4D  Compound Interest

When you deposit money in a bank, you are in effect loaning money to the bank. Most banks pay compound interest on deposits rather than simple interest. Under compound interest, interest is paid not only on your original deposit but also on interest already accumulated. To point out the difference between simple and compound interest, we look at a hypothetical example.

Let us suppose that Phoebe, having earned \$100 selling fresh milk, deposits this amount in a bank paying a 6% annual interest rate on savings. If this interest is simple, then each year the interest her deposit earns is

I = P · r · t = \$100 · .06 · 1 = \$6  .

Therefore, after one year her savings will have grown to \$106, after two years to \$112, after three years to \$118, and so on. But when you think about it, this arrangement is not totally fair because as the years go by her bank is keeping all the interest that she has earned, but it is paying new interest based only on her original deposit - the bank is enjoying free use of the money she has earned as interest!

Of course no one wants to be unfair to a little girl, so banks offer also compound interest savings plans. Under these plans, every once in a while, on a regular basis, the interest earned on an account is added to the principal to arrive at a new principal - and henceforth interest is based on the new principal instead of on the original amount deposited. If interest is compounded annually, then interest is added to the principal once a year. If it is compounded semiannually, then interest is added to the principal every six months, if compounded quarterly, then it is added every three months, etc. Interest compounded daily is added to the principal every day.

Now let us do some calculations on Phoebe's account assuming that her interest is compounded annually. After one year she has earned 6% of \$100 as interest, or \$6, so her new principal for the second year is 106% of \$100, or \$100 + \$6 = \$106. We can do this calculation in one step with the multiplication

106% · \$100 = 1.06 · \$100 = \$106  .

During the second year Phoebe earns interest on her new principal of \$106. She earns 6% of \$106, so after two years she will have 106% of \$106, or

106% · \$106 = 1.06 · \$106 = \$112.36  .

During the third year Phoebe earns interest on the new principal of \$112. 36, so after three years she will have 106% of \$112.36, or

106% · \$112.36 = 1.06 · \$112.36 ≈ \$119.10  .

Our conclusion after all these calculations is that each year Phoebe's money is multiplied by 1.06. The factor 1.06 comes from adding the annual interest rate 6% to 100%, to arrive at 106%. Alternatively, we could add 1 + .06 = 1.06. So if we want to know how much Phoebe has in her account say after 5 years, we begin with her original deposit of \$100 and multiply by 1.06 a total of 5 times; the result is

\$100 · 1.06 · 1.06 · 1.06 · 1.06 · 1.06 = \$100 · 1.065 ≈ \$133.82  .

(Such calculations are more easily done on a scientific calculator having a separate button for powers - first you compute 1.06 to the power 5, and then you multiply by 100.)

 Huh?

Next let us suppose that Phoebe's interest is compounded semiannually, so that now her interest is added to her principal every six months instead of every year. After the first six months, or 1/2 year, the amount of interest she has earned is

I = P · r · t = \$100 · .06 · ½ = \$100 · .03 = \$3  ,

and so her new principal after six months is \$100 + \$3 = \$103. We can look at it also this way - for a half-year period her interest percentage is cut in half, from 6% to 3%, so that the interest she earns is 3% of \$100. Thus her new principal after six months is 103% of \$100, or

103% · \$100 = 1.03 · \$100 = \$103  .

The same reasoning shows that Phoebe's principal gets multiplied by the factor 1.03 every half-year. After one year, two half-year periods have passed, so Phoebe's money has been multiplied by 1.03 twice; her amount after one year is

\$100 · 1.03 · 1.03 = \$100 · 1.032 = \$106.09  .

After two years, 4 half-year time periods have passed, so after two years Phoebe has

\$100 · 1.034 ≈ \$112.55  .

The interest rate 3% is called the periodic interest rate for Phoebe's loan; it represents the percentage of the principal her money earns over the time period between compoundings. We see that, for interest compounded semiannually, the periodic interest rate is half the annual interest rate.

If Phoebe's interest is compounded quarterly, then interest is compounded four times a year, so her periodic interest rate is 6%/4 = 1.5% = .015. Consequently, every three months her principal is multiplied by the factor 1 + .015 = 1.015. For instance, after two years, eight quarter-year time periods have passed, so her money has grown to

\$100 · 1.0158 ≈ \$112.65  .

