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Chapter 6 Branch-and-Bound

Chapter 6 Branch-and-Bound. 6.1 Illustrating Branch-and-Bound with the 0-1 Knapsack Problem 6.2 The Traveling Salesperson Problem. 1. An improvement on the backtracking algorithm. 2. Similar to backtracking, a state space tree is used .

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Chapter 6 Branch-and-Bound

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  1. Chapter 6Branch-and-Bound 6.1 Illustrating Branch-and-Bound with the 0-1Knapsack Problem 6.2 The Traveling Salesperson Problem Branch-and-Bound

  2. 1. An improvement on the backtracking algorithm. 2. Similar to backtracking, a state space tree is used. 3. Does not limit us to any particular way of traversing the tree. 4. It is used only for the optimization problem. 5. Branch-and-Bound algorithm computes a number (bound) at a node to determine whether the node is promising. P.S: the 0-1 Knapsack problem in Section 5.7 is actually a branch-and-bound algorithm. What is Branch-and-Bound? Branch-and-Bound

  3. 1. Best-first search with branch-and-bound pruning: A simple modification of the approach called breadth-first search with branch-and-bound pruning. Branch-and-Bound Pruning voidbreadth_first_tree_search(treeT) { queue_of_nodeQ; nodeu, v ; initialize(Q); v = root of T ; visit v; enqueue(Q, v); while (!empty(Q)) { dequeue(Q, v); for (each child u of v) { visit u; enqueue(Q, u); } } } Branch-and-Bound

  4. 7 1 3 8 6 5 2 13 14 10 12 15 4 11 9 16 Breadth-First Search of a Tree Branch-and-Bound

  5. 1. Breadth-First Search with Branch-and-Bound Pruning for the 0-1 Knapsack problem. 2. Best-First Search with Branch-and-Bound Pruning for the 0-1 Knapsack problem：an improvement of 1. A node is nonpromising if this bound is less than or equal tomaxprofit or weight  W. 6.1 Illustrating Branch-and-Bound with the 0-1Knapsack Problem Branch-and-Bound

  6. (0,0) $0 0 $115 (1,1) $40 2 $115 Example 6.1 Suppose thatn = 4, W=16, and we have the following ipiwipi/wi 1 $40 2 $20 2 $30 5 $6 3 $50 10 $5 4 $10 5 $2 Find the solutions. • Sol： • Setmaxprofit = 0 • Visit node(0,0) • profit = ? • weight = ? • 3. Visit node (1,1) ：profit = ? weight = ? bound = ? maxprofit = ? bound = ? Branch-and-Bound

  7. (0,0) $0 0 $115 (1,1) (1,2) $0 0 $82 $40 2 $115 (2,1) (2,2) (2,3) (2,4) $70 7 $115 $40 2 $98 $30 5 $82 $0 0 $60 (3,4) (3,1) (3,2) (3,3) (3,5) (3,6) $120 17 $0 $70 7 $80 $90 12 $98 $40 2 $50 $80 15 $82 $30 5 $40 (4,1) (4,2) (4,3) (4,4) $80 12 $80 $70 7 $70 $100 17 $0 $90 12 $90 Example 6.1 (Cont’d) n = 4, W=16, 4. Visit node(1,2) profit = ? weight = ? 5. Visit node (2,1) maxprofit = ? bound = ? Branch-and-Bound

  8. voidbreadth_first_branch_and_bound(state_space_treeT, number& best) { queue_of_nodeQ; nodeu, v ; initialize(Q); v = root of T ; enqueue(Q, v); best = value(v); while (!empty(Q)) {dequeue(Q, v); for (each child u of v) {if (value(u) is better than best) best = value(u); if (bound(u) is better than best) enqueue(Q, u); }} } Breadth-First Branch-and-Bound Branch-and-Bound

  9. voidknapsack2(intn, const intp[], const intw[], intW, int maxprofit) { queue_of_nodeQ; nodeu, v ; initialize(Q); v.level = 0 ; v.profit = 0 ; v.weight = 0 ; maxprofit = 0; enqueue(Q, v); while (!empty(Q)) {dequeue(Q, v); u.level = v.level+1 ; u.weight = v.weight +w[u.level] ; u.profit = v.profit+p[u.level] ; if (u.weight <=W && u.profit>maxprofit ) maxprofit = u.profit; if (bound(u) > maxprofit) enqueue(Q, u); u.weight = v.weight ; u.profit = v.profit ; if (bound(u) > maxprofi) enqueue(Q, u); } } Algorithm 6.1 Branch-and-Bound

  10. floatbound(nodeu) { indexj, k ; inttotalweight ; floatresult ; if (u.weight >=W) return 0; else {result = u.profit ; j = u.level+1 ; totweight = u.weight ; while (j <=n && totweight+w[j]<=W ) {totweight= totweight+w[j]; result= result+p[j]; j++; } k = j ; if (k <= n) result= result+(Wtotweight)p[k]/w[k]; return result ; } } Algorithm 6.1 (Cont’d) Branch-and-Bound

