Sunday, 18 November 2012

Math @Work: Importance of 'weightage' in finding averages

I have come across the concept of ‘score card’ for employees in an organization. It is an employee performance-assessment tool. Each employee’s work is categorized in to some parts and each part is given some weightage depending on its significance. The employee is allotted with a score in each part of the work depending on his performance in that work. Part-wise performance-percentages and overall performance-percentage are calculated and based on these parameters, his salary-hike is decided for the coming fiscal. Can we take the average of all the part-performance percentages to get the overall-performance figure?

Let us consider the following scenario:

The deciding parameters are Work Efficiency, Punctuality, Analysation skills and Problem solving skills each having equal weightage of 25%. And an employee has got the following scores:
 Parameter Weightage Employee score Part % Work efficiency 25 20 80% Punctuality 25 25 100% Analyzation skills 25 15 60% Problem solving skills 25 20 80%

The overall performance can be calculated straight-away like

Total score achieved by employee/ total max. Score = (20+25+15+20)/(25+25+25+25)

= 80%
I can also find it by taking average of part percentages:
(80+100+60+80)/4 = 320/4 = 80%
Now I change the weightages allotted to the parameters like this:
 Parameter Weightage Employee score Part % Work efficiency 50 40 80% Punctuality 20 20 100% Analyzation skills 10 6 60% Problem solving skills 20 16 80%
Overall performance % = (40+20+6+16)/(50+20+10+20) = 82%
But if I do it by taking average of individual percentages, I will get (80+100+60+80)/4 = 320/4 = 80%
Where has it gone wrong?
Here comes the concept of weightages. I need to take care of the weightage of each part. In the first problem, the weightage given to each part is 25 and hence weightage fraction for each part is 25/100 = ¼. So when I say that the overall average is (80+100+60+80)/4, it really meant this:
“(1/4)*80 +(1/4)*100 +(1/4)*60 +(1/4)*80” Here I multiply the part-percentages with the respective weightage.
Coming to the second problem, the weightage for the first part is 50/100 = ½, weightage for second part is 20/100 = 1/5, weightage for third part is 10/100 = 1/10 and weightage for fourth part is 20/100 = 1/5.
So the overall-performance = (1/2)*80 + (1/5)*100 + (1/10)*60 + (1/5)*80 = 40+20+6+16 = 82%