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

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Presentation on theme: "Chapter 6 Annual Cash Flow Analysis EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS."— Presentation transcript:

1 Chapter 6 Annual 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 Apply annual cash flow techniques in various situations in selecting the best alternative Develop and use spreadsheet in solving engineering economic problems Learning Objectives

4 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. Vignette: Lowest Prices on the Net! Buy Now! Why are there so many spam or junk e-mail selling ink or toner cartridges?

5 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?” Vignette: Lowest Prices on the Net! Buy Now!

6 Example 6-1 Annual Cash Flow 4 0 1 23 5 A A A A A 8 67 9 A A A A 10 A  0 P=1000 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 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%) 4 0 1 23 5 8 67 9 10  0 P=1000 10 S=200 Example 6-2 Annual Cash Flow A A A A A A A A A A

8 EUAC Formulas 0 P n S 4 0 1 23 A A A A n-1 A n A  (6-1) (6-3) (6-4)

9 4 0 1 23 5 A A A A A 8 67 9 A A A A 10 A  0 P=1000 10 S=200 Example 6-2 or i = 7%

10 Year Maintenance and Repair Cost 145 290 3180 4135 5225 4 01 23 5 90 45 180 225 135 PW Cost Example 6-3 i = 7%

11 Example 6-3by 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 Year Maintenance and Repair Cost 145 290 3135 4180 5225 Example 6-4 Annual Cash Flow i = 7% 4 01 23 5 90 45 180 225 135

13 Annual Cash Flow Analysis SituationCriterion Neither input nor output fixed: typical situation Maximize EUAW (Equivalent Uniform Annual Worth) EUAW=EUAB - EUAC Fixed input: amount of money or other input resources are fixed Maximize EUAB (Equivalent Uniform Annual Benefits) Fixed output: fixed task, benefit, or other outputs Minimize EUAC (Equivalent Uniform Annual Costs)

14 Device A 4 0 1 23 5 P=1000 A=300 Device B 4 0 1 23 5 P=1350 300 350 450 400 500 Example 6-5 Annual Cash Flow Which device should the company select? i = 7%

15 Device A 4 0 1 23 5 P=1000 A=300 Device B 4 0 1 23 5 P=1350 300 350 450 400 500 Example 6-5 Annual Cash Flow i = 7%

16 Plan APlan BPlan C Installed cost of equipment$15,000$25,000$33,000 Material and labor savings per year$14,000$9,000$14,000 Annual operating expenses$8,000$6,000 End-of-useful-life salvage value$1,500$2,500$3,300 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 Plan APlan BPlan C Installed cost of equipment$15,000$25,000$33,000 Material and labor savings per year$14,000$9,000$14,000 Annual operating expenses$8,000$6,000 End-of-useful-life salvage value$1,500$2,500$3,300 Plan APlan BPlan C Material and labor savings per year$14,000$9,000$14,000 Salvage value * (A/F, 8%, 10)104172228 EUAB =$14,104$9,172$14,228 Installed cost * (A/P, 8%, 10)$2,235$3,725$4,917 Annual Operating expenses8,0006,000 EUAC =$10,235$9,725$10,917 EUAW = EUAB – EUAC =$3,869-$553$3,311 Example 6-6 Annual Cash Flow i = 8% B B C C Based on maximizing EUAW, select Plan A.

18 Pump APump B Initial cost$7,000$5,000 End-of-useful-life salvage value$1,500$1,000 Useful life, in years126 If EUAC B was calculated over n = 12-year period Example 6-7 Annual Cash Flow i = 7% Calculate EUAC A for n=12 and EUAC B for n = 6:

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? DEVICE ADEVICE B ANNUAL COST FLOW ($800)($1000) $300$200 $300$250 $300 $350 $300$400 INITIAL COST ANNUAL COST SAVINGS

21 To maximize EUAW, select Device A. SOLUTION 1 DEVICE ADEVICE B ANNUAL COST FLOW ($800)($1000) $300$200 $300$250 $300 $350 $300$400 Question 1 contd

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

23 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 EQUIPMENTCOST SALVAGE VALUE USEFUL LIFE EQUIPMENT A$1500$2005 year EQUIPMENT B$1600$35010 year Analysis period = a common multiple of alternative lives

24 Question 2 Continues 0 1 2 3 4 5 $200 $1500 6 7 8 9 10 $1500 $200 Replacement Equipment A Investment Original Equipment A Investment EQUIPMENT A EQUIPMENTCOST SALVAGE VALUE USEFUL LIFE EQUIPMENT A$1500$2005 year EQUIPMENT B$1600$35010 year

25 Question 2 Continues 0 1 2 3 4 5 $1600 6 7 8 9 10 $350 EQUIPMENT B EQUIPMENTCOST SALVAGE VALUE USEFUL LIFE EQUIPMENT A$1500$2005 year EQUIPMENT B$1600$35010 year

