Time Value of Money (part 2)

12 years ago

Last time we discussed that money has a time component to it, that it changes value with time. We derived the basic formula for TVM (Time Value of Money):

\[ P_N = P_0(1 + i)^N \]

That is, the future value (FV) after N periods is equal to the present value (PV) multiplied by a geometric factor depending on the interest rate.

Let’s explore. Suppose you went to a bank and said you wanted a $1000 loan repayable in one payment after a year. The bank offers you 6% per annum. How much would your single repayment be? Using the above formula, it’s easy: $1000 × 1.06, or $1060. Let’s say now that you want a loan over two years, repayable in two equal payments, one after the first year and one after the second. How much should your payments be in that case? In these kinds of situations, it’s sometimes helpful to draw a cash flow diagram.

This diagram show that you receive (it’s positive) $1000 now and that you pay back (they’re negative) two payments of amount a, one after a year, the other after two. But how much is a? As I pointed out last time, when you deal with TVM and calculate values you have to rebase all your amounts to the same point in time. There are two basic times that are important: now, and the time when everything completes. For our supposed loan, those two times are now and two years hence.

Let’s rebase everything to now. The amount of the loan you receive is $1000 and you get it now, so its value is $1000. The first payment is one year hence and so its present value now is -$(a / 1.06) (it’s a payment so it’s negative). The second payment is two years hence and so its PV is -$(a / (1.06 × 1.06)). Since taking the loan and repaying it should be the same as not taking the loan at all (otherwise one of the counterparties is making more out of the arrangement than agreed – that is, the interest rate is higher or lower than was settled and favors one or the other), the sum of the three PVs (the net present value or NPV) should be zero.

So:

\[ 1000 - \frac{a} {1.06} - \frac{a} {1.06^2} = 0 \]

Which, with a bit of rearranging, gives us:

\[ a = \frac {1000} { \frac {1} {1.06} + \frac {1} {1.06^2}}  \]

Which, given a competent enough calculator, gives a = $545.44.

Let’s look at it from the other end of the cash flow diagram. Again, there should be no difference between taking the loan out and not taking it out so the net future value (NFV) of the arrangement should be zero. The FV of the loan amount is $1000 × 1.06 × 1.06, or $1123.60. The FV of the first payment is –$(a × 1.06). The FV of the second payment is just -$a. Since the NFV should be zero we have:

\[ (1000 \times 1.06^2) - a - (a \times 1.06) = 0 \]

Which, after rearranging and using our trusty calculator, gives us a = $545.44, just as before. In other words, viewing a financial transaction from either the start or the end gives the same answers. TVM works!

Let’s see if we can’t generalize our results. What if the loan terms was three years and involved three equal payments? First the cash flow diagram:

This shows us what we’d have as NPV:

\[ \begin{aligned} 1000 - \frac{a} {1.06} - \frac{a} {1.06^2} - \frac{a} {1.06^3} & = 0 \\ \text{or, } 1000 – a( \frac{1} {1.06} + \frac{1} {1.06^2} + \frac{1} {1.06^3} ) & = 0 \\ \text{or, } \frac {1000} {( \frac{1} {1.06} + \frac{1} {1.06^2} + \frac{1} {1.06^3} )} - a & = 0 \\ \text{giving, } \frac {1000} {( \frac{1} {1.06} + \frac{1} {1.06^2} + \frac{1} {1.06^3} )} & = a \end{aligned} \]

Which, using our trusty calculator, gives us a = $374.11.

I’m sure you can see the generalization now. Let r = (1 + i), and let there be N equal payments, then we have:

\[ \begin{aligned} P - a( \frac {1}{r} + \frac {1}{r^2}+ \frac {1}{r^3} + \ldots + \frac {1}{r^N} ) & = 0 \\ \text{or, } P - a \left( \frac {r^{N-1} + r^{N-2}+ r^{N-3} + \ldots + 1}{r^N} \right) & = 0 \\ \end{aligned} \]

This looks scary until we remember (or someone reminds us) that:

\[ \sum_{k=0}^{n-1} x_k = \frac {1-x^n}{1-x} \]

This insight gives us:

\[ P – \frac { a (1 - r^N) } { r^N (1 – r) } = 0 \]

Or

\[ a = P . r^N . \frac { (1 – r) } { (1 - r^N) } \]

Using this formula we can now do some simple mortgage calculations. So, $250K house, 10% down, borrow the rest over 30 years at 5% per annum, means how much as a monthly payment? (This is the first example from last time.)The accepted rule for converting an annual interest rate to a monthly one is to divide it by 12, so our monthly rate is 0.416667%. The principal is $225K, and there are 360 payment periods (12 months a year × 30 years). Plugging that into the formula gives us:

\[ a = 225000 \times \left( \frac { 1.00416667^{360} \times (1 – 1.00416667) } { (1 – 1.00416667^{360}) } \right) \]

Which is $1207.85 per month. Over 15 years instead with the same interest rate?

\[ a = 225000 \times \left( \frac { 1.00416667^{180} \times (1 – 1.00416667) } { (1 – 1.00416667^{180}) } \right) \]

Which is $1779.29 per month.

Next time we’ll look at irregular payments and car leases.