7  Equation of Value

7.1 Constructing the Equation of Value

Two or more amounts of money payable at different points in time, cannot be compared directly; we need all of them to be set in a common date. This is done by either accumulating or discounting the payments. The resulting equation is known as equation of value.

Example: Every Friday in February (the 7, 14, 21, and 28) Walt places a $1000 bet, on credit, with his off-track bookmaking service. The betting service charges a weekly effective interest rate of 8% on all credit extended. Unfortunately for Walt, he loses each bet and agrees to repay his debt to the bookmaking service in four installments, to be made on March 7, 14, 21, and 28. Walt pays $1100 on each of March 7, 14, and 21. How much must Walt pay on March 28 to completely repay his debt?

Solution: If we choose Feb 7th to be time \(t=0\), then the PV of the amount that Walt will be receiving is: \(1000(1+v+v^2+v^3)\).

The PV of the amount that he has to pay is: \(1100(v^4+v^5+v^6) + Xv^7\)

We have the equation of value formulated at time \(t=0\):

\[1000(1+v+v^2+v^3) = 1100(v^4+v^5+v^6) + Xv^7 \] Solving for \(X\) with \(v = (1+i)^{-1} = 1.08^{-1}\), we have \[X = \frac{1000(1+v+v^2+v^3) - 1100(v^4+v^5+v^6)}{v^7} = 2273.79\] If we choose Mar 28 (\(t=7\)) to be the time point of reference, the equation of value would be: \[1000[(1+i)^7 + (1+i)^6 + (1+i)^5 + (1+i)^4)] = 1100[(1+i)^3 + (1+i)^2 + (1+i)] + X\] Solving for \(X\) also results in \(X=2273.79\).

Remark: When a transaction involves only compound interest, an equation of value formulated at time \(t_1\), can be translated into an equation of value formulated at time \(t_2\) simply by multiplying the first equation by \((1+i)^{t_2-t_1}\).

7.2 Time Diagrams

One device which is often helpful in the solution of equations of value is the time diagram. A time diagram is not a formal part of a solution, but may be very helpful in visualizing the solution. Usually, they are very helpful in the solution of complex probems.

Example: In return for a payment of $1,200 at the end of 10 years, a lender agrees to pay $200 immediately, $400 at the end of 6 years, and a final amount at the end of 15 years. Find the amount of the final payment at the end of 15 years if the nominal rate of interest is 9% converted semiannually.

Solution: