Gomory Cuts. Chapter 5. Linear Arithmetic. Decision Procedures. An Algorithmic Point of View. Revision 1.0

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1 Chapter 5 Linear Arithmetic Decision Procedures An Algorithmic Point of View D.Kroening O.Strichman Revision 1.0

2 Cutting planes Recall that in Branch & Bound we first solve a relaxed problem (i.e., no integrality constraints). We now study a method for adding cutting planes constraints to the relaxed problem that do not remove integer solutions. Specifically, we will see Gomory cuts. Decision Procedures Gomory Cuts 2

3 Cutting planes, geometrically. satisfying assignments The dotted line is a cutting plane. Decision Procedures Gomory Cuts 3

4 Example: Gomory Cuts Suppose our input integer linear problem has... Integer variables x 1,..., x 3. Lower bounds 1 x 1 and 0.5 x 2. Decision Procedures Gomory Cuts 4

5 Example: Gomory Cuts Suppose our input integer linear problem has... Integer variables x 1,..., x 3. Lower bounds 1 x 1 and 0.5 x 2. After solving the relaxed problem: The final tableau of the general simplex algorithm includes the constraint x 3 = 0.5x x 2, (1) Decision Procedures Gomory Cuts 4

6 Example: Gomory Cuts Suppose our input integer linear problem has... Integer variables x 1,..., x 3. Lower bounds 1 x 1 and 0.5 x 2. After solving the relaxed problem: The final tableau of the general simplex algorithm includes the constraint x 3 = 0.5x x 2, (1)...and the solution α is {x , x 1 1, x 2 0.5} Decision Procedures Gomory Cuts 4

7 Example: Gomory Cuts Subtracting these values from (1) gives us x = 0.5(x 1 1) + 2.5(x 2 0.5). (2) Decision Procedures Gomory Cuts 5

8 Example: Gomory Cuts Subtracting these values from (1) gives us x = 0.5(x 1 1) + 2.5(x 2 0.5). (2) We now wish to rewrite this equation so the left-hand side is an integer: x 3 1 = (x 1 1) + 2.5(x 2 0.5). (3) Decision Procedures Gomory Cuts 5

9 Example: Gomory Cuts The two right-most terms must be positive because 1 and 0.5 are the lower bounds of x 1 and x 2, respectively. Since the right-hand side must add up to an integer as well, this implies that (x 1 1) + 2.5(x 2 0.5) 1. (4) Decision Procedures Gomory Cuts 6

10 Example: Gomory Cuts The two right-most terms must be positive because 1 and 0.5 are the lower bounds of x 1 and x 2, respectively. Since the right-hand side must add up to an integer as well, this implies that (x 1 1) + 2.5(x 2 0.5) 1. (4) This constraint is unsatisfied by α because α(x 1 ) = 1, α(x 2 ) = 0.5. Decision Procedures Gomory Cuts 6

11 Example: Gomory Cuts The two right-most terms must be positive because 1 and 0.5 are the lower bounds of x 1 and x 2, respectively. Since the right-hand side must add up to an integer as well, this implies that (x 1 1) + 2.5(x 2 0.5) 1. (4) This constraint is unsatisfied by α because α(x 1 ) = 1, α(x 2 ) = 0.5. Hence, this constraint removes the current solution. Decision Procedures Gomory Cuts 6

12 Example: Gomory Cuts The two right-most terms must be positive because 1 and 0.5 are the lower bounds of x 1 and x 2, respectively. Since the right-hand side must add up to an integer as well, this implies that (x 1 1) + 2.5(x 2 0.5) 1. (4) This constraint is unsatisfied by α because α(x 1 ) = 1, α(x 2 ) = 0.5. Hence, this constraint removes the current solution. On the other hand, it is implied by the integer system of constraints, and hence cannot remove any integer solution. Decision Procedures Gomory Cuts 6

13 Generalizing this example: Upper bounds. Both positive and negative coefficients. The description that follows is based on Integrating Simplex with DPLL(T) Technical report SRI-CSL Dutertre and de Moura (2006). Decision Procedures Gomory Cuts 7

14 There are two preliminary conditions for deriving a Gomory cut from a constraint: The assignment to the basic variable has to be fractional. The assignments to all the nonbasic variables have to correspond to one of their bounds. Decision Procedures Gomory Cuts 8

15 Consider the i-th constraint: x i = a ij x j, (5) where x i B. x j N Decision Procedures Gomory Cuts 9

16 Consider the i-th constraint: x i = a ij x j, (5) where x i B. x j N Let α be the assignment returned by the general simplex algorithm. Thus, α(x i ) = a ij α(x j ). (6) x j N Decision Procedures Gomory Cuts 9

17 Partition the nonbasic variables to those that are currently assigned their lower bound, and those that are currently assigned their upper bound J = {j x j N α(x j ) = l j } K = {j x j N α(x j ) = u j }. (7) Subtracting (6) from (5) taking the partition into account yields x i α(x i ) = j J a ij (x j l j ) j K a ij (u j x j ). (8) Decision Procedures Gomory Cuts 10

18 Let f 0 = α(x i ) α(x i ). As we assumed that α(x i ) is not an integer then 0 < f 0 < 1. We can now rewrite (8) as x i α(x i ) = f 0 + j J a ij (x j l j ) j K a ij (u j x j ). (9) Note that the left-hand side is an integer. Decision Procedures Gomory Cuts 11

19 We now consider two cases. (Case 1) If a ij (x j l j ) a ij (u j x j ) > 0 j K j J then, since the right-hand side must be an integer, f 0 + a ij (x j l j ) a ij (u j x j ) 1. (10) j J j K Decision Procedures Gomory Cuts 12

20 (Still in case 1) We now split J and K as follows: J + = {j j J a ij > 0} J = {j j J a ij < 0} K + = {j j K a ij > 0} K = {j j K a ij < 0} (11) Gathering only the positive elements in the left-hand side of (10) gives us: j J + a ij (x j l j ) or, equivalently, a ij (x j l j ) 1 f 0 j J + j K a ij (u j x j ) 1 f 0, (12) j K a ij 1 f 0 (u j x j ) 1. (13) Decision Procedures Gomory Cuts 13

21 (Case 2) If a ij (x j l j ) a ij (u j x j ) 0 j K j J then again, since the right-hand side must be an integer, f 0 + a ij (x j l j ) a ij (u j x j ) 0. (14) j J j K Decision Procedures Gomory Cuts 14

22 Eq. (14) implies that a ij (x j l j ) a ij (u j x j ) f 0. (15) j J j K + Dividing by f 0 gives us (End of case 2) j J a ij f 0 (x j l j ) + j K + a ij f 0 (u j x j ) 1. (16) Decision Procedures Gomory Cuts 15

23 Note that the left-hand side of both (13) and (16) is greater than zero. Therefore these two equations imply + j J + j K + a ij a ij (x j l j ) a ij (x j l j ) 1 f 0 f 0 j J f 0 (u j x j ) j K a ij 1 f 0 (u j x j ) 1. (17) Since each of the elements on the left-hand side is equal to zero under the current assignment α, then α is ruled out by the new constraint. In other words: the solution to the linear problem augmented with the constraint is guaranteed to be different from the previous one. Decision Procedures Gomory Cuts 16

Operations Research Lecture 6: Integer Programming

Operations Research Lecture 6: Integer Programming Notes taken by Kaiquan Xu@Business School, Nanjing University May 12th 2016 1 Integer programming (IP) formulations The integer programming (IP) is the