A company is evaluating Kansas City and Atlanta as alternative locations for a new plant to manufacture PC’s for small businesses. The following information has been collected. They want to use this information to compare the two locations. Different SME’s (subject matter experts) scored each of the non-economic factors. So, unfortunately, the scores for the non-economic factors are scaled differently. The scales are indicated on the table. On each scale, the highest score is always the best value. For example, for the scale 0-4, the best value is 4. For each of the Non-economic factors, it is possible to score the maximum.
Critical Success Factor |
Factor Weight |
Scale |
Kansas City |
Atlanta |
Cost per computer |
0.50 |
$3,900 |
$4,300 |
|
Non-economic factors |
||||
Cost of living |
0.10 |
0-2 |
1.2 |
1.4 |
Labor availability |
0.10 |
0-3 |
2.1 |
2.1 |
Union activities |
0.15 |
0-4 |
1.6 |
2.4 |
Proximity to similar industries |
0.10 |
0-2 |
1.4 |
1.0 |
Local transportation systems |
0.05 |
0-2 |
1.4 |
1.8 |
a) Using the factor scoring (rating) method as we learned in class, which site should be selected?
b) Suppose that the company wants to consider Omaha as a third site. Although the cost per computer for Omaha has not been determined yet, the company does know it will be at least $3,900. The scores for the qualitative factors have been determined and are the following:
Critical Success Factor |
Omaha |
Cost per computer |
??? |
Cost of living |
1.4 |
Non-economic factors |
|
Labor availability |
1.5 |
Union activities |
2.2 |
Proximity to similar industries |
1.0 |
Local transportation systems |
1.2 |
Given the information provided in the problem and the additional information provided in this part of the problem, what is the Cost per computer for Omaha that makes it factor rating value the same Kansas City’s value?
The following problem is solved by following method.
The factor rating method states that the factor rating which has the highest total score is the best alternative
so the site with highest total score must be selected as our site.
The Factor weight is weighted by the expert as their own measure.
The factor with most importance is weighted highest
The summation of Factor weight should be 1
Part 1
1. calculate the X score which is a product of factor weight (FW) and Kansas city scale (Kscale) of kansas city.
2.Similarly Calculate the X score which is a product of factor weight (FW) and Atlanta(Ascale) of Atlanta.
3. calculate the total score of each city by summation of Xscore as shown below
Critical Success Factor |
Factor Weight (FW) |
KansasCity (Kscale) |
X score (FW*kscale) |
Atlanta (Ascale) |
Xscore (FW*Ascale) |
Cost per computer | 0.5 | $3,900 | 0.5x3900=1950 | $4,300 | 0.5x4300=2150 |
Non-economic factors | |||||
Cost of living | 0.1 | 1.2 | 0.1x1.2=0.12 | 1.4 | 0.1x1.4=0.14 |
Labor availability | 0.1 | 2.1 | 0.1x2.1=0.21 | 2.1 | 0.1x2.1=0.21 |
Union activities | 0.15 | 1.6 | 0.15x1.6=0.24 | 2.4 | 0.15x2.4=0.36 |
Proximity to similar industries | 0.1 | 1.4 | 0.1x1.4=0.14 | 1 | 0.1x1=0.1 |
Local transportation systems | 0.05 | 1.4 | 0.05x1.4=0.07 | 1.8 | .05x1.8=0.09 |
Total Score | 1950.78 | 2150.9 |
Here we can see that Atlanta has the highest total score so we will select Atlanta as our site
Part 2
Critical Success Factor | Factor Weight(FW) | Ohmaha (Oscale) | xscore (FWxOscale) |
Cost per computer | 0.5 | 3900 | 0.5x3900=1950 |
Non-economic factors | |||
Cost of living | 0.1 | 1.4 | 0.1x1.4=0.14 |
Labor availability | 0.1 | 1.5 | 0.1x1.5=0.15 |
Union activities | 0.15 | 2.2 | 0.15x2.2=0.33 |
Proximity to similar industries | 0.1 | 1 | 0.1x1=0.1 |
Local transportation systems | 0.05 | 1.2 | 0.05x1.2=0.06 |
Total score=1950.78
Here the Summation of scale of Omaha is equal to that of kansas city
Above is the calculation of the Factor rating method of Omaha.
The Total scale summation of Omha = FWxOscale
The Total scale summation = 0.1x1.4+0.1x1.5+0.15x2.2+0.1x1+0.005x1.2
=0.78
The Total scale summation of kansas city =FWxKscale
The Total scale summation of kansas city=0.1x1.2+0.1x2.1+0.15x1.6+0.1x1.4+0.05x1.4
=0.78
As we can see that the Total scale summation of both kansas city and Omaha is equal we can infer that the cost per computer in Omaha is $3900. As Both Kansas city and Omaha has same scale score
So when we consider the cost per computer of Omaha with its factor weight we will get atleast $3900
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