3310/5310: ORIE Prelim 4 Solutions and grading scheme (problems 2, 3) 2(a) Give the cut developed by Gomory for the standard form integer programming problem with A, b, c integer-valued: max{cs | Ax = b, x 0, x integral} Solution. Consider an optimal basic solution of the LP-relaxation (where the solution is not integervalued). Let xBi be a basic variable in row i of the simplex tableau, with xBi = i , not an integer. b The row i equation has the following form: xBi +
xj nonbasic
aij xj = i . b
(1)
For all feasible solution x to the LP-relaxation, the following holds: xBi +
xj nonbasic
aij xj i . b
The following may not be satisfied by all solution x to the LP relaxation, but is satisfied by all integral-valued solution x to the LP relaxation: xB i +
xj nonbasic
aij xj i . b
(2)
By equations (1) and (2), we have: (ij - aij )xj i - i . a b b
xj nonbasic
(3)
[Total: 10 points. Seven points for correct cut/constraint as given in (3) above; 3 points for some explanation of the cut. If constraint (3) is not explicitly written but is explained well, then partial credit will be given, proportional to how clear the explanation is. An explanation is clear when it can very easily be seen that it describes inequality (3).] 2(b) Show that the slack variable for the cut from part (a) must also be integer-valued for any integral, feasible solution to the original problem. Why is this important? Solution. We add a slack variable s to our solution from part (a): b - (ij - aij )xj + s = -(i - i ). a b
xj nonbasic
Then, solving for s, we get: s =
xj nonbasic
(ij - aij )xj - (i - i ) a b b
xj nonbasic
xj nonbasic
aij xj
= i - b
aij xj - i - b
1
For any integral, feasible solution to the original problem, xj is integer, for all original variables xj . aij xj = xBi , which must be an This means that is integral. By equation (1), = i - b
xj nonbasic
integer. So, s must be integer-valued at any integral, feasible solution to the original problem. This is important since in future iterations, this slack variable could become basic. When optimal LP-relaxation solution is not integral, this allows us to choose any row of the current optimal simplex tableau with non-integral RHS to derive a new constraint, as opposed to only rows corresponding to original variables. Note that stating that this slack variable should be integral because we require integral solution in the end is not sufficient. You need to explain it further. [Total: 25 points. 20 points for proving that the slack variable is an integer at any feasible, integral solution; 5 points for giving an argument of why this fact is important. For importance: If you only state that this slack variable should be integral because we require integral solution in the end, you should only get 1 point. For the proof: other approaches would be given full or partial credit, depending on their correctness, but it needs to be very clear (you need to show that s is an integer algebraically). You may not assume the correctness of the algorithm to prove that the slack has to be integral (the point of the question is to show that slack the has to be an integer, which then show that the algorithm works). It is incorrect to state that (ij - aij ) and (i - i ) are integers. If a b b you used this to show that s is an integer, you should get at most 5 out of 20 points.] 3(a) Formulate the problem of determining a stable set of maximum cardinality in G as an IP problem. Your answer should include an explanation of the variables in your IP model. Solution. max
iV
xi
s.t. xi + xj 1 (i, j) E xj {0, 1} j V
[Total: 5 points. One point each for: correct objective function, edge constraints, and specifying that the xj 's are 0-1/binary variables; 2 points for brief explanation.] 3b Will the LP relaxation of your formulation in part (a) always been an optimal integer-valued solution? Solution. No. Consider the graph G = (V, E) with V = {1, 2, 3}, E = {(1, 2), (1, 3), (2, 3)} (i.e. this is a triangle with vertices labelled 1, 2, 3). The optimal LP-relaxation solution is x1 = x2 = x3 = 0.5, with value xi = 1.5. This solution is not integer-valued.
iV
[Total: 5 points. Three points for giving a graph example for which the optimal lp solution is non-integral; 2 points for giving a non-integral-valued optimal solution for this graph.] 3c Any set of vertices, say O V , of an odd cycle in G also determines a valid inequality for the formulation of part (a). What is the inequality? Solution.
iO
xi
|O|-1 2 .
2
[Total: 5 points.] 3d Denote by ST AB the maximum cardinality of a stable set, by COV the minimum size of a node cover, and by M AT CH the maximum size of a matching in G. In G, the complement V \S of a stable set S is a node cover. Why? Solution. Consider any edge (i, j) in the graph. We know that the vertices i and j cannot both be in S since S is a stable set. So, at least one endpoint of each edge (i, j) is in the set V \S. Hence, V form a node cover. [5 points] Thus ST AB + COV = |V |. Why? Solution. If S is the maximum-cardinality stable set and V \S is a node cover, then V \S must be a minimum-size node cover. Since S and V \S form a partition of the set of vertices V , we have: |V | = |S| + |V \S|| = ST AB + COV . [5 points] We also have that M AT CH COV . Why? Solution. Suppose M is a maximum matching in the graph G of cardinalty |M | = M AT CH. Then any node cover of G must include at least one endpoint of the edges in M . So, any node cover must have size at least MATCH. Hence, MATCH COV. [5 points] Hence ST AB + M AT CH |V | for any undirected graph G = (V, E). (From the previous two statements, we can deduce that ST AB + M AT CH ST AB + COV = |V |. Note that this is not a question, so no credit for this statement.) Give a class of graphs for which the latter relation is an equality and state a theorem which justifies your answer. Solution. Bipartite graphs satisfies ST AB + M AT CH = |V | since we know that for bipartite graphs, M AT CH = COV (the maximum cardinality of matching is equal to the minimum size of node cover, by Konig's theorem). So, ST AB + M AT CH = ST AB + COV = |V |. [10 points. Five points for a correct class of graph; 5 points for a correct theorem.] [Total: 25 points.]
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