Chap08 - 10/06/09 Chapter 8-Engineering Economics 1...

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Unformatted text preview: 10/06/09 Chapter 8-Engineering Economics 1 Definition of an Engineer An engineer is someone who can do for $1 what any fool can do for $2. 10/06/09 Chapter 8-Engineering Economics 2 Fundamental Problem of Engineering Economics Trade offs between initial costs and downstream costs 10/06/09 Chapter 8-Engineering Economics 3 Components of Life Cycle Costs ($10 3 ) Ajax Blaylock ($1000) ($1000) Initial Cost 30.00 20.00 Rebuilding at End of 3rd Year - 3.00 Salvage Value -4.00 - Maintenance (5 years) 5.00 10.00 Productivity Benefit (5 years) -2.50 - Electricity year 1 3.00 3.50 year 2 3.15 3.68 year 3 3.31 3.86 year 4 3.47 4.05 year 5 3.65 4.25 Total Costs 45.08 52.34 Table 8.1 10/06/09 Chapter 8-Engineering Economics 4 First Cost Decision Rule Ajax = 30.00 Blaylock = 20.00 Choose Blaylock since it has the least first cost 10/06/09 Chapter 8-Engineering Economics 5 Total Life-Cycle Cost Decision Rule Ajax = 45.08 Blaylock = 52.34 Choose Ajax since it has the least total life-cycle cost 10/06/09 Chapter 8-Engineering Economics 6 Annual Rate of Return Decision Rule Incremental Investment = Initial Cost of More Expensive Option - Initial Cost of Less Expensive Option = 30.00 - 20.00 = 10.00 Downstream Benefits = Avoidance of Rebuilding Cost + Income from Salvage + 5 Years Savings on Maintenance Costs + Productivity Enhancement + 5 Years Savings on Electricity Costs = 3.00 + 4.00 + (10.00 - 5.00) + 2.50 + (3.50 - 3.00) + (3.68 - 3.15) + (3.86 - 3.31) + (4.05 - 3.47) + (4.25 - 3.65)= 17.26 Average Annual Benefit = Downstream Benefits/Project Duration = 17.26/5 = 3.45 Annual Rate of Return = [Annual Benefits / Incremental Investment] 100 = 25 %. Choose Ajax unless another opportunity for investing $10,000 provides a greater rate of return. 10/06/09 Chapter 8-Engineering Economics 7 Payback Period (time to recover incremental investment) Benefits in 1st Year = 1.00 + 0.50 + 0.50 = 2.00 (maint) + (prod) + (elec) Total Benefits in 2 Years = 2.00 + 1.00 +0.50 +0.53 = 4.03 (yr 1) + (maint) + (prod) + (elec) Total Benefits in 3 Years = 4.03 + 1.00 +0.50 +0.55 + 3.0 = 9.05 (yr 2) + (maint) + (prod) + (elec)+ (rebuild) Total Benefits in 4 Years = 9.05 + 1.00 +0.50 +0.58 = 11.13 (yr 3) + (maint) + (prod) + (elec) The payback period is 3.5 years. Select Ajax unless another opportunity to invest the $10,000 provides a quicker payback. 10/06/09 Chapter 8-Engineering Economics 8 Trade-offs Between Capital Cost and Operating Cost: Ajax vs. Blaylock Motors ($10 3 )* Ajax Blaylock Initial Cost 30.00 20.00 Maintenance (annual cost each year for 5 years) 1.00 3.50 10/06/09 Chapter 8-Engineering Economics 9 Total Life-Cycle Cost Decision Rule* Ajax = 30 + 5(1.00) = 30 + 5 = 35 Blaylock = 20 + 5(3.5) = 20 + 17.5 = 37.5 Choose Ajax since it has the least total life-cycle cost 10/06/09 Chapter 8-Engineering Economics 10 Annual Rate of Return Decision Rule* Incremental Investment = Initial Cost of More Expensive Option - Initial Cost of...
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Chap08 - 10/06/09 Chapter 8-Engineering Economics 1...

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