Economics Module 3 - Present Worth Analysis PDF

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Economics Module 3 - Present Worth Analysis...

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Co-op Course Academic Component Engineering Economics Module 3 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:  Minimum Attractive Rate of Return  Present Worth Method or Present Value  Capitalized Worth Method or Capitalized Equivalency These concepts are used to evaluate the economic feasibility of an engineered solution. You should review Engineering Economics Module 1 and Module 2: Basic Concepts to make sure you are familiar with that material prior to continuing with this module. Minimum Attractive Rate of Return: The minimum attractive rate of return (MARR), also called the target rate, hurdle rate, cut-off rate or valuation rate, is the rate by which a project is evaluated when an interest rate is not known. It is usually chosen to maximize the economic viability of an organization when considering a project, and accounts for various risk factors. The MARR is typically established based on company policy and may be project specific. The MARR may take into account some of the following:  The amount of money available for investment, and the source and cost of these funds  The number of good projects available for investment and their purpose  The amount of perceived risk associated with investment opportunities available  The type of organization involved (e.g., government, public utility, private industry) (Sullivan, et.al. 2012). Present Worth Analysis Present worth (PW) is the concept of the equivalent worth of all cash flows at the end of year 0 – or, another way to look at it, at the beginning of year 1. All cash inflows and outflows are discounted to the present point in time at an interest rate (typically chosen as the MARR). Present worth as a function of interest rate for a series of cash flows is found as follows: n

PW (i%)   Fk (1  i) k

(1)

k 0

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 3: Present Worth - Plouff

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Where: i = effective interest rate, or MARR, per compounding period k = index for each compounding period (0≤k≤N) Fk = future cash flow at end of period k n = number of compounding periods in the study period This equation assumes there is a constant interest rate throughout the life of the project. This would have to be adjusted if the interest rate varied throughout the course of the project. To apply the present worth method to determine a project’s economic worthiness, use the following: If PW (i = MARR) ≥ 0, the project is economically justified

(2)

Example 1: 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? $15,000

$12,000

1

$12,000

2

$12,000

$12,000

3

4

$12,000

5

End of Year (i = 18%)

$50,000 PW = PW of cash inflows – PW of cash outflows Engineering Economics 3: Present Worth - Plouff

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PW(18%) = $12,000 x (P given A, 18%, 5) + $15,000 x (P given F, 18%, 5) - $50,000

PW (18%)  $12,000

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

PW = -$5,917.31 Since PW(18%) < 0, this equipment would not be economically justified

Example 2: A chemical company is going to install a new pipeline/pump system to connect a storage tank to a processing plant 500 m away, and is considering two options. The first system has an initial cost of $3,500, has a salvage value of $500, and costs $1.50 per hour to operate. The second system has an initial cost of $5,000, has a salvage value of $1,000, and costs $1.25 per hour to operate. Both systems have an estimated service life of 10 years and operate 800 hours per year. Assume that the company has a MARR of 12% for this project. Annual maintenance and pumping costs should be considered to be paid in their entireties at the end of the years in which their costs are incurred. Which of the two systems is the best investment using the approximate present worth of the piping/pump systems over the first 10 years of operation? System 1: End of Year (i = 12%) 1

$1,200

2

3

4

$1,200

$1,200

$1,200

….

$500 10

$1,200

$3,500

PW = PW of cash inflows – PW of cash outflows Engineering Economics 3: Present Worth - Plouff

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PW(12%) = $500 x (P given F, 12%, 10) - $1,200 x (P given A, 12%, 10) - $3,500

PW (12%)  $500

1 (1  0.12)10  1  $ 1 , 200  $3,500 (1  0.12) 10 0.12(1  0.12) 10

PW = -$10,119.3 System 2: End of Year (i = 12%) 1

$1,000

2

3

4

$1,000

$1,000

$1,000

….

$1,000 10

$1,000

$5,000

PW = PW of cash inflows – PW of cash outflows PW(12%) = $1,000 x (P given F, 12%, 10) - $1,000 x (P given A, 12%, 10) - $5,000

PW (12%)  $1000

1 (1  0.12)10  1  $ 1 , 000  $5,000 (1  0.12) 10 0.12(1  0.12) 10

PW = -$10,328.2 Since PW(12%) of System 1 [-$10,119.3] < PW(12%) of System 2 [-$10,328.2], System 1 should be selected as this equipment would cost less in present value terms. Notes: This is a simplistic approach to comparing two alternatives. Additional factors would typically be included in the analysis and will be discussed in future modules. Also, using the PW approach of comparing two alternatives works only for alternatives with equal time periods (equal n). Engineering Economics 3: Present Worth - Plouff

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Important assumptions of using the present worth (and the future worth and annual worth) model include:  the future is known with certainty (not really possible); and  money can be borrowed or lent at the same interest rate - i.e. capital markets are perfect (doesn’t usually happen in the real world). Although these models rely on these unrealistic assumptions, it is cost-beneficial to use them because the cost of using the simpler PW (or FW or AW) model is less than the benefits (incrementally improved results) of using more complex models. Rarely do more sophisticated models result in a reversal of decision of the outcomes of one of these simpler models. Capitalized Worth Analysis The capitalized worth (CW) analysis is a special case of the present worth method where all revenues and expenses are over an infinite time period. If only expenses are considered, the results obtained are sometimes called the capitalized cost. For this case in which there are ongoing end-of-period uniform payments, A, at interest rate, i, the outcome of P given A approaches 1/A as n goes to infinity. Therefore, the CW for a project is the annual equivalent of the project over its useful life divided by the interest rate, i (as a decimal). 1  CW (i %)  A   i

(3)

Example 3: Calculate the capitalized cost of a project which has an initial cost of $100,000. The annual operating cost is $5,000 for the first 3 years and $3,000 thereafter. There is a recurring $10,000 maintenance cost every 10 years. The interest rate is 10% per year.

1

2

End of Year (i = 10%) 3 4

$3,000

$3,000

$3,000

$2,000

$2,000 $2,000

$3,000

…….. 10 ……. 20 ….. $3,000

$10,000

$3,000

$10,000

$100,000 Engineering Economics 3: Present Worth - Plouff

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Consider $2,000 of the $5,000 annual operating cost for the first 3 years to be a one-time cost, leaving $3,000 annual operating costs forever. PW(10%) = - $2,000 x (P given A, 10%, 3) - $100,000

(1  0.10) 3  1  $100,000 = -$104,974 0.10(1 0.10)3 The recurring annual operating cost of $3,000 plus the equivalent annual cost of the $10,000 cost every 10 years is: PW (10%)  $2,000

A(10%) = -$3,000 - $10,000 x (A given F, 10%, 10) 0.10 i  $3,000  $10,000  $3,627.45 n 10 (1  i)  1 (1  0.10)  1 $3,627.45 CW (10%)  $104,974   $141,249 0.10 A(10%)  $3,000  $10,000

Engineering Economics 3: Present 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 3: Present Worth - Plouff

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