26 Question 2 Continues EQUIPMENT A EQUIPMENT B To minimize cost, we select EQUIPMENT B (closer to “ 0” value ). EQUIPMENTCOST SALVAGE VALUE USEFUL LIFE EQUIPMENT A$1500$2005 year EQUIPMENT B$1600$35010 year If we use n = 5 years for equipment A, we get exactly the same. EUAC A = - 1500 (A/P, 6%,5) + 200 (A/F, 6%,5) = - 1500 (0.2374) + 200 (0.1774) = - 320.62

27 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. MACHINE INITIAL COST Salvage Value at the End Of Useful Life USEFUL LIFE MACHINE A$40,000$8,0007 year MACHINE B$65,000$10,00013 year Analysis period for 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. MACHINE A MACHINE B 0 1 2 $65,000 $10,000 12 13 14 15 $65,000 25 26 $10,000 $65,000 90 91 $10,000 0 1 2 $8,000 6 7 $40,000 8 9 13 14 $8,000 $40,000 90 91 $8,000 MACHINEINITIAL COST Salvage Value USEFUL LIFE MACHINE A$40,000$8,0007 year MACHINE B$65,000$10,00013 year

29 Question 3 continues EUAW A = -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 0 1 2 3 4 5 $8,000 6 7 $40,000 7-year life MACHINE A MACHINEINITIAL COST Salvage Value USEFUL LIFE MACHINE A$40,000$8,0007 year MACHINE B$65,000$10,00013 year i = 10%

30 Question 3 continues 0 1 2 3 4 5 6 7 8 9 10 $10,000 $65,000 MACHINE B 11 12 13 13-year life To minimize the cost, we select MACHINE A MACHINEINITIAL COST Salvage Value USEFUL LIFE MACHINE A$40,000$8,0007 year MACHINE B$65,000$10,00013 year EUAW A = -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

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? MACHINE INITIAL COST (Million) Salvage Value at the End Of Useful Life USEFUL LIFE Tunnel$7$0permanent Pipeline$6$060 year

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. For fixed output, minimize EUAC. Select the pipeline. From Chapter 5 MACHINE INITIAL COST Salvage Value USEFUL LIFE Tunnel $7$0 permanent Pipeline $6$0 60 year

33 Think – Pair - Share An item 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 P=$500 i=6% 0 1 2 3 4 5 A= ?

34 Think – Pair - Share An item 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 P=$500 i=6% 0 1 2 3 4 5 A= $100.96

35 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? will be converted to i=6% 0 1 2 3 4 5 A=? i=6% 0 1 2 3 4 5

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

37 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. Other Analysis Period Think – Pair - Share

38 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? MACHINE INITIAL COST Salvage Value at the End Of Useful Life Terminal Value at the end of 10-year analysis period USEFUL LIFE MACHINE A$40,000$8,000$15,0007 year MACHINE B$65,000$10,000$15,00013 year Other Analysis Period Think – Pair - Share

39 Question 5 Continues $40,000 0 1 2 3 4 5 $8,000 6 7 8 9 10 $15,000 7-year life MACHINE A 11 12 13 14 $40,000 7-year life MACHINE INITIAL COST Salvage Value Terminal Value USEFUL LIFE MACHINE A$40,000$8,000$15,0007 year MACHINE B$65,000$10,000$15,00013 year

40 Question 5 Continues 0 1 2 3 4 5 6 7 8 9 10 $15,000 $65,000 MACHINE B 11 12 13 14 13-year life MACHINE INITIAL COST Salvage Value Terminal Value USEFUL LIFE MACHINE A$40,000$8,000$15,0007 year MACHINE B$65,000$10,000$15,00013 year

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 Pump APump B Initial cost$7,000$5,000 End-of-useful-life salvage value$1,500$1,000 Useful life, in years129 To minimize EUAC, select pump B. Assumption: Identical Replacements!

44 Example 6-9 Infinite Analysis Period TunnelPipeline Initial cost$5.5 million$5 million Maintenance00 Useful lifePermanent50 years Salvage value00

45 Example 6-10 Other Analysis Period AlternativesAlt. 1Alt. 2 Initial cost$50,000$75,000 Estimated salvage value at end of useful life$10,000$12,000 Useful life7 years13 years Estimated market value, end of 10 years$20,000$15,000

46 Another Example An item 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? P=$500 i=6% 0 1 2 3 4 5 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 paid495.94*48 – 20710.5 = $3,095.1

51 Problem 6-40 Solution Car cost: purchase price = $28000, resale for $11000 after 4 years annual operating expenses: $1200 plus $0.24/mile Pay $0.5/mile for driving salesperson’s own car. EAUC of driving personal car = EUAC of providing a company car 0.5x = (28000-11000)(A/P, 0.1, 4) + 11000(0.1) + 1200 + 0.24x 0.26x = 17000(0.3155) + 1100 + 1200 = 7663.5 x = 7663.5/0.26 = 29,475 miles/year Provide a company car if driving more than 29,475 miles/year. Do not provide a company car, otherwise.

52 End of Chapter 6


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