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Engineering Economics: Present Worth Analysis Techniques

Learn about Present Worth Analysis and its application in Engineering Economics. Understand NPW formula, mutually exclusive alternatives, and decision-making based on costs, benefits, and interest rates.

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Engineering Economics: Present Worth Analysis Techniques

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  1. Chapter 5Present Worth Analysis EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS

  2. Three Economic Analysis Methods There are three major economic analysis techniques: • Present Worth Analysis • Annual Cash Flow Analysis • Rate of Return Analysis This chapter discusses the first techniques

  3. Chapter Contents • Economic Criteria • Considering Project Life • Net Present Worth • Applying Present Worth Techniques • Useful Lives Equal the Analysis Period • Useful Lives Different from the Analysis Period • Infinite Analysis Period: Capitalized Cost • Multiple Alternatives • Spreadsheet Solution

  4. Economic Criteria • Depending on situation, the economic criterion should be chosen from one of the following 3: Engineering Economics

  5. Analysis Period • Specific time period, same for each alternative, called the analysis period, planning horizon, or project life • Three different analysis-period situations may be considered: • All alternatives have the same useful life: Set it as the analysis period. • Alternatives have different useful lives: Let the analysis period equal the least common multiple, or some realistic time (based on needs). • Infinite analysis period, n=∞ Engineering Economics

  6. Net Present Worth (NPW or PW) • Here is the basic NPW formula: PW = PW of benefits – PW of cost Engineering Economics

  7. Present Worth Techniques • Mutually exclusive alternatives: • Resolve their consequences to the present time. Engineering Economics

  8. Present Worth—Equal Useful Lives • Example:Consider two mechanical devices to install to reduce cost. Expected costs and benefits of machines are shown in the following table for each device. If interest rate is 6%, which device should be purchased? Engineering Economics

  9. A=$300 0 1 2 3 4 5 i=6% P= $1000 Example Continues Engineering Economics

  10. $500 $450 $400 $350 $300 0 1 2 3 4 5 i=6% P= $1350 Example Continues Engineering Economics

  11. A=$300 0 1 2 3 4 5 i=6% P= $1000 $500 $450 $400 $350 $300 0 1 2 3 4 5 i=6% P= $1350 Example Continues Work 5-4 DEVICE Bhas the larger present worth & is the preferred alternative Engineering Economics

  12. Present Worth—Equal Useful Lives • Example: Consider two investments with expected costs and benefits shown below for each investment. If investments have lives equal to the 5-year analysis period, which one should be selected at 10% interest rate? Engineering Economics

  13. Example Continues Engineering Economics

  14. Example Continues Engineering Economics

  15. Example Continues Salvage value is considered as another positive cash flow. Since criterion is to maximize PW (= present worth of benefits – present worth of costs), the preferred alterative is INVESTMENT1 Engineering Economics

  16. Alternatives with different Useful Lives • Example: Consider two new equipments to perform desired level of (fixed) output. expected costs and benefits of machines are shown in the below table for each equipment. If interest rate is 6%, which equipment should be purchased? Engineering Economics

  17. Example Continues • One method to select an analysis period is the least common multiple of useful lives. EQUIPMENT A $200 $200 Original Equipment A Investment Replacement Equipment A Investment 0 1 2 3 4 5 6 7 8 9 10 $1500 $1500 Engineering Economics

  18. Question Continues EQUIPMENT B $350 Original Equipment B Investment 0 1 2 3 4 5 6 7 8 9 10 $1600 Engineering Economics

  19. Question Continues EQUIPMENT A EQUIPMENT B For fixed output of 10 years of service of equipments, Equipment B is preferred because it has a smaller cost. Engineering Economics

  20. Present Worth-Useful Lives are Different from the Analysis Period • Example: Consider two alternative production machines with expected initial costs & salvage values shown below. If interest rate is 10%, compare these alternatives over a (suitable) 10-year analysis period (by using the present worth method)? Engineering Economics

  21. Example Continues $8,000 $15,000 $40,000 MACHINE A 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 7-year life 7-year life $40,000 Engineering Economics

  22. Example Continues $10,000 MACHINE B 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 13-year life $65,000 Engineering Economics

