IP Cut Homework from J and B Chapter 9: 14, 15, 16, 23, 24, 31 14. You wish to solve the IP below with a cutting plane technique. Maximize 4x 1 + 2x 2 + x 3 subject to 14x 1 + 10x 2 + 11x 3 32 10x 1 + 8x 2 + 9x 3 0 x 1, x 2, x 3 0 and integer After relaxing the integrality requirements and solving the resulting linear program, the following solution is obtained. Var. no. Name Value Status 1 x 1 1.208 BASIC 1 2 x 2 1.509 BASIC 2 3 x 3 0 ZERO 4 SLK 1(x 4 ) 0 ZERO 5 SLK 2(x 5 ) 0 ZERO Objective value = 7.849 The basis and basis inverse for this solution are B = 14 10 10 8 and B 1 = 1 8 10 212 10 14. a. What Gomory cuts can be derived from the current information? Express the cuts in terms of the nonbasic variables in the solution above. The Gomery cuts are: G1. 0.9906x 3 + 0.0377x 4 + 0.0472x 5 0.2075 G2. 0.1132x 3 + 0.0472x 4 + 0.934x 5 0.5094 b. Write the cuts found in part a in terms of the original structural variables. The Gomery cuts are: G1. -1x 1 + 0x 2 + 1x 3-1 or 1x 1 + 0x 2-1x 3 1 G2. -10x 1 + 7x 2 + 8x 3-1 or 10x 1-7x 2-8x 3 1 1
c. What Dantzig cut should be added to continue the cutting plane procedure? D. x 3 + x 4 + x 5 1 d. Write the Dantzig cut in terms of the original structural variables. D. -24x 1 2x 2-1x 3-31 e. Add the Gomery cut with the largest right-hand-side value to the original LP and use an LP code to find the solution. Report the values of the variables in the optimal solution. How has the objective value changed with the addition of the cut? The new LP solution Var. no. Name Value Status 1 x 1 1.182 BASIC 2 x 2 1.546 BASIC 3 x 3 0 ZERO 4 SLK 1(x 4 ) 0 ZERO 5 SLK 2(x 5 ) 0.546 BASIC 6 SLK Cut(x 6 ) 0 ZERO Objective value = 7.8182 The objective value has decreased. 2
15. The tableau below gives the LP solution to the relaxation of an integer programming problem with maximization objective. All a ij and b i coefficients in the original problem are integer. Row Basic Coefficients no. variables z x 1 x 2 x 3 x 4 x 5 x 6 RHS 0 z 1 0.1 0.3 0 0.2 0 0 23.1 1 x 6 0 1.3 0.3 0 1.0 0 1 5.3 2 x 3 0 0 1.1 1 0.4 0 0 1.6 3 x 5 0 0.8 0.2 0 0.5 1 0 3.7 a. Write out all Gomory cuts that can be derived from the tableau. Row 1: 0.3x 1 + 0.7x 2 0.3 Row 2: 0.1x 2 + 0.4x 4 0.6 Row 3: 0.2x 1 + 0.8x 2 + 0.5x 4 0.7 b. Write out the Dantzig cut that can be derived from the tableau. Dantzig: x 1 + x 2 + x 4 1 c. Add the Gomory cut with the largest right-hand-side value to the tableau and use the dual simplex method to find the new solution. Perform these computations by hand. From row 3: A: 0.2x 1 + 0.8x 2 + 0.5x 4 X 7 = 0.7 Row Basic Coefficients no. variables z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHS 0 z 1 0.1 0.3 0 0.2 0 0 0 23.1 1 x 6 0 1.3 0.3 0 1.0 0 1 0 5.3 2 x 3 0 0 1.1 1 0.4 0 0 0 1.6 3 x 5 0 0.8 0.2 0 0.5 1 0 0 3.7 4 x 7 0 0.2 0.8 0 0.5 0 0 1-0.7 Ratio 1/2 3/8 2/5 3
x 7 leaves the basis and x 4 enters Row Basic Coefficients no. variables z x1 x2 x3 x4 x5 x6 x7 RHS 0 z 1 0 0 0 0 22.96 1 x6 0 0 0 0 1 5.16 2 x3 0 0 1 0 0 1.46 3 x5 0 0 0 1 0 3.56 4 x7 0 1 0 0 0 0.875 Drop: d. Solve the problem with the cutting plane algorithm included in the Teach IP Excel Add-in. 4
16. Find the relaxed LP solution to the following integer program. Maximize 2x 1 + 5x 2 subject to x 1 + x 2 5 x 1 + x 2 2 x 1 x 2 2 x 1 + x 2 3 x 1, x 2 0 and integer Now identify which of the cuts below are valid for use in a cutting plane algorithm. The constraints are not cumulative so each part should be analyzed separately. Justify your conclusion in each case. a. x 1 3 b. x 2 4 c. x 1 + 3x 2 10 d. 1.75x 3 0.75x 4 0.25 (Note that x 3 and x 4 are respectively the slack variables for the first two constraints.) a. x 1 3 This is a valid inequality for the convex hull of the integer points. There are some points that are infeasible for the LP relaxation. It would be useful for a cutting plane algorithm b. x 2 4 This is a valid inequality, but it does not cut off any points in the LP relaxation. It would not be useful for a cutting plane. c. x 1 + 3x 2 10 This is a not a valid inequality for the convex hull of the integer solutions, so it would not be useful for a cutting plane algorithm d. 1.75x 3 0.75x 4 0.25 (Note that x 3 and x 4 are respectively the slack variables for the first two constraints.) Rewrite d in terms of the original variables: 1.75(5 x 1 x 2 ) 0.75(2 + x 1 x 2 ) 0.25 25/4-1.75x 1-1.75x 2 1.5 0.75x 1 + 0.75x 2 0.25 2.5x 1 + x 2 9/2 or 5x 1 + 2x 2 9 This is a not a valid inequality for the convex hull of the integer solutions, so it would not be useful for a cutting plane algorithm 5
