2. Valid Inequalities for the 0-1 Knapsack Polytope

Consider the constraint set of a 0-1 knapsack problem.

,

Where , +1 for , and +

1.

Assume for all . Thus dim(conv(S)) = n.

Assume .

is called an independent set if (: characteristic vector of Otherwise, called a dependent set (cover).

Prop 2.1: If C is a dependent set, then (cover inequality) is a valid inequality for S.

The extension of a minimal dependent set C is the set .

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Ex)

C1 = {1, 2, 3} (185) = {1, 2, 3} = C1.

cover inequality

C4 = {2, 3, 4, 5} (196) = {1, 2, 3, 4, 5}

Prop 2.2: If C is a minimal dependent set, then (2.3) (extended cover in-equality) is valid for S.

Pf) Suppose () and jR |C| so that . Now and by definition of we obtain ,

which contradicts .

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Prop 2.3: Let be a minimal dependent set with . If any of the following con-ditions holds, then (2.3) gives a facet of conv(S).

a.

b. and (i) (C\{j1, j2}){1} is independent.

c. C = E(C) and (ii) (C\{j1}){p} is independent, where p = min{j: j N\E(C)}

d. and (i) and (ii).

Pf) The following n independent sets satisfy (2.3) at equality.

1. for . There are |C| of these.

2. for . and is independent by (i) and . There are of these.

3. for . and is independent by (ii) and .

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J is a matrix of all ones. J-I is nonsingular. Ex-continued)

C1 = {1, 2, 3}, facet since C1 = E(C1) and (ii)

C4 = {2, 3, 4, 5} E(C4) = {1, 2, 3, 4, 5} facet since E(C4) = N and (i)

(

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C

E(C)\C

N\E(C)

Cor 2.4: If C is a minimal dependent set for S and , then gives a facet of conv, where .

Pf) is a minimal dependent set for

.

Now Prop 2.3 a) applies with and .

Prop 2.5: If C is a minimal dependent set for S and , then conv(S) has a facet represented by

,

where for all and where for all .

Pf) Lift starting with . facet of conv(S(C1, C2))

Do maximum lifting for each , and for each . (first lifting variable should be in N\C.)

Nonnegativity of the coefficients is implied by definitions.

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Prop 2.6: Let be a minimal dependent set with . Let for h = 1, … , r; also let and . Every valid inequality of the form (2.4) that represents a facet of conv(S) satisfies the following conditions:

I. If , then .

II. If , then (a) and (b) there is at least one facet of the form (2.4) with

Ex)

C = {6, 7, 8, 9}

, , , 35, 41, .

0, 0 aj 13, 0 or 1, aj = 14

1, 15 aj 25, 1 or 2, aj = 26

2, 27 aj 33, 2 or 3, aj = 34

2 35 aj 39

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Application to 0-1 IP

“Solving large-scale zero-one linear programming problems”, H. P. Crowder, E. L. Johnson, and M. W. Padberg, Operations Research 31, pp. 803-834, 1983

0-1 IP, max {cx: Ax b, x Bn}.

Consider each constraint individually, then a valid inequality for the con-straint and is also valid for S.

Let is a constraint of 0-1 IP

Let ,

Assume , for all . If not, set .

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After solving LP relaxation of IP, get optimal .

If , find violated cover inequality for .

Separation: Find a cover C such that ( b+1, assuming integral data) and * >

|C| - 1 *) < 1

Formulation: Let , if , and 0, otherwise.

= min *)

If < 1 z gives most violated cover inequality.

If 1 no violated cover inequality

Although there is no violated cover, lifting may identify a violated lifted cover inequality.

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For application of valid inequalities for 0-1 knapsack problem to 0-1 IP, see II.6.2.

For the separation of the violated cover inequality, we need to solve 0-1 knap-sack problem which is NP-hard. May use heuristics for separation. However, some sophisticated algorithms for the knapsack problem works very well computationally. Hence exact separation may be worth doing.

We also need to solve the 0-1 knapsack problem if we do lifting. However, for 01 lifting, it can be solved in polynomial time since the coefficients in the objective function is bounded by n. We can reverse the roles of the objective function and the constraint. See pp. 462, pp. 440-441 Prop 1.6. (The results are for general knapsack problem, but can be modified for 0-1 knapsack.)

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3. Valid Inequalities for the Symmetric Traveling Salesman Polytope

Symmetric TSP formulation: G = (V, E), |V| = m.

min

s.t for all

for all

and integer for all

Assume complete graph for facet results. Validity of inequalities considered holds for any graph. may be considered as part of the subtour elimination constraints.

Prop 3.5: The subtour elimination inequalities give facets of conv(S) for for all W with .

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2-matching inequalities: Let H V with and let be an odd set of disjoint edges, each of which has one end in H.

Use weights of ½ on the degree constraints for all , weights of ½ on for all , weights of ½ on for all , and rounding yields . ( )

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1/2

1/2

1/2

1/2

1/2

1/2

1

1

1

1

1

1

Fractional solution that can’t be cut off by subtour elimination (cut set) constraints (called envelope)

2-matching inequality can be restated as

i=1

k

Comb inequality:

Comb is a subgraph generated by a node set {H, W1 , … , Wk} with the follow-

ing properties1. | H Wi | 1, for i = 1, … , k.

2. | Wi \ H | 1, for i = 1, … , k.

3. 2 | Wi | m - 2, for i = 1, … , k.

4. Wi Wj = , for i j,

5. k is odd and at least 3

Comb inequality:

e E(H) xe + i=1k e E(Wi ) xe |H| + i=1

k (|Wi|– 1) – (k+1)/2 (3.7)

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Multiply degree constraints for all vH by ½ and sum them

e E(H) xe + ½ e (H) xe = |H|. (3.8)

add -½ xe 0, for all e (H)\ i=1k E(Wi) to (3.8)

e E(H) xe + ½ i=1k e (H)E(Wi ) xe |H|. (3.9)

Consider subtour elimination constraints for Wi , H Wi , Wi \H, respectively

e E(Wi) xe | Wi | - 1, for i = 1, … , k

e E(HWi) xe |HWi | - 1, for i = 1, … , k

e E(Wi\H) xe |Wi \ H| - 1, for i = 1, … , k.

multiply each of the above by ½, and add to (2.5)

( since E(Wi) = E(WiH) E(Wi\H) (E(Wi) (H)) )

e E(H) xe + i=1k e E(Wi ) xe |H| + ½ i=1

k [ (|Wi | – 1) + (|HWi | - 1)

+ (|Wi \ H| - 1) ]

= |H| + i=1k (|Wi | – 1) – k/2

since k is odd = |H| + i=1k (|Wi | – 1) – (k+1)/2Integer Programming 2009 13

There are polynomial time algorithms for separation of subtour elimination constraints and 2-matching inequalities. However, no polynomial time algo-rithm is known for the separation of more general comb inequalities. (use heuristics)

The comb inequalities can be generalized further generalized comb inequal-ities

Thm 3.7: The generalized comb inequalities give facets of conv(S).

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