  23. Example Continues MACHINE A MACHINE B For fixed output of 10 years of service of equipments, Machine A is preferred because it has a smaller cost. Engineering Economics

  24. Infinite Analysis Period (Capitalized Cost) • Capitalized cost is the present sum of money that is set aside now at a given interest rate to yield the funds (future interest earned) required to provide the service indefinitely. (5-2) Engineering Economics

  25. Infinite Analysis Period (Capitalized Cost) • Example: How much should one set aside to pay $1000 per year for maintenance on an equipment if interest rate is 2.5% per year and the equipment is kept in service indefinitely (perpetual maintenance)? Engineering Economics

  26. Multiple (3+) Alternatives • Question: Cash flows (costs and incomes) for three pieces of construction equipments are shown below. For 10% interest rate, which alternative should be selected? Engineering Economics

  27. Question Continues $1000 $850 $700 $550 $400 $400 $400 $400 0 1 2 3 4 5 6 7 8 EQUIPMENT 1 $2000 Engineering Economics

  28. Question Continues $700 $300 $300 $300 $300 $400 $500 $600 0 1 2 3 4 5 6 7 8 EQUIPMENT 2 $1500 Engineering Economics

  29. Question Continues $650 $500 $500 $550 $600 $700 $500 $500 0 1 2 3 4 5 6 7 8 To maximize NPW, choose EQUIPMENT 1 EQUIPMENT 3 $3000 Engineering Economics

  30. Question Continues (MS EXCEL) Use function: npv(rate, value range) - Return the net present value of a series of future cash flows “value range” at interest “rate”/period. rate = interest rate per period value range = the cash flow values Engineering Economics

  31. Question Continues (MS EXCEL) Engineering Economics

  32. Problem 5-15 Solutioni = 12% P = $980,000 purchase cost F = $20,000 salvage value after 13 years A = $200,000 annual benefit for 13 years PW = –P + A(P/A, 0.12, 13) + F(P/F, 0.12, 13) = –980000 + 200000(6.424) + 20000(0.2292) = $309,384 As PW > 0, purchase the machine. Or using MS EXCEL PW = -P + pv(0.12, 13, -200000, -20000) = $309,293.17 Terms A(P/A, 0.12, 13) and F(P/F, 0.12, 13) are combined! Engineering Economics

  33. Problem 5-23 Solutioni = 18%/12 = 1.5% per month A = $500 payment/month n = 36 payments P = ? price of a car she can afford P = A(P/A, 0.015, 36) = 500(27.661) = $13,831 What is P, if r = 6%? i = 6%/12 = 0.5% P = pv(0.005, 36, -500) = $16,435.51 Do Problems 5-24, 5-25, 5-26! Engineering Economics

  34. Problem 5-41 Outputs: 2000 lines for years 1~10 4000 lines for years 21~30 i = 10% per year, cables last for at least 30 yrs Option 1: 1 cable with capacity of 4000 lines Cost: $200k with $15k annual maintenance cost Option 2: 1 cable with capacity of 2000 lines now 1 cable with capacity of 2000 lines in 10 years Cost: $150k with $10k maintenance cost/year/cable (a) Which option to choose? (b) Will answer to (a) change if 2000 additional lines are needed in 5 years, instead of 10 years? Engineering Economics

  35. Problem 5-41 Solution (a) Present worth of cost for option 1 PW 0f cost = $200k + $15k(P/A, 10%, 30) = $341,400 Present worth of cost for option 2: PW of cost = $150k + $10k(P/A, 10%, 30) + $150k(P/F, 0.1, 10) + $10k(P/A, 0.1, 20)(P/F, 0.1, 10) = $334,900 Select option 2, as it has a smaller PW of cost. Engineering Economics

  36. Problem 5-41 Solution (b) Cost for option 1 will not change. PW 0f cost = $341,400 Present worth of cost for option 2: PW of cost = $150k + $10k(P/A, 10%, 30) + $150k(P/F, 0.1, 5) + $10k(P/A, 0.1, 25)(P/F, 0.1, 5) = $394,300 Therefore, the answer will change to option 1. Engineering Economics

  37. End of Chapter 5 Engineering Economics

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