  11. (0,0) $0 0 $115 (1,1) $40 2 $115 Example 6.2 Suppose thatn = 4, W=16, and we have the following ipiwipi/wi 1 $40 2 $20 2 $30 5 $6 3 $50 10 $5 4 $10 5 $2 Find the solutions. • Sol： • Setmaxprofit = 0 • Visit node(0,0) • profit = ? • weight = ? • 3. Visit node (1,1) ：profit = ? weight = ? bound = ? maxprofit = ? bound = ? Branch-and-Bound

  12. (0,0) $0 0 $115 (1,1) (1,2) $0 0 $82 $40 2 $115 (2,1) (2,2) $70 7 $115 $40 2 $98 (3,4) (3,1) (3,2) (3,3) $120 17 $0 $70 7 $80 $90 12 $98 $40 2 $50 (4,1) (4,2) $100 17 $0 $90 12 $90 Example 6.2 (Cont’d) n = 4, W=16, 4. Visit node(1,2) profit = ? weight = ? 5. Visit node (2,1) maxprofit = ? bound = ? Branch-and-Bound

  13. voidbest_first_branch_and_bound(state_space_treeT, number& best) { priority_queue_of_nodePQ; nodeu, v ; initialize(PQ); v = root of T ; insert(PQ, v); best = value(v); while (!empty(PQ)) {remove(PQ, v); if (bound(v) is better than best) for (each child u of v) {if (value(u) is better than best) best = value(u); if (bound(u) is better than best) insert(PQ, u); } } } Best-First Branch-and-Bound Branch-and-Bound

  14. voidknapsack3(intn, const intp[], const intw[], intW, int maxprofit) { priority_queue_of_nodePQ; nodeu, v ; initialize(PQ); v.level = 0 ; v.profit = 0 ; v.weight = 0 ; maxprofit = 0; insert(PQ, v); while (!empty(PQ)) {remove(PQ, v); if (v.bound > maxprofit) { u.level = v.level+1 ; u.weight = v.weight +w[u.level] ; u.profit = v.profit+p[u.level] ; if (u.weight <=W&&u.profit>maxprofit ) maxprofit = u.profit; u.bound = bound(u); if (u.bound > maxprofit) insert(PQ, u); u.weight = v.weight ; u.profit = v.profit ; u.bound = bound(u); if (u.bound > maxprofi) insert(PQ, u); } } } Algorithm 6.2 Branch-and-Bound

  15. How to find the optimal tour more efficient than using backtracking? Branch-and-Bound can be considered for searching the state space tree. Best-First Search：the best bound at present is searched first 6.2 The Traveling Salesperson Problem v1 v2 Optimal tour 10 4 7 v4 v3 7 2 v5 Branch-and-Bound

  16. 1.Starts at v1, the 2nd vertex can be v2, v3, v4, v5. v1：min{14,4,10,20}=4v2：min{14,7,8,7}=7 v3：min{4,5,7,16}=4 v4：min{11,7,9,2}=2 v5 ：min{18,7,17,4}=4 The minimum bound is 21. There can be no tour with a shorter length. 2. Suppose we have visited [v1, v2] v1：14 v2：min{7,8,7}=7 v3：min{4,7,16}=4 v4：min{11,9,2}=2 v5 ：min{18,17,4}=4 The minimum bound is 31. [1,2] [1] [1,3] [1,4] [1,2,3,5] [1,2,5] [1,2,4] [1,2,3] [1,2,3,4] [1,5] Computation of Bound for a Node Branch-and-Bound

  17. Visit node [1] ：bound=21, minlength= Visit node [1,2] ： bound=31, Visit node [1,3] ： bound=22, Visit node [1,4] ： bound=30, Visit node [1,5] ： bound=42, Visit node [1,3,2] ： bound=22, Visit node [1,3,4] ： bound=27, Visit node [1,3,5] ： bound=39, Visit node [1,3,2,4] ： bound=37, minlength=37 Visit node [1,3,2,5] ： bound=31, minlength=31 Best First Search Node Best First Search Node Example 6.3 Branch-and-Bound

  18. [1] Bnd=21 [1,5] Bnd=42 [1,2] Bnd=31 [1,4] Bnd=30 [1,3] Bnd=22 [1,3,2] Bnd=22 [1,3,4] Bnd=27 [1,3,5] Bnd=39 [1,4,2] Bnd=45 [1,4,3] Bnd=38 [1,4,5] Bnd=30 [1,3,4,5] Len=34 [1,3,2,5] Len=31 [1,3,2,4] Len=37 [1,3,4,2] Len=43 [1,4,5,2] Len=30 [1,4,5,3] Len=48 Example 6.3 (Cont’d) Branch-and-Bound

  19. for (all i such that 2i n &&i is not in v.path) {u.path = v.path; put i at the end of u.path; if(u.level== n2) { put index of only vertex not in u.path at the end of u.path ; put 1 at the end of u.path ; if(length(u)<minlength) {minlength=length(u); opttour = u.path ; } } else { u.bound = bound(u); if(u.bound<minlength) insert(PQ, u); } } }}} Algorithm 6.3 voidtravel2(intn, const numberW[][], ordered_set&opttour number&minlength) { priority_queue_of_node PQ; node u, v; initialize(PQ); v.level = 0; v.path = [1]; v.bound = bound(v); minlength = ; insert(PQ, v); while (!empty(PQ )) { remove(PQ, v); if(v.bound <minlength) { v.level = v.level +1; Branch-and-Bound

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