Cost Engineering

Dr. Nabil I El Sawalhi

Assistant Professor

Construction Management

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Estimating Methods

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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.

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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:

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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

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Estimating Methods

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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.

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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

• 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.

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2. Conference

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.

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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.

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• 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

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• 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

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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 band 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.

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• 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.

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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

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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.

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• 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

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• 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.

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Design Weight ,Ib

Ci

$

2 20

3 30

4 60

∑ 9 110

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The cost per pound is Ca = ∑ci/ ∑ni =110/9=$12.22

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

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• 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.

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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.

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• 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.

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• 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.

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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.

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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

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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.

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Cubic MethodExample (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

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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

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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

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