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Conversion Factor for Geometric Gradient Series

In document Engineering Economy (Page 86-90)

A geometric gradient series is a series of cash flows that increase or decrease by a constant percentage each period. The geometric gradient series may be used to model inflation or deflation, productivity improvement or degradation, and growth or shrinkage of market size, as well as many other phenomena.

In a geometric series, the base value of the series is A and the “growth” rate in the series (the rate of increase or decrease) is referred to as g. The terms in such a series are given by A, A(1  g), A(1  g)2, . . . , A(1  g)N–1 at the ends of periods 1, 2, 3, . . . , N, respectively. If the rate of growth, g, is positive, the terms are increas-ing in value. If the rate of growth, g, is negative, the terms are decreasincreas-ing. Figure 3.7 shows a series of receipts where g is positive. Figure 3.8 shows a series of receipts where g is negative.

The geometric gradient to present worth conversion factor, denoted by (P/A,g,i,N), gives the present worth, P, that is equivalent to a geometric gradient series where the base receipt or disbursement is A, and where the rate of growth is g, the interest rate is i, and the number of periods is N.

0 1 2 N⫺2 N

A

N⫹1 A(1⫹g) A(1⫹g)2

3 4 N⫺1

A(1⫹g)3 A(1⫹g)(N⫺3)

A(1⫹g)(N⫺2)

A(1⫹g)(N⫺1)

Figure 3.7 Geometric Gradient Series for Receipts With Positive Growth

The present worth of a geometric series is

where A the base amount g the rate of growth

i the interest rate N the number of periods

P the present worth Figure 3.8 Geometric Gradient Series for Receipts With Negative Growth

Estimating Growth Rates

Geometric gradient series can be used to model the effects of inflation, deflation, production rate change, and market size change on a future cash flow. When using a geometric gradient series, the relevant growth rate must be estimated.

The internet can be a useful research tool for collecting information such as expert opinions and statistics on trends for national and international activities by product type and industry. For exam-ple, a sales growth rate may be estimated by

considering a number of factors: economic condi-tion indicators (e.g., gross domestic product, employment, consumer spending), population growth, raw material cost, and even online shop-ping and business-to-business trading. Federal governments are keen to assess what has hap-pened in order to predict what will happen. In Canada, such information is collected by Statistics Canada and is available to the public online at www.statcan.gc.ca.

N E T V A L U E 3 . 1

We can define a growth-adjusted interest rate, i°, as

Then the geometric gradient series to present worth conversion factor is given by (P/A,g,i,N) (P/

Care must be taken in using the geometric gradient to present worth conversion factor. Four cases may be distinguished:

1. i > g > 0. Growth is positive, but less than the rate of interest. The growth-adjusted interest rate, i°, is positive. Tables or functions built into software may be used to find the conversion factor.

2. g > i > 0. Growth is positive and greater than the interest rate. The growth-adjusted interest rate, i°, is negative. It is necessary to compute the conversion factor directly from the formula.

3. g i > 0. Growth is positive and exactly equal to the interest rate. The growth-adjusted interest rate, i°, equals zero. As with any case where the interest rate is zero, the present worth of the series with constant terms, A/(1 g ), is simply the sum of all the N terms:

PN1Ag

4. g < 0. Growth is negative. In other words, the series is decreasing. The growth-adjusted interest rate, i°, is positive. Tables or functions built into software may be used to find the conversion factor.

EXAMPLE 3.7

Tru-Test is in the business of assembling and packaging automotive and marine testing equipment to be sold through retailers to “do-it-yourselfers” and small repair shops.

One of its products is tire pressure gauges. This operation has some excess capacity.

Tru-Test is considering using this excess capacity to add engine compression gauges to its line. It can sell engine pressure gauges to retailers for $8 per gauge and expect to be able to produce about 1000 gauges in the first month of production. It also expects that, as the workers learn how to do the work more efficiently, productivity will rise by 0.25 percent per month for the first two years. In other words, each month’s output of gauges will be 0.25 percent more than the previous month’s. The interest rate is 1.5 percent per month. All gauges are sold in the month in which they are produced, and receipts from sales are at the end of each month. What is the present worth of the sales of the engine pressure gauges in the first two years?

We first compute the growth-adjusted interest rate, i°:

We then make use of the geometric gradient to present worth conversion factor with the uniform cash flow A $8000, the growth rate g  0.0025, the growth-adjusted interest rate i° 0.0125, and the number of periods N  24.

P A(P/A,g,i,N)  A(P/1A,i°g,N) PV(i,N,A)/(1  g)

P 8000冢(P/A,11.0.2052%5,24)

From the interest factor tables we get P 8000冢210.0.602245

P 164 580

The present worth of sales of engine compression gauges over the two-year period would be about $165 000. Recall that we worked with an approximate growth-adjusted interest rate of 1.25 percent when the correct rate was a bit less than 1.25 percent. This means that $164 580 is a slight understatement of the present worth. ______________

EXAMPLE 3.8

Emery’s company, Dry-All, produces control systems for drying grain. Proprietary technology has allowed Dry-All to maintain steady growth in the U.S. market in spite of numerous competitors. Company dividends, all paid to Emery, are expected to rise at a rate of 10 percent per year over the next 10 years. Dividends at the end of this year are expected to total $110 000. If all dividends are invested at 10 percent interest, how much will Emery accumulate in 10 years?

If we calculate the growth-adjusted interest rate, we get 1

and it is natural to think that the present worth is simply the first year’s dividends multiplied by 10. However, recall that in the case where g i, the present worth is given by

PN1 Ag 101110.1000 1 000 000

Intuitively, dividing by (1 g) compensates for the fact that growth is considered to start after the end of the first period, but the interest rate applies to all periods. We want the future worth of this amount after 10 years:

F 1 000 000(F/P,10%,10)  1 000 000(2.5937)  2 593 700

Emery will accumulate $2 593 700 in dividends and interest. _____________________

In document Engineering Economy (Page 86-90)