How long will it take a money to one and one half times it self to 7.5 compounded bimontly

Simple Interest: A = P(1+rt)

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Simple Interest: A = P(1+rt)
Compound Interest:
Continuous Compound Interest:
Equivalent Value:
Present Value:
P = A. e-rt
4.3: The Growth, Decline Model:
Growth: P(t) = Po . ekt
Decline: P(t) = Po . e-kt
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How do you calculate how long it will take for money to double?
How much is $1000 worth at the end of 2 years if the interest rate of 6% is compounded daily?
How do you solve compounded semiannually?
How long will it take you to double your money if you invest it today and earn 10% per year?

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 rateThe 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:

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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

How do you calculate how long it will take for money to double?

The Rule of 72 is a calculation that estimates the number of years it takes to double your money at a specified rate of return. If, for example, your account earns 4 percent, divide 72 by 4 to get the number of years it will take for your money to double.

How much is $1000 worth at the end of 2 years if the interest rate of 6% is compounded daily?

Compound interest formulas Hence, if a two-year savings account containing $1,000 pays a 6% interest rate compounded daily, it will grow to $1,127.49 at the end of two years.

How do you solve compounded semiannually?

The formula for compounded interest is based on the principal, P, the nominal interest rate, i, and the number of compounding periods. The formula you would use to calculate the total interest if it is compounded is P[(1+i)^n-1].

How long will it take you to double your money if you invest it today and earn 10% per year?

At 10%, you could double your initial investment every seven years (72 divided by 10). In a less-risky investment such as bonds, which have averaged a return of about 5% to 6% over the same time period, you could expect to double your money in about 12 years (72 divided by 6).