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 | Compounded | Number of periods per year | Compound amount |
| annually | 1 | (1+1/1)1 = $2 |
| monthly | 12 | (1+1/12)12 = $2.6130 |
| daily | 360 | (1+1/360)360 = $2.7145 |
| hourly | 8640 | (1+1/8640)8640 = $2.71812 |
| each minute | 518,400 | (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. - If the interest rate is compounded continuously at an annual interest rate r, then
- Effective interest rate: = er - 1
- If the interest rate is compounded n times per year at an annual interest rate r, then
- Effective interest rate = (1+r/n)n - 1
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
- P(t): the new balance
- the new population
- the new price ...
- Po: the original balance
- the original population
- the original price
- k: the rate of change
- the growth, decline rate
- the interest rate
- t: the time (years, days...)
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
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