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First of all, I must say I'm in no way a data scientist or anything; I just happened to come across a problem, and I am attempting to use statistics to find the best solution.

The problem is about splitting each of six sets of items in two, with each combination having four parameters, and finding the split that minimizes four parameters.
Specifically, I'm trying to optimize the book schedule, so that my desk mate and I can bring the required schoolbooks to school with the least effort - that is, with

  • minimal total weight for me $W_m$,
  • minimal total weight for my desk mate $W_d$,
  • minimal swapped books for me $S_m$, and
  • minimal swapped books for my desk mate $S_d$.

How should I seek the best splitting, i.e. the combination that minimizes all four parameters?

My first thought was simple weighted Euclidean distance: simply assign a score $S = \sqrt{\alpha(W_m^2+W_d^2)+\beta(S_m^2+S_d^2)}$ to each combination, and then evaluate the $n$ lowest scores for some weights $(\alpha,\:\beta)$. However, my concern is that such a score would possibly give too much weight to suboptimal solutions, no matter the choice of weights. This seems especially true considering that [I read somewhere that] the Euclidean distance tends to favour closer points much more than farther ones.

Perhaps an improvement could be computing the average weight $W=\frac{W_m+W_d}2$ and the average swap number $S=\frac{S_m+S_d}2$, and then computing the weighted Euclidean distance between these, $S=\sqrt{\alpha W^2+\beta S^2}$.

Any thoughts?

whuber
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Giulio Muscarello
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    It is rare that one can simultaneously achieve four objectives optimally. You have to stipulate how to trade off improvements in some objectives against weakening of others. This is not a question of optimization, because it comes beforehand: it is a matter of your *values* and as such cannot be answered here. We can provide some guidance for how to discover and quantify your values: see *inter alia* http://stats.stackexchange.com/questions/9137 and http://stats.stackexchange.com/questions/9358. – whuber Oct 21 '14 at 19:00
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    In engineering, we do this all the time it is called [multiple objective optimization](http://en.wikipedia.org/wiki/Multi-objective_optimization). You would not be able to optimize all at once, and there will be several sub optimal solutions. The key is to find a [pareto optimal front](http://en.wikipedia.org/wiki/Pareto_efficiency) and then you could decide and pick an optimal solution. – forecaster Oct 21 '14 at 19:47
  • It depends, among other things, on how your preferences (you and your desk-mate's) weight up the various combinations of weight carried and books swapped. If your opposite number has to walk up a lot of hills, a little weight more or less may loom much larger than a substantial amount of swapping. If they come by car, a bit of weight may not matter as much as the effort of swapping. We can't tell you how you should prefer those tradeoffs. – Glen_b Oct 22 '14 at 00:23

1 Answers1

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Without a rubric there is no such thing as "best". Best has baggage.

The classic example is "what is the best path"? To answer it you have to know what is at the end. Are you walking to a Nobel prize or your execution?

Some bests oppose each other. Shortest time means maximum energy expenditure. Least energy means you take a while to get there.

The question behind the question seems to be about how weights combine, or how number of books combine.

Do you care about total weight? Is one book at one pound the same value as eight books at 2 ounces each? (small books have a lower price-to-purchase)

If I did this then I would make sure they fit in the bag, and I would account for total weight. I would account for purchase price of the book (I have to buy my own books) - this should reduce number of minima. I would account for unique configurations - I wouldn't want to have a unique switching for each day of the week.

But that is just me getting my nerd on.

EngrStudent
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