MER1601 MER Design of Thermal Fluid Systems Engineering Economics – Comparing Alternatives Professor Bruno Winter Term 2005

MER1602 (1) Comparing Alternatives (2) Group Meetings Today’s Plan

MER1603 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

MER1604 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)

MER1605 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 Use i = 10% and n = 10 yrs

MER1606 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: Which one should be selected based on an interest rate of 18% per year?

MER1607 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%?

MER1608 Annual Cost Method A A = 1800(A/P, 7%, 30) + 5 = 150 A B = 450(A/P, 7%, 10) + 20 = 84 Alternative B is better because it’s annual cost is lower than that of Alternative A.

MER1609 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

MER16010 Capitalized Cost Example Compare the following machines using i = 18% per year Machine XMachine Y First Cost$50,000$200,000 Annual Operating Costs 62,000 24,000 Salvage Value 10,000 0 Overhaul after 6 years ,000 Life, years 7 

MER16011 Benefit Cost Method Generally used for Public Projects B = Benefits = Cash Inflows C = Costs = Cash Outflows Want B-C > 0 or B/C > 1

MER16012 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.

MER16013 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?

MER16014 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?