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MIE 754 - Class #5 Manufacturing & Engineering Economics

MIE 754 - Class #5 Manufacturing & Engineering Economics. Concerns and Questions Quick Recap of Previous Class Today’s Focus: Chap 3 Comparing Alternatives with Different Useful Lives Chap 4 Rate of Return Methods Hmwk #3 Due in 1 Week: Chap 2 - 3, 4, 5, 6, 11, 12, 15,

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MIE 754 - Class #5 Manufacturing & Engineering Economics

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  1. MIE 754 - Class #5 Manufacturing & Engineering Economics • Concerns and Questions • Quick Recap of Previous Class • Today’s Focus: • Chap 3 Comparing Alternatives with Different Useful Lives • Chap 4 Rate of Return Methods • Hmwk #3 Due in 1 Week: • Chap 2 - 3, 4, 5, 6, 11, 12, 15, 16, 24, 30, 37

  2. Concerns and Questions?

  3. Quick Recap of Previous Class • Effective Interest • Comparison of Alternatives • Procedure • Defining Investment Alternatives • Useful Life versus Study Period

  4. Useful Life versus Study Period • Comparison must be over the same study period for ALL alternatives! • Useful Lives = Study Period • Useful Lives are Different Among Alters. • UL < SP • UL > SP

  5. Same Study Period Required!! • Use either: • Repeatability Assumption • Cotermination Assumption

  6. Example A B Capital Investment -$3,500 -$5,000 Annual Revenues 1,900 2,500 Annual Expenses -645 -1,020 Useful Life 4 6 Market Value 0 0

  7. Several In-class Examples

  8. The Effect of Compounding • Benjamin Franklin, according to the American Bankers Association, left $5,000 to the residents of Boston in 1791, with the understanding that it should be allowed to accumulate for a hundred years. By 1891 the $5,000 had grown to $322,000. A school was built, and $92,000 was set aside for a second hundred years of growth. In 1960, this second century fund had reached $1,400,000. As Franklin put it, in anticipation: "Money makes money and the money that money makes makes more money." • Question: What average interest rate per year was earned from 1791 to 1891?

  9. The “Ben Franklin” Problem Solution Given: P=$5,000 N=100 F = $322,000 Find: i'% F = P(F|P, i'%, 100) $322,000 = $5000(F|P, i'%, 100) therefore, (F|P, i'%, 100) = 64.4 From tables, (F|P, 4%, 100) = 50.5049 (F|P, 5%, 100) = 131.501 F = P(1+i')N therefore, 322 = 5(1+i')100 64.4 = (1+i')100 or (1+i') = 1.0425 so that i' = 4.25%

  10. Chapter 4 - Rate of Return Methods • Compare against minimum standard of desirability - minimum attractive rate of return (MARR) • Internal Rate of Return (IRR) Method Solves for the interest rate that equates the equivalent worth of a project's cash outflows (expenditures) to the equivalent worth of cash inflows (receipts or savings).

  11. IRR is Like a “Break-Even” Problem • Find i' such that FW(neg, i') = FW(pos, i') • FW = 0 = FW(pos, i') - FW(neg, i') • Can use any one of the EW methods for IRR: • PW(i' %) = 0 • AW(i' %) = 0 • FW(i' %) = 0 • Why?

  12. Can Solve for i'by: • Trial and error • Linear interpolation • An equation solver • Computer program • You will need to know how to interpolate in this course!

  13. Evaluating Project with IRR • Compare IRR to MARR to determine whether or not the project is acceptable with respect to profitability. • IRR = i'  MARR acceptable • IRR = i' < MARR unacceptable (reject)

  14. Difficulties with the IRR Method • The IRR Method assumes that recovered funds are reinvested at the IRR rather than the MARR • Possible multiple IRRs • Why should you learn the RR Methods? • The majority of companies favor the RR methods for evaluating investment projects

  15. Example Problem Using IRR Cost/Revenue Estimates Initial Investment: $50,000 Annual Revenues: $20,000 Annual Operating Costs: $2,500 Salvage Value @ EOY 5: $10,000 Study Period: 5 years MARR 20%

