1. Engineering Economic Analysis
Chapter 6
Annual Cash Flow Analysis
Donald G. Newnan
San Jose State University
Ted G. Eschenbach
University of Alaska Anchorage
Jerome P. Lavelle
North Carolina State University
Neal A. Lewis
University of New Haven
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2. Chapter Outline Annual Cash Flow Calculations
Annual Cash Flow Analysis
Analysis Period
Analysis Period Equal to the Alternative Lives
Analysis Period Different from the Alternative Lives
Infinite Analysis Period
Analyzing Loans
Annuity Due
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3. Learning Objectives Define equivalent uniform annual cost & benefit
Express problem as annual cash flow equivalent
Conduct equivalent uniform annual worth analysis
Compare alternatives
With lives that are equal, a common multiple, or infinite
With a fixed study period
Develop & use spreadsheets to analyze loans
Use annuity due for beginning of period cash flows
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4. Vignette: Energy Star Appliances
US Department of Energy goal: reduce home’s energy use for appliances & electronics by 20%. New appliances meeting energy efficiency criteria earn ENERGY STAR label.
Adoption slow:
Steady electricity prices (only 1 – 3% increase/year)
Many investment decisions are made by home builders, landlords, & property managers—not consumers
Relative short period of residency (US average < 8 years)
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5. Vignette: Energy Star Appliances
Typical American household
400 loads of laundry per year
40 gallons of water per load
ENERGY STAR  reduce water & energy consumption by 40%
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6. Vignette: Energy Star Appliances
Major appliances last 7 to 15 years; refrigerators longest
For major appliances what should be considered in keep & move vs. leave or sell when moving?
Which appliance replacement decision for a clothes washer or a refrigerator is more sensitive to electricity price changes?
How would you explain this to someone not skilled in engineering economics?
EnergyGuide labels are not required for TVs, ranges, ovens, & clothes dryers.
Why do you think this is true?
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7. Example 6-1 Annual Cash Flow Calculations
Machine costs $30,000; O&M = $2000; saves $10,000 in labor costs. Salvage value at 5 years = $7000. What is EUAW at 7%?
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8. Example 6-2 Annual Cash Flow Calculations
Consider only capital costs from Example 6-1. Machine costs $30,000. Salvage value at 5 years = $7000.
What is EUAC or capital recovery cost at 10%?
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9. Annual Cash Flow CalculationsP n S 4 1 2 3 A
n-1 n 
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)
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10. Example 6-3 Annual Cash Flow Calculations
Year
Maintenance & Repair Cost 1
545 2
590 3
635 4
680 5
725
i = 7%
First find PW using the NPV function.
Then convert to EUAC using the PMT function.
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11. Example 6-4 Add an overhaul in year 3
Maintenance & Repair Cost 1
545 2
590 3 4
680 5
725
i = 7%
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12. EAUC for a simple overhaul
A machine requires a $100K overhaul in year 5 of 8 year life. EAUC of overhaul at i = 10%?
$62.09K
$133.10K
$11.64K
$18.74K
I don’t know how to do this C
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13. EAUC for a simple overhaul
A machine requires a $100K overhaul in year 5 of 8 year life. EAUC of overhaul at i = 10%?
