How do I quanitify the number of people who did the work?
August 31, 2010 10:01 AM Subscribe
How do I accurately quantify the number of people who did the work?
Let's say you have 100 units of work. If 5 people each do 20 units of work, then it's fair to say 5 people did the work. However, if 1 person does 96 units, and 4 people do 1 unit each, then it's fair to say ~1 person did the work... but what's the formula?
My current thinking is to sum the amount done by each person divided by the amount done by the person who did the most. So the math would look like:
Case 1: 20/20 + 20/20 + 20/20 + 20/20 + 20/20 = 5
Case 2 96/96 + 1/96 + 1/96 + 1/96 + 1/96 = ~1
That algorithm doesn't feel right, but it seems to give me what I want. Am I missing something obvious?
Thank you!
Let's say you have 100 units of work. If 5 people each do 20 units of work, then it's fair to say 5 people did the work. However, if 1 person does 96 units, and 4 people do 1 unit each, then it's fair to say ~1 person did the work... but what's the formula?
My current thinking is to sum the amount done by each person divided by the amount done by the person who did the most. So the math would look like:
Case 1: 20/20 + 20/20 + 20/20 + 20/20 + 20/20 = 5
Case 2 96/96 + 1/96 + 1/96 + 1/96 + 1/96 = ~1
That algorithm doesn't feel right, but it seems to give me what I want. Am I missing something obvious?
Thank you!
The problem you're having is what metric you're trying to get.
If you're looking for man days, then report in man days (or man months, or man years). You would use this where the pure level of resourcing was important. Or, if you got more granular you could report man days by function, type of person etc.
If you're trying to quantify how many people got involved, which would be an indication of project complexity, or some feature of the project that meant you had to pull in lots of people then use that.
In the first example, if a man month is, say, 21 days and your project took 105 days then it took 5 man months. This doesn't distinguish on the basis of individual input.
posted by MuffinMan at 10:12 AM on August 31, 2010
If you're looking for man days, then report in man days (or man months, or man years). You would use this where the pure level of resourcing was important. Or, if you got more granular you could report man days by function, type of person etc.
If you're trying to quantify how many people got involved, which would be an indication of project complexity, or some feature of the project that meant you had to pull in lots of people then use that.
In the first example, if a man month is, say, 21 days and your project took 105 days then it took 5 man months. This doesn't distinguish on the basis of individual input.
posted by MuffinMan at 10:12 AM on August 31, 2010
Response by poster: I'm using "people" in my example, but the real problem is estimating how many resources were used to get a certain amount of output from two different systems. Basically, I need to know what to divide the output by in order to get apples vs apples.
posted by mpls2 at 10:31 AM on August 31, 2010
posted by mpls2 at 10:31 AM on August 31, 2010
Best answer: Your index is really just the total amount of work done, divided by the amount of work done by the person who did the most.
There are lots of fancy mathematical ways to quantify this. They are called Renyi entropies or diversity indices. As you can see, there are a lot of them both for technical reasons and because they seem to measure slightly different things. (Note that any of the indices in those articles will need some tweaking to get it into the units of "number of people".) Your index is the reciprocal of the Berger-Parker index.
One that is fairly easy to understand and has some of the properties you want, I think, is derived from the Simpson index. This is used in ecology, and is just the probability that two individuals in a population which comes from many different species are of the same species. In what you're doing, you could imagine it as the probability that two "pieces of work" picked at random from the whole job were done by the same person.
To compute it, you just take the sum of the squares of the fractions of work done by each person. So in the first case, you get 5*(20/100)^2 = 0.2, and in the second (.96)^2 + 4*(.01)^2 = .922.
Instead of giving these numbers, I'd give their reciprocals: 5 and ~1.085, respectively. If there are n people who did equal amounts of work, you'll get n; you'll never get a number from this procedure that's more than the number of people.
posted by madcaptenor at 10:37 AM on August 31, 2010 [3 favorites]
There are lots of fancy mathematical ways to quantify this. They are called Renyi entropies or diversity indices. As you can see, there are a lot of them both for technical reasons and because they seem to measure slightly different things. (Note that any of the indices in those articles will need some tweaking to get it into the units of "number of people".) Your index is the reciprocal of the Berger-Parker index.
