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IENG 217Cost 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
1m0 exponent,g correlatinm
size designreference Q
equipmentofsizedesign
Qsize designreference for cost
Qdesignnew for cost
r
r
c
c
r
m
r
cr
Q
C
C
where
Q
QCC
Power Law and Sizing
design of tindependen items of costC
designnew to coupledindex inflationI
designreference to coupledindex inflation
i
c
r
ir
c
m
r
cr
I
where
CI
I
Q
QCC
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
CER1
2
.
.
C KQ
C C CQ
Q
N
m
f vc
r
m
s
FHG
IKJ
K empirical constant
Q capacity expressed as design dimension
m=correlatingcoefficient
C fixed costs
C variable costs
3. C=KQ
f
v
m
m
r
cvf Q
QCCC
Factor Method Uses a ratio or percentage approach; useful for
plant and industrial construction applications
C C f C f
C selected major equipment
f factors for estimatingmajor items
f factor for estimating indirect
c ii
n
e I
e
i
I
( )( )
11
where
C cost of design
cost of
expenses
Factor Method
Basic Item Cost
Factor
Adjustment for Inflation
C CI
I
where
C t of itemat benchmark time
C t of itemincurrent time
I index of benchmark year
I index for current time
r cr
c
r
c
r
c
FHG
IKJ
cos
cos
c
rcr I
ICC
Example; Plant Project
Example; Plant Project
Ir = 100.0Ic = 114.1
Major Process Design Equipment Current Cost
Benchmark Cost
Rising Film Reactor $2,900,000 $2,541,630Ozonation Reactor $700,000 $613,497 Total $3,600,000 $3,155,127
Example; Plant Project
4.1
1.7
1.1
Example; Plant Project
Item Current Cost Current Index
Major Item Bench
fi Benchmark Cost
Project Strt Index
Project Strt Cost
Midwest Index
Plant Strt Cost
Equipment $3,600,000 114.1 $3,155,127 1.0 $3,155,127 134.0 $4,227,870 $4,227,870Major Process Item Bld Erection 1.7 $5,363,716 134.0 7,187,379 1.000 7,187,379 Direct Materials 4.1 $12,936,021 134.0 17,334,268 1.000 17,334,268 Eng. Costs 1.1 $3,470,640 134.0 4,650,657 1.000 4,650,657Building Site 1,250,000Process Building 2,100,000Railroad Spur 40,000Utilities 650,000 Total $37,440,175
Other Project Methods
Expected Value Range Percentile Simulation
Expected Value
Suppose we have the following cash flow diagram.
NPW = -10,000 + A(P/A, 15, 5)
1 2 3 4 5
A A A A A
10,000
MARR = 15%
Expected Value
Now suppose that the annual return A is a random variable governed by the discrete distribution:
A
p
p
p
2 000 1 6
3 000 2 3
4 000 1 6
, /
, /
, /
Expected Value
A
p
p
p
2 000 1 6
3 000 2 3
4 000 1 6
, /
, /
, /
For A = 2,000, we have
NPW = -10,000 + 2,000(P/A, 15, 5)
= -3,296
Expected Value
A
p
p
p
2 000 1 6
3 000 2 3
4 000 1 6
, /
, /
, /
For A = 3,000, we have
NPW = -10,000 + 3,000(P/A, 15, 5)
= 56
Expected Value
A
p
p
p
2 000 1 6
3 000 2 3
4 000 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
p(NPW) 1/6 2/3 1/6
Expected Value
E[Return] = (1/6)-3,296 + (2/3)56 + (1/6)3,409
= $56
A 2,000 3,000 4,000
p(A) 1/6 2/3 1/6
NPW -3,296 56 3,409
p(NPW) 1/6 2/3 1/6
