MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 3 - Comparing Alternatives Professor Anderson Spring 2012 1

Choosing Between Alternatives

5 Methods (that we will cover) (1) Present Worth Method (2) Annual Cost Method (3) Capitalized Cost Method (4) Benefit Cost Ratio Method (5) Rate of Return 2

Present Worth Method (Equal Lives)

Two uses:  comparison of alternatives  placing valuation on prospective receipts (What would you be willing to pay now for the earnings you expect to make later?) Calculate the PW of each alternative:  the greatest PW (investments)  smallest PW (costs) 3

Present Worth Example

Three ideas to reduce labor costs are proposed A - Continue with current method B - Build Equipment to reduce labor costs C - Purchase New Equipment Annual Disbursements Initial Costs Salvage Value A $ 9,200.00

$ $ Use i = 10% and n = 10 yrs B $ 6,400.00

$ 15,000.00

$ C $ 5,625.00

$ 25,000.00

$ 5,000.00

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Unequal Lives

PW method requires that the lives of all of the options are equal .

If the alternatives have different lives assume costs are repeated. Repeat until Lowest Common Multiplier of the unequal lives.

Two excavating machines are available: Annual Disbursements Initial Costs Salvage Value Life Machine A $ 3,500 $ 11,000 $ 1,000 6 Machine B $ 3,100.00

$ 18,000.00

$ 2,000.00

9 Which one should be selected based on an interest rate of 18% per year?

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Annual Cost Method

Alternatives that have unequal lives are more easily compared using the annual cost method.

Assumes each alternative will be replaced by an identical twin at the end of its life. (infinite renewal). EXAMPLE: Which of the following alternatives is superior over a 30 year period if the IR is 7%?

Alt. A Alt. B Life, years 30 10 Cost O&M per year $1800 $5 $450 $20 6

Annual Cost Method

Life, years Cost O&M per year Alt. A 30 $1800 $5 Alt. B 10 $450 $20 A A A B = 1800(A/P, 7%, 30) + 5 = 150 = 450(A/P, 7%, 10) + 20 = 84 Alternative B is better because it’s annual cost is lower than that of Alternative A.

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Capitalized Cost Method

   The present worth of a project with an infinite life is called the capitalized or life cycle cost.

It is the amount of money required at time zero to perpetually support the project on the earned interest only.

CC = Initial Cost + Annual Costs/i 8

Capitalized Cost Example

Compare the following machines using 18% per year i = First Cost Annual Operating Costs Salvage Value Overhaul after 6 years Life, years Machine X $50,000 62,000 10,000 --- 7 Machine Y $200,000 24,000 0 4,000  9

Benefit Cost Method

Generally used for Public Projects B = Benefits = Cash Inflows C = Costs = Cash Outflows Want B-C > 0 or B/C > 1 10

Rate of Return Method

It is often useful to calculate the rate of return on an investment.

Solve for the interest rate that equates the worth of cash inflows to that of cash outflows ==> Result is IRR - Internal Rate of Return Set up PW or AW = 0 and solve for i.

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Rate of Return Example

If you pay me $15,000 now and I promise to pay you back in 10 end of year payments of $1,250 what is your rate of return?

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Rate of Return Example

Okay here’s a better deal: If you pay me $15,000 now and I promise to pay you back in 10 end of year payments of $2,800 what is your rate of return?

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