Exponential of that sum |
| Linear: \(\delta(u)=\alpha+\beta u\) | \(\alpha t+\tfrac{1}{2}\beta t^2\) | \(e^{\alpha t+\frac{1}{2}\beta t^2}\) |
The integrated force is dimensionless when the force is measured per year and time is measured in years. A time stated in months must first be converted to years if the force is per year. For example, \(18\) months is \(1.5\) years, not \(18\), when multiplying by an annual force.
Why the Accumulation Function Matters
The accumulation function is the basic link between an interest model and a fund value at a later time. For a constant force, a single exponential expression is sufficient. For a time-varying force, the integral accounts for how long each force level applies. In either case, accumulated value comes from the initial amount multiplied by the factor for the correct time interval.
A useful final check is to ask whether the integrated force reflects the entire path of the force over the interval. For a piecewise model, verify each duration and boundary. For a linear model, integrate the function rather than using one endpoint. These checks are particularly valuable when an effective annual rate is requested: the equivalent rate must reproduce the same total accumulation over the stated number of years.
Check Your Understanding
For each problem, assume a deterministic force of interest, no additional deposits or withdrawals, and annual time units. Round dollar amounts to the nearest cent and rates to three decimal places.
- A deposit of $4,000 earns a constant force of \(0.025\) per year for \(6\) years. Find its accumulated value and the equivalent effective annual rate. (Answer: $4,647.34; 2.532%)
- A $7,500 investment earns force \(0.02\) for the first \(3\) years and \(0.05\) for the next \(2\) years. Find its value after \(5\) years. (Answer: $8,801.33)
- A $12,000 fund earns \(\delta(t)=0.01+0.004t\) for \(5\) years. Find its accumulated value and the equivalent effective annual rate over the interval. (Answer: $13,262.04; 2.020%)
- A $9,000 investment earns force \(0.04\) for \(1.5\) years and \(0.08\) for the next \(2.5\) years. Find its value after \(4\) years and its equivalent effective annual rate. (Answer: $11,672.37; 6.716%)
- A $5,000 investment earns \(\delta(t)=0.03+0.005t\) from time zero through year \(6\). Find the correct value at year \(6\), then find the value incorrectly obtained by applying the year-6 force throughout all \(6\) years and the resulting overstatement. (Answer: Correct value $6,549.82; incorrect value $7,166.65; overstatement $616.83)