Why does PERT work?

Question

e is the limit of (1+1/n)^n. So how come we do Pe^rt to calculate continuously compounded interest? The regular formula for compound interest is (1+r/n)^tn, with rate being part of the base, not the exponent. How come with regards to continuously compounded interest, r is moved to the exponent?

Answer 1

We wish to evaluate $$\lim_{n\to\infty}\left(1+\frac rn\right)^n.$$ Let $m=\frac nr$. Note $m\to\infty$ if and only if $n\to\infty$. So the limit becomes

$$\lim_{m\to\infty}\left(1+\frac1m\right)^{rm}=e^r.$$

Answer 2

I think it might be useful to a more general audience to also discuss the motivation of continuous compounding from first principles.

Suppose a fund accumulates interest at an effective annual rate of $i$ (per year); that is to say, for an amount $K$ deposited into the fund at the beginning of the year, the total amount of interest added to the fund at the end of the year is $Ki$, and the accumulated value of the fund at the end of the first year therefore is $K + Ki = K(1+i)$. In such a case, because the interest rate is effective, all that matters is that the fund in effect accumulates $Ki$ in interest in one year, no matter how it is valued at some intermediate point within that year.

Now suppose that instead, interest is compounded twice per year; e.g., the fund has a nominal annual rate $i^{(2)} = i$ compounded semiannually, from which the effective six-month interest is $i^{(2)}/2$. That is to say, every six months, the fund earns interest at a rate of $i^{(2)}/2$. So after six months, the fund has earned $Ki^{(2)}/2$ in interest and has value $K\left(1+\frac{i^{(2)}}{2}\right)$. At the end of the year, the fund has value $K\left(1+\frac{i}{2}\right)\left(1+\frac{i}{2}\right) = K\left(1+\frac{i^{(2)}}{2}\right)^2$. Note that $$K\left(1+\frac{i^{(2)}}{2}\right)^2 = K\left(1+i^{(2)}+\frac{(i^{(2)})^2}{4}\right) > K(1+i^{(2)}),$$ which makes sense because when interest is compounded twice per year, interest on the second half of the year is earned on the previously earned interest in the first half of the year.

So, what if we compound the interest more frequently, say on a monthly basis? In other words, if the same nominal rate is given but is now compounded 12 times a year, the effective monthly interest rate is now $i^{(12)}/12$, and at the end of the year, the accumulated value is $$K\left(1 + \frac{i^{(12)}}{12}\right)^{12} = K\left(1 + i^{(12)} + \tfrac{11}{24}(i^{(12)})^2 + \tfrac{55}{432}(i^{(12)})^3 + \cdots \right) > 1 + i^{(12)} + \frac{(i^{(12)})^2}{4} > 1 + i^{(12)}.$$ In fact, it is not too difficult to see that if $n > m$, then for the same nominal rate $i^{(n)} = i^{(m)} = i$, $$\left( 1 + \frac{i^{(n)}}{n} \right)^{n} > \left( 1 + \frac{i^{(m)}}{m} \right)^{m},$$ which is just another way of saying that the more frequently that interest is compounded, the greater the accumulated value at the end of one year.

But if we compound interest continuously, do we get an infinite amount of money? No. This is because $$\lim_{n \to \infty} \left( 1 + \frac{i^{(n)}}{n} \right)^n = e^\delta,$$ where $\delta = \lim_{n \to \infty} i^{(n)}$ is the "nominal rate of interest compounded continuously," and $n$ is the number of compounding periods per year (the proof of which is explained in the other answer). So even if we compound "continuously," the effective annual rate remains finite, because as we compound more frequently, the rate per period is also proportionally smaller. The value $\delta$ is called the force of interest. Loosely speaking, it is a measure of how rapidly interest is accrued over an instantaneous period of time.

In more sophisticated time value of money calculations, the force of interest can be regarded as a function of time, rather than as a constant as we have done here. In such a case we write $\delta(t)$ and the accumulated value at a time $t = n$ of a single payment $K_0$ at time $t = 0$, is expressed in the form $$K(n) = K_0 \exp \left( \int_{t=0}^n \delta(t) \, dt \right).$$