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I have a question:

In the branch and bound algorithm, if you reach two leaf nodes...do you have to go back and solve all the other problems?

It seems to cumbersome.

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Typically in a branch and bound, you explore the leaf nodes one by one, in a branching process. Along the way, you keep track of the optimal value seen so far as well as a way of bounding the value that can be seen in any child leaf nodes.

You never "have to" solve the upper-bound. You do so only to save running time. You can be as lazy as you want in terms of how to update the minimum value seen or how much work you want to expend in your bounds.