Target Average Price

• Apr 3rd 2009, 12:17 PM
squelch
Target Average Price
If I have 100 apples at different weights, how do I share them out between a given number of people to a given ratio, so that each has as close as possible the same average weight per apple.

For example - 3 people require 20, 30 and 50 apples respectively. There are 100 apples of various known weights.

The average weight of each persons apples needs to be as close as possible.

I am afraid I don't know which branch of maths this falls under. Any help appreciated!
• Apr 9th 2009, 08:41 PM
TKHunny
I'm going with "applesauce".

You need more information, I think. There are far too many solutions.
• Apr 10th 2009, 01:06 PM
squelch
Hi TKHunny

Good thought - but the apples cannot be cut up! There are many permutations I know, but this is the only information you get:
• The number of apples
• The weight of each apple
• The number of people
• The number of apples each person requires
When the variation in apple weight is realistic, I can get the correct result. However, if I have a couple of really (ridiculously) heavy apples the result can be bad.
• Apr 10th 2009, 04:42 PM
TKHunny
Quote:

Originally Posted by squelch
I can get the correct result.

I continue to struggle with this concept. I think it highly unlikely that there is "the" correct result. I takes relatively few apples and people before there are many appropriate solutions. If a situation truly has an optimal solution, there must first be a definition of "optimal". Given a definition, the apples sizes also probably are derived to accomodate.
• Apr 11th 2009, 12:42 AM
CaptainBlack
Quote:

Originally Posted by squelch
Hi TKHunny

Good thought - but the apples cannot be cut up! There are many permutations I know, but this is the only information you get:
• The number of apples
• The weight of each apple
• The number of people
• The number of apples each person requires
When the variation in apple weight is realistic, I can get the correct result. However, if I have a couple of really (ridiculously) heavy apples the result can be bad.

One algorithm might be:

Assume there are sufficient apples.

Give the first person the number of apples they require choosen to be the lightest apples.

Now repeat for the second person etc.