Question

I’m in the process of learning about simulated annealing algorithms and have a few questions on how I would modify an example algorithm to solve a 0-1 knapsack problem.

I found this great code on CP:

http://www.codeproject.com/KB/recipes/simulatedAnnealingTSP.aspx

I’m pretty sure I understand how it all works now (except the whole Bolzman condition, as far as I’m concerned is black magic, though I understand about escaping local optimums and apparently this does exactly that). I’d like to re-design this to solve a 0-1 knapsack-“ish” problem. Basically I’m putting one of 5,000 objects in 10 sacks and need to optimize for the least unused space. The actual “score” I assign to a solution is a bit more complex, but not related to the algorithm.

This seems easy enough. This means the Anneal() function would be basically the same. I’d have to implement the GetNextArrangement() function to fit my needs. In the TSM problem, he just swaps two random nodes along the path (ie, he makes a very small change each iteration).

For my problem, on the first iteration, I’d pick 10 random objects and look at the leftover space. For the next iteration, would I just pick 10 new random objects? Or am I best only swapping out a few of the objects, like half of them or only even one of them? Or maybe the number of objects I swap out should be relative to the temperature? Any of these seem doable to me, I’m just wondering if someone has some advice on the best approach (though I can mess around with improvements once I have the code working).

Thanks!

Mike

Answer 1

With simulated annealing, you want to make neighbour states as close in energy as possible. If the neighbours have significantly greater energy, then it will just never jump to them without a very high temperature — high enough that it will never make progress. On the other hand, if you can come up with heuristics that exploit lower-energy states, then exploit them.

For the TSP, this means swapping adjacent cities. For your problem, I’d suggest a conditional neighbour selection algorithm as follows:

1. If there are objects that fit in the empty space, then it always puts the biggest one in.
2. If no objects fit in the empty space, then pick an object to swap out — but prefer to swap objects of similar sizes.

That is, objects have a probability inverse to the difference in their sizes. You might want to use something like roulette selection here, with the slice size being something like (1 / (size1 – size2)^2).