EML4550 - Engineering Design Methods Engineering-Economics Introduction Economic

EML4550 - Engineering Design Methods Engineering-Economics Introduction Economic

EML4550 - Engineering Design Methods Engineering-Economics Introduction Economic Decision Rules Hyman: Chapter 8 EML4550 2007 1 Overview Engineering economics: The application of economic principles to engineering problems. Decision making of engineering design must balance trade-offs among cost (economic feasibility) and performance (technical accomplishments) in the most economical manner. A sound economical analysis should involve structured process and mathematical modeling techniques in order to meet design needs. EML4550 -- 2007 Economic Decision Criteria (from the simplest to the comprehensive) Simple criteria 1. Lowest initial cost (minimize capital cost) 2. Lowest overall life-cycle cost (no time value of money, regardless of when these costs occur) 3. Average annual rate of return (investment decision) 4. Payback period

Time value of money? Present Value Analysis Present value Annualized cost Unequal lifetimes analysis o o Common multiple of lifetimes Annualized cost of ownership and operation EML4550 -- 2007 Example: Ajax vs. Blaylock motors (Hyman Sec. 8.2) Need to decide between two brands/types of motors Ajax will cost $30k and require $1k/yr to maintain (high quality) Blaylock will cost $20k and require $2k/yr to maintain (lower quality) Furthermore, Blaylock requires an additional $3k to rebuild after 3 years and has no resale value after 5 years, Ajax can be re-sold after 5 years for $4k (salvage value) Ajax being a better motor, will decrease wear and tear on other systems, so factory will accrue $500/yr in savings on other systems (gain in productivity) Ajax uses $3k/yr in electricity, Blaylock uses $3.5/yr in electricity Electricity rates increase by 5% each year (inflation) Ajax or Blaylock, which one should I buy? EML4550 -- 2007 Example: Ajax vs. Blaylock motors

(Hyman Sec. 8.2) ETU Ajax Blaylock Initial Cost 30 20 Rebuild at end of year 3 -- 3 Salvage value -4 -- 5 10 - 2.5 --

Year 1 3 3.5 Year 2 3.15 3.65 Year 3 3.31 3.86 Year 4 3.47 4.05 Year 5 3.65 4.25 45.08 52.34 Maintenance (5 years)

Productivity benefit (5 yrs) Electricity (O&M): Totals (No time value): EML4550 -- 2007 Criterion 1: Lowest capital cost Buy Blaylock (20 < 30) Cash-starved, no financing Unsophisticated buyer Ajax Blaylock Initial Cost 30 20 Rebuild at end of year 3 -- 3 Salvage value -4 -- Maintenance (5 years) 5

10 - 2.5 -- Year 1 3 3.5 Year 2 3.15 3.65 Year 3 3.31 3.86 Year 4 3.47 4.05 Year 5 3.65

4.25 45.08 52.34 Productivity benefit (5 yrs) Electricity (O&M): EML4550 -- 2007 Totals (No time value): Criterion 2: Lowest life-cycle cost No time value of money (simple addition of life-cycle costs) Choose Ajax (45.08 < 52.34) Slightly more sophisticated buyer Understanding of total cash flow Lacks understanding of Net Present Value (NPV) Ajax Blaylock Initial Cost 30 20 Rebuild at end of year 3 -- 3

Salvage value -4 -- Maintenance (5 years) 5 10 - 2.5 -- Year 1 3 3.5 Year 2 3.15 3.65 Year 3 3.31 3.86 Year 4

3.47 4.05 Year 5 3.65 4.25 45.08 52.34 Productivity benefit (5 yrs) Electricity (O&M): EML4550 -- 2007 Totals (No time value): Criterion 3: Average annual rate of return Assume the lowest cost option is the base case Calculate the % return on investment of paying extra money to buy the better equipment Is the % return high enough? EML4550 -- 2007 Criterion 3: Average annual rate of return Incremental investment = Initial cost (expensive) - Initial cost (cheap) = 30 - 20 = $10k Downstream benefits of Ajax motor:

Avoidance of rebuilding Income from salvage Savings on maintenance Increased in productivity revenue Savings in electricity: Year 1 Year 2 Year 3 Year 4 Year 5 +3 +4 (10-5)=+5 +2.5 (3.5-3)=+0.5 (3.65-3.15)=+0.5 (3.86-3.31)=+0.55 (4.05-3.47)=+0.58 (4.25-3.65)=+0.6 Sum of all benefits: $17.23k EML4550 -- 2007 Criterion 3: Average annual rate of return

Total benefits from extra investment (no time value of money): $17.26k Average annual benefit over five years = 17.23k/5 = $3.45k Average annual rate of return = 3.45/10 = 34.5% (assume $10k initial investment by choosing the more expensive Ajax motor) Is this attractive? Compare with corporate hurdle rate (a rate of return above which an investment makes sense and below which it does not. Also called required rate of return or profit margin.) Can the additional expenses be invested in other options to achieve higher profit margin (>34.5%)? Need benchmark data for comparison. EML4550 -- 2007 Criterion 4: Payback period Extra investment of $10k, when do I get it back? Benefits: Maintenance + Productivity + Electricity Year 1: 1 + 0.5 + 0.5 = 2 Years 1+2: 2 + 1 + 0.5 + 0.53 = 4.03 Years 1+2+3: 4.03 + 1 + 0.5 + 0.55 +3 (rebuild cost) = 9.05 Years 1+2+3+4: 9.05 + 1 + 0.5 + 0.58 = 11.13 Payback period ~ 3.5 years Attractive? Not attractive? Soon enough? A benchmark is needed to justify the economical value. EML4550 -- 2007

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