Sanderson M. Smith

THE ALGEBRA OF LOANS

"Goodness is the only investment that never fails."
Henry David Thoreau

Money must circulate in a capitalistic society. Many business organizations, such as banks, keep money circulating by providing loans to businesses and to the general public. As businesses, these organizations must themselves make money, so they impose a charge for the privilege of using their money. This charge is called interest..

The purpose of this article is to illustrate the mathematics (algebra) involved in loan transactions. There are four basic algebraic concepts used.

(1) If a is not zero, then the linear equation ax = b has solution x = b/a.

Example: 8x = 12 ==> x = 12/8 = 3/2 = 1.5.

(2) If w is not zero, then w-1 = 1/w.

Examples: (a) 5-1 = 1/5 = 0.2; (b) 1/13 = 13-1.

(3) If an amount A is invested for n periods of time and earns an interest rate i per period, then the accumulation after n time periods is A(1+i)n.

Example: If \$1,000 is invested for 12 years at 9% per year, then the accumulation after 12 years is \$1,000(1.09)12 = \$2,812.66.

(4) If r is not equal to 1, then the sum of the geometric series a + ar + ar2 + ar3 + ... arn-1 =

a(1 - rn)/(1-r)

Example: The series 3 + 6 + 12 + 24 + 48 is geometric with a = 3 and r = 2. Using the formula, the sum of the 5 terms is 3(1 - 25)/(1 - 2) = 3(-31)/(-1) = 93.

 For those who are interested, here is the derivation for the sum of n terms in a geometric series: If Sn represents the sum of the first n terms of a geometric series, then Sn = a + ar + ar2 + ar3 + ... + arn-1 Now rSn = ar + ar2 + ar3 +ar4 + ... + arn-1 + arn Hence, if r is not 1, then Sn - rSn = Sn(1-r) = a - arn ==> Sn = a(1 - rn)/(1-r)

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EXAMPLE #1:

Suppose I lend you \$10,000 at an annual interest rate of 9%, and you agree to pay me back with three equal end-of-year payments. If x is the amount of the annual payment, then the loan display is as follows:

 Loan \$10,000 Payments x x x Year 0 1 2 3

What is the value of x? Remember, I want to earn a true annual rate of 9%. If A1 is the amount I must invest at 9% to accumulate to x after one year, then

A1(1.09) = x ==> A1 = x/(1.09) = x[1/(1.09)] = x(1.09)-1.

If A2 is the amount needed right now to accumulate to x after two years, then

A2(1.09)2 = x ==> A2 = x/(1.09)2 = x(1.09)-2.

Defining A3 is a similar manner, we have

A3(1.09)3 = x ==> A3 = x/(1.09)2 = x(1.09)-3.

In this situation, \$10,000 = A1 + A2 + A3. Hence

\$10,000 = x(1.09)-1 + x(1.09)-2 + x(1.09)-3 = x[(1.09)-1 + 1.09)-2 + (1.09)-3]

This is a linear equation. Using a calculator to compute the value in the brackets, we have

\$10,000 = x[2.531294666] ==> x = \$3,950.55.

Let's set up an installment payment schedule:

 End of Year Payment Interest portion of payment Principal portion of payment Principal outstanding after payment 0 --- --- --- \$10,000.00 1 \$3,950.55 \$900.00 \$3,050.55 \$6,949.45 2 \$3,950.55 \$625.45 \$3,325.10 \$3,624.35 3 \$3,950.55 \$326.20 \$3,624.35 \$0.00 TOTALS \$11,851.65 \$1,185.65 \$10,000.00 ---

A brief explanation: You have borrowed \$10,000 from me at a 9% rate. Each year you will pay me 9% of what you owe me as interest. At the end of the first year, you owe me (\$10,000)(.09) = \$900 in interest. So, your payment of \$3,950.55 includes \$900 in interest, and the amount of the loan you have paid off is \$3,950.55 - \$900.00 = \$3,050.55. After one year you still owe me \$10,000 - \$3,050.55 = \$6,949.45. Hence, for the next year, the interest portion of your payment is (\$6,949.45)(.09) = \$625.45, and the amount of the loan you are repaying (the principal portion of the payment) is \$3,950.55 - \$625.45 = \$3,325.10. The amount of the outstanding loan after the second payment is \$6,949.45 - \$3,325,10 = \$3,624.35. Doing these computations is a recursive process. You should note that the schedule does "balance out." That is, after your last payment, you owe \$0.00, as should be expected. You pay me \$1,185.65 in interest over the three-year life of this loan.

