1 Chapter 6 Annual Cash Flow Analysis EGN 3615ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS1
2 Chapter Contents Annual Cash Flow Calculations Annual Cash Flow AnalysisAnalysis periodAnalysis period equal to alternative livesAnalysis period = a common multiple of alternative livesAnalysis period for continuing requirementInfinite analysis periodOther analysis periodUsing Spreadsheets
3 Learning ObjectivesApply annual cash flow techniques in various situations in selecting the best alternativeDevelop and use spreadsheet in solving engineering economic problems
4 Vignette: Lowest Prices on the Net! Buy Now! Why are there so many spam or junk selling ink or toner cartridges?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.
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 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%?P=100041235A867910
7 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%)P=100010S=20041235867910AAAAAAAAAA
9 Example 6-2 i = 7% i = 7% or or P=1000 10 S=200 4 1 2 3 5 A 8 6 7 9 P=100010S=200i = 7%41235A867910i = 7%oror
10 Maintenance and Repair Cost Example 6-3i = 7%YearMaintenance and Repair Cost145290318041355225412359045180225135PWCost
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) = $ Step 2: Find the EUAC: EUAC = pmt(0.07, 5, ) = $129.51
12 Example 6-4 Annual Cash Flow i = 7%YearMaintenance and Repair Cost145290313541805225412359045180225135
13 Annual Cash Flow Analysis SituationCriterionNeither input nor output fixed: typical situationMaximize EUAW (Equivalent Uniform Annual Worth)EUAW=EUAB - EUACFixed input: amount of money or other input resources are fixedMaximize EUAB (Equivalent Uniform Annual Benefits)Fixed output: fixed task, benefit, or other outputsMinimize EUAC (Equivalent Uniform Annual Costs)
14 Example 6-5 Annual Cash Flow Device ADevice B41235P=135030035045040050041235P=1000A=300i = 7%i = 7%Which device should the company select?
16 Example 6-6 Annual Cash Flow Plan APlan BPlan CInstalled cost of equipment$15,000$25,000$33,000Material and labor savings per year$14,000$9,000Annual operating expenses$8,000$6,000End-of-useful-life salvage value$1,500$2,500$3,300Each 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%Plan APlan BPlan CInstalled cost of equipment$15,000$25,000$33,000Material and labor savings per year$14,000$9,000Annual operating expenses$8,000$6,000End-of-useful-life salvage value$1,500$2,500$3,300CBCBPlan APlan BPlan CMaterial and labor savings per year$14,000$9,000Salvage value * (A/F, 8%, 10)104172228EUAB =$14,104$9,172$14,228Installed cost * (A/P, 8%, 10)$2,235$3,725$4,917Annual Operating expenses8,0006,000EUAC =$10,235$9,725$10,917EUAW = EUAB – EUAC =$3,869-$553$3,311Based on maximizing EUAW, select Plan A.
18 Example 6-7 Annual Cash Flow i = 7%Pump APump BInitial cost$7,000$5,000End-of-useful-life salvage value$1,500$1,000Useful life, in years126Calculate 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?DEVICE ADEVICE BANNUAL COST FLOW($800)($1000)$300$200$250$350$400INITIAL COSTANNUAL COST SAVINGS
21 Question 1 contd SOLUTION 1 To maximize EUAW, select Device A. DEVICE BANNUAL COST FLOW($800)($1000)$300$200$250$350$400Question 1 contdSOLUTION 1To 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 5EQUIPMENTCOSTSALVAGEVALUEUSEFULLIFEEQUIPMENT A$1500$2005 yearEQUIPMENT B$1600$35010 year
24 Replacement Equipment A Investment COSTSALVAGEVALUEUSEFULLIFEEQUIPMENT A$1500$2005 yearEQUIPMENT B$1600$35010 yearQuestion 2 Continues12345$200$1500678910Replacement Equipment A InvestmentOriginal Equipment AInvestmentEQUIPMENT A
25 Question 2 Continues $350 $1600 EQUIPMENT B EQUIPMENT COST SALVAGE VALUEUSEFULLIFEEQUIPMENT A$1500$2005 yearEQUIPMENT B$1600$35010 yearQuestion 2 Continues12345$1600678910$350EQUIPMENT B
