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

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

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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.

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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?”

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=1000 4 1 2 3 5 A 8 6 7 9 10 

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

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

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

10. 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

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%
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

13. 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)

14. 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?

15. 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%

16. 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?

17. 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.

18. 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

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

21. 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.

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

24. 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

25. 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

26. 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).

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.
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

28. 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

29. 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, =

30. 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

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 $0
permanent
Pipeline $6
60 year

32. 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.

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

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

36. 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

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?
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

39. 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

40. 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

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!
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.

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

45. 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

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=$500
i=6% 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, ) = $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
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

50. 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

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 = ( )(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.

52. End of Chapter 6