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There was no high-powered math involved in the computations above. The most difficult computation was summing the series (1.09)-1 + 1.09)-2 + (1.09)-3. Note that this is a geometric series with a = (1.09)-1 and r = (1.09)-1. While summing term by term isn't too tedious with just three terms (in this case, three payments), it could become cumbersome if a loan involves many payments. So, let's find a way around this:

Suppose now that you borrow an amount A, and will pay it off with year-end payments for n years at an annual interest rate, i. Using the loan stated at the beginning of this article as a model, the amount, x, of the annual loan payment would be the solution to the equation

A = x[(1+i)-1 + (1+i)-2 + (1+i)-3 + ... + (1+i)-n]

Note that the expression in brackets is a geometric sequence with a = (1+i)-1 and r = (1+i)-1. Using the geometric series sum formula, the sum in the brackets is

(1+i)-1[1 - (1+i)-n]/[1-(1+i)-1] = [1 - (1+i)-n]/(1+i)/[1-(1+i)-1] = [1 - (1+i)-n]/[(1+i)-1] = [1 - (1+i)-n]/i.

(The simplification isn't as bad as it seems. Try it!)

Hence, we have A = x [1 - (1+i)-n]/i ==> x = (Ai)/ [1 - (1+i)-n]

Here is a summary of what we have just done:

 Given a loan of amount A at interest rate i that is to be repaid by equal payments over n time periods of equal length, then the amount, x, of each payment is x = (Ai)/ [1 - (1+i)-n] Use of this formula assumes that the interest rate is a true rate for the time interval between payments.

In Example #1, we can calculate the payment x using the formula.

x = (\$10,000)(.09)/[1 - (1.09)-3] = \$3,950.55

Now, let's apply the formula to a fairly realistic situation.

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EXAMPLE #2:

A loan of \$30,000 will be paid off with twelve end-of-month payments over a period of one year. The interest rate is 18% per annum, compounded monthly. We will produce the payment schedule.

It is important to note that a typical loan, such as a home mortgage, involves monthly payments. The interest rate is usually an annual rate, compounded monthly. In this example, the true monthly rate is 18%/12 = 1.5%. Hence, the interest portion of the first payment is (\$30,000)(.015) = \$450.

Using the payment formula, we have x = (\$30,000)(.015)/[1 - (1.015)-12] = \$2,750.40.

 End of Month Payment Interest portion of payment Principal portion of payment Principal outstanding after payment. 0 ---- --- --- \$30,000.00 1 \$2,750.40 \$450.00 \$2,300.40 \$27,699.60 2 \$2,750.40 \$415.49 \$2,334.91 \$25,364.69 3 \$2,750.40 \$380.47 \$2,369.93 \$22,994.77 4 \$2,750.40 \$344.92 \$2,405.48 \$20,589.29 5 \$2,750.40 \$308.84 \$2,441.56 \$18,147.73 6 \$2,750.40 \$272.22 \$2,478.18 \$15,669.54 7 \$2,750.40 \$235.04 \$2,515.36 \$13,154.19 8 \$2,750.40 \$197.31 \$2,553.09 \$10,601.10 9 \$2,750.40 \$159.02 \$2,591.38 \$8,009.72 10 \$2,750.40 \$120.15 \$2,630.25 \$5,379.46 11 \$2,750.40 \$80.69 \$2,669.71 \$2,709.75 12 \$2,750.40 \$40.65 \$2,709.75 \$0.00 TOTALS \$33,004.80 \$3,004.80 \$30,000.00 ---

If computations are done correctly, a loan payment schedule will "balance out" in the end. If the payment is calculated incorrectly, the schedule will not balance out and you will know an error has been made. It should also be noted that use of a spreadsheet makes the construction of a loan payment schedule quite easy.

