Economics Module 4 - Future and Annual Worth Analysis PDF

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Economics Module 4 - Future and Annual Worth Analysis...

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Co-op Course Academic Component Engineering Economics Module 4 Evaluating the Economic Feasibility of an Engineered Solution1 The goal of this module is to continue the introduction of some basic concepts of engineering economics. Concepts that will be discussed include:  Future Worth Method or Future Value  Annual Worth Method and Equivalent Uniform Annual Cost These concepts are used to evaluate the economic feasibility of an engineered solution. You should review Engineering Economics Module 1, Module 2, and Module 3 to make sure you are familiar with that material prior to continuing with this module. Future Worth Analysis The objective in all time value of money methods is to maximize future wealth, and therefore, the future worth (FW) is very useful in capital investment decision situations. The FW of a series of cash flows is its equivalent value at the end of year, n, at an interest rate that is typically the MARR. Future worth of a project is equivalent to its present worth as follows: FW(%) = PW x (F given P, i%, n)

(1)

Furthermore, n

FW (i%)   Fk (1  i) n k

(2)

k 0

Finally, If FW (i = MARR) ≥ 0, then the project is economically justified

(3)

1

Content of this module is largely excerpted from Chapter 5 of Engineering Economy (fifteenth edition) by Sullivan, Wicks and Koelling (2012) and Chapters 6 and 7 of Engineering Economics by Sepulveda, Souder and Gottfried (1984).

Engineering Economics 4: Future and Annual Worth - Plouff

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Example 1: Referring back to Example 1 of Engineering Economics Module 3, a piece of new equipment has been proposed to increase the productivity of a manual process. The investment cost is $50,000, and the equipment will have a market value of $15,000 at the end of five years. Increased productivity attributable to the equipment will amount to $12,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the company’s MARR is 18% per year, what is the future worth of the project? $15,000

$12,000

1

$12,000

2

$12,000

$12,000

3

4

$12,000

5

End of Year (i = 18%)

$50,000 FW = FW of cash inflows – FW of cash outflows FW(18%) = $12,000 x (F given A, 18%, 5) - $50,000 x (F given P, 18%, 5) + $15,000 FW (18%)  $12,000

(1  0.18) 5  1  $50,000(1  0.18) 5  $15,000 0.18

FW = -$13,537.4 Since FW(18%) < 0, this equipment would not be economically justified. This agrees with the present worth analysis in Example 1. Also, per Equation 4, FW should equal PW*(F given P, i%, n): Engineering Economics 4: Future and Annual Worth - Plouff

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FW (18%)  $5,917.31(1  0.18) 5  $13,537.4

Example 2: Determine the future worth of the following series of cash flows, based on an interest rate of 15% per year, compounded annually: $0 (end of year 0), $2,000 (end of year 1), $4,000 (end of year 2), $6,000 (end of year 3), $8,000 (end of year 4), $10,000 (end of year 5), $10,000 (end of year 6), $10,000 (end of year 7). Notice that end of year 1 through year 5 is a gradient series and end of year 6 through year 7 is a uniform series.

$10,000 $10,000 $10,000 $8,000 $6,000 $4,000 $2,000

1

2

3

4

5

6

7

End of Year (i = 15%)

FW(15%) = (F given G, i%, n) + (F given A, i%, n)

FW (15%)  A

(1 i ) n  1 1  (1  i) n  1  (1  i ) n  1 G   n  A i i i i 

FW (15%)  $2,000

(1  0.15) 5  1 1  (1  0.15) 5 1  (1  0.15) 2  1  $2,000   5 $ 10 , 000  0.15 0.15  0.15 0.15 

FW(15%) = $58,216.51

Engineering Economics 4: Future and Annual Worth - Plouff

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Annual Worth Analysis The annual worth (AW) method establishes an equal annual series of dollar amounts, for a stated period of time, and is equivalent to the cash inflows and outflows at a given interest rate that is typically the MARR. AW is the annual equivalent revenues or savings, R, minus annual equivalent expenses, E, less the annual equivalent capital recovery, CR. AW(i%) = R – E – CR(i%)

(4)

Also, the AW of a project is equivalent to its PW and FW, such that: AW(i%) = PW (A given P, i%, n) and AW(i%) = FW (A given F, i%, n) If AW (i = MARR) ≥ 0, then the project is economically justified.

