LCM 7 Parametric

8/10/2019 LCM 7 Parametric

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Project Cost & Finance Management


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Parametric Estimating

Parametric cost estimating is the use of historical cost dataand statistical techniques to predict future costs

Statistical techniques are used to develop cost estimating relationships(CERs) that tie the cost or price of an item (e.g., a product, good,service, or activity) to one or more independent variables (i.e., costdrivers)

Parametric models are used in the early design stages to get an idea of

how much the product (or project) will cost, on the basis of a fewphysical attributes (such as weight, volume, and power)

The output of the parametric models (an estimated cost) is used togauge the impact of design decisions on the total cost


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Various statistical and other mathematical techniquesare used to develop the CERs

For example, simple linear regression and multiplelinear regression models, which are standard statisticalmethods for estimating the value of a dependent variable(the unknown quantity) as a function of one or more

independent variables, are often used to develop estimatingrelationships


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Various statistical and other mathematical techniquesare used to develop the CERs

For example, simple linear regression and multiplelinear regression models, which are standard statisticalmethods for estimating the value of a dependent variable(the unknown quantity) as a function of one or more

independent variables, are often used to develop estimatingrelationships


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Examples of Cost Drivers Used inParametric Estimating

Product Cost Driver (Independent Variable)Construction Floor space, roof surface area, wall surface

areaTrucks Empty weight, gross weight, horsepower

Passenger car Weight, wheel base, passenger space,horsepower

Turbine engine Maximum thrust, cruise thrust, specific fuelconsumption

Mechanical Engine Piston displacement, compression ratio,

horsepowerAircraft Empty weight, speed, wing areaPower plants KiloWattsMotors HorsepowerSoftware Number of lines of code


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Parametric Estimating Techniques

Unit Technique

Factor Technique

Power Sizing Technique

Cost Estimating Relationship


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Parametric Estimating Techniques

Unit Technique

Theunit techniqueinvolves using a per unit factor that canbe estimated effectively

Examples are as follows:1.Capital cost of plant per kilowatt of capacity2.Revenue per mile

3.Capital cost per installed telephone4.Revenue per customer served5.Temperature loss per 1,000 feet of steam pipe6.Operating cost per mile

7.Construction cost per square foot


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Unit Technique-Example

Suppose that we need a preliminary estimate of thecost of a particular house. Using a unit factor of, say,$95 per square foot and knowing that the house is

approximately 2,000 square feet, we estimate its cost tobe

$95 2,000= $190,000

While the unit technique is very useful for preliminaryestimating purposes, such average values can bemisleading.In general, more detailed methods will result in greater

estimation accuracy.


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In 2008, your multinational firm needs to acquire a total of 50,000 squarefeet of additional office space in London, Hong Kong, and New York City.

You have been tasked with developing a cost estimate for 20,000 square

feet of space in London, 10,000 square feet of space in Hong Kong, and20,000 square feet of space in New York City.

While researching this topic, you find an article inUSA Todaythat providesthe following annual cost estimates per square foot of rental space in the

three cities: London, $233.57; Hong Kong, $201.29; and New York City,$175.00.

How much should your firm expect to pay per year for the 50,000 squarefeet of office space?



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The rental expense per year will be approximately 20,000$233.57 in London plus 10,000$201.29 in Hong Kong plus20,000$175.00 in New York City.

The total expense will be $10,184,300 per year for theadditional office space.



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The factor technique is an extension of the unit method inwhich we sum the product of several quantities or componentsand add these to any components estimated directly.

That is,

C= Cd+ fmUm,d m

WhereC= cost being estimated;Cd= cost of the selected componentestimated directly;fm= cost per unit of componentm;

Um= number of units of componentm.

Factor Technique


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As a simple example, suppose that we need a slightly refinedestimate of the cost of a house consisting of 2,000 squarefeet, two porches, and a garage.

Using a unit factor of $85per square foot, $10,000per porch,and $8,000 per garage for the two directly estimatedcomponents,

We can calculate the total estimate as

Total Cost = Cost of Porches + Cost of Garage + Cost ofHouse

= $10,000 2+ $8,000+ $85 2,000

= $198,000

Factor Technique - Simple Example


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Thepower-sizing technique, which is sometimes referred to asanexponential model, is frequently used for developing capitalinvestment estimates for industrial plants and equipment.

This CER recognizes that cost varies as some power of thechange in capacity or size.

That is;

CA/ CB= ( SA/ SB)X

CA= CB( SA/ SB)X


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CA= CB( SA/ SB)X

Where:

CA= cost for plant A CB = cost for plant B (both in $ as of the point in time for which the estimate is

desired);

SA = size of plant A SB = size of plant B (both in same physical units); x= cost-capacity factorto reflect economies of scale

The value of the cost-capacity factor will depend on the type of plant or equipmentbeing estimated.

For example, X = 0.68 for nuclear generating plants and 0.79 for fossil-fuelgenerating plants.x< 1indicates decreasing economies of scale (each additional unit of capacity costsless than the previous unit),x> 1 indicates increasing economies of scale (each additional unit of capacity costsmore than the previous unit), and

x= 1indicates a linear cost relationship with size.


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Power Sizing Technique-Example

Suppose that an aircraft manufacturer desires to make a preliminaryestimate of the cost of building a 600-MWfossil-fuel plantfor the assemblyof its new long- distance aircraft. It is known that a 200-MW plantcost $100million 20 years ago when the approximate cost index was 400, and thatcost index is now 1,200. The cost-capacity factor for a fossil-fuel power plant

is 0.79.


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Power Sizing Technique-Example

Before using the power-sizing model to estimate the cost of the600-MW plant(CA), we must first use the cost indexinformation to update the known cost of the 200-MW plant 20years ago to a current cost.