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EXAMPLE #3:

A typical 30 year mortgage: You borrow \$200,000 at 7.5%.

A mortgage rate is an annual rate, compounded monthly. In this case the true monthly rate is 0.075/12 = 0.00625. There would be (30)(12) = 360 payments over the life of this loan. Each payment, x, would be

x = (\$200,000)(0.00625)/[1 - (1.00625)-360] = \$1,398.43.

The total of all payments would be (\$1,398.43)(360) = \$503,434.45.

The total amount of interest paid would be \$503,434.45 - \$200,000.00 = \$303,434.45.

One could use a spreadsheet to construct a complete loan schedule of the 360 payments. The interest portion of the payments would be of importance since such interest is a tax deductible item. What follows is just an analysis of the first three payments under this mortgage.

 End of month Payment Interest portion of payment Principal portion of payment Principal outstanding after payment 0 --- --- --- \$200.000.00 1 \$1,398.43 \$1,250.00 \$148.43 \$199,851.57 2 \$1,398.43 \$1,249.07 \$149.36 \$199,702.21 3 \$1,398.43 \$1,248.14 \$150.29 \$199,551.92

This is basically how mortgages work: That is, in the beginning the payments are mostly interest. For instance, in this mortgage, the first three payments which total to \$4,195.29 contain an amount of interest equal to \$3,747.21. The amount of the loan is reduced by only \$448.08 with these payments.

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EXAMPLE #4:

I have read that it is not unusual to find an American family that has \$8,000 worth of credit card debt. Credit card rates around 18% per annum (true rate of 1.5% a month) are common. If you have this debt, you basically have a loan of \$8,000 from the credit card company. Since the interest rate is very high, this is definitely not the type of loan you want to have. Your best best would be to repay the \$8,000 immediately if you could afford to do so.

If you can't pay it off immediately, you now have the MATH POWER to do some computations. Suppose you wonder what monthly payments would be required to pay off the \$8,000 in one year. Well, simply do the computation:

x = (\$8,000)(0.015)/[1 - (1.015)-12] = \$733.44.

Under this play, you would pay a total of (\$733.44)(12) = \$8,801.28, and this would include \$801.28 in interest.

Unfortunately, people who have substantial credit card debt often pay what is just the minimum required amount each month. Assuming choice is involved, this is not a wise financial decision. Let's demonstrate why.

Assume you own \$8,000, the annual rate is 18% compounded monthly, and you take the option of paying just \$150 a month. What follows is a summary of the first year under this plan:

 End of month Payment Interest required Principal in payment Principal outstanding after payment 0 --- --- --- \$8000.00 1 \$150.00 \$120.00 \$30.00 \$7,970.00 2 \$150.00 \$119.55 \$30.45 \$7,939.55 3 \$150.00 \$119.09 \$30.91 \$7,908.64 4 \$150.00 \$118.63 \$31.37 \$7,877.27 5 \$150.00 \$118.16 \$31.84 \$7,845.43 6 \$150.00 \$117.68 \$32.32 \$7,813.11 7 \$150.00 \$117.20 \$32.80 \$7,780.31 8 \$150.00 \$116.70 \$33.30 \$7,747.01 9 \$150.00 \$116.21 \$33.79 \$7,713.22 10 \$150.00 \$115.70 \$34.30 \$7,678.92 11 \$150.00 \$115.18 \$34.82 \$7,644.10 12 \$150.00 \$114.66 \$35.34 \$7,608.76 Totals for 12 months \$1,800.00 \$1,408.76 \$391.24 ---

After just one year of paying the minimum required amount of \$150 per month, You have paid a total of \$1,800, you still owe \$7,608.76, and you have paid \$1,408.76 in interest. Scary, isn't it? There are plenty of folks who charge lots of expensive items on credit cards and pay off just the minimum balance required each month. Can you now see why they never seem to get out of credit card debt?

 For another example of the credit card trap, see item #42 under WRITINGS AND REFLECTIONS

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"It's good to have money and the things money can buy, but it's good, too, to check up once in a while and make sure that you haven't lost the things money can't buy."
George Horace Lorimer