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h6sol - CO 227 Homework assignment 6 Fall `10 Page 1 CO 227...

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CO 227 - Homework assignment 6 Fall ‘10 Page 1 CO 227 - Fall ‘10 Homework assignment #6: (Due at 5pm, on Monday, Dec 6th) Instructions: Please show all your work and justify your answers. Answers without proper justification will not be considered. Please list who you collaborated with. Please staple your homework and write your name and student id in pen. Total value = 90 points Question 1 - Branch-and-bound (30 points) Consider the following integer program: max z = x 1 +5 x 2 s.t. 4 x 1 +3 x 2 6 3 x 1 +2 x 2 18 x 1 , x 2 0 and integer (IP) The picture below shows the branch-and-bound tree in the middle of solving this problem (only the nodes that were already explored are shown). Subp. 1 z * = 28 . 94 x * 1 = 2 . 47 x * 2 = 5 . 29 Subp. 2 z * = 25 . 5 x * 1 = 3 x * 2 = 4 . 5 Subp. 4 INFEASIBLE Subp. 5 z * = 23 . 33 x * 1 = 3 . 33 x * 2 = 4 Subp. 6 z * = 19 x * 1 = 4 x * 2 = 3 Subp. 7 z * = 23 x * 1 = 3 x * 2 = 4 Subp. 3 z * = 25 . 33 x * 1 = 2 x * 2 = 4 . 67 x 1 3 x 1 2 x 2 5 x 2 4 x 1 4 x 1 3 (a) (3 points) Suppose that we know z 2 which is the value of the optimal value to the subproblem 2 and z 3 which is the optimal value to the subproblem 3. How would you express the optimal value to (IP) in terms of z 2 and z 3 ?
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CO 227 - Homework assignment 6 Fall ‘10 Page 2 [CAREFUL: the optimal solution to the subproblems is NOT the optimal solution to the LP relax- ations of the subproblems. It does take into account the integrality constraints] Solution: The original problem (IP) is divided into two subproblems 2 and 3. Therefore the optimal value for (IP) is given by: max { z 2 , z 3 } (b) (4 points) Solve the following IP: (show all your work) max z = x 1 +5 x 2 s.t. 4 x 1 +3 x 2 6 3 x 1 +2 x 2 18 x 1 3 x 1 , x 2 0 and integer Solution: Notice that this problem is exactly the subproblem 2 in the branch-and-bound tree. From the tree, we notice that all the nodes that correspond to solving subproblem 2 have been explored and there are no more nodes to branch on. Hence, the optimal solution to subproblem 2 is the best solution found in that subtree, which is: x = (3 , 4) T with objective value 23. (c) (5 points) Suppose you stopped branching now, and just used the best integral solution you have found so far. Give a bound on how far from the optimal value for (IP) that your best integral solution’s objective value could be. Show all your work. Solution: The best solution found in the given branch-and-bound tree has value 23. Now suppose that z IP is the optimal value for (IP). We know that: 23 z IP = max { z 1 , z 2 } But, from what we’ve seen in the previous question, z 1 = 23. Moreover, since the LP relaxation of IP 2 has value 25.33, we know that z 2 25 . 33. In particular, since z 2 must be an integer number, we know that z 2 25. Hence 23 z IP = max { z 1 , z 2 } ≤ max { 23 , 25 } = 25 So our best integral solution’s objective value is at most 2 units away from the optimal value.
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