Equation of value: formula sheet

Key formulas

Simple interest accumulation

A(t) = 1 + it

Interest is on the original capital only. Used for short terms.

Compound interest accumulation

A(t) = (1 + i)^t

i is the effective rate per unit time and t is in the same unit.

Discount factor

v = 1 ÷ (1 + i), so PV = v^t × payment

v^t = (1 + i)^(-t).

Nominal to effective interest

1 + i = (1 + i^(m) ÷ m)^m

i^(m) is paid m times a year at i^(m) ÷ m per period.

Effective rate of discount

d = i ÷ (1 + i) = iv = 1 − v

Also i − d = id and i = d ÷ (1 − d).

Nominal discount rate

1 − d = (1 − d^(p) ÷ p)^p

Also (1 + i^(m)/m)^m = (1 − d^(p)/p)^(−p).

Force of interest from i

δ = ln(1 + i), so 1 + i = e^δ

Constant force. d = 1 − e^(−δ).

Accumulation with varying force

A(t) = exp(∫ from 0 to t of δ(s) ds)

Discount factor over the same period is the reciprocal.

Limits of nominal rates

lim of i^(m) as m → ∞ = δ = lim of d^(p) as p → ∞

Ordering for i > 0: d < d^(p) < δ < i^(m) < i.

Accumulation factor, constant effective rate

A(t1,t2) = (1 + i)^(t2 − t1)

Needs i to be an effective rate per the same unit of time as t.

Discount factor, constant effective rate

v^n = (1 + i)^(−n), where v = 1 ÷ (1 + i)

Present value of ₹1 due in n years.

Accumulation factor from force of interest

A(t1,t2) = exp( ∫ δ(s) ds from t1 to t2 )

Works for any varying δ(s). Integrate first, then exponentiate.

Discount factor from force of interest

v(t1,t2) = exp( −∫ δ(s) ds from t1 to t2 )

Reciprocal of the accumulation factor.

Constant force of interest

A(0,n) = e^(δn), and δ = ln(1 + i)

So e^δ = 1 + i.

Consecutive periods

A(t0,t2) = A(t0,t1) × A(t1,t2)

Valid for t0 < t1 < t2. Use it when rates change in steps.

Varying effective annual rates

A(0,n) = (1+i1)(1+i2)…(1+in)

i_k is the effective rate in year k.

Payment accumulation and discounting

Value at t2 = Payment at t1 × A(t1,t2); PV at t1 = Payment at t2 ÷ A(t1,t2)

Keep the dates straight before choosing a factor.

Equation of value (discrete, constant i)

Σ Aⱼ × (1 + i)^(t − tⱼ) over income = Σ Bₖ × (1 + i)^(t − sₖ) over outgo

Aⱼ is income at time tⱼ, Bₖ is outgo at time sₖ, t is the valuation date. A factor (1 + i)^(t − time) accumulates if the cashflow is before t and discounts if it is after t.

Discount factor

v = 1 ÷ (1 + i), and v^n = (1 + i)^(−n)

Using time 0 as the valuation date gives PV of income = PV of outgo with factors v^t.

Net cashflow form

Σ cₜ × v^t = 0

cₜ is the net cashflow at time t (income positive, outgo negative). This is the same equation, written as net present value equal to zero.

Continuous cashflows

∫ ρ(s) × exp(−∫₀^s δ(r) dr) ds over income = same expression over outgo

ρ(s) is the rate of payment at time s and δ(r) is the force of interest. With constant δ, the discount factor is e^(−δs).

Equation of value for life contingencies

EPV of income = EPV of outgo

This is the equivalence principle. Each cashflow is multiplied by its probability of being paid and by its discount factor.

Independence of valuation date

Multiplying both sides by (1 + i)^k leaves the solution unchanged

Holds for a constant rate or a consistent discount function. Do not mix different bases on the two sides.

Equation of value

Σ (payments in) × v^t = Σ (payments out) × v^t, where v = 1 ÷ (1 + i)

Use any single comparison date, but value every cash flow at that same date and at the same rate.

