project cost estimation

1. Cost Engineering
Dr. Nabil I El Sawalhi
Assistant Professor
Construction Management
CE - L8 1

2. Estimating Methods
CE - L8 2

3. CE - L8 3

4. CE - L8 4

5. Algorithms
• Algorithms are generally applied in a
sequential series of steps to estimate various
elements of the total estimate. In detail unit
cost estimating, the steps are usually:
• 1) estimate the direct field cost,
• 2) the field indirect cost,
• 3) the office costs, and finally,
• 4) the profit and contingency.
CE - L8 5

6. Estimate General Form
• The algorithm in effect transforms project
technical and programmatic descriptive
information into cost terms.
• These estimating algorithms are often referred
to as cost estimating relationships (CERs).
• In a very simple, but common form, a CER may
appear as:
CE - L8 6

7. Cost resource = factor x parameter
where:
• Factor = $ (total, labor, or material, etc.), or
time (labor hours, equipment rental hours,
etc.) a unit cost factor in terms of resource
unit of measure
• Parameter = units of measure of the estimate
item
CE - L8 7

8. Estimating Methods
CE - L8 8

9. 1. Opinion
• In the absence of data and shortage of time,
there may be no way to evaluate designs rather
than opinion.
• The key to opinion estimating is human ware.
• The individual is selected to a job because of his
experience, common sense, and knowledge.
• Opinion estimating is done collectively in
conferences.
CE - L8 9

10. 2. Conference
• Non quantitative technique which provides single
value.
• It is done in a round table fashion.
• A mediator provides a question “ what is the labor
and material cost for this design.
• A hidden card gambit(strategy, maneuver) has each
of the committee expert reveal a personal value to
question.
• This could provide a consensus.
CE - L8 10

11. • If agreement is not reached, discussion and
presentation are permitted ( called estimate-
talk- estimate or E-T-E).
• The hidden card idea prevent any
brainstorming session which generally gives
optimistic estimate.
• A ranking scheme along the “good- better-
best” approach can be applied.
CE - L8 11
2. Conference

12. 2. Conference
• The lack of analysis and a trial of verifiable
facts leading from the estimate to the
governing situation are major drawback to the
conference method.
• Although there is little faith in the method
accuracy, these factors seldom prevent its use.
CE - L8 12

13. 3. Comparison (similarity or analogy)
• It use formal logic in estimating.
• If we faced with difficult design estimate
we designate it as design a and construct a
simpler design problem for which an
estimate can be found.
CE - L8 13

14. • The simpler problem is b.
• Ca (Da) <= Cb (Db)
• Where Ca,b = value of the estimate for design a
& b dollars.
• Da,b = design a or b
CE - L8 14

15. • If we have additional known design Dc which
have lower estimate than Ca, then the logic can
be expanded to
• Cc (Dc) <= Ca (Da) <= Cb (Db)
• Assume that design b and c generally satisfy the
technical requirements as nearly as possible.
• Comparison apply to any complexity of design.
• The comparison method is some time called
similarity or analogy
CE - L8 15

16. Example
• Three exterior wall systems for residence are
given as shown in the following figure . The
design that has the unknown cost is a, while b
and c have been explored for costs and
standards are available, which indicate that for
a 10-ft-long section Cb= $5.16/ft2 and Cc =
$4.93/ft2.
CE - L8 16

17. • The inspectable judgment for this method
initially requires that the unknown design be
squeezed between two designs that are above
and below the unknown cost. Does the reader
believe that design a satisfies this
requirement? . If this initial premise is
accepted, then the cost is between these two
numbers.
CE - L8 17

18. CE - L8 18

19. 4. Unit method
• The unit method use historical & quantitative
evidence and leads to a cost driver easily
understood.
• Other names of unit method are:
• -average, order of magnitude;
• -Lump sum
• -Function
• -Parameter.
• -Module estimating
CE - L8 19

20. Unit
continued..
• It involves various refinements.
• Extensions of this method lead to the factor
estimating method.
• Examples:
• Cost of house per square foot,
• Cost of electrical transmission per mile,
• Construction cost per hospital bed,
• Chemical plant cost per barrel of oil capacity.
CE - L8 20

21. • The most common unit cost estimate for
building is the cost per square meter.
• A similar one is the cost per cubic meter.
• Another types of unit cost estimate is:
• Cost per unit or trade unit estimate (the
building is broken down into basic trades and
cost and percentage factors are given.
• Ca = ∑ci/ ∑ni
CE - L8 21

