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 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 Definitions Always be clear what your definitions are. Following Maple's notation, we will use the symbol := to indicate a definition. t := time (units are months) yt := account balance in month t (units are $) D := monthly deposit (units are $) I := annual interest rate Note that time here 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 that all months are not the same number of days. (This is a modeling assumption.) Translation 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. Here the balance this month is a function of the balance last month. The subscript t is a variable, so it can take on different values. 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 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 to do by hand but straightforward with a computer. The iterative method is not as powerful as the Algebraic method 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 = a2*yt-1 + b*(a + 1) = a*(a*(a*yt-2 + b) + b) + b= a3*yt-2 + b*(a2 + a + 1) etc. until, following the pattern, we are all the way back to y0 = at+1*y0 + b*(at + ... + a + 1) The last term can be rearranged by a clever trick to give yt+1 = at+1*y0 + b*(1-at+1)/(1-a) or, if you prefer yt in terms of t, 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. Applying this result 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? Model 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 algebraic formula. Implementation in Excel Open a spreadsheet. We will keep a parameter table in the spreadsheet so that we can easily see the parameters, and change them if we want to try different scenarios. In column A we will put what month it is. Label the column and make a series from 0 to however many months you want. In column B we will put the balance owed, computed by the iterative method. Note that in cell B3 we have used B2 to represent yt-1, which should be different in every row where we are computing yt. (relative cell referencing). But we have used $F$3 and $E$3 to refer to parameters, and we don't want Excel to look in a different row every time we are computing a new yt, so the $ go into the cell references (absolute cell referencing). In column C, we will put the balance computed by the formula from the theorem. Be careful typing the formula! There is a lot of opportunity for error. Copy the formula in B3 down to fill in the rest of column B. Likewise copy down the formula in C3. The numbers in columns B and C should be the same even though they were computed by different methods. A 1 t 2 0 B yt (iterative) 3000 3 1 =(1+$E$3/12)*B2 + $F$3 4 5 6 7 8 9 C D yt (formula) 3000 =(1+$E$3/12)^A3*$C$2 + $F$3*(1(1+$E$3/12)^A3)/(-$E$3/12) E F G parameters I D 0.03 150 2 3 4 5 6 7 When the formulas are entered and copied down the numbers should look like A 1 2 3 4 5 6 7 8 9 t 0 1 2 3 4 5 6 7 B yt (iterative) 3000.00 3157.50 3315.39 3473.68 3632.37 3791.45 3950.93 4110.80 C D yt(formula) 3000.00 3157.50 3315.39 3473.68 3632.37 3791.45 3950.93 4110.80 E parameters I 0.03 F D 150 and we clearly don't have the $10,000 until 43 months! What should the monthly deposit be in order to get the car in 24 months? We could try replacing the D value in the spreadsheet with various different values to try scenarios. It is more direct and more generally useful 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 10000 = (1 + .03/12)24*3000 + D*(1-(1+ 0.03/12)24)/(-0.03/12) 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 errorprone) or we can use Maple to take care of the algebra for us. G Implementation in Maple We see that saving $150 a month won't buy the car in 24 months. We have to save $275.87 per month (rounded to the nearest penny)