Page 261 - Analysis, Synthesis and Design of Chemical Processes, Third Edition

We illustrate the use of these equations in Examples 9.4, 9.5, and 9.6. The letters p.a. following the
interest refers to per year (per annum).

Example 9.4

For an investment of $500 at an interest rate of 8% p.a. for 4 years, what would be the future value of this
investment, assuming compound interest?
From Equation (9.5) for P = 500, i = 0.08, and n = 4 we obtain
n
4
F = P(1 + i) = 500(1 + 0.08) = $680.24
4
Note: Simple interest would have yielded F = 500(1 + (4)(0.08)) = $660 ($20.24 less).
4

Example 9.5

How much would I need to invest in a savings account yielding 6% interest p.a. to have $5000 in five
years’ time?
From Equation (9.6) using F = $ 5000, i = 0.06, and n = 5 we get
5
n
5
P = F / (1 + i) = 5000/(1.06) = $3736.29
n
If we invest $3736.29 into the savings account today, we will have $5,000 in five years’ time.

Example 9.6

I need to borrow a sum of money (P) and have two loan alternatives.
a. I borrow from my local bank, which will lend me money at an interest rate of 7% p.a. and pay
compound interest.
b. I borrow from “Honest Sam,” who offers to lend me money at 7.3% p.a. using simple interest.

In both cases, I need the money for three years. How much money would I need in three years to pay off
this loan? Consider each option separately.
Bank: From Equation (9.5) for n = 3 and i = 0.07 we get
3
F = (P)(1 + 0.07) = 1.225 P
3
Sam: From Equation (9.4) for n = 3 and i = 7.3 we get
F = (P)(1 + (3)(0.073)) = 1.219 P
3

Sam stated a higher interest rate, and yet it is still preferable to borrow the money from Sam because
1.219P < 1.225P. This is because Sam used simple interest, and the bank used compound interest.