Taking a Weighted Average

Date: 03/20/2003 at 21:11:54
From: Matt
Subject: Two Different Ways to Take a Weighted Average

I am currently doing a study of truck availability and Use of 
Availability for a work project. It involves taking and combining the 
following data to get the total number of hours the truck fleet 
operates per day:
              
Truck A: Availability = 80%, Use of Availability = 90%
Truck B: Availability = 90%, Use of Availability = 70%
Truck C: Availability = 95%, Use of Availability = 80%

To determine how many hours each truck operates per day the equation 
is (24 * Availability * Use of Availability).

I can't figure out which is the best way to get the average operating 
hours for the fleet. Should I find the hours each truck operates and 
sum the values or can I find the average Availability and Use of 
Availability for the Fleet and get the answer that way?  I tried both 
ways and get slightly different numbers.  Is one method incorrect or 
is it just a question of significant digits?

Trial #1 - Find individual hours then add together:

Truck A = (24 * 0.8 * 0.9) = 17.28 hours
Truck B = (24 * 0.9 * 0.7) = 15.12 hours
Truck C = (24 * 0.95 * 0.8) = 18.24 hours
Truck Fleet (A + B + C) = (17.28+15.12+18.24) = 50.64

Trial #2 - Find Fleet averages then calculate hours:

Truck Fleet Availability = (0.8*0.9*0.95)/3 = 0.8833
Truck Fleet Use of Avail = (0.9*0.7*0.8)/3 = 0.8

Truck Fleet Hours = (3 * 24 * 0.8833 * 0.8) = 50.88 hours

Thank you!

Date: 03/21/2003 at 10:59:23
From: Doctor Douglas
Subject: Re: Two Different Ways to Take a Weighted Average

Hi, Matt,

Thanks for submitting your question to the Math Forum.

How to properly take averages in different situations usually comes 
down to what question is being asked, or what quantity you want to 
estimate. Sometimes this is a tricky business to get right.

Because you asked for "the total number of hours the truck fleet 
operates per day," you can simply add up the individual trucks, as you 
did in method 1.  This is certainly a "weighted average," because each 
truck's use is being weighted by its availability. I think method 1 is 
correct.

Now, what is method 2? The average availability is 0.8833, and the 
average use (irrespective of availability) is 0.8. You could multiply 
these two together, and the answer would provide an estimate of hours 
based on an *imaginary* truck whose availability is the average 
availability and whose use is the average use. This is probably a more 
complicated statistic than you need. It doesn't take into account 
correlations between the availability and the use, which could be 
important (in fact they probably are - see the example below). And 
most importantly, method 2 doesn't answer your question as directly as 
method 1 - you directly compute the total number of hours (by 
computing the hours for each truck) in method 1, not just estimate a 
fleet average based on the three imaginary trucks.

To see how the "correlations" I mentioned could be significant,
imagine you have the following three trucks:

   availability       use
       98%            10%
       50%             0%
        2%           100%

Method 1 gives the total number of hours for all three trucks as  
24(.98*10 + .50*0 +  0.02*1) = 2.832 hours.  But method 2 gives the
following:  24*(.98 + .5 + .02)*(.1 + 0 + 1)/3 = 13.2 hours. How can 
this be right? None of the trucks manages more than 24*0.98*0.10 = 
2.352 hours individually. The answer lies in how the use and 
availability "correlate." The average availability is 50%, and the 
average use is a shade above one-third. The use average has been 
"skewed" by the high use number of the third truck, whose availability 
is only 2% in the first place.  If there had been a fourth truck with 
availability 1% and use 100%, this problem would have been even more 
dramatic.

The bottom line is that it is best to use the data when available, and 
to be careful when doing too much manipulation of "average" values.  

I hope this helps.  Please write back if you have more questions 
about this!

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/