Level annuity-immediate

a_n = (1 − v^n) ÷ i

Payments of 1 at the end of each period for n periods. For payment X, the present value is X × a_n.

Linear interpolation for the rate

i ≈ i1 + (i2 − i1) × (T − f(i1)) ÷ (f(i2) − f(i1))

f(i) is the value at rate i and T is the target value. Choose i1 and i2 close together with T between f(i1) and f(i2).

Newton-Raphson iteration

i_(k+1) = i_k − f(i_k) ÷ f′(i_k)

Here f(i) is the equation written as f(i) = 0. Repeat until two successive values agree to the required accuracy.

Unknown term of a level annuity

n = −ln(1 − i × A ÷ P) ÷ ln(1 + i)

From A = P × a_n, where A is the present value, P the payment and i the rate per period. Check the answer is sensible.

Discount factor

v = 1 ÷ (1 + i); d = i ÷ (1 + i) = iv

i is the effective rate per payment period. Always match the rate period to the payment period.

Annuity-immediate (payments at end of each period)

a(n) = (1 − vⁿ) ÷ i; accumulation s(n) = ((1 + i)ⁿ − 1) ÷ i

a(n) is the value one period before the first payment. s(n) is the value at the time of the last payment.

Annuity-due (payments at start of each period)

ä(n) = (1 − vⁿ) ÷ d = (1 + i) × a(n)

Value at the time of the first payment. Also ä(n) = 1 + a(n−1).

Perpetuities

a(∞) = 1 ÷ i; ä(∞) = 1 ÷ d

Valid for i > 0. The first is paid at year end, the second at the start.

Deferred annuity

Value now of a(n) deferred m periods = vᵐ × a(n) = a(m+n) − a(m)

Payments run from time m+1 to m+n. Check the first payment date carefully.

Increasing annuity-immediate (payments 1, 2, ..., n)

(Ia)(n) = (ä(n) − n × vⁿ) ÷ i

Payment of t at time t. Value is at time 0. For (Iä)(n) use (ä(n) − n vⁿ) ÷ d.

Decreasing annuity-immediate (payments n, n−1, ..., 1)

(Da)(n) = (n − a(n)) ÷ i

Payment of n at time 1, down to 1 at time n.

Perpetuity increasing by 1 each period

(Ia)(∞) = 1 ÷ (i × d) = (1 + i) ÷ i²

Payments 1, 2, 3, ... at times 1, 2, 3, ...

Loan outstanding balance

Bₜ = X × a(n−t) at rate i, for level instalment X at year ends

Prospective method. Interest in instalment t+1 = i × Bₜ; capital repaid = X − i × Bₜ.

m-thly payments

a⁽ᵐ⁾(n) = (1 − vⁿ) ÷ i⁽ᵐ⁾, where i⁽ᵐ⁾ = m × ((1 + i)^(1/m) − 1)

This is for total payment 1 a year paid in m instalments of 1/m each at the end of each m-th period.

Net present value

NPV(i) = Σ C(t) × v^t, where v = 1 ÷ (1 + i)

C(t) is net cash flow at time t (inflow positive, outflow negative). For continuous flows use ∫ ρ(t) e^(-δt) dt.

Internal rate of return

NPV(i) = 0 solved for i

There may be no solution, or several, when cash flows change sign more than once.

Linear interpolation for IRR

i ≈ i₁ + (i₂ − i₁) × NPV(i₁) ÷ (NPV(i₁) − NPV(i₂))

Choose i₁ and i₂ close together with NPVs of opposite sign. The answer is an estimate.

Discounted payback period

smallest t such that Σ (s ≤ t) C(s) × v^s ≥ 0

Use the same rate i. State whether you give the time of the cash flow or an interpolated time.

Level annuity-immediate

a(n) = (1 − v^n) ÷ i

Useful to value level loan repayments or level project inflows.

Loan equation of value

Amount received = PV of repayments at the effective rate i

Solve for i to get the effective rate of return on the loan.