Now let us consider the formula for the general situation. Suppose you deposit a principal P, at a certain annual interest rate, but compounded after every time period of some specified length. How much is your new amount A after N time periods? We let R denote the periodic interest rate - that is, R is the percentage interest your money earns each time period. Then after each time period your money is multiplied by (1 + R) (where R is first expressed as a decimal). After N compoundings, your money has been multiplied by (1 + R) a total of N times, and your amount is

A = P · (1 + R)N  .

example 1

A paperboy, saving for his education, deposited \$500 at an annual interest rate of 8%, compounded quarterly. What will his investment be worth in 7½ years, when he enters college at U.H.?

The principal is P = \$500. As interest is compounded four times a year, the periodic interest rate is R = 8% ÷ 4 = 2%. The number of compoundings in 7½ years is N = 4 · 7½ = 30. At that time the paperboy's investment will be worth

A = P · (1 + R)N = \$500 · (1 + .02)30 = \$500 · 1.0230 ≈ \$905.68  .

Note that to compute the periodic interest rate you divide the annual interest rate by the number of compoundings each year. However, to compute the number of time periods you multiply the number of years by the number of compoundings each year. If we let

 n = # of compoundings per year, r = annual interest rate, R = periodic interest rate, t = # of years of the loan, N = # of compoundings of the loan,

then we have the relations

R = r/n      ,      N = n · t  .

Here is the equation for the amount A after t years with beginning principal P:

 COMPOUND INTEREST FORMULA A = P · (1 + R)N = P · (1 + r/n)n·t  .

The formula is not hard to remember if you understand how it is derived - at each compounding the amount is multiplied by (1 + R), so that after N compoundings the original principal P has been multiplied by (1 + R)N.

example 2

The Teradas, upon their marriage 40 years ago, invested \$10,000 at an annual interest rate of 5%. We compute the amount in their account today, assuming that interest in the account has been compounded

1. annually,
2. semiannually,
3. quarterly,
4. monthly,
5. weekly,
6. daily.

The principal is P = \$10,000, the annual interest rate is r = 5% = .05, and the number of years is t = 40. The numbers of compoundings per year, in the six parts of the question, are n = 1, 2, 4, 12, 52, and 365, respectively. In each case we substitute these values into the compound interest formula to arrive at the amounts

 a)  A = \$10,000 · (1 + .05/1)1 · 40 = \$70,399.89 b)  A = \$10,000 · (1 + .05/2)2 · 40 = \$72,095.68 c)  A = \$10,000 · (1 + .05/4)4 · 40 = \$72,980.21 d)  A = \$10,000 · (1 + .05/12)12 · 40 = \$73,584.17 e)  A = \$10,000 · (1 + .05/52)52 · 40 = \$73,819.59 f)  A = \$10,000 · (1 + .05/365)365 · 40 = \$73,880.44

Observe that as you increase the number of compoundings per year you tend to get diminishing returns. The difference over 40 years between annual and semiannual compounding is nearly \$1700, whereas the difference between weekly and daily compounding is only about \$61.

Calculator Tip

In calculations like those of Example 2, you should not round off anything until you get to the final answer. To understand why, look again at part d) of Example 2. If we carry out the calculations to 10 decimal places (which is about the upper limit for most calculators), we find that

\$10,000 · (1 + .05/12)12 · 40 = \$10,000 · 1.0041666667480 ≈ \$73,584.17  .

On the other hand, if we choose to round off the term 1.0041666667 to just 1.0042, then the calculations give the answer

\$10,000 · 1.0042480 ≈ \$74,766.01  .

Our rounding led to a mistake of almost \$1200! That is because, even though the difference between 1.0041666667 and 1.0042 is very small, this difference gets magnified tremendously when we multiply the rounded term by itself 480 times. (One error leads to 480 errors, which accumulate on top of each other; and then at the end, when we multiply by 10,000, the error is multiplied by 10,000.) The best way to perform the whole computation on your calculator is to

 1)  divide .05 by 12 (without rounding the answer), 2)  add 1 (without rounding the answer), 3)  take this amount to the power 480 (without rounding the answer), 4)  multiply by \$10,000 (rounding your final answer to the nearest penny).

example 3

The Thomas family has just been blessed with a new baby. If they invest money now, at an 8% annual interest rate compounded semiannually, what principal will they need in order that the investment be worth \$50,000 when the child enters college 18 years later?

