Chapter 4 Class Handout

Simple Interest: A = P(1+rt)

P: the principal, the amount invested
A: the new balance
t: the time
r: the rate, (in decimal form)

Ex1: If $1000 is invested now with simple interest of 8% per year. Find the new amount after two years.
P = $1000, t = 2 years, r = 0.08.
A = 1000(1+0.08(2)) = 1000(1.16) = 1160

Compound Interest:

P: the principal, amount invested
A: the new balance
t: the time
r: the rate, (in decimal form)
n: the number of times it is compounded.
Ex2:Suppose that $5000 is deposited in a saving account at the rate of 6% per year. Find the total amount on deposit at the end of 4 years if the interest is:

P =$5000, r = 6% , t = 4 years

a) simple : A = P(1+rt)
A = 5000(1+(0.06)(4)) = 5000(1.24) = $6200

b) compounded annually, n = 1:
A = 5000(1 + 0.06/1)(1)(4) = 5000(1.06)(4) = $6312.38

c) compounded semiannually, n =2:
A = 5000(1 + 0.06/2)(2)(4) = 5000(1.03)(8) = $6333.85

d) compounded quarterly, n = 4:
A = 5000(1 + 0.06/4)(4)(4) = 5000(1.015)(16) = $6344.93

e) compounded monthly, n =12:
A = 5000(1 + 0.06/12)(12)(4) = 5000(1.005)(48) = $6352.44

f) compounded daily, n =365:
A = 5000(1 + 0.06/365)(365)(4) = 5000(1.00016)(1460) = $6356.12

Continuous Compound Interest:

Continuous compounding means compound every instant, consider investment of 1$ for 1 year at 100% interest rate. If the interest rate is compounded n times per year, the compounded amount as we saw before is given by: A = P(1+ r/n)nt
the following table shows the compound interest that results as the number of compounding periods increases:

P = $1; r = 100% = 1; t = 1 year

CompoundedNumber of periods per yearCompound amount
annually1(1+1/1)1 = $2
monthly12(1+1/12)12 = $2.6130
daily360(1+1/360)360 = $2.7145
hourly8640(1+1/8640)8640 = $2.71812
each minute518,400(1+1/518,400)518,400= $2.71827

As the table shows, as n increases in size, the limiting value of A is the special number

e = 2.71828

If the interest is compounded continuously for t years at a rate of r per year, then the compounded amount is given by:

A = P. e rt

Ex3: Suppose that $5000 is deposited in a saving account at the rate of 6% per year. Find the total amount on deposit at the end of 4 years if the interest is compounded continuously. (compare the result with example 2)
P =$5000, r = 6% , t = 4 years
A = 5000.e(0.06)(4) = 5000.(1.27125) = $6356.24

Ex4: If $8000 is invested for 6 years at a rate 8% compounded continuously, find the new amount:
P = $8000, r = 0.08, t = 6 years.
A = 8000.e(0.08)(6) = 8000.(1.6160740) = $12,928.60

Equivalent Value:

When a bank offers you an annual interest rate of 6% compounded continuously, they are really paying you more than 6%. Because of compounding, the 6% is in fact a yield of 6.18% for the year. To see this, consider investing $1 at 6% per year compounded continuously for 1 year. The total return is:
A = Pert = 1.e(0.06)(1) = $1.0618
If we subtract from $1.618 the $1 we invested, the return is $0.618, which is 6.18% of the amount invested.
The 6% annual interest rate of this example is called the nominal rate
The 6.18% is called the effective rate.
Ex5: Which yield better return: a) 9% compounded daily or b) 9.1% compounded monthly?
a) effective rate = (1+0.09/365)365 - 1 = 0.094162
b) effective rate = (1+0.091/12)12 - 1 = 0.094893
the second rate is better.

Ex6: An amount is invested at 7.5% per year compounded continuously, what is the effective annual rate?
the effective rate = er - 1 = e 0.075 - 1 = 1.0079 - 1 = 0.0779 = 7.79%

Ex7: A bank offers an effective rate of 5.41%, what is the nominal rate?
er - 1 = 0.0541
er = 1.0541
r = ln 1.0541 then r = 0.0527 or 5.27%

Present Value:

If the interest rate is compounded n times per year at an annual rate r, the present value of a A dollars payable t years from now is:

If the interest rate is compounded continuously at an annual rate r, the present value of a A dollars payable t years from now is

P = A. e-rt

Ex8: how much should you invest now at annual rate of 8% so that your balance 20 years from now will be $10,000 if the interest is compounded
a) quarterly: P = 10,000.(1+0.08/4)-(4)(20)= $ 2,051.10
b) continuously: P = 10,000.e-(0.08)(20) = $2.018.97

4.3: The Growth, Decline Model:

Same formulas will be applied for population, cost ...:

Growth: P(t) = Po . ekt

Decline: P(t) = Po . e-kt

For compounded continuously, the time T it takes to double the price, population or balance using k as the rate of change, the growth rate or the interest rate is given by:

===>

Note: the time it takes to triple it is T = ln3/k and so on..., (only if it is compounded continuously).

Ex9: The growth rate in a certain country is 15% per year. Assuming exponential growth :
a) find the solution of the equation in term of Po and k.
b) If the population is 100,000 now, find the new population in 5 years.
c) When will the 100,000 double itself?
Answer: a) Po. e 0.15t; b) 211,700; c) 4.62 years

Ex10: If an amount of money was doubled in 10 years, find the interest rate of the bank.
Answer: 6.93%

Ex11: In 1965 the price of a math book was $16. In 1980 it was $40. Assuming the exponential model :
a) Find k (the average rate) and write the equation.
b) Find the cost of the book in 1985.
c) After when will the cost of the book be $32 ?
Answer: a) 6.1%; b) $ 54.19; c) T = 11.36 years

Ex12: How long does it take money to triple in value at 6.36% compounded daily?
Answer: 17.27 years

Ex13: A couple want an initial balance to grow to $ 211,700 in 5 years. The interest rate is compounded continuously at 15%. What should be the initial balance?
Answer: $100,000

Ex14: The population of a city was 250,000 in 1970 and 200,000 in 1980 (Decline). Assuming the population is decreasing according to exponential-decay, find the population in 1990.
Answer: 160,000