# SI and CI

Tag:

Simple Interest
Let the annual rate of interest be i (as a fraction, that is 100i percent), 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 4% per year simple interest 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

Compound Interest
Let the annual rate of interest be i (as a fraction, that is 100i percent), 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)]).

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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 4% per year compound interest 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.

Annual Percentage Rate (APR)
The annual percentage rate (APR) is the rate r of simple interest for 1 year, which will be equal to the actual amount of compound interest for that year at rate i compounded q times per year. 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%.

Installment Loans
An installment loan is a form of an arrangement called an annuity. In this scenario, a lender loans a borrower a principal amount P, on which the borrower will pay an interest rate of i per year for n years, with q equal payments per year, and the borrower also repays 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 last part of the principal and interest owed. 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 an 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.

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))

Question 1
The March, 1974 edition of _Scientific American_ said that the rate of energy consumption increased 3.1 percent per year between 1947 and 1971. If the coal reserves would last 600 years at the 1972 consumption rate, how long would they last if consumption continued to increase at the rate of 3.1 percent per year?

Answer log(600(0.031) + 1)/log(1.031) =97 solve it.

NOTES
The time period for principal doubling itself is given by the approximation formula n=(72/r) n=69/r+0.35
r is rate of interest for one time period.

The above empirical formulae would be valid for most of the common rate of interest and time periods. The error would increase if r decreases below 5%. Both the formulae give very close values