Actuarial Mathematics for Modelling · Present value and accumulated value of cashflow streams

Continuous Cashflow Streams and Force of Interest

Updated 11 October 2026 · Fact-checked

A continuous cashflow stream pays money at rate ρ(t) per year at every instant. Its present value is ∫ ρ(t) v(t) dt, where v(t) = exp(−∫₀ᵗ δ(s) ds). Its accumulated value at time T is ∫ ρ(t) exp(∫ₜᵀ δ(s) ds) dt. Integrate over the payment period.

Understand Continuous Cashflow Streams and Force of Interest

Most cashflows in exams arrive at fixed dates. Some are better modelled as flowing all the time, such as premium income from many policies or rent received daily. For these we use a rate of payment ρ(t), measured in rupees per year at time t.

In a very short interval from t to t + dt, the amount paid is about ρ(t) dt. So the total paid between times a and b is ∫ ρ(t) dt from a to b. If ρ(t) is constant at ρ, the total is ρ(b − a).

The force of interest δ(t) is the instantaneous rate of interest at time t. It links to the discount function by v(t) = exp(−∫₀ᵗ δ(s) ds). A ₹1 payment at time t has present value v(t) at time 0. If δ is constant, v(t) = e^(−δt).

To value a stream, treat each tiny payment ρ(t) dt as a single payment at time t. Discount it, then add up all the pieces by integrating. That is why present value becomes an integral. Accumulated value works the same way, but you grow each piece forward to the end date instead.

The key skill is setting up the integral correctly: the limits, the rate ρ(t), and the discount or accumulation factor. After that it is calculus.

Key rules to remember

Total payment: Total paid from a to b = ∫ₐᵇ ρ(t) dt; No interest involved. For constant ρ this is ρ(b − a).
Discount factor from force of interest: v(t) = exp(−∫₀ᵗ δ(s) ds); Gives the present value at time 0 of ₹1 paid at time t.
Accumulation factor between two times: A(t₁, t₂) = exp(∫ from t₁ to t₂ of δ(s) ds); Growth of ₹1 from time t₁ to time t₂, where t₁ < t₂.
Present value of a continuous stream: PV = ∫ₐᵇ ρ(t) exp(−∫₀ᵗ δ(s) ds) dt; Payments run from time a to time b, valued at time 0.
Accumulated value of a continuous stream: AV at time T = ∫ₐᵇ ρ(t) exp(∫ₜᵀ δ(s) ds) dt; Here T ≥ b. Each payment grows from t to T.
Constant ρ and constant δ: PV = ρ ∫₀ⁿ e^(−δt) dt = ρ (1 − e^(−δn)) ÷ δ; This equals ρ ā_n̅|. Here ā_n̅| = (1 − v^n) ÷ δ with δ = ln(1 + i).
Accumulated value, constant case: AV at time n = ρ (e^(δn) − 1) ÷ δ; This equals ρ s̄_n̅|.

How to solve Continuous Cashflow Streams and Force of Interest questions

Use the same routine for any continuous stream question, whether δ is constant or varies with time.

1Write down ρ(t) and the period over which payments are made, from a to b.
2Write down δ(t), or convert the given interest rate to δ using δ = ln(1 + i) or δ = −ln(1 − d).
3Decide the valuation date. Time 0 means present value. A later date T means accumulated value, or a value at some other time.
4Build the factor. For PV at time 0 use exp(−∫₀ᵗ δ(s) ds). For value at time T use exp(∫ₜᵀ δ(s) ds).
5Evaluate the inner integral of δ first, giving a function of t, then simplify the exponential.
6Set up the outer integral of ρ(t) times that factor, with the correct limits, and integrate.
7Check that the answer is sensible: PV should be less than total paid if δ > 0, and AV should be more.
8State the answer with units in rupees, to the accuracy asked.

Quickest way: Use the constant-rate shortcut, or split the period

When to use it: Use when ρ and δ are both constant over a period, or when they change at fixed times.

If ρ and δ are constant, use PV = ρ (1 − e^(−δn)) ÷ δ directly. Do not integrate again.
If either changes at set times, split the integral at those times.
Value each piece at its own start. Then discount that value back to time 0 using the accumulated force up to that start.
Add the pieces.
For accumulated value, either use AV = PV × exp(∫₀ᵀ δ(s) ds), or work forwards piece by piece.