In this problem we are given the desired amount A = \$50,000, and we wish to know the amount of principal required to attain that amount. The annual interest rate is r = 8%, the number of compoundings per year is n = 2, and the number of years is t = 18. The periodic interest rate is

R = r/n = 8%/2 = 4% =.04  ,

and over the life of the loan the number of compoundings is

N = n · t = 2 · 18 = 36  .

We substitute into

A = P · (1 + R)N

and find that

\$50,000 = P · (1 + .04)36 = P · 1.0436  ,

P = \$50,000 ÷ 1.0436 ≈ \$12,183.44  .

The Thomas family must invest about \$12,183.

It is not always easy to compare compound interest rates and decide right away which is the better bargain. What would be the better rate - 10% compounded semiannually or 9.5% compounded monthly? Even though one rate is higher, the other is compounded more often, so it is not obvious which works out better in the end. An objective way to compare different compounding plans is to calculate, under each plan, by what percentage an investment will grow in one year. This percentage yearly growth is called the effective annual yield, or just annual yield, of the plan. If money is compounded more often than once a year, then the annual yield will be somewhat higher than the annual percentage rate - that is because you gain a little when interest is paid on your interest after the compounding.

One simple method of computing the annual yield is to assume you invest \$100 in the plan, and then calculate by how many dollars your money grows in one year. Since a percentage measures parts per hundred, this number of dollars increase will be the percentage increase in your investment.

example 4

We compute the annual yield on an investment offering

(a)  10.5% annual interest, compounded annually,

(b)  10% annual interest, compounded semiannually,

(c)  10% annual interest, compounded daily,

(d)  10.2% annual interest, compounded monthly.

(a)  Under annual compounding your interest is not compounded during the year, but only at the end of the year. Thus, after one year, your money has grown by 10.5%, the same as the annual interest rate. The annual yield is 10.5%, the annual interest rate.

(b)  If you invest a principal P = \$100 at 10% interest, compounded semiannually, then there are n = 2 compoundings in the first year, and the annual interest rate is r = 10% =.10. After one year the time is t = 1, and your investment will have grown to the amount

A = P · (1 + r/n)n · t = \$100 · (1 + .10/2)2 · 1 = \$100 · 1.052 = \$110.25

Since your \$100 has grown by \$10.25, the percentage increase is 10.25%. This percentage is the annual yield.

(c)  Under daily compounding at 10% annual interest, the variables are n = 365, r = .10, t = 1, P = \$100; after one year the \$100 investment has grown to

A = P · (1 + r/n)n · t = \$100 · (1 + .10/365)365 · 1 ≈ \$110.52

The \$100 investment has increased by \$10.52, and so the annual yield is 10.52%.

(d)  The variables in this case are n = 12, r = 10.2% = .102, t = 1, P = \$100; after one year under monthly compounding the \$100 investment has grown to

A = P · (1 + r/n)n · t = \$100 · (1 + .102/12)12 · 1 = \$110.69

The increase is \$10.69, and hence the annual yield is 10.69%.

Notice that plan (d), offering the highest annual yield among the four plans, is the best bargain of the four.

Inflation is related to compound interest, in that the formulas involved are the same. The annual rate of inflation measures the percentage that prices grow each year. If the rate of inflation is 3% a year, then, on the average, prices of goods and services are increasing at about 3% a year. Prices on some things may be increasing faster than 3%, while other prices may be increasing more slowly than 3% or even decreasing - but on average over all goods and services, the increase will be 3%. Thus, for a typical item having a price P today, under a rate of inflation of 3% the item will after one year cost 3% more, or 103% of P. Thus a year from now the item will cost

A = P · 1.03  .

Every year the price is multiplied by 1.03, so after N years, if the rate of inflation remains at 3%, the item will cost

A = P · 1.03N

More generally, if the annual rate of inflation remains constant at R, then a typical item having a price P today will after N years cost the amount

A = P · (1 + R)N

(But you must convert the rate R from a percent to a decimal before using this formula.)

example 5

Suppose that the annual rate of inflation for the next 5 years will remain at about 3%. If a loaf of bread costs \$2.25 today, about how much would you expect a similar loaf to cost 5 years from today?