(5)

The AW method is often used to compare alternatives with unequal project time periods. It also can be used to rank alternatives according to their desirability. The alternatives must be mutually exclusive and repeatedly renewed up to the duration of the longest time period alternative. The calculated annual cost is often called the equivalent uniform annual cost (EUAC) and the equivalent uniform annual series (EUAS) for cash inflows. The CR amount for a project is the equivalent uniform annual cost of the capital invested. It is an annual amount that covered the loss in value of the asset, and interest on invested capital. One method of determining the CR is: CR(i%) = I(A given P, i%, n) – S(A given F, i%, n) Where:

(6)

I = the initial investment for the project S = the salvage (or market) value at the end of the project time period

In some case, the initial investment, I, is spread over several periods in which case the PW of all investment amounts should be used.

Engineering Economics 4: Future and Annual Worth - Plouff

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Example 3: Machine A and Machine B are being considered for a 10-year service in an industrial facility. The MARR is 8%. What are the equivalent uniform annual costs of each machine, respectively, and which is the better economic choice?

Initial cost Salvage Value Annual Operating Cost Annual Repair Cost

Machine A $5,000 $0 $1,500 $750

Machine B $10,000 $2,000 $1,000 $500

Machine A

1

2

$2,250

$2,250

End of Year (i = 8%) 3

$2,250

4

$2,250

….

$2,250

10

$2,250

$5,000 EUAC(i%) = E + [I(A given P, i%, n) – S(A given F, i%, n)] 0.08(1  0.08) 10   i (1  i )n  EUAC (8%)  $1,500  $750  $5,000    0  $2,250  $5,000 n 10  (1  0.08)  1   (1 i)  1  EUAC(8%) = $2,995.15

Engineering Economics 4: Future and Annual Worth - Plouff

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Machine B

1

2

$1,500

$1,500

End of Year (i = 8%) 3

$1,500

$2,000 4

$1,500

….

$1,500

10

$1,500

$10,000 EUAC(i%) = E + [I(A given P, i%, n) – S(A given F, i%, n)]

 i(1  i) n    i EUAC (8%)  $1,000  $500  $10,000   $2,000    n n  (1  i)  1 (1  i) 1   0.08(1  0.08)10 EUAC (8%)  $1,500  $10,000 10  (1  0.08) 1 EUAC(8%) = $2,852.24

   0.08   $2,000   10  (1 0.08)  1  

EUACMachine B [$2,852.24] < EUACMacine A [$2,995.15] Therefore, choose Machine B because it has a smaller EUAC.

Example 4: Referring back to Example 1 of Engineering Economics Module 3, a piece of new equipment has been proposed to increase the productivity of a manual process. The investment cost is $50,000, and the equipment will have a market value of $15,000 at the end of five years. Increased productivity attributable to the equipment will amount to $12,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the company’s MARR is 18% per year, is this proposal a good one considering AW?

Engineering Economics 4: Future and Annual Worth - Plouff

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$15,000

$12,000

1

$12,000

2

$12,000

$12,000

3

4

$12,000

5

End of Year (i = 18%)

$50,000

AW(i%) = R – E – CR(i%) AW(18%) = $12,000 – [ I x (A given P, 18%, 5) - S x (A given F, 18%, 5)]

 i (1  i ) n    i  $15,000 AW (18%)  $12,000  $50,000   n n  (1  i)  1  (1  i)  1     0.18(1  0.18) 5  0.18 AW (18%)  $12,000  $50,000  $15,000   5 5  (1  0.18) 1   (1  0.18) 1  AW(18%) = -$1,892.2 Since AW(18%) < 0, this equipment would not be economically justified. This agrees with the present worth analysis in Example 1 of Engineering Economics Module 3 and the future worth analysis in Example 1 of this module.

Engineering Economics 4: Future and Annual Worth - Plouff

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Summary of Formulas for Cash Flow Patterns and Interest Factors Table 1 provides a summary of the formulas used for cash flow patterns and interest factors discussed throughout this module.

To Find F

Table 1: Formulas for Cash Flow Patterns and Interest Factors Given Formula P

P

F

A

P

P

A

A

F

F

A

A

G

F

G

P

G

Engineering Economics 4: Future and Annual Worth - Plouff

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