One that is fairly easy to understand and has some of the properties you want, I think, is derived from the Simpson index. This is used in ecology, and is just the probability that two individuals in a population which comes from many different species are of the same species. In what you're doing, you could imagine it as the probability that two "pieces of work" picked at random from the whole job were done by the same person.
To compute it, you just take the sum of the squares of the fractions of work done by each person. So in the first case, you get 5*(20/100)^2 = 0.2, and in the second (.96)^2 + 4*(.01)^2 = .922.
Instead of giving these numbers, I'd give their reciprocals: 5 and ~1.085, respectively. If there are n people who did equal amounts of work, you'll get n; you'll never get a number from this procedure that's more than the number of people.
posted by madcaptenor at 10:37 AM on August 31, 2010 [3 favorites]
oh, great, I killed this question with math.
posted by madcaptenor at 11:03 AM on August 31, 2010
posted by madcaptenor at 11:03 AM on August 31, 2010
That's pretty cool, madcaptenor. You can infer the following interesting (?) property about your modified Simpson index: suppose you had one person do x of the work and the other n-1 people split the remaining 1-x amount of work evenly. If the first person did 0 of the work, the index gives n-1, as expected. However, the index also returns n-1 if the first person did 2/n of the work. For example, if you have one person doing half the work and three other people doing 1/6 of the work, the index says that 3 people did the work (i.e.: 1 / [ (1/2)^2 + 3*(1/6)^2 ] = 3).
On the flip-side, we might ask how much work the first person must do in order to take credit for all the work. Well, the index is equal to one if and only if one person does all the work, so that's not illuminating. However, we can ask instead how much work does the first person have to do in order to get the index below 1.5 (i.e.: rounded to the nearest integer, it's one). Well, messing around with Wolfram Alpha seems to indicate that if the first person does ~80% of the work (it varies with the total number of people), they get to take full credit, more or less.
posted by mhum at 12:27 PM on August 31, 2010
On the flip-side, we might ask how much work the first person must do in order to take credit for all the work. Well, the index is equal to one if and only if one person does all the work, so that's not illuminating. However, we can ask instead how much work does the first person have to do in order to get the index below 1.5 (i.e.: rounded to the nearest integer, it's one). Well, messing around with Wolfram Alpha seems to indicate that if the first person does ~80% of the work (it varies with the total number of people), they get to take full credit, more or less.
posted by mhum at 12:27 PM on August 31, 2010
Alright, I'll bite at the question implicit in what mhum said: say there are n+1 people, person 1 does x total work, and each other person does (1-x)/n. Then my modified index comes out to n/(x^2*n+1-2x*x^2). Setting this equal to 3/2 and solving the quadratic, you get (3+sqrt(6n^2-3n))/(3(n+1)). As n goes to infinity this approaches sqrt(6)/3, or sqrt(2/3), or about 0.816.
This is exactly what you'd expect, now that I think about it. The sum of the squares of the amounts of work done by everybody needs to be no more than 2/3 to get the index equal to 1.5. And if there are lots of other people then all their contributions are small and they don't count much, so all the contribution comes from the first person.
posted by madcaptenor at 1:03 PM on August 31, 2010
This is exactly what you'd expect, now that I think about it. The sum of the squares of the amounts of work done by everybody needs to be no more than 2/3 to get the index equal to 1.5. And if there are lots of other people then all their contributions are small and they don't count much, so all the contribution comes from the first person.
posted by madcaptenor at 1:03 PM on August 31, 2010
Response by poster: Thanks, madcaptenor! You've been very helpful.
posted by mpls2 at 1:57 PM on August 31, 2010
posted by mpls2 at 1:57 PM on August 31, 2010
This thread is closed to new comments.
I'd say in both examples five people did the work, just that in the first example they split the effort evenly, while in the second example one did a majority of it with small contributions from each of the others.
In most contexts there is a sort of threshold where people don't mind if their small contribution is not formally recognized, but there is no formula for this. In part this is because all work is not equal. When I go to the dentist a majority of the work is done by the receptionist and hygenist, but I wouldn't leave out the dentist himself, even though his contribution of units of work to my case might be quite small.
posted by meinvt at 10:10 AM on August 31, 2010