Transcript Annual Equivalent Worth Criterion

Lecture No.19
Chapter 6
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th edition, © 2010
Chapter Opening Story - Drinking Water from Ocean
Build a $300 million
water-desalination
plant in Carlsaba, CA
Produce 50 million
gallons of drinking
water a day
Supply about
100,000 homes
Cost 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?
Contemporary Engineering Economics, 5th edition © 2010
Annual Worth Analysis
Principle: Measure an
 Annual Equivalent Conversion
investment worth on
annual basis
Benefits: By knowing
the annual equivalent
worth, we can:
Seek consistency
of report format
Determine the
unit cost (or unit
profit)
Facilitate the
unequal project
life comparison
Contemporary Engineering Economics, 5th edition © 2010
Fundamental Decision Rules
Single Project Evaluation:
 If AE(i) > 0, accept the
investment.
 If AE(i) = 0, remain
indifferent to the
investment
 If AE(i) < 0, reject the
investment
Comparing Mutually
Exclusive Alternatives:
 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 = $1,650,000
 Service life = 25 years
 Salvage value = 0
 Expected improvement in fuel efficiency = 1%
 Fuel cost = $0.05kWh
 Load 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 Savings
Required input power before
adding the second unit:
150,000kW
 272,727kW
0.55
Required input power after
adding the second unit:
 Annual Fuel Savings
Afuel savings  (reduction in fuel requirement)  (fuel cost)
:
150,000kW
 267,857kW
0.56
.
Reduction in energy
consumption: $4,870kW
Annual operating hours:
(operating hours per year)
=(4,870kW)  ($0.05/kWh)
 7,446 hours/year 
=$1,813,101/year
Operating hours = (365)(24)(0.85)
=7,446 hours/year
Contemporary Engineering Economics, 5th edition © 2010
Ex. 6.1 Annual Worth Calculations
 (a) with constant fuel price:
PW (12%) = -$1,650,000 + $1,813,101 (P/A, 12%, 25) = $12,570,403
AE(12%) = $12,570,403 (A/P, 12%, 25) = $ 1,602,726
 (b) with escalating fuel price:
A1 =$1,813,101
PW (12%)  $1,650,000  $1,813,101(P / A1 ,4%,12%,25)
 $17,459,783
AE (12%)  $17,459,783(A / P ,12%,25)
 $2,226,122
Contemporary Engineering Economics, 5th edition © 2010
Ex. 6.2 Annual Equivalent Worth - Repeating
Cash Flow Cycles
 First Cycle
PW(12%)  $1,000,000
[($800,000  $100,000(A G ,12%,4)](P A ,12%,4)
 $1,000,000  $2,017,150.
 $1,017,150.
 Repeating Cycles
AE 12% = $1,017,150  A P , 12%, 4 
= $334,880.
Contemporary Engineering Economics, 5th edition © 2010
Example 6.3 Comparing Mutually Exclusive
Alternatives
Contemporary Engineering Economics, 5th edition © 2010
Required Assumptions
 The 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 performance
 Model A:
PW 15% = $22,601
 
AEC 15%  = $22,601  A P , 15%, 3
= $9,899.
 Model B:
PW 15%  = $25,562
AEC 15%  = $25,562  A P , 15%, 4 
= $8,954.
Contemporary Engineering Economics, 5th edition © 2010
Annual equivalent cost
= Capital cost + Operating costs
Contemporary Engineering Economics, 5th edition, © 2010
Annual Equivalent Cost
involved, the AE method is
called the annual equivalent
cost.
 Revenues must cover two
kinds of costs: Operating
costs and capital costs.
Annual Equivalent Costs
 When only costs are
Contemporary Engineering economics, 5th edition, © 2010
Capital
costs
+
Operating
costs
Capital (Ownership) Costs
 Def: 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:
S
0
N
I
0 1
CR(i)  I(A / P , i , N)  S(A / F , i , N)
 (I  S)(A / P , i , N)  iS
Contemporary Engineering economics, 5th edition, © 2010
2
3
CR(i)
N
Capital (Ownership) Costs Associated with
Various Vehicles
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
Contemporary Engineering economics, 5th edition, © 2010
Example - Capital Cost Calculation for Mini
Cooper
 Capital recovery Cost
 Given:




I = $19,800
N = 3 years
S = $12,078
i = 6%
 Find: CR(6%)
$12,078
0
1
2
3
$19,800
CR(i) = (I -S) (A/P , i , N) + iS
CR(6%) = ($19,800 - $12,078) (A/P , 6%, 3)
+ (0.06)$12,078
= $3,614
Contemporary Engineering economics, 5th edition, © 2010
Example 6.4 Required Annual Revenue
to Justify an Investment
 Cost of Owning & Operating
 Given:
 I = $20,000
$4,000
Capital cost
 S = $4,000
0
 N = 5 years
1
2
3
4
5
 i = 10%
$20,000
+
 Find: See if an annual
revenue of $5,000 is large
enough to. cover both the
capital and operating costs
0
1
2
3
4
$500
Operating cost
Contemporary Engineering economics, 5th edition, © 2010
5
Ex. 6.4 Solution:
Need additional
revenue in the
amount of $120.76
to justify the
Investment
Contemporary Engineering economics, 5th edition, © 2010