  16. Solution using IRR method Find i'% such that the PW(i'%) = 0 0=-50,000+17,500(P|A, i'%,5)+10,000(P|F, i'%,5) PW(20%) = 6354.50 tells us that i' > 20% PW(25%) = 339.75 > 0, tells us that i'% > 25% PW(30%) = -4,684.24 < 0, tells us that i'% < 30% 25% < i' < 30% Use linear interpolation to estimate i'% or Use “short-cut” to draw a conclusion

  17. Linear Interpolation example i% PW 25 339.75 i' 0 30 -4684.24 i' = 25.3% > MARR, accept

  18. Comparing Mutually Exclusive Alternatives (MEAs) with RR Methods • Fundamental Purpose of Capital Investment: • Obtain at least the MARR for every dollar invested. • Basic Rule: • Spend the least amount of capital possible unless the extra capital can be justified by the extra savings or benefits. (i.e., any increment of capital spent above the minimum must be able to pay its own way)

  19. Why not select the investment opportunity that maximizes IRR? See example below ABB-A() Investment -$100 -$10,000 -$9,900 Lump-Sum $1,000 $15,000 $14,000ReceiptNext Year IRR 900% 50% 41.4% If MARR = 20%, would you rather have A or B if comparable risk is involved?

  20. Comparing MEAs - using the IRR method cont'd. If MARR = 20%, PWA = $733 and PWB = $2,500 * Never simply maximize the IRR. * Never compare the IRR to anything except the MARR. IRRA->B: PWA->B = 0 = -9,900 + 14,000(P|F, i'%, 1) 9,900/14,000 = (P|F, i'%, 1) i' = 41.4% > MARR

  21. Rate of Return Methods for Comparing Alternatives MUST use an Incremental Approach! Step 1. Rank order alternatives from least to greatest initial investment. Step 2. Compare current feasible alternative with next challenger in the list Step 3. Compute RR (IRR or ERR) and compare with MARR. If RR < Marr choose the least initial investment alternative. If RR  MARR choose the greater initial investment alternative Step 4. Remove rejected alternative from list. Continue with next comparison

  22. Example Problem: Given three MEAs and MARR = 15% 123 Investment (FC) -28,000 -16,000 -23,500 Net Cash Flow/yr 5,500 3,300 4,800 Salvage Value 1,500 0 500 Useful Life 10 yrs 10 yrs 10 yrs Study Period 10 yrs 10 yrs 10 yrs Use the IRR procedure to choose the best alternative.

  23. Example Problem Cont. Step 1. DN -> 2 -> 3 -> 1 Step 2. Compare DN -> 2  cash flows  Investment -16,000 - 0 = -16,000  Annual Receipts 3,300 - 0 = 3,300  Salvage Value 0 - 0 = 0 Compute  IRRDN->2 PW(i') = 0 = -16,000 + 3,300(P|A, i'%, 10) i'DN->2 15.9%

  24. Step 3. Since i' > MARR, keep alt. 2 (higher FC) as current best alternative. Drop DN from further consideration. Step 4. Next comparison: 2 -> 3  Investment -23,500 - (-16,000) = -7,500  Annual Receipts 4,800 - 3,300 = 1,500  Salvage Value 500 - 0 = 500 Computing  IRR2->3 PW(i') = 0 0= -7,500 + 1,500(P|A, i'%, 10) + 500(P|F, i'%, 10)

  25. i'2->3 15.5% Since i' > MARR, keep Alt. 3 (higher FC) as current best alternative. Drop Alt. 2 from further consideration. Next comparison: 3 -> 1  cash flows  Investment -28,000 - (-23,500) = -4,500  Annual Receipts 5,500 - 4,800 = 700  Salvage Value 1,500 - 500 = 1,000 Compute  IRR3->1 PW(i') = 0 0= -4,500 + 700(P|A, i'%, 10) + 1,000(P|F, i'%, 10) i'3->1 10.9%

  26. Since i' < MARR, keep alt. 3 (lower FC) as current best alternative. Drop alt. 1 from further consideration. Step 5. All alternatives have been considered. Recommend alternative 3 for investment.

  27. Graphical Interpretation of Example

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