22. • Ca = average dollar cost per unit of design
• Ci = value of design in dollars
• Ni = design i unit.
• Example:
• Consider the cost of field installed gate valve
with carbon steel piping system. There are
three historical observations of total material
and labor cost.
CE - L8 22

23. Design
Weight ,Ib
Ci
$
2 20
3 30
4 60
∑ 9 110
CE - L8 23
The cost per pound is Ca = ∑ci/ ∑ni =
110/9=$12.22

24. Example. For a concrete roadway:
• Given:
• A unit cost model of the roadway with a
modifying parametric algorithm: One all-in,
line-item algorithm (concrete) of many in the
given model is:
• $ of concrete = $500 per cubic meter x (N)
cubic meters
CE - L8 24

25. • Where the corresponding parametric algorithm for
quantity parameter N is:
• (N) = (road width in meters) x (road length in
meters) x (road loading in kg/cm2) x (1 cm/kg)/(
100 cm/m)
• and the following road design parameters are
given:
• Road width = 10 m , Road length = 100 m
• Road loading = 12 kg/cm2
• $ of concrete=$500x [(1O)x(lOO)x(12)x(1)/(1OO)]
• = $60,000
• plus remaining algorithms of all other part items of
the model.
CE - L8 25

26. Area Estimate
• This is prepared depending on the basis of the
area of building, the rate being deducted from
the cost of similar building having similar
specification, heights and construction in the
locality.
• It is calculated by finding the area of the
building and multiplying by the unit area rate.
The area is calculated from the covered area
by taking external dimension of the building at
the floor level.
CE - L8 26

27. • The method builds a sorely needed bridge
between nuts-and-bolts unit cost estimating
and esoteric (unclear) parametric analysis.
• This method is most heavily used in trade
specific or general commercial building
construction and plant applications. The
models and parametric relationships are
prebuilt in some systems and left to the user
in others.
CE - L8 27

28. • Courtyard and other open areas should not be
included in the area.
• For multistory buildings there are two
methods to use:
A) All floors including basement floor and roof
have an equal cost
B) Basement and roof have different cost.
CE - L8 28

29. Example 1.
• A building of size 20x35m consist of :
• Basement, g floor,1st floor, 2nd floor and roof.
And the total cost of building is 630,000 $.
Calulate the cost of m2 assuming that:
• A) all floors including basement and roof have
the same cost
• B) cost of basement is 60% of cost of the rest
of floors and cost of roof is 40% of cost of rest
floors.
CE - L8 29

30. a) Assumption 1
• Basement and other floor area = 700x5=3500m2
• Cost per m2 = 630,000/3500= 180$
b) Assumption 2
• basement area = 700 x .6 = 420m2
• Roof area = 700x.4 = 280m2
• Other floors = 700x 3 = 2100
• Total area = 2800m
• Cost of floor =630000/2800 =225$
• Cost of basement = 225x0.6 = 135$
• Cost of roof = 225x0.4 =$90
CE - L8 30

31. Cube Rate Estimate for Building
• It is an approximate estimate and it is
prepared based on cubical contents of the
building.
• The cube rate being deducted from the cost of
similar building having similar specifications
and construction, in the locality.
• Cubic rate is more accurate than the area
because it considers the height of the
building.
CE - L8 31

32. Cubic Method
Example (2)
• A building of size 40x60m and consist of basement,
g floor, 1st floor , 2nd floor. The height of the
buildings are :
• 2.8m for basement. 3.0m for all other floors. The
total cost is $2,250,000. estimate the cost of cubic
meter using the following two assumptions:
• A) the cost of all floors are the same
• B) the cost of cubic meter of basement is %60 of
other floors
CE - L8 32

33. a) assumption 1
• Volume of basement = 40x60x2.8 =
6720m3
• Volume of any other floor = 40x60x3 =
7200m3
• Total volume of all floors = 6720x1+7200x3=
28,320 m3
• Cost of cubic meter = 2,250,000/28320= $
79,450
CE - L8 33

34. b) assumption 2
• Volume of basement = 40x60x2.8x0.60=
4032m3
• Volume of any other floor = 40x60x3= 7200m3
• Total volume of all floors = 4032x1+7200x3 =
25,632 m3
• Cost of cubic meter 2,250,000/22,632= $ 87,780
CE - L8 34