Common mistakes in Continuous Cashflow Streams and Force of Interest

Using v^t = (1 + i)^(−t) with a time-varying force and ignoring the integral.
Students are used to constant interest rates and forget δ now depends on t.
Fix: Always compute ∫ δ(s) ds first. Only use e^(−δt) when δ is truly constant.
Integrating δ from 0 to t when finding accumulation from t to T.
The PV formula is memorised and applied blindly.
Fix: Check the limits. For growth from t to T the inner integral runs from t to T.
Forgetting to include ρ(t) inside the outer integral when it is not constant.
Students copy the constant case, ρ × factor, into a variable-rate question.
Fix: If ρ depends on t, it stays inside the integral and is multiplied by the discount factor.
Treating the given interest rate i as δ.
The words 'rate of interest' and 'force of interest' sound alike.
Fix: Convert with δ = ln(1 + i). Only use δ directly when the question states it.
Using wrong limits when the stream starts after time 0.
Students integrate from 0 by habit.
Fix: Integrate payments only over the stated payment period, but keep the discount factor measured from time 0.
Dropping the minus sign in the exponent for present value.
Mixing up the accumulation and discount factors.
Fix: PV uses exp(−∫δ). If your PV exceeds the total paid and δ > 0, recheck the sign.

Worked examples

Example 1

A continuous payment stream is paid at a constant rate of ₹12,000 per year for 5 years. The force of interest is constant at 6% per year. Find the present value, to the nearest rupee.

Show the solution

ρ = 12,000, δ = 0.06, n = 5.
PV = ρ (1 − e^(−δn)) ÷ δ.
δn = 0.06 × 5 = 0.3.
e^(−0.3) = 0.740818.
1 − 0.740818 = 0.259182.
0.259182 ÷ 0.06 = 4.31970.
PV = 12,000 × 4.31970 = 51,836.4.

Answer: The present value is about ₹51,836.

Example 2

A stream is paid continuously for 4 years at rate ρ(t) = 1,000 + 200t per year at time t. The force of interest is δ(t) = 0.05 per year for all t. Find the accumulated value at time 4, to the nearest rupee.

Show the solution

AV = ∫₀⁴ (1,000 + 200t) e^(0.05(4 − t)) dt.
Factor out e^(0.2): AV = e^(0.2) ∫₀⁴ (1,000 + 200t) e^(−0.05t) dt.
Find the PV integral I = ∫₀⁴ (1,000 + 200t) e^(−0.05t) dt.
Part 1: 1,000 ∫₀⁴ e^(−0.05t) dt = 1,000 (1 − e^(−0.2)) ÷ 0.05.
e^(−0.2) = 0.818731, so 1 − 0.818731 = 0.181269, and ÷ 0.05 = 3.62538. Part 1 = 3,625.38.
Part 2: 200 ∫₀⁴ t e^(−0.05t) dt. Use ∫ t e^(−kt) dt = −t e^(−kt) ÷ k − e^(−kt) ÷ k².
With k = 0.05: at t = 4, −4 × 0.818731 ÷ 0.05 − 0.818731 ÷ 0.0025 = −65.4985 − 327.4924 = −392.9909.
At t = 0: −1 ÷ 0.0025 = −400.
The integral = −392.9909 − (−400) = 7.0091.
Part 2 = 200 × 7.0091 = 1,401.82.
I = 3,625.38 + 1,401.82 = 5,027.20.
AV = e^(0.2) × 5,027.20 = 1.221403 × 5,027.20 = 6,140.2.

Answer: The accumulated value at time 4 is about ₹6,140.

Exam tips

Write the integral before any numbers. Marks are usually given for the correct set-up even if the arithmetic slips.
For a constant ρ with piecewise δ, split at the change times. This is faster and less error-prone than one big integral.
In computer-based Paper B, check your hand formula against an R or Excel numerical integration to catch limit errors.
Keep at least five decimal places in exponentials until the final step, as rounding early can shift the rupee answer.
Show the notation clearly: ρ(t), δ(t), v(t). State the valuation date in your working.

Practice questions from Present value and accumulated value of cashflow streams

Continuous Cashflow Streams and Force of Interest: frequently asked questions

What is the difference between rate of payment and force of interest?

The rate of payment ρ(t) is how fast money is paid, in rupees per year. The force of interest δ(t) is how fast money grows, as an annual rate at each instant. You combine them inside one integral.

How do I find the present value with a variable force of interest?

First compute v(t) = exp(−∫₀ᵗ δ(s) ds). Then integrate ρ(t) v(t) over the payment period. Split the integral if δ or ρ changes form at a certain time.

How is δ related to the effective annual rate i?

With constant interest, δ = ln(1 + i), so 1 + i = e^δ. Discount factors then become v^t = e^(−δt).

Is ā_n̅| the same as a_n̅|?

No. a_n̅| assumes discrete payments, while ā_n̅| assumes continuous payment at rate 1 per year. For constant δ, ā_n̅| = (1 − v^n) ÷ δ, and it is slightly larger than a_n̅|.