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Annual Cash Flow Analysis: Maximizing Profitability in Engineering Projects

Learn how to calculate and analyze annual cash flows for engineering projects, select the best alternatives, and use spreadsheets effectively. Explore different analysis periods and make informed decisions to maximize profits.

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Annual Cash Flow Analysis: Maximizing Profitability in Engineering Projects

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  1. Chapter 6Annual Cash Flow Analysis EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS

  2. Chapter Contents • Annual Cash Flow Calculations • Annual Cash Flow Analysis • Analysis period • Analysis period equal to alternative lives • Analysis period = a common multiple of alternative lives • Analysis period for continuing requirement • Infinite analysis period • Other analysis period • Using Spreadsheets

  3. Learning Objectives • Apply annual cash flow techniques in various situations in selecting the best alternative • Develop and use spreadsheet in solving engineering economic problems

  4. Vignette: Lowest Prices on the Net! Buy Now! • It is a common tactic for manufacturers to sell inkjet or laser printers at very low prices. Then take advantage at the time when the ink or toner cartridges need to be replaced. • Why are there so many spam or junk e-mail selling ink or toner cartridges?

  5. Vignette: Lowest Prices on the Net! Buy Now! • King Camp Gillette, inventor of the safety razor, gave his razor away free of charge. But his business revenue soared. Why? • Can Gillette’s strategy work with other products? Why or Why not? • What ethical issues do producers and marketers face in designing and selling their products? Is it true that “anything goes in business” and “caveat emptor?”

  6. 8 4 6 7 2 3 1 9 0 5 10 0 A A A A A A A A A A P=1000 Example 6-1 Annual Cash Flow A student bought $1000 worth of furniture. What is the equivalent uniform annual cost (EUAC) if it is expected to last 10 years and the interest rate is 7%?

  7. S=200  8 4 6 7 2 3 0 1 9 0 5 10 10 P=1000 Example 6-2 Annual Cash Flow A student bought $1000 worth of furniture. What is the equivalent uniform annual cost (EUAC) if it is expected to last 10 years and can be sold for $200? (i = 7%) A A A A A A A A A A

  8. S  n n n-1 4 2 3 0 1 0 A A A A A A P EUAC Formulas (6-1) (6-3) (6-4)

  9. S=200  8 4 6 7 2 3 0 1 9 0 5 10 10 A A A A A A A A A A P=1000 Example 6-2 i = 7% i = 7% or or

  10. PWCost 45 90 135 180 225 4 2 3 0 1 5 Example 6-3 i = 7%

  11. Example 6-3 by spreadsheet npv(rate, value range) - rate = interest/period - value range = cash flow values Step 1: Find the PW of the costs: PW of cost = npv(b2, b3:b7) = $531.01 Step 2: Find the EUAC: EUAC = pmt(0.07, 5, -531.01) = $129.51

  12. 45 90 135 180 225 4 2 3 0 1 5 Example 6-4 Annual Cash Flow i = 7%

  13. Annual Cash Flow Analysis

  14. 500 450 400 350 A=300 300 P=1000 P=1350 4 4 2 2 3 3 1 1 0 0 5 5 Example 6-5 Annual Cash Flow Device B Device A i = 7% i = 7% Which device should the company select?

  15. 500 450 400 350 A=300 300 P=1000 P=1350 4 4 2 2 3 3 1 1 0 0 5 5 Example 6-5 Annual Cash Flow Device B Device A i = 7% i = 7%

  16. Example 6-6 Annual Cash Flow Each of Plans A, B, and C has a 10-year life. If interest is 8%, which plan should be adopted?

  17. Example 6-6 Annual Cash Flow i = 8% C B C B Based on maximizing EUAW, select Plan A.

  18. Example 6-7 Annual Cash Flow i = 7% Calculate EUACA for n=12 and EUACB for n = 6: If EUACB was calculated over n = 12-year period

  19. 5 Types of Analysis Periods • There are 5 kinds of analysis-period situations in Annual Cash Flow analysis (illustrated by questions 1-5). - Analysis period equal to alternative lives (QUESTION 1) - Analysis period a common multiple of alternative lives (QUESTION 2) - Analysis period for continuing requirement (QUESTION 3) - Infinite analysis period (QUESTION 4) - Another analysis period (QUESTION 5)

  20. Analysis period equal to alternative lives • Question 1: There are two devices which have useful lives of 5 years with no salvage value. the below table shows initial costs and annual cost savings for each item. with interest 12%, which device should be chosen? INITIAL COST ANNUAL COST SAVINGS

  21. Question 1 contd SOLUTION 1 To maximize EUAW, select Device A.

  22. QUESTION 1 CONTINUES SOLUTION 2 (by present worth analysis) To maximize PW, select Device A – the same conclusion!

  23. Analysis period = a common multiple of alternative lives • Question 2: Considering 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? please note that this is the same example discussed in chapter 5

  24. Question 2 Continues $200 $200 Original Equipment A Investment Replacement Equipment A Investment 0 1 2 3 4 5 6 7 8 9 10 EQUIPMENT A $1500 $1500

  25. Question 2 Continues $350 0 1 2 3 4 5 6 7 8 9 10 EQUIPMENT B $1600

  26. Question 2 Continues EQUIPMENT A EQUIPMENT B If we use n = 5 years for equipment A, we get exactly the same. EUACA = - 1500 (A/P, 6%,5) + 200 (A/F, 6%,5) = - 1500 (0.2374) + 200 (0.1774) = - 320.62 To minimize cost, we select EQUIPMENT B (closer to “0” value).

