Lecture No.19 Chapter 6 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010
Jan 12, 2016
Lecture No.19Chapter 6
Contemporary Engineering EconomicsCopyright © 2010
Contemporary Engineering Economics, 5th edition, © 2010
Chapter Opening Story - Drinking Water from Ocean
Build a $300 million water-desalination plant in Carlsaba, CAProduce 50 million gallons of drinking water a daySupply about 100,000 homesCost the plant $1.10 in electricity per 1,000 gallons of water
At Issue: What would be the production cost per gallon of water from ocean?
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Annual Worth AnalysisPrinciple: Measure an investment worth on annual basisBenefits: By knowing the annual equivalent worth, we can:
Seek consistency of report formatDetermine the unit cost (or unit profit)Facilitate the unequal project life comparison
Annual Equivalent Conversion
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Fundamental Decision RulesSingle Project Evaluation: Comparing Mutually
Exclusive Alternatives:If AE(i) > 0, accept the investment.
If AE(i) = 0, remain indifferent to the investment
If AE(i) < 0, reject the investment
Service projects: select the alternative with the minimum annual equivalent cost (AEC).
Revenue projects: select the alternative with the maximum AE(i).
Contemporary Engineering Economics, 5th edition © 2010
Example 6.1 Economics of Installing A Feedwater Heater Install a 150MW unit:
Initial cost = $1,650,000Service life = 25 yearsSalvage value = 0Expected improvement in fuel efficiency = 1%Fuel cost = $0.05kWhLoad factor = 85%
Determine the annual worth for installing the unit at i = 12%.
If the fuel cost increases at the annual rate of 4%, what is AE(12%)?
Contemporary Engineering Economics, 5th edition © 2010
Ex. 6.1 Calculation of Annual Fuel SavingsRequired input power before adding the second unit:
Required input power after adding the second unit:
Reduction in energy consumption: $4,870kW
Annual operating hours:
Annual Fuel Savings
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:
.
150,000kW272,727kW
0.55
150,000kW267,857kW
0.56
Operating hours = (365)(24)(0.85) =7,446 hours/year
Ex. 6.1 Annual Worth Calculations (a) with constant fuel price:PW (12%) = -$1,650,000 + $1,813,101 (P/A, 12%, 25) = $12,570,403AE(12%) = $12,570,403 (A/P, 12%, 25) = $ 1,602,726
(b) with escalating fuel price:
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Ex. 6.2 Annual Equivalent Worth - Repeating Cash Flow Cycles
First Cycle
Repeating Cycles
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PW(12%) $1,000,000[($800,000 $100,000( ,12%,4)]( ,12%,4)
$1,000,000 $2,017,150.$1,017,150.
A G P A
AE 12% = $1,017,150 , 12%, 4
= $334,880.
A P
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Required AssumptionsThe service life of the selected alternative is required on a
continuous basis.Each alternative will be replaced by an identical asset that
has the same costs and performanceModel A:
Model B:
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PW 15% = $22,601
AEC 15% = $22,601 , 15%, 3
= $9,899.
A P
PW 15% = $25,562
AEC 15% = $25,562 , 15%, 4
= $8,954.
A P
Annual equivalent cost = Capital cost + Operating costs
Contemporary Engineering Economics, 5th edition, © 2010
Contemporary Engineering economics, 5th edition, © 2010
Annual Equivalent CostWhen only costs are
involved, the AE method is called the annual equivalent cost.
Revenues must cover two kinds of costs: Operating costs and capital costs.
Capital costs
Operating costs
+
Annu
al E
quiv
alen
t Cos
ts
Contemporary Engineering economics, 5th edition, © 2010
Capital (Ownership) CostsDef: Owning an equipment
is associated with two transactions—(1) its initial cost (I) and (2) its salvage value (S).
Capital costs: Taking these items into consideration, we calculate the capital costs as:
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SEGMENT BEST MODELS ASKING PRICE
PRICE AFTER 3 YEARS
Compact car Mini Cooper $19,800 $12,078
Midsize car Volkswagen Passat
$28,872 $15,013
Sports car Porsche 911 $87,500 $48,125
Near luxury car BMW 3 Series $39,257 $20,806
Luxury car Mercedes CLK $51,275 $30,765
Minivan Honda Odyssey
$26,876 $15,051
Subcompact SUV Honda CR-V $20,540 $10,681
Compact SUV Acura MDX $37,500 $21,375
Full size SUV Toyota Sequoia $37,842 $18,921
Compact truck Toyota Tacoma $21,200 $10,812
Full size truck Toyota Tundra $25,653 $13,083
Example - Capital Cost Calculation for Mini Cooper
Given: I = $19,800 N = 3 years S = $12,078 i = 6%
Find: CR(6%)
Capital recovery Cost
Contemporary Engineering economics, 5th edition, © 2010
Example 6.4 Required Annual Revenue to Justify an Investment
Given: I = $20,000 S = $4,000 N = 5 years i = 10%
Find: See if an annual revenue of $5,000 is large enough to cover both the capital and operating costs
Cost of Owning & Operating
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.
Ex. 6.4 Solution:Need additional revenue in the amount of $120.76 to justify the Investment
Contemporary Engineering economics, 5th edition, © 2010