Chapter 6 Annual Cash Flow Analysis EGN 3615

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

Chapter Contents Annual Cash Flow Calculations Annual Cash Flow 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

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

Vignette: Lowest Prices on the Net! Buy Now! • Why are there so many spam or junk e-mail 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.

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? ”

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%? 0 P=1000 0 1 2 3 4 5 6 7 8 9 10 A A A A A

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%) S=200 0 P=1000 10 0 1 2 3 4 5 6 7 8 9 10 A A A A A

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

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

Example 6 -3 i = 7% Year 1 2 3 Maintenance and Repair Cost 45 90 180 4 5 135 225 0 1 2 3 4 5 45 90 PWCost 135 180 225

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(b 2, b 3: b 7) = $531. 01 Step 2: Find the EUAC: EUAC = pmt(0. 07, 5, -531. 01) = $129. 51

Example 6 -4 Annual Cash Flow i = 7% Year 1 2 3 4 5 Maintenance and Repair Cost 45 90 135 180 225 0 1 2 3 4 5 180 225 45 90 135

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

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

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

Example 6 -6 Annual Cash Flow Installed cost of equipment Material and labor savings per year Annual operating expenses End-of-useful-life salvage value Plan A $15, 000 $14, 000 $8, 000 $1, 500 Each of Plans A, B, and C has a 10 -year life. If interest is 8%, which plan should be adopted? Plan B $25, 000 $9, 000 $6, 000 $2, 500 Plan C $33, 000 $14, 000 $6, 000 $3, 300

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

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

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)

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 DEVICE A DEVICE B ANNUAL COST FLOW ($800) ($1000) $300 $200 $300 $250 $300 $350 $300 $400

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

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

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 EQUIPMENT COST SALVAGE VALUE USEFUL LIFE EQUIPMENT A $1500 $200 5 year EQUIPMENT B $1600 $350 10 year

EQUIPMENT COST SALVAGE VALUE USEFUL LIFE EQUIPMENT A $1500 $200 5 year EQUIPMENT B $1600 $350 10 year Question 2 Continues $200 0 $1500 1 Original Equipment A Investment 2 3 EQUIPMENT A 4 5 $1500 Replacement Equipment A Investment 6 7 8 9 $200 10

EQUIPMENT COST SALVAGE VALUE USEFUL LIFE EQUIPMENT A $1500 $200 5 year EQUIPMENT B $1600 $350 10 year Question 2 Continues 0 $1600 1 2 3 EQUIPMENT B 4 5 6 $350 7 8 9 10

EQUIPMENT COST SALVAGE VALUE USEFUL LIFE EQUIPMENT A $1500 $200 5 year EQUIPMENT B $1600 $350 10 year Question 2 Continues EQUIPMENT A Bget exactly the same. If we use n = 5 years for. EQUIPMENT equipment A, we 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).

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. Salvage Value at the End Of Useful Life USEFUL LIFE MACHINE A $40, 000 $8, 000 7 year MACHINE B $65, 000 $10, 000 13 year MACHINE INITIAL COST

Question 3 continues MACHINE INITIAL COST Salvage Value USEFUL LIFE MACHINE A $40, 000 $8, 000 7 year MACHINE B $65, 000 $10, 000 13 year � 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 MACHINE A 0 1 2 $40, 000 6 7 8 9 13 $40, 000 $10, 000 14 90 $40, 000 $10, 000 91 $10, 000 MACHINE B 0 1 $65, 000 2 12 13 14 $65, 000 15 25 26 $65, 000 90 91

Question 3 continues MACHINE INITIAL COST Salvage Value USEFUL LIFE MACHINE A $40, 000 $8, 000 7 year MACHINE B $65, 000 $10, 000 13 year $8, 000 MACHINE A 0 1 2 3 4 7 -year life 5 6 7 i = 10% $40, 000 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

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

Infinite Analysis Period Question 4: 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 $0 permanent Pipeline $6 $0 60 year

Question 4 continues MACHINE INITIAL COST Salvage Value USEFUL LIFE Tunnel $7 $0 permanent Pipeline $6 $0 60 year 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.

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? 0 P=$500 1 i=6% 2 A= ? $100 3 4 5

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? 0 P=$500 1 i=6% 2 $100 3 4 A= $100. 96 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? 0 1 2 3 4 5 i=6% will be converted to 0 1 2 3 A=? 4

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

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.

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? MACHINE INITIAL COST Salvage Terminal Value at the end of 10 End Of -year analysis Useful Life period USEFUL LIFE MACHINE A $40, 000 $8, 000 $15, 000 7 year MACHINE B $65, 000 $10, 000 $15, 000 13 year

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

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

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.

Additional Examples

Example 6 -8 Analysis Period for a Continuing Requirement Assumption: Identical Replacements! Initial cost End-of-useful-life salvage value Useful life, in years To minimize EUAC, select pump B. Pump A Pump B $7, 000 $5, 000 $1, 500 $1, 000 12 9

Example 6 -9 Infinite Analysis Period Initial cost Maintenance Useful life Salvage value Tunnel $5. 5 million 0 Permanent 0 Pipeline $5 million 0 50 years 0

Example 6 -10 Other Analysis Period Alternatives Initial cost Estimated salvage value at end of useful life Useful life Estimated market value, end of 10 years Alt. 1 $50, 000 $10, 000 7 years $20, 000 Alt. 2 $75, 000 $12, 000 13 years $15, 000

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? 0 P=$500 1 i=6% 2 3 4 A = $118. 70 5

Some Exercise Problems

Problem 6 -7 Solution Given r = 15%, n = 500 months, and F = $1 M, 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

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 0 1 $18, 700. 00 2 $18, 361. 29 3 $17, 020. 60 interest principal $109. 08 $107. 11 $105. 12 $338. 71 $340. 69 $342. 67 ending $18, 700. 00 $18, 361. 29 $18, 020. 60 $17, 677. 93

Problem 6 -19 - continued With taxes and fees consideration Tag & fees: Taxes at 7%: $600 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

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. 5 x = (28000 -11000)(A/P, 0. 1, 4) + 11000(0. 1) + 1200 + 0. 24 x 0. 26 x = 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.

End of Chapter 6