Transcript Chapter 5 Present Worth Analysis - Help-A-Bull

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
 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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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
0
1
2
3
4
5
6
7
8
9
10
A
A
A
A
A
A
A
A
A
A
P=1000
EUAC  1000(A/P,7 %,10)
 $142.40
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
10

0
1
2
3
4
5
6
7
8
9
10
A
A
A
A
A
A
A
A
A
A
P=1000
EUAC  1000(A/P,7%,10) - 200(A/F,7%,10)
EUAC  1000(0.1424) - 200(0.0724)
 $127.92
EUAC Formulas
S
0
n

0
1
2
3
4
n-1
n
A
A
A
A
A
A
P
EUAC  P(A/P, i, n) - S(A/F, i, n)
(6-1)
EUAC  (P - S)(A/F, i, n)  P(i)
(6-3)
EUAC  (P - S)(A/P, i, n)  S(i)
(6-4)
Example 6-2
S=200
0
10
i = 7%
i = 7%

0
1
2
3
4
5
6
7
8
9
10
A
A
A
A
A
A
A
A
A
A
P=1000
EUAC  P(A/P, i, n) - S(A/F, i, n)
EUAC  1000(A/P,7%,10) -200(A/F,7%,10)  $127.92
or
EUAC  (P - S)(A/P, i, n)  S(i)
EUAC  (1000-200)(A/P,7%,10)  200(7%)  $127.92
or
EUAC  (P - S)(A/F, i, n)  P(i)
EUAC  (1000-200)(A/F,7%,10)  1000(7%)  $127.92
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
135
PWCost
180
225
PWCost  45( P F , 7%,1)  90 ( P F , 7%, 2)  180( P F , 7%,3)
 135( P F , 7%, 4)  225( P F , 7%,5)  $531
EUAC  531(A/P,7%,5)  $130
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) = $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
EUAC  45  45(A/G,7%,5)  $129
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
5
1
0
350 400
2
450
500
4
5
3
i = 7%
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
5
0
1
350 400
2
450
500
4
5
3
i = 7%
i = 7%
P=1350
EUAWA  - 1000(A/P,7 %,5)  300
 - 1000(0.243 9)  300  $56.1
EUAWB  - 1350(A/P,7 %,5)  300  50(A/G,7%, 5)
 - 1350(0.243 9)  300  50(1.865)  $64.0
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
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
i = 7%
Pump B
$5,000
$1,000
6
Calculate EUACA for n=12 and EUACB for n = 6:
EUAC A  (P - S)(A/P, i, n)  S(i)
 (7000 - 1500)(A/P, 7%, 12)  1500(7%)  $797
EUACB  (5000 - 1000)(A/P, 7%, 6)  1000(7%)  $909
If EUACB was calculated over n = 12-year period
EUACB  5000  (5000 - 1000)(P/F,7%,6)(A/P, 7%, 12)  1000(7%)
 $909
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
ANNUAL COST
FLOW
($800)
($1000)
$300
$200
$300
$250
$300
$300
$300
$350
$300
$400
Question 1 contd
SOLUTION 1
DEVICE A
DEVICE B
ANNUAL COST
FLOW
ANNUAL COST
FLOW
($800)
($1000)
$300
$200
$300
$250
$300
$300
$300
$350
DEVICE A
$300
EUAW  800(A/P,12%,5)  300
 800(0.2774
)  300  78.08
$400
DEVICE B
EUAW  1000(A/P,12%,5)  200 50(A/G,12%,5)
 1000(0.277
4)  200 50(1.775) 11.35
To maximize EUAW, select Device A.
QUESTION 1 CONTINUES
SOLUTION 2 (by present worth analysis)
DEVICE A
PW  800 300(P/A,12%,5)
 800 300(3.605) 281.50
DEVICE B
PW  1000 200(P/A,12%,5)  50(P/G,12%,5)
 1000 200(3.605) 50(6.397) 40.85
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
1
$1500
Original Equipment A
Investment
2
3
EQUIPMENT A
4
5
Replacement Equipment A
Investment
6
7
8
9
$200
10
$1500
EUAWA  [(1500 200( P / F , 6%, 5)
 1500( P / F , 6%, 5)  200( P / F , 6%,10)](A/P,6%,10)
 [1500 200(0.7473)
 1500(0.7473)  200(0.5584)](0.1359)
 320.70
EQUIPMENT
COST
SALVAGE
VALUE
USEFUL
LIFE
EQUIPMENT A
$1500
$200
5 year
EQUIPMENT B
$1600
$350
10 year
Question 2 Continues
0
1
2
3
4
5
6
$350
7
8
EQUIPMENT B
$1600
EUAWB  P(A/P,i, n)  S(A/F, i, n)
