# Integer prog 64 ie301 operations research i spring

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Integer Prog. 64 IE301 Operations Research I Spring 2015 TREE Subproblem 1 Z=165/4 X1=15/4 X2=9/4 Subproblem 2 Subproblem 3 t=1 x1≥4 x1≤ 3 arc node chronological order in which the problems are solved
9.3 Branch and Bound Method for Solving Pure Integer Programming Problems EXAMPLE 9 Branch-and-Bound Method Every point is included in either region Feasible regions have no points in common. x1 x2 1 1 2 2 3 3 4 5 6 IP feasible point 4 5 6 7 8 9 Subproblem 2 Figure 12 Feasible region for subprblems 2 and 3 x1 x2 1 1 2 2 3 3 4 5 6 IP feasible point Subproblem 3 4 5 6 7 8 9 Optimal LP solution to subproblem 1 X1=3.75 X2=2.25 Figure 12 Feasible region for subprblems 2 and 3 A B C F E D G Integer Prog. 65 IE301 Operations Research I Spring 2015
9.3 Branch and Bound Method for Solving Pure Integer Programming Problems EXAMPLE 9 Branch-and-Bound Method We now choose any subproblem that has not been solved as an LP Choose subproblem 2 Z=41, x1=4, x2=9/5 (Point C) The subproblem solution to subproblem 2 did not yield an all integer solution. This means we may need to branch further under subproblem 2. Our accomplishments are summarized in figure 13 Subproblem 1 Z=165/4 X1=15/4 X2=9/4 Subproblem 2 Z=41 X1=4 X2=9/5 Subproblem 3 t=1 t=2 x1≥4 x1≤ 3 Subproblem 2 did not yield an all integer solution Figure 13 Telfa subproblems 1 and 2 Solved Integer Prog. 66 IE301 Operations Research I Spring 2015
9.3 Branch and Bound Method for Solving Pure Integer Programming Problems EXAMPLE 9 Branch-and-Bound Method From figure 12 we find that the optimal solution to the subproblem 3 is point F: z=39, x1=x2=3 Subproblem 3 yield an all integer solution with z-value=39. A solution obtained by solving a subproblem in which all variables have integer values is a candidate solution . So, there is no need to go further in Subproblem 3 branch. Because the candidate solution may be optimal, we must keep it, until a better feasible solution to the IP is found. The z-value (z=39) for the candidate solution is a lower bound on the original IP. x1 x2 1 1 2 2 3 3 4 5 6 IP feasible point Subproblem 3 4 5 6 7 8 9 Optimal LP solution to subproblem 1 X1=3.75 X2=2.25 Subproblem 2 Figure 12 Feasible region for subprblems 2 and 3 A B C F E D G Integer Prog. 67 IE301 Operations Research I Spring 2015
9.3 Branch and Bound Method for Solving Pure Integer Programming Problems EXAMPLE 9 Branch-and-Bound Method The subproblem solution to subproblem 2 did not yield an all integer solution. But subproblem 2 still has the potential to give an integer solution better than the current lower bound=39 (found by Subproblem 3) So we choose to use subproblem 2 to create two new subproblems. We choose fractional valued variable in the optimal solution to subproblem 2 and then branch on that variable.