26 Question 2 Continues EQUIPMENT A EQUIPMENT B COSTSALVAGEVALUEUSEFULLIFEEQUIPMENT A$1500$2005 yearEQUIPMENT B$1600$35010 yearQuestion 2 ContinuesEQUIPMENT AEQUIPMENT BIf we use n = 5 years for equipment A, we get exactly the same.EUACA = (A/P, 6%,5) (A/F, 6%,5)= (0.2374) (0.1774)=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.MACHINEINITIAL COSTSalvageValue at the End Of Useful LifeUSEFULLIFEMACHINE A$40,000$8,0007 yearMACHINE B$65,000$10,00013 year
28 Question 3 continues $8,000 $40,000 $10,000 $65,000 MACHINE A INITIAL COSTSalvageValueUSEFULLIFEMACHINE A$40,000$8,0007 yearMACHINE B$65,000$10,00013 yearQuestion 3 continuesUnder 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.12$8,00067$40,0008913149091MACHINE A12$65,000$10,0001213141525269091MACHINE B
31 Infinite Analysis Period Question4: The water will be carried via two alternatives:A tunnel through mountainA pipeline which goes around mountainIf there is a permanent need for an aqueduct, which option should be selected at 6% interest rate?MACHINEINITIAL COST (Million)SalvageValue at the End Of Useful LifeUSEFULLIFETunnel$7$0permanentPipeline$660 year
32 MACHINEINITIAL COSTSalvageValueUSEFULLIFETunnel$7$0permanentPipeline$660 yearQuestion 4 continuesUnder 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 5For fixed output, minimize EUAC. Select the pipeline.
33 Think – Pair - Share $100 i=6% P=$500 A= ? 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?$100P=$500i=6%12345A= ?
34 Think – Pair - Share $100 i=6% P=$500 A= $100.96 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?$100P=$500i=6%12345A= $100.96
35 Think – Pair - Share i=6% A=? 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 toi=6%12345A=?
37 Other Analysis Period Think – Pair - Share 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 Other Analysis Period Think – Pair - Share MACHINE A $40,000 $8,000 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?MACHINEINITIAL COSTSalvageValue at the End Of Useful LifeTerminal Value at the end of 10-year analysis periodUSEFULLIFEMACHINE A$40,000$8,000$15,0007 yearMACHINE B$65,000$10,00013 year
43 Example 6-8 Analysis Period for a Continuing Requirement Assumption: Identical Replacements!Pump APump BInitial cost$7,000$5,000End-of-useful-life salvage value$1,500$1,000Useful life, in years129To minimize EUAC, select pump B.
44 Example 6-9 Infinite Analysis Period TunnelPipelineInitial cost$5.5 million$5 millionMaintenanceUseful lifePermanent50 yearsSalvage value
45 Example 6-10 Other Analysis Period AlternativesAlt. 1Alt. 2Initial cost$50,000$75,000Estimated salvage value at end of useful life$10,000$12,000Useful life7 years13 yearsEstimated market value, end of 10 years$20,000$15,000
46 Another Example i=6% P=$500 A = $118.70 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=$500i=6%12345A= $118.70
48 Problem 6-7Solution Given r = 15%, n = 500 months, and F = $1M, find A. A = F(A/F, 15%/12, 500) = MPT(15%/12, 500, 0, ) = $25.13 What is A, if r = 10%? A = PMT(10%/12, 500, 0, ) = $ What is A, if r = 6%? A = PMT(6%/12, 500, 0, ) = $450.17
49 Problem 6-19 month beginning interest principal ending 0 $18,700.00 SolutionFind A, for r = 7%, n = 48, & P = – 2350 – 850 = $18,700.A = P(A/P, 7%/12, 48)= PMT(7%/12, 48, ) = $447.79month beginning interest principal ending$18,700.001 $18, $ $ $18,361.292 $18, $ $ $18,020.603 $17, $ $ $17,677.93
50 Problem continuedWith taxes and fees consideration Tag & fees: $600 Taxes at 7%: 0.07(21900– ) = $1,410.5 P = = $20,710.5 A = P(A/P, 7%/12, 48) = PMT(7%/12, 48, ) = $ vs $ Interest paid *48 – = $3,095.1
51 Problem 6-40Solution 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 = ( )(A/P, 0.1, 4) (0.1) x 0.26x = 17000(0.3155) = x = /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.