Transcript IENG 492 Accounting for Engineers

IENG 217
Cost Estimating for Engineers
Project Estimating
Hoover Dam
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U.S. Reclamation Service opened debate 1926
Six State Colorado River Act, 1928
Plans released 1931, RFP
Bureau completed its estimates
3 bids, 2 disqualified
Winning bid by 6 companies with bid price at
$48,890,955
Winning bid $24,000 above Bureau estimates
Project Methods
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Power Law and sizing CERs
Cost estimating relationships
Factor
Power Law and Sizing
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In general, costs do not rise in strict proportion to
size, and it is this principle that is the basis for the
CER
m
Qc
C  Cr
Qr
F
I
G
J
HK
where
C  cost for new design Qc
Cr  cost for reference design size Qr
Qc  design size of equipment
Qr  reference design size
m  correlating exponent , 0  m  1
Power Law and Sizing
F
Q I I
CC GJ
HQ KI
m
c
c
r
r
r
 Ci
where
Ir  inflation index coupled to reference design
Ic  inflation index coupled to new design
Ci  cost items independent of design
Power Law and Sizing
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Ten years ago BHPL built a 100 MW
coal generation plant for $100 million.
BHPL is considering a 150 MW plant of
the same general design. The value of
m is 0.6. The price index 10 years ago
was 180 and is now 194. A substation
and distribution line, separate from the
design, is $23 million. Estimate the
cost for the project under consideration.
Class Problem
CER
1. C  KQ m
K  empirical constant
Q  capacity expressed as design dimension
m=correlating coefficient
F
Q I
C GJ
HQ K
m
2. C  C f
c
v
r
C f  fixed costs
Cv  variable costs
3. C= KQ m N s
Factor Method
Uses a ratio or percentage approach; useful for
plant and industrial construction applications
n
C  (Cc   fi Ce )( fI  1)
i 1
where
C  cost of design
Ce  cost of selected major equipment
fi  factors for estimating major items
fI  factor for estimating indirect expenses
Factor Method
Factor
Basic Item Cost
Adjustment for Inflation
F
I I
C  C GJ
HI K
r
r
c
c
where
Cr  cos t of item at benchmark time
Cc  cos t of item in current time
Ir  index of benchmark year
Ic  index for current time
Example; Plant Project
Example; Plant Project
Ir =
Ic =
Major Process
Design Equipment
Rising Film Reactor
Ozonation Reactor
Total
Current Cost
$2,900,000
$700,000
$3,600,000
100.0
114.1
Benchmark
Cost
$2,541,630
$613,497
$3,155,127
Example; Plant Project
4.1
1.7
1.1
Example; Plant Project
Item
Current Cost
Equipment
Major Process Item
Bld Erection
Direct Materials
Eng. Costs
Building Site
Process Building
Railroad Spur
Utilities
Total
$3,600,000
Current
Index
114.1
Major Item
Bench
$3,155,127
fi
Project Strt
Index
134.0
Project Strt
Cost
$4,227,870
Midwest
Index
1.0
Benchmark
Cost
$3,155,127
Plant Strt
Cost
$4,227,870
1.7
4.1
1.1
$5,363,716
$12,936,021
$3,470,640
134.0
134.0
134.0
7,187,379
17,334,268
4,650,657
1.000
1.000
1.000
7,187,379
17,334,268
4,650,657
1,250,000
2,100,000
40,000
650,000
$37,440,175
Other Project Methods
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Expected Value
Range
Percentile
Simulation
Expected Value
Suppose we have the following cash flow
diagram.
A
A
A
A
A
MARR = 15%
1
2
3
4
10,000
NPW = -10,000 + A(P/A, 15, 5)
5
Expected Value
Now suppose that the annual return A is a
random variable governed by the discrete
distribution:
2,000
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A  3,000
4,000
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p  1/ 6
p 2 / 3
p1/ 6
Expected Value
2,000
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A  3,000
4,000
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p  1/ 6
p 2 / 3
p1/ 6
For A = 2,000, we have
NPW = -10,000 + 2,000(P/A, 15, 5)
= -3,296
Expected Value
2,000
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A  3,000
4,000
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p  1/ 6
p 2 / 3
p1/ 6
For A = 3,000, we have
NPW = -10,000 + 3,000(P/A, 15, 5)
=
56
Expected Value
2,000
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A  3,000
4,000
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p  1/ 6
p 2 / 3
p1/ 6
For A = 4,000, we have
NPW = -10,000 + 4,000(P/A, 15, 5)
= 3,409
Expected Value
There is a one-for-one mapping for each
value of A, a random variable, to each
value of NPW, also a random variable.
A
2,000
3,000
4,000
p(A)
1/6
2/3
1/6
NPW
-3,296
56
3,409
1/6
2/3
1/6
p(NPW)
Expected Value
A
2,000
3,000
4,000
p(A)
1/6
2/3
1/6
NPW
-3,296
56
3,409
1/6
2/3
1/6
p(NPW)
E[Return] = (1/6)-3,296 + (2/3)56 + (1/6)3,409
= $56