IENG 492 Accounting for Engineers
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Transcript IENG 492 Accounting for Engineers
IENG 217
Cost Estimating for Engineers
Project Estimating
Hoover Dam
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
Power Law and sizing CERs
Cost estimating relationships
Factor
Power Law and Sizing
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
CC 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
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
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
A 3,000
4,000
p 1/ 6
p 2 / 3
p1/ 6
Expected Value
2,000
A 3,000
4,000
p 1/ 6
p 2 / 3
p1/ 6
For A = 2,000, we have
NPW = -10,000 + 2,000(P/A, 15, 5)
= -3,296
Expected Value
2,000
A 3,000
4,000
p 1/ 6
p 2 / 3
p1/ 6
For A = 3,000, we have
NPW = -10,000 + 3,000(P/A, 15, 5)
=
56
Expected Value
2,000
A 3,000
4,000
p 1/ 6
p 2 / 3
p1/ 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