5 x2 C1 4 C2 Cb 3 Ca 2 1 C3 Cc C4 0 1 2 3 4 5 x2 Cd 6
23. (Branch and cut) { XE "Branch and cut" }{ XE "Integer programming:branch and cut" }How could a cutting plane technique be incorporated into an implicit enumeration technique for solving the pure integer programming problem? a. Write out the steps. The primary change involves the Bound procedure: If the relaxed solution fails to find a feasible (integer) solution or fails to discover infeasibility, we may add cuts to the LP relaxation in an attempt to obtain a better bound or find infeasibility. In the backtrack procedure we must eliminate all cuts added at lower levels of the tree. b. Under what circumstances could the cuts be used at each node of the search tree? Cuts are only valid for the whole tree if they are added at node 0. 7
24. (Generalized Gomory cut) { XE "Gomory cuts:generalized" }Suppose we wish to use a cutting plane technique to solve a pure integer program. Assume that the LP relaxation has been solved and the ith constraint in the simplex tableau is x B(i) + - a ij x j = b - i j Q where x B(i) is the ith basic variable and Q is the set of nonbasic variables. Multiply this equation by any rational number h. Now, using the same approach outlined in Section 8.4 to derive the basic Gomory fractional cut given in Eq. (16), derive from the equation above a more generalized version of Eq. (16). x B( i) + a ij x j = b i (1) Multiplying the expression by a rational number h: hx B i (2) ( ) + ha ij x j = hb i At the optimal integer solution all variables are nonnegative integers. Replacing the coefficients on the left side by the next smaller integer we obtain the inequality h x B( i) + ha ij x j hb i (3) Since the left side must now be integer, the right side may be replaced by the next lower integer. h x B( i) + ha ij x j hb i (4) Multiplying (1) by the integer part of h and subtracting (4) we have h x B( i) + h a ij x j = h b i ( ha ij h a ij )x j hb i h b i or ( h a ij ha ij )x j h b i hb i The last expression is the desired cut. It is valid for every h. 8
31. Shown below is an integer programming model and the optimal tableau of its linear programming relaxation. Maximize z = 2x 1 + x 2 3x 3 + 5x 4 subject to 3x 1 x 2 + x 3 + 2x 4 8 x 1 + x 2 + 4x 3 x 4 6 2x 1 + 3x 2 x 3 + x 4 10 x 1 + x 3 + x 4 7 x j 0 and integer, j = 1,,4 Row Basic Coefficients no. variables z x 1 x 2 x 3 x 4 x s1 x s2 x s3 x s4 RHS 0 z 1 6 0 4 0 2 0 1 0 26 1 x 4 0 11/7 0 2/7 1 3/7 0 1/7 0 34/7 2 x s2 0 17/7 0 33/7 0 4/7 1 1/7 0 64/7 3 x 2 0 1/7 1 3/7 0 1/7 0 2/7 0 12/7 4 x s4 0 4/7 0 5/7 0 3/7 0 1/7 1 15/7 a. Write the Dantzig cut that can be derived from the tableau. x 1 + x 3 + x s1 + x s3 1 b. From the set of possible Gomory cuts, write the one with the greatest right-handside value. The Gomery cut with the greatest RHS is from row 1. (4/7)x 1 + (2/7)x 3 + (3/7)x s1 + (1/7)x s3 6/7 c. Write the Gomory cut that can be derived from row 0. Since the coefficients in row 0 are all integer, no Gomery cut can be derived from the row. d. Write the Gomory cut that can be obtained from row 4. Add it to the tableau and reoptimize using the dual simplex method. The cut from row 4. (3/7)x 1 + (5/7)x 3 + (4/7)x s1 + (6/7)x s3 1/7 9
Objective: Max Press to setup. Objective Value: 25.833 LP Status: Optimal IP Status: Variables I- 1 I- 2 I- 3 I- 4 Name: X 1 X 2 X 3 X 4 Values: 0 1.6667 0 4.8333 Linear Obj. Coef.: 2 1-3 5 Variable Values : 0 0 0 0 Variable Values : 9999 9999 9999 9999 straints Constraint Bounds Name Value Value Value Constraint Coefficients Cut 1 13-9999 13 3 2-1 2 Con 1 8-9999 8 3-1 1 2 Con 2-3.167-9999 6 1 1 4-1 Con 3 9.8333-9999 10 2 3-1 1 Con 4 4.8333-9999 7 1 0 1 1 e. Solve the problem with the cutting plane algorithm embodied in the Teach IP Excel Add-in. When cut 4 is added, an integer solution is immediately obtained. Objective: Max Objective Value: 22 LP Status: Optimal IP Status: Variables I- 1 I- 2 I- 3 I- 4 Name: X 1 X 2 X 3 X 4 Values: 0 2 0 4 Linear Obj. Coef.: 2 1-3 5 Variable Values : 0 0 0 0 Variable Values : 9999 9999 9999 9999 straints Constraint Bounds Name Value Value Value Constraint Coefficients Cut 1 4-9999 4 1-6E-17 0 1 Con 1 6-9999 8 3-1 1 2 Con 2-2 -9999 6 1 1 4-1 Con 3 10-9999 10 2 3-1 1 Con 4 4-9999 7 1 0 1 1 10