  27. Analysis period for continuing requirement • Question 3: Considering two alternative production machines with expected initial costs and salvage values of machines are shown below for each machine. If interest rate is 10%, compare these alternatives as continuing requirement.

  28. Question 3 continues • Under the assumption of identical replacement of equipments at the end of their useful lives (continuing requirement), EUAC of machine A will be compared to EUAC of machine B without taking the least common multiple of useful lives (shown below) into consideration. $8,000 $8,000 $8,000 MACHINE A 0 1 2 6 7 8 9 13 14 90 91 $10,000 $10,000 $10,000 $40,000 $40,000 $40,000 MACHINE B 0 1 2 12 13 14 15 25 26 90 91 $65,000 $65,000 $65,000

  29. Question 3 continues EUAWA = -P(A/P,i,n) + S(A/F,i,n) = -40,000(A/P,10%,7) + 8,000(A/F,10%,7) = -40,000(0.2054) + 8,000(0.1054) = -8,216 + 840.80 = -7375.20 i = 10% $8,000 MACHINE A 0 1 2 3 4 5 6 7 7-year life $40,000

  30. Question 3 continues EUAWA = -P(A/P,i,n) + S(A/F,i,n) = -65,000(A/P,10%,13) + 10,000(A/F,10%,13) = -65,000(0.1408) + 10,000(0.0408) = -9,152 + 408 = -8,744 $10,000 MACHINE B 0 1 2 3 4 5 6 7 8 9 10 11 12 13 13-year life $65,000 To minimize the cost, we select MACHINE A

  31. Infinite Analysis Period • Question4: The water will be carried via two alternatives: • A tunnel through mountain • A pipeline which goes around mountain If there is a permanent need for an aqueduct, which option should be selected at 6% interest rate?

  32. Question 4 continues • Under the assumption of the continual identical replacement of the limited life alternative, The EUAC for the infinite analysis period will be equal to the EUAC computed for limited life. From Chapter 5 For fixed output, minimize EUAC. Select the pipeline.

  33. 0 1 2 3 4 5 Think – Pair - Share Anitem was purchased for $500. If the item’s expected life is 5 years with salvage value $100 at EOY 5 what will be the equivalent uniform annual cost (EUAC) be at 6 % interest rate? $100 i=6% P=$500 A= ?

  34. 0 1 2 3 4 5 Think – Pair - Share Anitem was purchased for $500. If the item’s expected life is 5 year with salvage value $100 at EOY 5 what will be the equivalent uniform annual cost (EUAC) be at 6 % interest rate? $100 i=6% P=$500 A= $100.96

  35. 0 1 2 3 4 5 Think – Pair - Share Question: An item was purchased. The annual costs which will occur at EOY 1, EOY 2, EOY 3, EOY 4 , and EOY 5 are $50, $100, $150, $200, and $250, respectively. What will the equivalent uniform annual cost (EUAC) be at 6 % interest rate? i=6% i=6% 0 1 2 3 4 5 A=? will be converted to

  36. QUESTION CONTINUES 0 1 2 3 4 5 $50 $100 $150 $200 $250 = 0 1 2 3 4 5 $50 $50 $50 $50 $50 + 0 1 2 3 4 5 0 $50 $100 $150 $200

  37. Think – Pair - Share Other Analysis Period • An analysis period may be equal to - the life of the shorter-life alternative; - the life of the longer-life alternative; or - something entirely different, based on the actual/realistic need.

  38. Think – Pair - Share Other Analysis Period • Question 5: Consider two alternative production machines with expected initial costs and salvage values of machines shown below. If interest rate is 10%, which alternative should be selected for an analysis period of 10 years by using EUAC?

  39. Question 5 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

  40. Question 5 Continues $15,000 MACHINE B 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 13-year life $65,000

  41. Question 5 Continues For fixed output of 10 years of service of equipments, Machine A is preferred, because its EUAC is closer to “0” cost.

  42. Additional Examples

  43. Example 6-8 Analysis Period for a Continuing Requirement Assumption: Identical Replacements! To minimize EUAC, select pump B.

  44. Example 6-9 Infinite Analysis Period

  45. Example 6-10 Other Analysis Period

  46. 0 1 2 3 4 5 Another Example Anitem was purchased for $500. If the item’s expected life is 5 years with no salvage value, what will be the equivalent uniform annual cost (EUAC) at 6 % interest rate? i=6% P=$500 A = $118.70

  47. Some Exercise Problems

  48. Problem 6-7 Solution Given r = 15%, n = 500 months, and F = $1M, find A. A = F(A/F, 15%/12, 500) = MPT(15%/12, 500, 0, -1000000) = $25.13 What is A, if r = 10%? A = PMT(10%/12, 500, 0, -1000000) = $133.56 What is A, if r = 6%? A = PMT(6%/12, 500, 0, -1000000) = $450.17

  49. Problem 6-19 Solution Find A, for r = 7%, n = 48, & P = 21900 – 2350 – 850 = $18,700. A = P(A/P, 7%/12, 48) = PMT(7%/12, 48, -18700) = $447.79 month beginning interest principal ending 0 $18,700.00 1 $18,700.00 $109.08 $338.71 $18,361.29 2 $18,361.29 $107.11 $340.69 $18,020.60 3 $17,020.60 $105.12 $342.67 $17,677.93

  50. Problem 6-19 - continued With taxes and fees consideration Tag & fees: $600 Taxes at 7%: 0.07(21900–2350+600) = $1,410.5 P = 18700 + 600 + 1410.5 = $20,710.5 A = P(A/P, 7%/12, 48) = PMT(7%/12, 48, -20710.5) = $495.94 vs$447.79 Interest paid 495.94*48 – 20710.5 = $3,095.1

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