Using Indexing, we find that the current cost of a 200-MWplant is

CB= $100 million( 1,200/ 400) = $300 million.

We obtain the following estimate for the 600-MW plant:CA= $300 million( 600-MW / 200-MW)

0.79CA= $300 million 2.38= $714 million.


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Cost Estimating Relationship (CER)

A CER is a mathematical model that describes the cost of an engineeringprojectas a function of one or more design variables.

CERs are used to make early design decisions that are cost-effective in

addition to meeting technical requirements.

There are four basic steps in developing a CER:

1. Problem definition

2. Data collection and normalization

3. CER equation development

4. Model validation and documentation


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A CER is a mathematical model that describes the cost of an engineeringproject as a function of one or more design variables.

CERs are used to make early design decisions that are cost-effective inaddition to meeting technical requirements.

There are four basic steps in developing a CER:

1. Problem definitionMust be defined clearly and explicitly

2. Data collection and normalizationCollection: Reliability of Data for accurate resultsNormalization: Data is normalized to account for differences due to inflation,

geographical location, labor rates, and so on.


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3. CER equation development

Type of Relationship Generalized Equation

Linear Cost = b0+ b1x1+ b2x2+ b3x3+.

Power Cost = b0+ b1x1b11+ b2x2

b12 + .

Logarithmic Cost = b0+ b1(log x1)+ b2(log x2)+.

Exponential Cost = b0+ b1eb11

x1+ b2e

b22

x2+


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Linear Relationship

Type of Relationship Generalized Equation

Linear Cost= b0+ b1x1+ b2x2+ b3x3+ .

All of the equation forms can easily be transformed into alinear form.

The above equations can be used to calculate the values of

the coefficientsb0andb1in the simple linear equation

y= b0+b1x


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Linear Relationship -Example

A company has fixed costs of $7,000 for plant andequipment and variable costs of $600for each unit ofoutput. What is total costat varying levels of output?


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Linear Relationship -Example

Let x= units of output

Let C= total cost

C= fixed cost + variable cost

= 7,000 + 600 x


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Linear Relationship

y= b0+ b1x

n n n n n

b1 = [nxiyixiyi] / [nxi2

(xi)2

]i = 1 I = 1 I = 1 I =1 I =1

n n

b0 = [yi- b1xi ] / nI=1 I=1


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Linear Relationship-ExampleCost Estimating Relationship (CER) for a Spacecraft

y= b0+ b1x

In the early stages of design, it is believed that the cost of a Martian roverspacecraft is related to its weight. Costand weightdata for six spacecraft havebeen collected and normalized and are shown in the table. A plot of the data

suggests a linear relationship.

Use a linear relationship model to determine the values of the coefficients for theCER.

Spacecraft Weight (lb) Cost ($ million)

i xi yi1 400 2782 530 4143 750 5574 900 6895 1,130 740

6 1,200 851


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1 400 278 111200 160000 77284

2 530 414 219420 280900 171396

3 750 557 417750 562500 310249

4 900 689 620100 810000 474721

5 1130 740 836200 1276900 547600

6 1200 851 1021200 1440000 724201

4910 3529 3225870 4530300 2305451

b1= 0.659b0= 48.282

Costi= 48.282 + 0.659xi

wherexrepresents the weight of the spacecraft in pounds, and 400x 1,200

=

xi yi xi yi xi2 yi

2


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Model Validation determines how well the CER can predict cost (i.e., modelvalidation) and document the development and appropriate use of the CER

Validation can be accomplished with statisticalgoodness of fit measures such asStandard Error (SE) and the Correlation Coefficient.

The SE is calculated by;

n

SE= [(yiCosti) 2 /n2]I=1

where Costi is the cost predicted by using the CER with the independent variablevalues for data setiandyiis the actual cost.

A small value of SE is preferred.


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The correlation coefficient(R) measures the closeness of the actual data points tothe regression line (y= b0+ b1x).

It is simply the ratio of explained deviation to total deviation.

n n n

R= (xix)(yi - y) / [ (xi x) 2 ] [ (yi -y) 2 ]i =1 i =1 i =1

n n

wherex= 1/n xi andy= 1/n yii =1 i =1

The sign (+/-) ofRwill be the same as the sign of the slope (b1) of the regression

line


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=


xi Cost yi (yi - Costi)2400 311.882 278 1147.989

530 397.552 414 270.536

750 542.532 557 209.323900 641.382 689 2267.473

1130 792.952 740 2803.914

1200 839.082 851 142.038

= 6841.276

SE = SQRT (6841.276/4)

SE = 41.356


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129753.0556 16835855426

50218.05556 2521853104

2129.722222 4535716.744

8234.722222 67810650.08

47321.38889 2239313846

100314.7222 10063043495

337971.6667

(xi x)(yi - y)

(xi x) (yi - y)

400 278 -418.3333333 -310.166667 175002.7778 96203.3611

530 414 -288.3333333 -174.166667 83136.11111 30334.0278

750 557 -68.33333333 -31.1666667 4669.444444 971.361111

900 689 81.66666667 100.833333 6669.444444 10167.3611

1130 740 311.6666667 151.833333 97136.11111 23053.36111200 851 381.6666667 262.833333 145669.4444 69081.3611

512283.3333 229810.83

xi yi (xi x)2 (yi -y)

2

(xi x)2(yi -y)

2

=

=



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0.98500839SE = 41.35 R =

Standard Error (SE) & The correlation coefficient(R)


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Problem Statement

Reef Office Supplies is interested in estimating the relationship between customer

service costs and sales. The following data are available:

Customer

Month Service Cost Sales

May $6,000 $100,000

June $6,500 $140,000July $7,300 $170,000

August $10,200 $200,000

September $10,800 $225,000

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LCM 7 Parametric