IEOR 269, Spring 2010 Integer Programming and Combinatorial ...

IEOR269 notes, Prof. Hochbaum, 2010 31
such a tour and we can verify in polynomial time that 1) it is a valid tour and 2) its total distance
is ≤ M. However if the answer to our problem is “no”, then (as far as we know) the only way to
verify this would be to check every possible tour (this is certainly not a poly-time computation).
Lec5
17 Addendum on Branch and Bound
In general, the idea of Branch and Bound is implicit enumeration. Consequently, this can create
serious problems for general IP.
Example 17.1. Consider the following 0-1 Knapsack problem.
max 2x 1 + 2x 2 + · · · + 2x 100
subject to 2x 1 + 2x 2 + · · · + 2x 100 ≤ 101 (10)
x i ∈ {0, 1} for i = 1, . . . , 100 (11)
One optimal solution to this problem is x 1 = x 2 = · · · = x 50 = 1 and x 51 = · · · = x 100 = 0. In fact,
setting any subset of 50 variables to 1 and the remaining to 0 yields an optimal solution.
For a 0-1 Knapsack problem of 100 variables, there are 2 100 possible solutions. The naive way to
find the optimal solution would to enumerate all of them, but this would take way too much time.
Although Branch and Bound is an enumeration algorithm, the idea is to enumerate the solutions
but discarding most of them before they are even considered.
Let us consider the LP-based Branch and Bound algorithm on the problem defined in Example
17.1. At the top node of the Branch and Bound tree, we don’t make any assumptions about the
bounds. The LP relaxation of the problem gives an objective value of 101, which is obtained by
setting any subset of 50 variables to 1, another 49 variables to 0, and the remaining variable to
1
2
; 101 is now the upper bound on the optimal solution to the original problem. Without loss of
generality, let the solution to the LP relaxation be
⎧
x ⎪⎨ 1 = x 2 = · · · = x 50 = 1
x 51 = 1 ⎪⎩ 2
x 52 = x 53 = · · · = x 100 = 0.
Branching on x 51 (the only fractional variable) gives no change in the objective value for either
branch. In fact, using Branch and Bound we would have to fix 50 variables before we found a
feasible integer solution. Thus, in order to find a feasible solution we would need to search 2 51 − 1
nodes in the Branch and Bound tree.
Although this is a pathological example where we have to do a lot of work before we get a lower
bound or feasible solution, it illustrates the potential shortfalls of enumeration algorithms. Further,
it is important to note that, in general, the rules for choosing which node to branch on are not well
understood.