Interest Calculations[Simple Interest] [Compound Interest] [Annual Percentage Rate (APR)] [Installment Loans] [Regular Deposits]
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.
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.
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
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 byA = P(1+ni).
To find the interest rate i, useP = A/(1+ni).
To determine how many compounding periods are needed to reach a given amount,i = ([A/P]-1)/n.
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
Compound InterestLet 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
Then the present value is given byA = P(1+[i/q])nq.
To find the interest rate i, useP = A(1+[i/q])-nq.
To determine how many years are needed to reach a given amount,i = q([A/P]1/nq - 1).
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 formn = log(A/P)/(q log[1+(i/q)]).
Here e = 2.718281828459... is the base of natural logarithms.A = Pein, P = Ae-in, i = log(A/P)/(n log[e]), n = log(A/P)/(i log[e]).
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
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
To find the interest rate i given the APR r, user = (1+[i/q])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.i = q[(1+r)1/q-1].
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
Installment LoansIn 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
The amount of the fixed payment is determined byA = P(1+[i/q])k - Mq([1+(i/q)]k-1)/i, = (P-Mq/i)(1+[i/q])k + Mq/i.
The amount of principal that can be paid off in n years isM = Pi/[q(1-[1+(i/q)]-nq)].
The number of years needed to pay off the loan isP = M(1-[1+(i/q)]-nq)q/i.
The total amount paid by the borrower is Mnq, and the total amount of interest paid isn = -log(1-[Pi/(Mq)])/(q log[1+(i/q)]).
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 principalThe 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,
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.i3 = (i1[P2-P]+i2[P-P1])/(P2-P1),
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
Regular DepositsSuppose 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
This can, of course, be solved for M or n algebraically:P = M([1+(i/q)]nq-1)(q/i)
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.)M = iP/[q([1+(i/q)]nq-1)] n = (1/q) log(1 + [(iP)/(Mq)])/log(1 + [i/q])
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
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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