$62.09K
$8.74K
$11.64K
$18.74K
I don’t know how to do this
=PV(10%,5,0,-100) = $62.09K
=PMT(10%,8,-62.09) = $11.64K C
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14. Annual Cash Flow Analysis
Situation
Criterion
Neither input nor output fixed
Maximize EUAW (Equivalent Uniform Annual Worth)
EUAW = EUAB − EUAC
Fixed input
Maximize EUAB (Equivalent Uniform Annual Benefits)
Fixed output
Minimize EUAC (Equivalent Uniform Annual Costs)
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15. Example 6-5
i = 7%
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16. Example 6-6, i = 8% Plan A Plan B Plan C Installed equipment cost
$15,000
$25,000
$33,000
Material & labor savings/year
$14,000
$9,000
Annual operating expenses
$8,000
$6,000
Salvage value
$1,500
$2,500
$3,300
Plan A
Plan B
Plan C
Material & labor savings/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
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17. Example 6-6, Spreadsheet Solution
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18. Example 6-7 Pump A Pump B Initial cost $7000 $5000 Salvage value $1200
$1000
Useful life, in years 12 6
Annual cost for 6 years for Pump B is
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19. Example 6-7 The same as the 6-year analysis
The annual cost for 12 years for Pump B is
The same as the
6-year analysis
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20. Annual Cash Flow & Analysis Period
Analysis period = alternative lives
Use that analysis period
Analysis period = common multiple of lives
or continuing requirement
or infinite
Assume identical replacements & each annual equivalent based on that alternative’s life
Some other analysis period
Estimate terminal values for all alternatives at end of analysis period
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21. Example 6-8 Analysis Period for a Continuing Requirement at 7%
Pump A
Pump B
Initial cost
$7,000
$5,000
Salvage value
$1,500
$1,000
Useful life (years) 12 9
𝐸𝑈𝐴𝐶 𝐴 = 7000− 𝐴 𝑃 ,7%, % =$797
𝐸𝑈𝐴𝐶 𝐵 = 5000− 𝐴 𝑃 ,7%, % =$684
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22. Pumps: Brass, 3 years, $12K Stainless, 4 years, $15K
EAC of brass vs stainless at 8%
Brass = $3.62K Stainless = $5.82K
Brass = $4.53K Stainless = $4.66K
Brass = $4.66K Stainless = $4.53K
None of the above
I don’t know
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23. Pumps: Brass, 3 years, $12K Stainless, 4 years, $15K
EAC of brass vs stainless at 8%
Brass = $3.62K Stainless = $5.82K
Brass = $4.53K Stainless = $4.66K
Brass = $4.66K Stainless = $4.53K
EAC (brass) = PMT(8%,3,-12) = $4.66
EAC (stainless) = PMT(8%,4,-15) = $4.53
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24. Example 6-9 Infinite Analysis Period
Tunnel
Pipeline
Initial cost
$5.5 million
$5 million
Maintenance
Useful life
Permanent
50 years
Salvage value
𝐸𝑈𝐴𝐶 𝑇𝑢𝑛𝑛𝑒𝑙 =𝑃 𝑖 =5.5𝑚𝑖𝑙𝑙𝑖𝑜𝑛(6%)=$330,000
𝐸𝑈𝐴𝐶 𝑃𝑖𝑝𝑒𝑙𝑖𝑛𝑒 =5.5𝑚𝑖𝑙𝑙𝑖𝑜𝑛( 𝐴 𝑃, 6%,50)=$317,000
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25. Example 6-10 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-year
$20,000
$15,000
𝑁𝑃𝑊 1 =−50, ,000−50,000 𝑃 𝐹 ,8%,7 +20,000 𝑃 𝐹 ,8%,10
=−$64,076
𝐸𝑈𝐴𝐶 1 = 𝑁𝑃𝑊 1 𝐴 𝑃 ,8%,10 =64, =$9,547
𝑁𝑃𝑊 2 =−75,000+15,000 𝑃 𝐹 ,8%,10 =−$69,442
𝐸𝑈𝐴𝐶 2 = 𝑁𝑃𝑊 2 𝐴 𝑃 ,8%,10 =69, =$10,347
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26. Example 6-11 Analyzing Loan with Spreadsheet: Amortization Schedule
Loan $2400, 6% nominal annual rate, 6 monthly payments
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27. Example 6-12 Finding a Balance Due
How much owed ½ way through a 48-month loan of $15,000 at 9% nominal?
Monthly Payment
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28. Example 6-14 Annuity Due for a Lease
Find PW for year & equivalent monthly cost; lease payments = of $1200/month; interest = 1%/month
Use Type = 1
EU monthly Cost = 1200(1+0.01) = $1212
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