Math 132

Savings Plans and First-Order Finite Difference Models

 by S. R.. Lubkin, based on work by R. White

Application

This lesson deals with the time value of money and compound interest. We will examine interest-bearing savings accounts with regular monthly deposits. How long will it take to save a target amount? If we want a target amount by some deadline, how much do we need to save each month?

Math Model

Definitions

Always be clear what your definitions are. We will use the symbol := to indicate a definition, since that is Maple's notation.

t := time (units are integers)
yt :=  account balance in month t (units are $)
D := monthly deposit (units are $)
I := annual interest rate

Note that time is not a continuous quantity, but discrete. Months are counted by integers. All months are considered to be the same length, even though we know all months are not. (This is a modeling assumption.)

Dynamical System

Let's begin with our word equation. The dynamics of the savings account are as follows: The amount you have each month is based on the amount you had in the previous month, plus interest, plus the amount you have deposited. In a word equation this looks like:
(balance this month) = (balance last month) + (interest on last month's balance) + (monthly deposit)
We need to make a further model of one of the terms, the interest, noting that the interest is based on what you owed:
(interest on last month's balance) = (balance last month)*(annual interest rate)/12
Now we can translate. We replace our words with symbols, and combine the two word equations above to get
yt = yt-1 + yt-1*I/12 +D
Or, simplifying,
yt = (1 + I/12)*yt-1 + D
The above equation is called a finite difference model. It is an example of a dynamical system, which is a mathematical model of some process in time.

Notice that the finite difference model could easily be rewritten as

yt+1 = (1 + I/12)*yt + D
just by adding 1 to t, that is, going one month ahead.

Solution of the Model

What is a "solution" of a finite difference model? So far, we have yt as a function of yt-1, but what we want is yt as a function of t.
There are two main ways to solve the model, iteration and algebra.

Iterative method:

We begin at the beginning. If we start with an initial balance of y0 at time 0, then we can find the balance in month 1 by plugging y0 into our finite difference equation. Then we can find y2 by plugging y1 into the finite difference equation, and so on.
We have y0. Then
y1 = (1 + I/12)*y0 + D
gives us y1.
y2 = (1+ I/12)*y1 + D
gives us y2. Repeat as necessary. You could see that this might be tedious unless we used a computer. Unfortunately, the iterative method is not as powerful as the

Algebraic method:

This is kind of interesting. We want to find some formula which will give us yt in terms of t directly, without having to compute each monthly balance in between. Let's begin with a simpler finite difference model,
yt+1 = a*yt + b
Then
yt+1  = a*yt + b
          = a*(a*yt-1 + b) + b
          = a*(a*(a*yt-2 + b) + b) + b
etc. until we are all the way back to y0
          = at+1*y0 + b*(at + ... + a + 1)
or, rearranging,
yt+1 = at+1*y0 + b*(1-at+1)/(1-a)
or, if you prefer yt in terms of t, which was what we were trying to find in the first place,
yt = at*y0 + b*(1-at)/(1-a)
This is nice, because we only have one formula to compute to find, say, y100, instead of 100 formulas with the iterative method.
The tidy result we just derived can be stated in the form of a theorem.
 
Theorem: For a finite difference equation of the form
yt+1 = a*yt + b
the solution yt is given by
yt = at*y0 + b*(1-at)/(1-a)
Applying the theorem to the specific example of our savings plan means that we have a = (1 + I/12) and b = D. That gives
yt = (1 + I/12)t*y0 + D*(1-(1+ I/12)t)/(-I/12)

Example

Suppose we want to buy a car. Dear Aunt Martha sends a gift of $3000 to help us get that car. But we want a $10,000 car. We open a savings account with an annual interest rate of 3%. If we deposit $150 a month into the savings account, how long will we have to wait for the car? For that matter, suppose we want to have the car in exactly 2 years. What does our deposit need to be each month in order to buy the car in 24 months?

Solution:

Using our finite difference model for the example above gives us a finite difference equation of
yt = (1 + .03/12)*yt-1 + 150
with an initial condition of y0 = 3,000. We will implement the iterative method in Excel and check it against the formula from the theorem.

Implementation in Excel:

Open a spreadsheet. What should the monthly deposit be in order to get the car in 24 months? We need to use the algebraic method. The algebraic solution in general is
yt = (1 + I/12)t*y0 + D*(1-(1+ I/12)t)/(-I/12)
and in our case, our numbers give since at t=24 we want yt to be 10,000. How do we solve this for D? We can do it by hand (tedious and error-prone) or we can use Maple.

Implementation in Maple:

In the linked worksheet we see that $150 a month really won't buy the car in 24 months. We have to save $275.87 per month.