Effective rate from a monthly rate

(1 + i) = (1 + j)^12, where j is the effective monthly rate

Convert before comparing with annual rates.

Quick revision

• Effective annual rate i: (1 + i) = (1 + i⁽ᵐ⁾/m)ᵐ for nominal rate i⁽ᵐ⁾ convertible m times a year.
• Force of interest δ = ln(1 + i), so 1 + i = e^δ.
• Discount rate d = i ÷ (1 + i) = 1 − v, where v = 1 ÷ (1 + i).
• Equation of value: PV of inflows = PV of outflows, at the same date and rate.
• Any valuation date gives the same answer if the rate is constant; pick the one that simplifies the algebra.
• a⌉ₙ = (1 − vⁿ) ÷ i for payments of 1 at the end of each year for n years.
• ä⌉ₙ = (1 − vⁿ) ÷ d for payments at the start of each year, and ä⌉ₙ = (1 + i) × a⌉ₙ.
• s⌉ₙ = ((1 + i)ⁿ − 1) ÷ i, and s⌉ₙ = (1 + i)ⁿ × a⌉ₙ.
• A deferred annuity starting after k years has value vᵏ × a⌉ₙ for payments at end of years k+1 to k+n.
• IRR is the rate that makes NPV equal to zero; check for sign changes in the cashflows.
• Loan balance after t payments equals the PV of the remaining payments at the loan rate.
• Interest in a loan payment = rate × opening balance; capital repaid = payment − interest.

Common mistakes

• Treating a nominal rate as an effective rate, for example using 12% convertible monthly as 12% a year. Fix: Always read the conversion frequency. Use i^(m) ÷ m per period, or convert to i first.
• Using d = 1 − i or d = i ÷ (1 − i). Fix: Remember d = i ÷ (1 + i) = 1 − v. Check with i = 10%: d = 0.1 ÷ 1.1 ≈ 9.09%, which is less than i.
• Using the integral of δ(s) from 0 to t when the payment dates are not from time 0. Fix: Use the actual limits t1 and t2 of the period you are valuing over. Draw the timeline first.
• Forgetting to exponentiate the integral and using ∫ δ(s) ds as the factor. Fix: The integral is the exponent. The factor is exp(integral).
• Mixing up income and outgo, or leaving out a payment. Fix: Draw the timeline first. Label every flow as in or out from one person's view. Count the flows on the timeline against the question.
• Discounting a cashflow that should be accumulated, or the reverse. Fix: Use (1 + i)^(t − time). If the cashflow is after the valuation date, the exponent is negative and you discount. If before, you accumulate.
• Valuing different cash flows at different dates Fix: Fix one comparison date before you write anything. Apply that date to every term.
• Interpolating with values that do not straddle the target Fix: Check that the target lies between f(i1) and f(i2). If not, you are extrapolating, so try a better pair of rates.
• Using the annual rate when payments are monthly or half-yearly. Fix: Convert first. Use the effective rate per payment period, and set n as the number of payments.
• Treating a deferred annuity as starting m periods after time 0 when it starts m+1 periods after. Fix: For payments in arrears, deferred m years means first payment at time m+1. The value is vᵐ a(n). Check with the timeline.

Exam tips

• Start each rate question by writing which type of rate you have and its frequency. This avoids the most common lost marks.
• Show the conversion formula in standard notation before you substitute numbers. Method marks are given for this.
• Keep the sanity check d < δ < i in mind. If your answer breaks the order, recheck it.
• For varying force of interest, set up the integral explicitly and state the limits before evaluating.
• In Paper B, store the effective rate as a variable and use log() and exp() to move between rates, so you can reuse it.
• In written questions, state the formula in standard notation before substituting, for example A(t1,t2) = exp(∫ δ(s) ds). Method marks depend on it.
• Check the limits of integration against the payment dates. This is the most common source of lost marks.
• Keep at least 5 significant figures in intermediate steps and round only the final answer.