Actuarial Mathematics for Modelling · Annuity and accumulation functions

Force of Interest and Accumulation with Varying Rates

Updated 11 October 2026 · Fact-checked

The force of interest δ(t) is the instantaneous rate of growth of money at time t: δ(t) = A'(t) ÷ A(t) = d/dt ln A(t). To get the accumulation from time s to t, integrate: A(s,t) = exp(∫ from s to t of δ(r) dr). Discount with the reciprocal.

Understand Force of Interest and Varying Rates

Interest rates such as i measure growth over a whole period, for example a year. The force of interest δ(t) measures growth at a single instant. Think of it as a speedometer reading for your money.

Let A(t) be the accumulation function, the value at time t of ₹1 invested at time 0, so A(0) = 1. The rate of growth at time t, relative to the current value, is A'(t) ÷ A(t). This is the definition of δ(t). Since A'(t) ÷ A(t) is the derivative of ln A(t), we can integrate both sides.

Integrating from 0 to t and using ln A(0) = 0 gives ln A(t) = ∫ δ(r) dr from 0 to t. So A(t) = exp(∫ from 0 to t of δ(r) dr). For money invested at time s and valued at time t, A(s,t) = exp(∫ from s to t of δ(r) dr). The discount factor from t back to s is the reciprocal, v(s,t) = exp(−∫ from s to t of δ(r) dr).

If δ is constant, the integral is simply δ(t − s). Then A = e^(δ(t−s)). Comparing with compound interest (1 + i)^(t−s) gives e^δ = 1 + i, so δ = ln(1 + i). The IAI exam often gives δ as a function of time, such as a linear or piecewise function. You then integrate it, piece by piece if needed. When rates are quoted as effective annual rates that differ by year, you multiply the yearly growth factors instead.

Key rules to remember

Definition of force of interest: δ(t) = A'(t) ÷ A(t) = d/dt [ln A(t)]; A(t) is the accumulation of ₹1 from time 0 to t. Holds when A is differentiable.
Accumulation from force of interest: A(s,t) = exp( ∫ from s to t of δ(r) dr ); Value at time t of ₹1 invested at time s. Use s = 0 for A(t).
Discount factor: v(s,t) = exp( −∫ from s to t of δ(r) dr ); Present value at time s of ₹1 due at time t.
Constant force of interest: δ = ln(1 + i), 1 + i = e^δ, v = e^(−δ); Applies only when the effective rate i is the same for all periods.
Varying effective annual rates: A(0,n) = (1 + i₁)(1 + i₂)…(1 + iₙ); iₖ is the effective rate in year k. Present value is the reciprocal.
Link to discount rate: δ = ln(1 + i) = −ln(1 − d); Useful for converting between i, d and δ when rates are constant.

How to solve Force of Interest and Varying Rates questions

Use this method for any question that gives δ(t) or varying rates and asks for an accumulation, present value or equivalent rate.

1Write down the time interval clearly: start time s and end time t, and whether you need an accumulation or a present value.
2Identify how the rate is given: a function δ(t), piecewise δ values, or effective annual rates for each year.
3If δ(t) is given, set up the integral ∫ from s to t of δ(r) dr. Split it at any point where the formula for δ changes.
4Evaluate the integral carefully, using the right limits for each piece. Do not integrate from 0 if the money starts at time s.
5Exponentiate to get the accumulation factor exp(integral). For a present value, use exp(−integral).
6Multiply by the amount invested, or the payment, to get the monetary answer.
7If asked for an equivalent effective annual rate, solve (1 + i)^n = accumulation factor, so i = factor^(1/n) − 1. State the result to a sensible accuracy.
8Check reasonableness: positive δ must give a factor above 1 for accumulation.

Quickest way: Integrate, exponentiate, then multiply

When to use it: Use when the exam gives δ(t) as a simple polynomial or piecewise constant and you need a single accumulation or present value.

Find the antiderivative of δ once, call it F(t), so the integral is F(t) − F(s).
For piecewise δ, add up (rate × length) for constant pieces. No integration is needed.
Compute the exponent first as one number, then use e^ on your calculator once.
For present values, change the sign of the exponent. Do not take the reciprocal of a rounded number.
For a constant-rate check, test your answer against e^(δ × time).

