MATP6620/ISYE6760 Combinatorial Optimization & Integer Programming
Homework 3.
Solutions

Due: Thursday, March 4, 2021, by end of day.
Penalty for late homeworks: 10% for each day or part of a day.

The following graph G = (V,E) is used in question 1. 1. Consider a node packing problem on the above graph, with each vertex having weight equal to 2 more than its degree. The LP relaxation includes the clique constraints vCxv 1 for each maximal clique C in the graph. The point xA = 0.4, xB = 0.6, xC = 0.4, xD = 0.5, xE = 0.4, xF = 0.5, xG = 0.1, xH = 0.4, and xJ = 0.1 is feasible in the LP relaxation.
1. Show that the given point is not in the convex hull of feasible solutions, by giving a valid constraint that is violated by this point.
2. Find an optimal solution to the node packing problem for this graph. Prove your solution is optimal.

Solution:

1. Vertices A,B,C,H,F constitute an odd hole, giving the valid constraint The left hand side of this constraint evaluates to 2.3 for the given point.

2. If xG = 1 then the largest possible value for the left hand side of the constraint in part (a) is 1, so the constraint can be lifted to (1)

The vertices D,E,J form a clique, so we also have the valid constraint Thus, any packing can use at most 3 vertices.

(We could also argue this using the odd hole constraint formed by {C,D,E,F,H}, which can be lifted with both G and J, so at most 2 of the 7 vertices {C,D,E,F,G,H,J} can be used. Also, at most one of {A,B} can be used.)

The vertices and their degrees are:

 degree vertices 5 C 4 F,G,H,J 3 A,D,E 2 B

We claim the node packing A,C,E with value 17 is optimal.

First note, any packing with just 2 vertices has value at most 13. So the optimal packing must contain 3 vertices.

Vertices C and F constitute a maximal packing, so they cannot both be part of an optimal packing. Vertex C is adjacent to all of G, H, and J, so any packing containing C with 3 nodes cannot have value larger than 17.

F, G, H form a clique, so no packing can contain more than 2 of the vertices of degree 4.

So no packing has value larger than 17.

Alternatively:

Solving the LP relaxation of the node packing problem with the lifted odd hole constraint (1) and the clique inequalities gives the integral solution. Here are the AMPL model and data files, and the AMPL output.

2. Consider the constraints 1. By considering the different possibilities for x, show that t1 + t2 1.
2. The constraints can be modeled equivalently as Show that the valid constraint t1 + t2 1 has Chvatal rank equal to 2.

Solution:

1. Four choices for x, with the minimum possible corresponding choices for t: 2. Rank is no larger than 2:

Write constraints as: 0.5(2) + 0.5(4) implies (6)

0.5(3) + 0.5(5) implies (7)

In the second round, 0.5(6) + 0.5(7) implies as required.

Rank is at least 2:

The point x = (0.5,0.5) and t = (0,0) satisfies (2)–(5). This point is a convex combination of the integer points x = (0,0), t = (0,0) and x = (1,1), t = (0,0). So any valid linear combination of (2)–(5) must be satisfied by at least one of these points, and so the rounded version must also be satisfied by at least one of the points.

Alternatively, since the point x = (0.5,0.5) and t = (0,0) is feasible in the LP relaxation, the maximum value of -t1 - t2 in the LP relaxation is 0. Thus, by the proposition in the notes, the best rank one inequality is only -t1 - t2 0, so the desired inequality has CG rank at least two.

3. The optimal tableau to the linear programming relaxation of the integer program where x3 and x4 are the slack variables in the two constraints. Find the Gomory and strong Gomory cutting planes implied by the two constraints. Express these constraints in terms of the original variables x1 and x2 and draw them on a graph of the feasible region.

Solution:

First constraint: Fractional parts: f3 = , f4 = , f0 = .

Gomory cut: In terms of the original variables, x3 = 14 - x1 - 2x2 and x4 = 8 + x1 - 6x2, so we have or (8)

Mixed integer Gomory cut: or equivalently In terms of the original variables, we have or (9)

Second constraint: Fractional parts: f3 = , f4 = , f0 = .

Gomory cut: In terms of the original variables, we have or (10)

Mixed integer Gomory cut: Since f3 f0 and f4 f0, the mixed integer Gomory cut is identical to the original Gomory cut.

Graph: 4. Let x Bn satisfy the constraints (11)

Show that the constraint (12)

is valid. Give a fractional point with 0 x e that satisfies the original n(n - 1)2 constraints but violates the new constraint. Show that the new constraint has Chvatal rank no larger than O(log n).

Solution:

• Any binary vector that violates (12) must have at least two components with value 0, say xp = xq = 0, with 1 p < q n. Then this point violates the corresponding constraint of form (11).
• Each xi = 0.5 satisfies (11) and violates (12).
• We use induction to show the constraint has rank no more than O(log(n)). Let k be an integer with 2 < k < n. Assume we know (13)

for all subsets C ⊆{1,,n} with |C| = k. Let J ⊆{1,,n} with k < |J| < 2k - 1. We show the inequality (14)

can be derived in one step. We need two parameters: We add together the α constraints of type (13) for every subset of size k of J, and give the constraints weight 1∕β. This gives the constraint Since this is a constraint and since all the coefficients on the left hand side are integral, we can round up the right hand side. Since k < |J| < 2k - 1, rounding gives (14) as required.

Thus by induction the inequality (13) has Chvatal rank as follows:

 k Chvatal rank 2 0 3 1 4 or 5 2 6,7,8,9 3 10,…,17 4 …

This can be expressed compactly by saying the rank is no larger than log 2(k - 1)for k = 2,3,

If |J|≥ 2k then (14) cannot be obtained in one step from all the inequalities (13). In particular, if we set each xi = 1 - 1∕k then all of the inequalities (13) are satisfied, and the left hand side of (14) evaluates to Thus, rounding cannot give (14) in one step.

A proof that the rank is at least O(log(k)) is a consequence of Theorem 4.12 in .

5. The AMPL model of the LP relaxation of a random weighted node packing problem with 15 nodes is contained in the file
http://www.rpi.edu/~mitchj/matp6620/hw3/nodepack.mod

The initial model contains only the adjacency constraint that just one endpoint of an edge can appear in the node packing. Pick a seed and then solve the problem using a cutting plane algorithm:

1. Solve the LP relaxation.
2. If the solution is integral, STOP.
3. If necessary, add one or more valid inequalities to the LP. These inequalities can be clique inequalities or odd hole inequalities.

(It is highly likely that you will need to use both clique inequalities and odd hole inequalities, and that these inequalities will be sufficient to solve the problem.)

(Hint: The graph consists of the cycle 1 - 2 - 3 - 4 -- 14 - 15 - 1, plus some extra edges. You might be able to see the structure by displaying adjacency.)

Solution:

Here is the model file with the seed 3242. Solving this problem required the addition of two clique constraints and 4 odd hole constraints, with each clique containing 3 nodes and each odd hole containing 5 nodes. Here is the output file with the seed 3242.

6. The Project:
Along with your solutions to this homework, hand in a brief description of what you would like to do for the project part of this course.

### References

   R. Müller and A. S. Schulz. Transitive packing: a unifying concept in combinatorial optimization. SIAM Journal on Optimization, 13(2):335–367, 2002.

 John Mitchell Amos Eaton 325 x6915. mitchj at rpi dot edu Office hours: Monday and Thursday 1pm–2pm. webex: https://rensselaer.webex.com/meet/mitchj