W7_IBRAHEEM_PREFERRED ESTIMATING TECHNIQUE

1.0 Problem definition:  Preferred Estimating Technique (problem 3-21, chapter 3. Engineering Economics)

The cost of building a supermarket is related to the total area of the building. Data for the last 10 supermarket built for Regork, Inc., are shown in the accompanying table from problem 3-21 on chapter 3 of engineering economics.

Table 1.0

2.0 Development of feasible alternative:

The following cost estimating models can be used for cost for building a 23,000 square feet supermarket. Two parametric equations were singled-out from numerous alternatives:

1. Quantitative techniques from selected estimating models

This is inclusive of indexes, unit and factor techniques

2. Parametric Cost estimating techniques

This is inclusive of power sizing techniques and CER.

For the purpose of this analysis, Unit Technique will be used from selected estimating models while CER will be selected from parametric cost estimating techniques.

The two scenarios will be analyzed and the best alternative selected based on degree of accuracy and details involved at conceptual.

3.0 Develop the outcome for each alternative:

Alternative 1 (Unit Factor Method)

Table 1.0 above does not have a single unit factor for multiplication. Calculating the Mean as a common factor

=             [(800000/14500) + (825000/15000) + (875000/17000) +nth value]/n

Where n= 10 in this case

=             539.5099/10

=             $53.95 per square foot. This will be used as the unit factor for cost estimation.

Calculating cost of project

=             $53.95*23,000

=             $1,240,850. 00

Alternative 2 (using CER)

Developing a Cost estimating relationship for the construction of supermarket using linear relation

y= b₀+b₁x

b₀ and b₁ are co-efficient while x is design variable (in this case size in square foot)

b₁=         n ∑x₁*yi – (∑x₁)*(∑y₁)/ n ∑ x₁² – (∑x₁)²

b₀=         (∑y₁ – b₁ ∑ x₁) / n

Substituting variables into relevant equation

n ∑x₁*yi– (∑x₁)*(∑y₁) =  142173450

n ∑ x₁² – (∑x₁)²=                                2761402500

b₁=                                         0.051485957

and substituting for value of b₀

(∑y₁ – b₁ ∑ x) / n =            508.3104238/10

b₀                              =           50.83104

Below is the spreadsheet for general overview of produced result

Table 2.0

Simple linear equation Cost = 50.83104 + 0.051485x

Where x = space area in square- feet.

Substituting 23,000 into the equation, cost equals $1,079.7 (in thousands)

Model validation

Finding the Standard deviation SE from known values

SE= SQRT (∑ y₁ – COST₁)² / n ₁-2

    = 66.79

Solving for correlation coefficient R

Table 3.0

R=∑(x₁ – mean)*( y₁ – mean) /SQRT (∑( x₁ – mean)²*(∑( y₁ – mean²))

  = 0.976475252

Closeness of values to 1 indicates a strong linear relationship between dependent and independent variables.

4.0 Selection of acceptable Criteria:

Acceptable criteria will be based on justification on which of the two alternatives produces a reasonable degree of accuracy in terms of technical requirement of cost estimation as well as strong linear relationship with variables.

5.0 Analysis and Comparison of alternatives:

Unit factor method provides a simple but not so detailed method of cost estimation. Regression method on CER produced a more detailed step-by-step process.

6.0 Selection of preferred alternative:

The most preferred alternative is alternative 2

7.0 Performance Monitoring & Post Evaluation of result:

Other available data will be subjected to cost estimation models to appreciate the different outcomes. Monitoring measures will be put in place for effective future comparison.

References:

Sullivan, W. G., Wicks, E.M., & Koelling, C.P. (2012). Engineering Economy (15^th ed.) (pp 89-122) New Jersey, NJ. Pearson Higher Education, Inc.

  

Purdue OWL APA style. (2011). APA formatting and style guide. Retrieved from http://owl.english.purdue.edu/owl/resource/560/19/

MOJEKWU, J.N (1996): Business Statistics (2^nd  ed.) (pp 87-100.) Ogun, NG. IPS Educational Press