Common mistakes in Force of Interest and Varying Rates

Integrating δ(t) from 0 to t when the money is invested at time s > 0.
Students memorise A(t) and forget the general form A(s,t).
Fix: Always write the limits as s to t first. Use 0 only when the investment starts at time 0.
Forgetting to exponentiate and giving the integral as the accumulation factor.
The integral looks like a final answer, especially when it is a clean number.
Fix: Write A = exp(integral) as a separate line. The integral alone is ln of the factor.
Taking δ equal to i, or using δ = i in a calculation.
Both are called rates, and for small values they are numerically close.
Fix: Use δ = ln(1 + i). For example, i = 8% gives δ = ln 1.08, which is about 7.70%, not 8%.
Using one formula for δ across a change point in a piecewise δ(t).
Students integrate the first formula over the whole period to save time.
Fix: Split the integral at each change point and add the parts, each with its own limits.
Averaging varying annual rates instead of multiplying growth factors.
Averaging seems natural for 'average rate', but compounding is multiplicative.
Fix: Multiply (1 + iₖ) for each year. Convert to an equivalent single rate only at the end using the nth root.
Sign error in the present value, using exp(+integral).
Students reuse the accumulation expression without reversing it.
Fix: Discounting goes backwards. Use exp(−integral), which equals 1 ÷ the accumulation factor.

Worked examples

Example 1

The force of interest at time t (in years) is δ(t) = 0.02 + 0.01t. Find (a) the accumulation at time 4 of ₹1,00,000 invested at time 0, and (b) the present value at time 0 of ₹1,00,000 payable at time 4.

Show the solution

The integral from 0 to 4 is ∫(0.02 + 0.01r) dr = [0.02r + 0.005r²] from 0 to 4.
At r = 4: 0.02 × 4 = 0.08, and 0.005 × 16 = 0.08. The sum is 0.16.
At r = 0 the value is 0. So the integral equals 0.16.
(a) Accumulation factor = e^0.16. Now e^0.16 ≈ 1.17351.
Accumulation = 1,00,000 × 1.17351 = ₹1,17,351 (to the nearest rupee).
(b) Discount factor = e^(−0.16) ≈ 0.85214.
Present value = 1,00,000 × 0.85214 = ₹85,214 (to the nearest rupee).

Answer: (a) About ₹1,17,351. (b) About ₹85,214.

Example 2

The force of interest is 4% per year for the first 3 years, then 6% per year for the next 2 years. Find (a) the accumulation at time 5 of ₹50,000 invested at time 0, and (b) the equivalent constant effective annual rate over the 5 years.

Show the solution

The force of interest is piecewise constant, so the integral is the sum of rate × length.
From 0 to 3: 0.04 × 3 = 0.12. From 3 to 5: 0.06 × 2 = 0.12. Total = 0.24.
(a) Accumulation factor = e^0.24 ≈ 1.27125.
Accumulation = 50,000 × 1.27125 = ₹63,562.5, about ₹63,562 to the nearest rupee if rounded down from the exact figure 63,562.4, so state ₹63,562.
(b) Let the equivalent constant effective rate be i, so (1 + i)^5 = e^0.24.
Then 1 + i = e^(0.24 ÷ 5) = e^0.048 ≈ 1.04917.
So i ≈ 4.92% per year.

Answer: (a) About ₹63,562. (b) An equivalent effective annual rate of about 4.92%.

Exam tips

Write the limits of integration explicitly every time. Marks are often given for correct setup before any arithmetic.
For piecewise δ, draw a quick timeline marking where the rate changes. This avoids wrong lengths.
Keep the exponent unrounded until the final step. Rounding the integral early can shift the rupee answer.
In written questions, show the definition δ(t) = A'(t) ÷ A(t) when asked to derive or prove a relationship. Then integrate.
In the computer-based paper, you may be asked to code the integral or accumulate yearly factors. State your assumption on time units and how the rate is applied.

Practice questions from Annuity and accumulation functions

Force of Interest and Varying Rates: frequently asked questions

What is the formula for the force of interest δ(t)?

δ(t) = A'(t) ÷ A(t), which equals d/dt of ln A(t). It is the instantaneous proportional rate of growth of the accumulation function. For compound interest at effective rate i it is the constant ln(1 + i).

How do I derive the accumulation function from the force of interest?

Start with δ(t) = d/dt ln A(t). Integrate from 0 to t and use A(0) = 1 to get ln A(t) = ∫ δ(r) dr. Then A(t) = exp(∫ from 0 to t of δ(r) dr). For an investment starting at s, change the lower limit to s.

Is the force of interest the same as the nominal rate?

Not in general. It is the limit of the nominal rate convertible m times a year as m tends to infinity. It equals ln(1 + i) when interest is compound at effective rate i.

How do I handle different interest rates in different years?

Multiply the accumulation factors for each year, (1 + i₁)(1 + i₂) and so on. If you are given δ for each period, add rate × length for each period and then exponentiate. Be careful to use the correct time for each rate.