Exam FMMeasurement of interestFree to read

Effective interest and discount rates

The effective rate i measures interest paid at the end of a period and the effective discount rate d measures it at the start; v = 1 − d = 1/(1+i) converts between them.

The formulas

Accumulation

$$a(t)=(1+i{)}^t$$

value at time t of 1 invested now

Discount factor

$$v=\frac{1}{1+i}=1-d$$

Discount rate

$$d=\frac{i}{1+i}=iv$$

Interest from discount

$$i=\frac{d}{1-d}$$

The identity worth memorising

$$i-d=id$$

Where it comes from

1. Interest of $i$ on 1 is paid at the END of the period, so 1 grows to $1+i$.
2. The same interest measured at the START of the period is $d$, and 1 at the end is worth $1-d$ now, so $v=1-d$.
3. Setting the two views of the same growth equal: $v=\frac{1}{1+i}$, hence $d=1-\frac{1}{1+i}=\frac{i}{1+i}$.
4. Multiplying $d=\frac{i}{1+i}$ through by $1+i$ gives $d+di=i$, that is $i-d=id$.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• Discount is interest paid in advance, so d is always slightly SMALLER than i, and i − d = id is the gap.
• v = 1 − d and v = 1/(1+i) are the same statement — whichever you remember gives you the other rate.

Traps

• Treating d as i and discounting one period too few.
• Using i = d(1+d) instead of i = d/(1 − d).

Related

• Nominal rates convertible m-thly
• Force of interest
• Equations of value and the time line

Drill this: the Exam FM question bank has original questions on this topic, and today’s free round is open to everyone.