The price today is P = \$2.25, and the rate of inflation is R = 3% = .03. Each year the price is multiplied by 1.03, so after N = 5 years the bread should cost about

A = P · (1 + R)N = \$2.25 · (1 + .03)5 = \$2.25 · 1.035 ≈ \$2.61  .

example 6

A mathematics professor now earns a salary of \$75,000 a year. If the rate of inflation remains at 2% per year, what salary will he need in 10 years in order to maintain his present extravagant lifestyle?

It now costs \$75,000 a year to pay for the professor's current needs, and this cost will increase by 2% each year. So each year his salary will have to be multiplied by 102% = 1.02. After 10 years he will need a salary in the amount of

A = \$75,000 · 1.0210 ≈ \$91,425  .

EXERCISES 4D

1. Suppose you loan \$800 to your penniless cousin, at a 12% annual interest rate compounded monthly. If he makes no payments, how much money will your cousin owe you after

1. one month?
2. two months?
3. three months?
4. one year?
5. two and a half years?

 Miss Wilkes deposited \$12,000 from an inheritance at an annual interest rate of 4%. How much will be in her account in 3 years if interest is compounded annually? semiannually? quarterly? monthly? weekly? daily?

1. A salesman, trying to interest Mr. Barnes in an investment scheme, promised him that his investment would increase in value by 10% every 9 months. If Mr. Barnes agrees to invest \$20,000, and if the salesman is correct in his prediction, what will be the value of this investment in six years?

2. Suppose the salesman offers Mr. Barnes another deal, where he may invest money at a 7% annual interest rate, compounded semiannually.
1. If Mr. Barnes likes this deal and invests \$30,000, how much will his investment be worth when he retires in 20 years?
2. Suppose that Mr. Barnes wishes this investment to be worth \$250,000 when he retires in 20 years. How much would he have to invest today to meet this goal?

 Mr. Fenwick is fortunate, in that the day he was born his father invested \$1000 for him in an interest bearing account, at 5% annual interest compounded quarterly. Today robust Mr. Fenwick turns 100 years old and, saving this money for a rainy day, he has not yet touched it. How much is his father's gift to him worth now? If his father had wanted Mr. Fenwick to be a millionaire on his 100th birthday, how much money would he have had to deposit on the day he was born? By coincidence, Mr. Fenwick's seventh wife delivered a baby boy on his 100th birthday. If Mr. Fenwick signs over his account to his new son as a birth gift, how much will this new son have in the account when he himself reaches 100 birthdays?

1. In calling around to various banks, Jill found the following available interest plans for three year certificates of deposit:

1. 5.8% yearly interest, compounded annually,
2. 5.75% yearly interest, compounded semiannually,
3. 5.7% yearly interest, compounded monthly,
4. 5.65% yearly interest, compounded daily.

Determine the annual yield for each of these plans. In which plan would you recommend that Jill invest her money?

 Suppose the annual rate of inflation will remain at about 4% for the next 10 years. If a light bulb today costs \$1.29, how much would you expect a comparable bulb to cost in 10 years?

1. During the six calendar years from the start of 1990 to the end of 1995, the annual rate of inflation in Hungary was about 26%. Estimate, say at the beginning of 1996, the price in Hungary of
1. a refrigerator costing 9000 forint at the beginning of 1994,
2. a mug of beer costing 50 forint at the beginning of 1992,
3. a haircut and shave costing 25 forint at the beginning of 1990.
4. If a worker in Budapest were making 80,000 forint per year at the beginning of 1990, what yearly salary would this worker require in 1995 in order to maintain a standard of living consistent with how he lived in 1990?

2. Estimate the future cost
1. in three years of a \$40 shirt, under an annual rate of inflation of 6.5%,
2. in 12 years of a \$25 facial massage, under an annual rate of inflation of 1.75%.

3. In Uzbekistan, the annual rate of inflation from the beginning of 1992 to the end of 1995 averaged about 560%. How much would a worker, making say 500 som per year in 1992, need to earn in 1995 in order to maintain her standard of living?