# 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

### Comments for Decay and Growth Factors

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

 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