# Chapter 6 Annual Cash Flow Analysis

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

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

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

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

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=1000 10 S=200 4 1 2 3 5 8 6 7 9 10 A A A A A A A A A A

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

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

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

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

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

Annual Cash Flow Analysis
Situation Criterion 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)

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

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

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

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

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

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

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

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

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

Question 2 Continues EQUIPMENT A EQUIPMENT B
COST SALVAGE VALUE USEFUL LIFE EQUIPMENT A \$1500 \$200 5 year EQUIPMENT B \$1600 \$350 10 year Question 2 Continues EQUIPMENT A EQUIPMENT B If 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).

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. MACHINE INITIAL COST 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

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

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

MACHINE INITIAL COST Salvage Value USEFUL LIFE MACHINE A \$40,000 \$8,000 7 year MACHINE B \$65,000 \$10,000 13 year Question 3 continues 1 2 3 4 5 6 7 8 9 10 \$10,000 \$65,000 MACHINE B 11 12 13 13-year life 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, = -8,744 To minimize the cost, we select MACHINE A

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 \$0 permanent Pipeline \$6 60 year

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

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

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? \$100 P=\$500 i=6% 1 2 3 4 5 A= \$100.96

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 to i=6% 1 2 3 4 5 A=?

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

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.

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? 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 \$15,000 7 year MACHINE B \$65,000 \$10,000 13 year

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

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

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.

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

Example 6-9 Infinite Analysis Period
Tunnel Pipeline Initial cost \$5.5 million \$5 million Maintenance Useful life Permanent 50 years Salvage value

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

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=\$500 i=6% 1 2 3 4 5 A = \$118.70

Some Exercise Problems

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, ) = \$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

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

Problem continued With 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

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 = ( )(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.

End of Chapter 6

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