Loans and Interest

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## Interest Calculations

[Simple Interest]  [Compound Interest]  [Annual Percentage Rate (APR)]  [Installment Loans]  [Regular Deposits]

### Introduction

Many financial calculations involve interest. Examples are loans, deposits, and annuities.

There are two kinds of interest: simple and compound. We will illustrate the difference with an interest-earning deposit.

A person makes a deposit with a financial institution, which promises a certain rate of interest per year, paid after specified intervals of time.

Note: The interest rate may be expressed as a percentage per year (yearly rate), or as an annual percentage rate (APR). These sound similar, but they are different, and it's important to know which one you're dealing with.

With simple interest, the amount of the deposit remains the same, and the amount of interest is paid to the depositor at the end of each interval of time. With compound interest, the amount of the deposit rises because the interest is added to the deposit at the end of each interval of time.

For example, suppose the deposit is \$1000, the yearly rate of interest is 6 percent, and the payment intervals are quarterly.

• If this is simple interest, the financial institution will pay the depositor \$15 at the end of each quarter, for a total of \$60 interest earned for the year (6 percent of \$1000). The total assets of the depositor after one year will be \$1060.

• If this is compound interest, the payment will still be \$15 at the end of the first quarter, but the interest will be added to the deposit, making the deposit now \$1015.

At the end of the second quarter, the interest will be calculated using this larger amount and will come to \$15.225, which will be added to the deposit, making the new total \$1030.225.

The interest paid at the end of the third quarter will be calculated using the second-quarter total, and will come to \$15.453375, which will again be added to the deposit, for a total of \$1045.678375.

At the end of the last quarter, the interest will be calculated based on the third-quarter amount, and will come to \$15.685175625. Thus the total assets of the depositor at the end of the year will be \$1061.363550625.

Of course, financial institutions do not keep track of these fractions of cents, and interest payments are rounded to the nearest cent. This means that the actual amounts paid are not the numbers shown above. Instead, the second interest payment will be \$15.23, the third will be \$15.45, the fourth will be \$15.69, and the total assets will be \$1061.37.

Notice that with compound interest, the depositor's assets at the end of the year are \$1.37 more than with simple interest. This is because during the last three quarters of the year, the depositor is earning interest on the interest previously earned.

Note: All the formulas below assume that interest earned is computed exactly, and not rounded at all. The effect of rounding is usually an extremely small amount over the course of many payments.

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### Simple Interest

Let the yearly interest rate be i (as a fraction, e.g. a rate of 6% would correspond to i=0.06), the amount of the principal be P, the number of years be n, and the amount after n years be A. Then

A = P(1+ni).
If you want to know what principal to deposit in order to have an amount A after n years at interest rate i, that principal is called the present value, and is given by
P = A/(1+ni).
To find the interest rate i, use
i = ([A/P]-1)/n.
To determine how many compounding periods are needed to reach a given amount,
n = ([A/P]-1)/i.
Example: Suppose you deposit \$6000 in a bank and receive simple interest at a yearly rate of 4% for 7 years. Then the parameters will be principal P = \$6000, interest rate per period i = 0.04, and number of periods n = 7. The amount of interest you will have received by the end of 7 years will be Pni = (\$6000)(7)(0.04) = \$1680, so you will have A = \$7680.

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### Compound Interest

Let the yearly rate of interest be i (as a fraction, e.g. a rate of 6% would correspond to i=0.06), the amount of the principal be P, the number of years be n, the number of times per year that the interest is compounded be q, and the amount after n years be A. Then
A = P(1+[i/q])nq.
Then the present value is given by
P = A(1+[i/q])-nq.
To find the interest rate i, use
i = q([A/P]1/nq - 1).
To determine how many years are needed to reach a given amount,
n = log(A/P)/(q log[1+(i/q)]).
Interest may be compounded quarterly, monthly, weekly, daily, or even more frequently. As the frequency of compounding increases, the amount A increases, but ever more slowly -- in fact it approaches a limit with continuous compounding. The formulas for this situation are found by taking the limit of the formulas above as q increases without bound. They take the form
A = Pein,
P = Ae-in,
i = log(A/P)/(n log[e]),
n = log(A/P)/(i log[e]).
Here e = 2.718281828459... is the base of natural logarithms.

Example: Suppose you deposit \$6000 in a bank and receive interest at a yearly rate of 4% for 7 years, compounded monthly. The parameters will be principal P = \$6000, annual interest rate i = 0.04, number of years n = 7, and number of periods per year q = 12. Over these 7 years the principal will grow over to the amount A, where
A = P(1+[i/q])nq,
= (\$6000)(1+[0.04/12])(7)(12),
= (\$6000)(1.003333333...)84,
= \$7935.08.

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### Annual Percentage Rate (APR)

If a yearly interest rate i is compounded q times per year, where q is greater than 1, this will be equivalent to a higher rate, r, of simple interest paid at the end of the year. The latter rate is called the annual percentage rate (APR). It is given by the formula
r = (1+[i/q])q-1.
To find the interest rate i given the APR r, use
i = q[(1+r)1/q-1].
The APR is mainly used to compare loans with different interest rates and payment intervals. The lower the APR, the lower the cost of the loan to the borrower.
Example: Suppose your credit card charges 18% interest per year, but you have to pay the interest due monthly. What is the annual percentage rate? Here the parameters are rate i = 0.18 and the number of compounding periods q = 12. Then the annual percentage rate (APR) r is given by
r = (1+[i/q])q-1,
= (1+[0.18/12])12-1,
= (1.015)12-1,
= 0.195618...,
or an APR of 19.5618%.

