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Using Weighted Criteria to Make Decisions

Date: 07/18/2008 at 19:37:32
From: AS
Subject: Finding a result by assigning weights

Hypothetical Example: 
I want to award my employee for highest attendance and learning new 
skills.  I have 3 employees who worked the following days in a month:

Employee 1: 10 days
Employee 2: 15 days
Employee 3: 20 days

Employee 1: Learned 2 new skills
Employee 2: Learned 2 new skills
Employee 3: Learned 1 new skill

Weighting: 50% for attendance and 50% for new skills learned therefore 
best employee is employee 3

Employee 1 score is .5(10)+.5(2) = 6.5
Employee 2 score is .5(15)+.5(2)= 8.5
Employee 3 score is .5(20)+ .5(1)= 10.5

Therefore Employee 3 gets that best employee award.  I somehow feel 
this is not the right method of calculation.

Am I able to calculate averages as above where skill and attendance 
are apples and oranges (different)?

I looked up weighted average but that only shows calculating weights 
for one type of quantity.

Date: 07/19/2008 at 22:34:47
From: Doctor Wilko
Subject: Re: Finding a result by assigning weights

Hi AS,

Thanks for writing to Dr. Math!

This is a very practical question in business and government.  Often a 
decision maker has to prioritize many alternatives (picking which 
employee gets the quarterly award), where the candidates can be ranked 
by multiple criteria (attendance record, skills learned, etc.).

If you were prioritizing your employees strictly based on attendance,
Employee 3 gets the reward because he has the highest attendance.

But, you actually have two criteria to rank your employees against:
attendance record and skills learned.  Now it's not so clear if 
Employee 3 still deserves the award.  He does have the highest
attendance, but he only learned one new skill.  Employee 2 on the 
other hand learned two new skills, but worked five less days.  Now 
it's getting a little fuzzy on which employee should be rewarded.  Now
suppose you have 20 employees and there are four or five criteria to
pick an award winner; now the selection is almost impossible to do
fairly.  This is where the decision maker needs some objective method
to help him make a fair decision.

Your decision is to pick the award winner from the three employees,
based on two criteria, attendance record and skills learned.  If we
use a decision tree to model your problem, it looks as follows:

                     Attendance  Skills
                       Days      Learned 
             / E1        10        2   
  Decision ----- E2      15        2 
             \ E3        20        1

There are two main components we'll look at to model your decision.
  1. You have to measure all criteria on similar numerical scales (to
     compare apples to apples), and

  2. You have to assign your importance (weights) to those criteria
     that will be used to rank the employees.

Once you have these two components, you can calculate a weighted
average for each employee.  This final weighted average will allow you
to rank your employees in order to determine who gets the reward.

Let's do these two steps now.

1. Measure all Criteria on Similar Numerical Scales
The first thing we'll do is use the same scale to measure both 
criteria, say a 0-100 scale, where 0 is the worst or least desired
outcome for each criterion and 100 is the best or most desired outcome
for each criterion.

Criteria #1, Attendance:

Assume for attendance, 5 days is the worst outcome and 30 days is the
best outcome for your employees.  5 days of attendance gets mapped to
the score 0 and 30 days of attendance gets mapped to a score of 100.

To get the scores for the intermediate days, you could use 
"proportional scoring" as one technique.  For example, Employee 1 is
20% of the way from the lowest to the highest value 
[(10-5)/(30-5)=0.20, and 100*0.20 = 20%], so Employee 1 gets a score
of 20 for his 10 days of attendance.  Likewise, Employee 2's attendance
gets mapped to a score of 40, and Employee 3's attendance gets mapped
to a score of 60.

Remember, we're trying to map days of attendance onto a scale from
0-100.  The mapping looks as follows:

  E1, 10 days --> score of 20
  E2, 15 days --> score of 40
  E3, 20 days --> score of 60

Criteria #2, Skills Learned:

Now, let's map the number of skills learned onto the same 0-100 scale
using the same proportional weighting technique.  First set your
endpoints of the scale.  0 skills learned gets a score of 0 and say 4
skills learned gets a score of 100.  Now, using proportional weighting
the mapping looks as follows:

  E1, 2 skills --> score of 50
  E2, 2 skills --> score of 50
  E3, 1 skill  --> score of 25

Are you seeing how we can compare apples to apples?  I have taken two
completely different criteria and put them on the same scale so
comparing them makes sense.

Now, if I re-draw my decision tree from above, it looks like:

                     Attendance  Skills
                       Score      Score
             / E1        20        50   
  Decision ----- E2      40        50 
             \ E3        60        25

2. Weight the Criteria

Next, you have to decide what's more important between attendance
record and skills learned.  Is a score of 50 on the attendance scale
the same as a score of 50 on the skills scale?  It might be, but this
is where you as the decision maker assign your importance to the

Importance is assigned to the criteria through weights.  These weights
will allow you to calculate a weighted average using the two criteria
to get an overall score for each employee.  Then you can rank the
employees by the overall score to determine who gets the award.

It is your call concerning the relative importance of the two 
criteria.  It is important however to consider the ranges of the
criteria.  The weights should reflect the relative value of going from
best to worst on each scale.  For example, if improving attendance 
from 5 days to 30 is three times more important than going from zero
skills learned to four skills learned, then this implies you weight
attendance = 0.75 and skills = 0.25.  But to go with your proposal
that the criteria are equal, you'd weight attendance = 0.50 and skills
= 0.50.

Now with your weights chosen, you can finally calculate the weighted
averages to get the overall score of each employee.  The decision tree
will look as follows:

                        0.50      0.50
                     Attendance  Skills                  Overall
                       Score      Score                   Score
             / E1        20        50  = .50(20)+.50(50) = 35
  Decision ----- E2      40        50  = .50(40)+.50(50) = 45
             \ E3        60        25  = .50(60)+.50(25) = 42.5

Now you can rank your employees according to the final weighted
average to see who shall receive the award:

  E2 = 45   (highest overall score)
  E3 = 42.5
  E1 = 35   (lowest overall score)

I'd say before accepting this as the final answer, especially when you
are building the model the first time, you'd step back and look at the
scales you constructed and the weights you applied to the criteria. 
This is called "Sensitivity Analysis".  It is easy to see that if
Skills was weighted higher, Employee 3 may have come out the winner. 
Also, if the Skills score was a little higher than 25, Employee 3
might have received the highest overall score.  Just be aware of this
as you construct your scales for the criteria and as you assign
weights to the criteria.

Model building is usually an iterative process, but once you've tested
your model and you feel confident it has captured your priorities, you
can use it objectively to rank all your employees to see who gets the
award.  I think this is where the value of this method lies; in the
end you've built a transparent, standardized, and repeatable process
for selecting employees for the award.

If you're interested in this topic, try looking up articles related to
decision analysis, value-focused thinking, and multiple-objective
decision making to name a few.

Does this help?  Please write back if you still have questions.

- Doctor Wilko, The Math Forum 
Associated Topics:
High School Number Theory
High School Statistics

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