Taking a Weighted AverageDate: 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/ |
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