Fast Formulas #6: Finding the Time

Florence Cathedral 24-Hour Clock. Credit: Wikimedia Commons

Florence Cathedral 24-Hour Clock: Wikimedia Commons.

 

The Fast Formulas on this blog provide shortcut calculations for financial instruments such as mortgages and bonds. Most formulas involve the symbol n to represent the number of periods (e.g. years or months) until the end of the instrument. Usually n is already known: for example, you might know how long the mortgage is, but you need to find the monthly payment. Sometimes, though, you don’t know the timespan, and you need to figure it out from other information.

Time to Required Future Value

Let’s start with a simple example. Suppose you are able to invest some of your money at 5% per year, compounded annually. You wish to grow your $50,000 investment to $100,000.  How long will that take? In other words, what is n in the time value of money[1] equation,

FV=(1+r)^{n}PV,                                                                                                      (Formula 6.1)

where PV , the present value, is $50,000; FV , the future value, is $100,000; r, the interest rate, is 5%; and n is the unknown number of compounding periods?

To find n, divide both sides by PV and then take the natural logarithm[2] of both sides:

\ln \left( \dfrac{FV}{PV}\right)=\ln((1+r)^{n}).

Because logarithms have the nice property that \ln (a^{n})= n \ln (a), we can move the exponent n to the outside of the logarithm:

\ln \left( \dfrac{FV}{PV}\right)=n \ln(1+r),

leading to the Fast Formula for the number of periods in the time value of money equation:

n=\dfrac {\ln \left( \dfrac{FV}{PV}\right) }{\ln (1+r)}.                                                                                                            (Formula 6.2)

In can use this formula to calculate the number of periods in our example:

n=\dfrac{\ln(2)}{\ln(1.05)}\approx\dfrac{0.6931}{0.0488}\approx14.2067.

So it will take a little more than fourteen years to double your money at 5%.[3]

You could also apply Formula 6.2 to problems with single-payment securities such as zero coupon bonds, T-Bills, or commercial paper. For example, suppose you know that a zero-coupon bond has been priced at 70% to yield 4%, but you don’t know the maturity. Apply Formula 6.1 with semiannual compounding (which is typical for bonds) to find that:

n=\dfrac {\ln\left( \dfrac {100}{70}\right)} {\ln \left( 1+\dfrac {0.04}{2} \right)}\approx\dfrac {0.3567}{0.0198}\approx18.0115.

So the bond will mature in about nine years (eighteen semiannual periods).

Time to Pay Off Mortgage

Let’s apply the logarithm method to another problem. Suppose you have just taken out a $100,000 30-year mortgage at 4.2%. As we know from Formula 1.3, your monthly payment is given by

L=rPD_{n}.                                                                                                                   (Formula 1.3)

where

L is the monthly payment;
r is the monthly interest rate (0.35% in our case);
P is the principal amount;
n is the term of the mortgage, in months; and
D_{n}=\dfrac {1} {1-d^{n}} , where d=\dfrac {1} {1+r} .

Applying Formula 1.3 gives us the monthly payment of $489.02.

Now suppose you wanted to accelerate your mortgage by paying $1,000 a month. How much sooner than thirty years would the mortgage expire?

To answer this question, let’s work with Formula 1.3 to extract and isolate n from Formula 1.3.  Begin by replacing D_{n} with its definition:

L=\dfrac {rP}{1-d^{n}}.

After a few steps, you can isolate d^{n}:

d^{n}=1-\dfrac {r}{L}P.

Now take logarithms of both sides:

\ln (d^{n})=\ln(1-\dfrac {r}{L}P,

and, as before, move the exponent to the outside of the logarithm:

n\ln(d)=\ln(1-\dfrac {r}{L}P,

and then solve for n:

n=\dfrac {\ln\left( 1-\dfrac {r}{L}P\right)}{\ln(d)}.                                                  (Formula 6.3)

Since you decided to pay $1,000 a month, we change L to $1,000.  Now we find that

n=\dfrac{\ln\left(1-\dfrac{0.0035}{1,000}\times 100,000 \right)}{\ln\left( \dfrac {1}{1.0035} \right)}\approx\dfrac {-0.4308}{-0.0035}\approx123.2961.

The mortgage can be paid off in about ten years if you pay $1,000 a month for 123 months and a smaller payment in the 124th month.  An approximate doubling of the payment cuts the repayment time by about two-thirds.

In General

In the cases we examined, time to future value and time to mortgage payoff, the variable for the number of periods appeared exactly once in the original formula.  In such cases, you should always be able to isolate n and solve for it.  If n appears as an exponent, you will need to use logarithms to find the number of periods.  However, if n appears more than once in the formula, most likely there will not be a direct way to solve for it, but iteration should work.

 

[1]I always thought that it made more sense to say “money value of time” than “time value of money.”.

[2]You could use another logarithm but the natural logarithm is a better choice since it is essential to many financial calculations.

[3] You could have estimated this very easily by using the Rule of 72.

Copyright 2015.  All rights reserved.

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This entry was posted in Bonds, Fast Formulas, Math and tagged , , , , . Bookmark the permalink.

2 Responses to Fast Formulas #6: Finding the Time

  1. Don Pistulka says:

    Win,
    Because I am not a mathematician (plus I’m lazy) I usually take the easy way out and use Excel’s NPER() function. However, I always appreciate knowing how the math works.
    Thanks

    Like

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