An iteration of Branch and Bound One iteration of Branch and Bound consists of the following four steps: Some definitions. Branch and Bound.

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1 ranch and ound xamples and xtensions epartment of Management ngineering Technical University of enmark ounding ow do we get ourselves a bounding function? Relaxation. Leave out some constraints. Keep the same objective function f. So now we maximize f over a larger solution space P. If all original constraints are satisfied we have an incumbent. hange the objective function f to g. ere gx) fx) for all x S. Optimality of g does not automaticly guarantee optimality of f. Use the two above at the same time. 3 Some definitions branching is a division of the original feasible set S into subsets S1... SK. branching is valid if K i=1 S i = S It is possible to determine and optimize over each set Si branching is called partitive if the sets Si are disjoint. 2 n iteration of ranch and ound One iteration of ranch and ound consists of the following four steps: 1. Selection of the node to be processed. 2. ound calculation. 3. Pruning? 4. ranching. 4 1

2 Strategy for branching In many situations there is not only one variable/candidate for branching. The question is which to choose. ranch on most fractional variable. ranch on least fractional variable. stimate the cost of forcing xj to become integer. These are all heuristic strategies, there is no guaranteed best strategy. Mostly mentioned strategies S: epth irst Search strategy: select among the active nodes one of those with the largest level. S: readth irst Search strategy: select among the active nodes one of those with the lowest level. es: est irst Search strategy: select among the active nodes one of those with the lowest bound. (sometime also called global best node selection). 7 Strategy for selecting the next subproblem In choosing a search strategy we might consider two different goals: Minimizing overall solution time. inding a good feasible solution to our original problem. 6 The est irst Search strategy One way to minimize overall solution time is to try to minimize the size of the search tree. subproblem (& node) is critical if the bound exceeds the value of the optimal solution. critical node must always be examined. Only nodes with values larger than or equal to the optimum will be searched. No superfluous calculations take place in est irst Search after the optimum has been found. 8 5

3 ranch and ound for TSP Input: distance matrix for a digraf = V ) with n vertices (so dij may be different from dji). If the edge i j) is in then dij equals the distance from i to j, otherwise dij equals. Output: The length d) of a shortest directed circuit in visiting each vertex exactly once, and an n-vector of edges v1 v2)... vn 1 vn)). Method: &: ounding function, search and branching strategy is described below. TSP xample Mathematical formulation of the TSP min st. n n dijxij i=1 n j=1i=j n j=1 i=1i=j ij S xij = 1 i 1... n} xij = 1 j 1... n} xij S 1 S V xij 0 1} i j 1... n} 10 TSP Lower bound calculation

4 TSP - & omponents I ounding function: Perform row and column reductions as for the assignment problem subtract the smallest element in each row from all row elements and then the smallest element in each column of the resulting matrix from all column elements. The sum of the subtracted elements constitutes a lower bound for the length of the optimal TSP tour. 13 TSP - & omponents III Search strategy: epth first: xamine the subproblem with the included edge first. ecause i j) is most expensive to leave out, there is a fair chance that the other subproblems is pruned in a later iteration. 15 TSP - & omponents II ranching: ind that 0-position i j) in the reduced matrix, for which the sum of the next-smallest elements in row i and column j is largest ( the most expensive edge to leave out in a new solution ). onstruct two subproblems: one in which i j) is forced into the solution (remove row i and column j from the matrix and set dji to ), and one in which i j) is excluded from the solution (set dij equal to ). 14 TSP ranching on 6 3) 1-branch 0-branch

5 TSP numeration tree with solution L=81 +(6,3) (6,3) L=81 L=129 +(6,3); +(4,6) +(6,3); (4,6) L=81 L=113 +(6,3); +(4,6); +(2,1) +(6,3); +(4,6); (2,1) L=84 L=101 +(6,3); +(4,6); +(2,1); +(1,4) +(6,3); +(4,6); +(2,1); (1,4) L=112 Solution with value 104! 17 ounds One way to identify a bound for the TSP is by relaxing constraints. This could be to allow subtours. This bound is although known to be rather weak. n alternative is the 1 tree relaxation. 19 The Symmetric TSP min s.t. ij) j ij S dijxij xij = 2 i n} xij S 1 S V xij 0 1 i j) 18 The 1-tree bound Identify a special vertex 1 (this can be any vertex of the graph). 1 and all edges incident with 1 are removed from. or the remaining graph determine the minimum spanning tree T. Now the two smallest edges e1 and e2 incident with 1 are added to T producing T1 (called a 1 tree) 20

6 Why is T1 a bound? We need to convince ourselves that the total cost of T1 is a lower bound of the value of an optimal tour. Note that a amiltonian tour can be divided into two edges e 1 and e 2 that are incident with 1 and the rest of the tour (let us call it T ). So the set of amiltonian tours is a subset of 1-trees of. Since e1 e2 are the two smallest edges incident to 1 + d d de e2 e + d e. urthermore as T is a 2 tree dt ) dt ). 21 TSP of ornholm So the cost of T1 is less that or equal to the cost of any amiltonian tour. In the case T1 is a tour we have found the optimal solution and can prune by bounding. otherwise we need to bound tree bound of ornholm Tree in rest of dge left out by Kruskal s MST algorithm 1 tree edge ost of 1 tree = 97 24

7 Strengthening the bound Idea: Vertices of T1 with high degree are incident with too many attractive edges. Vertices of degree 1 have on the other hand too many unattractive edges. efine πi as the degree of vertex i minus 2. Note that i V π i equals 0 since T1 has n edges and therefore the degree sum is 2n. or each edge i j) we transform the cost to d ij = d ij + πi + πj. 25 ow do we branch? Observe that in the case our 1-tree is not a tour at least one vertex has degree 3 or more. So choose a vertex v with degree 3 or more. or each edge ui v) generate a subproblem where ui v) is excluded from the set of edges. 27 Strengthen the bound Modified distance matrix: ost of 1 tree = ranching on ornholm xclude (,) xclude (,) xclude (,) ost of 1 tree = 98 The optimal tour! ost of 1 tree =

8 Using & as a heuristic rom the root-node of the &-tree we have a (global) upper bound u. When we compute an initial solution at root node or bound by optimality we have a global lower bound l. This gives us the opportunity to estimate the worst-case deviation from the optimum. 29 Keys to success ave a good bounding function. If several are available the stronger the better even if it is computationally harder to solve. et a good incumbent early (never underestimate the value of a good initial solution). xperiment, experiment, experiment. 31 Optimization-based heuristics Stop the & when the estimate on the deviation is below a certain threshold. on t investigate all branches in the &-tree (especially applied within inary integer programming) eam Search: Usefull technique if memory is limited. Set an upper limit α on active nodes that are stored. If this limit is reached only keep the α best. 30

Methods and Models for Combinatorial Optimization Exact methods for the Traveling Salesman Problem

Methods and Models for Combinatorial Optimization Exact methods for the Traveling Salesman Problem L. De Giovanni M. Di Summa The Traveling Salesman Problem (TSP) is an optimization problem on a directed