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Analysis & Design of Algorithms (CSCE 321)

Analysis & Design of Algorithms (CSCE 321). Prof. Amr Goneid Department of Computer Science, AUC Part 13. Branch & Bound. Branch & Bound Algorithms. Introduction General Idea Example: The 0/1 Knapsack Problem. Introduction.

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Analysis & Design of Algorithms (CSCE 321)

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  1. Analysis & Design of Algorithms(CSCE 321) Prof. Amr Goneid Department of Computer Science, AUC Part 13. Branch & Bound Prof. Amr Goneid, AUC

  2. Branch & Bound Algorithms • Introduction • General Idea • Example: The 0/1 Knapsack Problem Prof. Amr Goneid, AUC

  3. Introduction • Backtracking uses depth-first search with pruning to traverse the (virtual) state space. • We can achieve better performance for many problems using a breadth-first search with pruning. This approach is known as Branch-and-Bound. • Branch and bound is a systematic method for solving optimization problems Prof. Amr Goneid, AUC

  4. Introduction • B&B is a rather general optimization technique that applies where the greedy method and dynamic programming fail. • However, it is generally slower. • On the other hand, if applied carefully, it can lead to algorithms that run reasonably fast on average. Prof. Amr Goneid, AUC

  5. The General Idea • The general idea of B&B is a BFS-like search for the optimal solution • Not all nodes get expanded (i.e., their children generated). Rather, a carefully selected criterion determines which node to expand and when, and another criterion tells the algorithm when an optimal solution has been found. Prof. Amr Goneid, AUC

  6. The General Idea • Compared to backtracking, B&B requires two additional items: • A bound on any further additions, • The value of best solution obtained so far. Prof. Amr Goneid, AUC

  7. Example: The 0/1 Knapsack • Consider a Knapsack of capacity W. Find the most valuable subset of (n) items that can fit into the knapsack. • Item (i) has a weight wi and a value vi, the value density is di = vi / wi • We arrange the items according to the value density, d1 ≥ d2 ≥ … ≥ dn Prof. Amr Goneid, AUC

  8. The 0/1 Knapsack • The state space is like backtracking. Each item is represented in a layer in a binary tree, with levels 0 … n. • The root of the tree at level 0 indicates that no item is yet selected. • A left branch indicates selecting an item, a right branch indicates no selection. Prof. Amr Goneid, AUC

  9. The 0/1 Knapsack • Consider w = total weight of items selected so far, v = total value obtained so far. • Consider dbest = best value density among remaining items. • The upper bound on the value so far is computed as: ub = v + (W – w) dbest • We branch to the next level from the more promising node (higher ub). Prof. Amr Goneid, AUC

  10. The 0/1 Knapsack • We terminate at the current node for one of the following 3 reasons: • The ub of node is not more promising than what is obtained so far. • The node violates the constraint w ≤ W • No further choices. Prof. Amr Goneid, AUC

  11. Example • W = 10, n = 4 Prof. Amr Goneid, AUC

  12. B & B State Space Numbers in each node are for W, V, UB 0 0,0,100 1 No 1 yes 1 2 4,40,76 0,0,60 2 No Inferior to node 8 2 yes 3 4 W = 11 4,40,70 3 yes 3 No 6 5 9,65,69 4,40,64 4 yes 4 No Inferior to node 8 7 8 W = 12 9,65,65 Optimal Solution Prof. Amr Goneid, AUC

  13. The 0/1 Knapsack • Level (0): Node (0) w = 0 v = 0 dbest = 10 ub = 0 + (W - 0)*10 = 100 • Level (1): • Node (1) left: with item (1) w = 4 v = 40 ub = 40 + (W - 4)*6 = 76 • Node (2) right: without (1) w = 0 v = 0 ub = 0 + (W – 0)*6 = 60 • Node (1) is more promising, Branch from that node. • Level (2): • Node (3) left: with item (2) w = 4 + 7 = 11 Not feasible • Node (4) right: without (2) w = 4 v = 40 ub = 40 + (W – 4)*5 = 70 Prof. Amr Goneid, AUC

  14. The 0/1 Knapsack • Level (3): Branching from node (4) • Node (5) left: with item (3) w = 9 v = 65 ub = 65 + (W – 9)* 4 = 69 • Node (6) right: without (3) w = 4 v = 40 ub = 40 + (W – 4) *4 = 64 • Node (5) is more promising, Branch from that node • Level (4): Branching from node (5) • Node (7) left: with item (4) w = 9 + 3 = 12 Not feasible • Node (8) right: without (4) w = 9 v = 65 ub = 65 • Node (8) is the optimal solution • The final solution is: (item 1 + item 3): total weight is 9 total value is 65 Prof. Amr Goneid, AUC

  15. Exercise • Solve the following instance of the knapsack problem by the branch-and-bound algorithm: W = 16, n = 4 Prof. Amr Goneid, AUC

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