Compound Growth Help
Introduction to Compound Growth
A quantity (such as a population, amount of money, or radiation level) changes exponentially if the growth or loss is a fixed percentage over a period of time. To see how this works, we will see how the value of an account grows over four years if $100 is deposited and earns 5% interest, compounded annually. Compounded annually means that the interest earned in the previous year earns interest.
After one year, $100 has grown to 100 + 0.05(100) = 100 + 5 = $105. In the second year, the original $100 earns 5% plus the $5 earns 5% interest: 105 + (105)(0.05) = $110.25. Now this amount earns interest in the third year: 110.25 + (110.25)(0.05) = $115.76. Finally, this amount earns interest in the fourth year: 115.76 + (115.76)(0.05) = $121.55. If interest is not compounded, that is, the interest does not earn interest, the account would only be worth $120. The extra $1.55 is interest earned on interest.
Compound growth is not dramatic over the short run but it is over time. If $ 100 is left in an account earning 5% interest, compounded annually, for 20 years instead of four years, the difference between the compound growth and noncompound growth is a little more interesting. After 20 years, the compound amount is $265.33 compared to $200 for simple interest (noncompound growth). A graph of the growth of each type over 40 years is given in Figure 9.1. The line is the growth for simple (noncompounded) interest, and the curve is the growth with compound interest.
We can use a formula to compute the value of an account earning compounded interest. If P dollars is invested for t years, earning r interest rate, then it will grow to A dollars, where A = P (1 + r ) t.
Find the compound amount.
- $5000, after three years, earning 6% interest, compounded annually
We will use the formula A = P (1 + r ) t . P = 5000, r = 0.06, and t = 3. We want to know A , the compound amount.
A = 5000(1 + 0.06) 3 = 5000(1.06) 3 = 5000(1.191016)
The compound amount is $5955.08.
- $10,000 after eight years, interest, compounded annually
A = 10,000(1 + 0.0725) 8 = 10,000(1.0725) 8 ≈ 10,000(1.7505656)
The compound amount is $17,505.66
Many investments pay more often than once a year, some paying interest daily. Instead of using the annual interest rate, we need to use the interest rate per period, and instead of using the number of years, we need to use the number of periods. If there are n compounding periods per year, then the interest rate per period is and the total number of periods is nt . The compound amount formula becomes
Find the compound amount.
- $5000, after three years, earning 6% annual interest
(a) compounded semiannually
(b) compounded monthly
For (a), interest compounded semiannually means that it is compounded twice each year, so n = 2.
The compound amount is $5970.26.
For (b), interest compounded monthly means that it is compounded 12 times each year, so n = 12.
The compound amount is $5983.40.
- $10,000, after eight years, earning annual interest, compounded weekly Interest that is paid weekly is paid 52 times each year, so n = 52.
The compound amount is $17,853.17.
The more often interest is compounded per year, the more interest is earned. $1000 earning 8% annual interest, compounded annually, is worth $1080 after one year. If interest is compounded quarterly, it is worth $1082.43 after one year. And if interest is compounded daily, it is worth $1083.28 after one year. What if interest is compounded each hour? Each second? It turns out that the most this investment could be worth (at 8% interest) is $1083.29, when interest is compounded each and every instant of time. Each instant of time, a tiny amount of interest is earned. This is called continuous compounding. The formula for the compound amount for interest compounded continuously is A = Pe rt , where A , P , r , and t are the same quantities as before. The letter e stands for a constant called Euler’s number. It is approximately 2.718281828. You probably have an e or e x key on your calculator. Although e is irrational, it can be approximated by numbers of the form
where m is a large rational number. The larger m is, the better the approximation for e . If we make the substitution and use some algebra, we can see how is very close to e rt , for large values of n . If interest is compounded every minute, n would be 525,600, a rather large number!
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