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### Installment Loans

In an installment loan, a lender loans a borrower a principal amount P, on which the borrower will pay a yearly interest rate of i (as a fraction, e.g. a rate of 6% would correspond to i=0.06) for n years. The borrower pays a fixed amount M to the lender q times per year. At the end of the n years, the last payment by the borrower pays off the loan.

After k payments, the amount A still owed is

A = P(1+[i/q])k - Mq([1+(i/q)]k-1)/i,
= (P-Mq/i)(1+[i/q])k + Mq/i.
The amount of the fixed payment is determined by
M = Pi/[q(1-[1+(i/q)]-nq)].
The amount of principal that can be paid off in n years is
P = M(1-[1+(i/q)]-nq)q/i.
The number of years needed to pay off the loan is
n = -log(1-[Pi/(Mq)])/(q log[1+(i/q)]).
The total amount paid by the borrower is Mnq, and the total amount of interest paid is
I = Mnq - P.

Example: Suppose you finance your car with a loan of \$12000 at a yearly interest rate of 11% for four years, and make equal payments monthly. How much will your payments have to be? Here the parameters are principal P = \$12000, interest rate i = 0.11, number of years n = 4, and number of periods per year q = 12. Then the monthly car payment M is given by
M = Pi/[q(1-[1+(i/q)]-nq)],
= (\$12000)(0.11)/[(12)(1-[1+(0.11/12)]-(4)(12))],
= \$110/(1-1.009166666...-48),
= \$310.15.
How much will be owed at the end of two years? Here we have a monthly payment of M = \$310.15, and the number of periods k = 2q = 24, and compute the amount due A at that time.
A = (P-Mq/i)(1+[i/q])k + Mq/i,
= (\$12000-[\$310.15][12]/0.11)(1+[0.11/12])24 +
(\$310.15)(12)/0.11,
= -\$27180.264935 + \$33834.545454,
= \$6654.28.
How much interest will have been paid in total?
I = Mnq - P,
= (\$310.15)(4)(12) - \$12000,
= \$14887.20 - \$12000,
= \$2887.20.

Example: You purchase a new home for \$250,000, and pay \$50,000 as a down payment. You take out a mortgage for the remaining \$200,000 at 8% interest for 30 years, compounded monthly, with equal monthly payments. What is your monthly mortgage payment? Here the parameters are principal P = \$200000, interest rate i = 0.08, number of years n = 30, and number of periods per year q = 12. Then your monthly payment M is given by

M = Pi/[q(1-[1+(i/q)]-nq)],
= (\$200000)(0.08)/[(12)(1-[1+(0.08/12)]-(30)(12))],
= (\$1333.333333...)/(1-[1.006666666...]-360),
= \$1467.53.

The interest rate i can't be solved for algebraically, and must be found numerically. One way to do this is as follows.

Guess a value i1 for i. A reasonable guess that will be too high is the interest rate for simple interest, i1 = Mnq/P - 1. Using that, compute the principal P1 for that rate. If P1 < P, then the interest rate i1 > i, but if P1 > P, then i1 < i. Using that fact, try another interest rate i2 and compute the corresponding principal value P2. Then try the new rate,
i3 = (i1[P2-P]+i2[P-P1])/(P2-P1),
which is gotten by linear interpolation. Replace the worse of the two starting interest rates with this new rate i3. Repeat this, always using the two interest rates with the corresponding principals closest to P as i1 and i2. Continue until you have found an interest rate such that the corresponding principal when rounded to the nearest cent gives P. Then i can be taken equal to that interest rate.

Example: What annual interest rate would you be paying if you paid off a principal of \$1000 in two years with monthly payments of \$50? Here P = \$1000, n = 2, q = 12, and M = \$50, and you want to find i. Here is a table of guesses for i and values obtained from them by interpolation, and the corresponding principal amounts.
i                P       P - \$1000
0.20            \$982.40    -\$17.60
0.18           \$1001.52      \$1.52
0.18159         \$999.98     -\$0.019221
0.181570145    \$1000.00     -\$0.000017576
Thus the interest rate would be 18.157% per year.

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### Regular Deposits

Suppose you opened an account at a bank which was paying an annual interest rate of i (a fraction, equivalent to 100i%). You make a deposit of M at the end of each of q equal time periods each year (including the end of the last period). The interest is compounded once per period. Then the value P of the account at the end of n years is given by
P = M([1+(i/q)]nq-1)(q/i)
This can, of course, be solved for M or n algebraically:
M = iP/[q([1+(i/q)]nq-1)]
n = (1/q) log(1 + [(iP)/(Mq)])/log(1 + [i/q])
The interest rate i can't be solved for algebraically, and must be found numerically. (See elsewhere on this page for methods for doing this.)

Example: Suppose you deposit \$100 in a bank at the end of every month, and receive 6% per year compound interest for 4 years, compounded monthly. The parameters will be payment M = \$100, annual interest rate i = 0.06, number of years n = 4, and number of periods per year q = 12. Over these 4 years the principal will grow to the amount P, where
P = M([1+(i/q)]nq-1)q/i
= (\$100)([1+(0.06/12)](4)(12)-1)12/0.06
= (\$100)(1.00548-1)200
= \$5409.78

Note that if you make the deposits at the start of each period, you get an extra month of interest, which you can compute by multiplying the total by (1+(i/q)):

P = M([1+(i/q)]nq-1)(q/i)(1+(i/q))
= M([1+(i/q)]nq-1)(1+(q/i))

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From the Dr. Math archives:

From the Internet:

• The Rule of 72, contributed by Chuck Cilek and Richard Alpert to the Investment FAQ, by Christopher Lott.
• Interest Tutorial on interest computations, from Introduction to Scientific Programming by Joseph L. Zachary, with an interactive applet that can be used to compare different scenarios graphically.