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Decay and Growth Factors

by Lee Van Beek
(Cincinnati, Ohio)










































I am taking a course in managerial math and I can't seem to get the basics of finding a growth or decay facor for a change in data

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Mar 18, 2012
Decay and Growth Factors
by: Staff


Part I

Question:

by Lee Van Beek
(Cincinnati, Ohio)


I am taking a course in managerial math and I can't seem to get the basics of finding a growth or decay factor for a change in data


Answer:

When the rate of change in your data from one period to the next is a constant percentage (%), the function which describes that change is call an exponential function.


It has the form:

f(t) = (initial value)*(1 + rate of change)^t


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Definitions:


t = time (the time could be the number of days, the number of quarters, the number of years, etc.)


Growth Factor:

(1 + rate of change) is called the “growth factor” if its value is greater than 1.


Decay Factor:

(1 + rate of change) is called the “decay factor” if its value is less than 1, but greater than 0.



(1 + rate of change)^t = (1 + rate of change) raised to the “t” power


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Example using a “growth factor”:

Since you are currently enrolled in a course emphasizing managerial math, you will be applying the exponential function to problems which involve compound interest.

As an example, suppose you deposit $250 in a savings account which earns 5% annual interest that is compounded yearly.

How much money will be in the savings account at the end of 5 years?

The exponential function states:

f(t) = (initial value)*(1 + rate of change)^t


However, when interest rate problems are involved, the function is normally written as follows:

A = P*(1 + r)^t

A = final balance in the savings account
P = principle (the initial deposit in the bank)
r = decimal form of the interest rate compounded each time period
t = time periods

P = $250
r = .05 (this is the decimal form of 5%. It is = 5%÷100)
t = 5 years


A = P*(1 + r)^t

A = $250*(1 + .05)^5


>>> Note: (1 + .05) = 1.05. 1.05 is greater than 1. Therefore, 1.05 is called the “growth factor”.


A = $250*(1.05)^5

A = $250*(1.2762815625)

A = $319.070390625

A = $319.07 (rounded to the nearest penny)

The final balance in the savings account is: A = $319.07


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Mar 18, 2012
Decay and Growth Factors
by: Staff


------------------------------------------------------------

Part II



Example using a “decay factor”:


Suppose you purchase a new piece of manufacturing equipment for $10,000.

If the value of the equipment depreciates at a rate of 12% per year (accelerated depreciation), how much will the equipment be worth after 5 years?


The exponential function states:

f(t) = (initial value)*(1 + rate of change)^t


However, you may see this equation rewritten as follows:

V = P*(1 + R)^t

V = future value of equipment
P = present value of equipment
R = depreciation rate
t = time periods

P = $10,000
R = -.12 (this is the decimal form of 12%. It is = 12%÷100)
t = 5 years


V = P*(1 + R)^t

V = $10,000*(1 - .12)^5


>>> Note: (1 - .12) = 0.88. 0.88 is less than 1. Therefore, 0.88 is called the “decay factor”.


V = $10,000*(.88)^5

V = $10,000*(0.5277319168)

V = $5,277.319168

V = $5,277.32 (rounded to the nearest penny)

The value of the equipment after 5 years is: V = $5,277.32



Thanks for writing.

Staff
www.solving-math-problems.com



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