 1600(A/P, 6%,10)  350(A/F, 6%,10)
 1600(0.1359)  350(0.0759)  190.88
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
EUAWA  [(1500 200( P / F , 6%, 5)
 1500( P / F , 6%, 5)  200( P / F , 6%,10)](A/P ,6%,10)
 [1500 200(0.7473)
 1500(0.7473)  200(0.5584)](0.1359)
 320.70
If we use n = 5 years forEQUIPMENT
equipment A, weBget exactly the same.
EUAW
(A/P (A/P,
,i, n) 6%,5)
S(A/F,+i,200
n) (A/F, 6%,5)
EUACBA = -P1500
1500(A/P
(0.2374)
+ )200
(0.1774)
= - 1600
, 6%,10
 350
(A/F, 6%,10)
320.62
= - 1600
(0.1359)  350(0.0759)  190.88
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
$8,000
$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
 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
$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.
EUAC
Tunnel
 Pi  $7 million (0.06)  420,000
From Chapter 5
EUAC
Pipeline
 P(A/P, 6%, 60)  6 million (0.0619)
 371,400
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?
$100
0
P=$500
1
i=6%
2
A= ?
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?
$100
0
P=$500
1
i=6%
2
3
4
5
A= $100.96
EUAC  P (A/P, 6%, 5)  S (A/F, 6%, 5)
 500 (0.2374)  100(0.1774 )  118.70  17.74
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%
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 $50 $50
+
0
1
0
2
3
4
5
$50 $100 $150
$200
Equivalent Uniform
AnnualCost (EUAC)
 50  50 (A/G, 6%, 5)
 50  50(1.884) 50(2.884)
 144.20
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
Value at the
at the end of
End Of
10-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
EUAWMACHINE
9
10
11
12
13
14
7-year life
$40,000
A
 [40,000  (40,000 8,000) (P /F,10%,7)
 15,000(P /F,10%,10)](A/P ,10%,10)
 [40,000  32,000(.05132) 15,000( 0.3855)](0.1627)
 $8,239.11
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
9
10
11
12
13
13-year life
$65,000
EUAWMACHINE
B
 [65,000  15,000(P/F,10%,10)](A/P,10%,10)
 [65,000  15,000( 0.3855)](0.1627)
 9634.69
14
Question 5 Continues
EUAWMACHINE
A
 [40,000  (40,000 8,000) (P /F,10%,7)
 15,000(P /F,10%,10)](A
/P ,10%,10)
 [40,000  32,000(.05132) 15,000( 0.3855)](0.1627)
 8,239.11
EUAWMACHINE
B
 [65,000  15,000(P/F,10%,10)](A/P,10%,10)
 [65,000  15,000( 0.3855)](0.1627)
 9634.69
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
Pump A Pump B
$7,000
$5,000
$1,500
$1,000
12
9
EUAC A  (7000 - 1500)(A/P, 7%, 12)  1500(7%)  $797
EUACB  (5000 - 1000)(A/P, 7%, 9)  1000(7%)  $684
To minimize EUAC, select pump B.
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
EUACTunnel  P(i)  5.5 million(6% )  $330,000
EUACPipeline  5 million(A/ P, 6%, 50)  $317,000
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
NPW1  - 50,000  (10,000 - 50,000)(P F , 8%,7)  20,000(P F , 8%,10)
 $64,076
EUAC1  NPW1( A P , 8%,10)  64,076(0.1 490)  $9,547
NPW2  - 75,000  15,000(P F , 8%,10)  $68,052
EUAC2  NPW2 ( A P , 8%,10)  69,442(0.1 490)  $10,142
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?
i=6%
0
P=$500
1
2
3
4
5
A = $118.70
 Equivalent 


 Uniform   A  P (A/P ,6%, 5)  500 (0.2374)
 AnnualCost 


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, -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
0
1
2
3
beginning
interest
principal
$18,700.00
$18,361.29
$17,020.60
$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
0.5x = (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.
End of Chapter 6