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1 Lecture 4 Maximal Flow Problems Set Covering Problems.

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Presentation on theme: "1 Lecture 4 Maximal Flow Problems Set Covering Problems."— Presentation transcript:

1 1 Lecture 4 Maximal Flow Problems Set Covering Problems

2 2 Agenda  maximal flow problems  set covering problems

3 3 Maximal Flow Problems

4 4  arcs labeled with their capacities  question: LP formulation of the total maximal total maximal flow from the sources to the sinks

5 5 Maximal Flow Problems  obvious maximum: 9 units  LP formulation, let  x be the flow from node 0  x ij be the flow from node i to node j 5 4 8 1 2 0

6 6 Maximal Flow Problem  let x ij be the flow from node i to node j

7 7 Maximal Flow Problem S T     

8 8 S T      

9 9 Comments for the Maximal Flow Problem  special structure of network flow  integral solutions for integral capacities

10 10 Further Comments for Network Flow Problems  network components in many practical problems  easier to solve with packages  more likely to have integral optimal solutions  many practical LP problems being dual of network flow problems  optimal integral solutions

11 11 Set Covering Problems

12 12 Set Covers  a set S = {1, 2, 3, 4, 5}  a collection of subsets of S,  = {{1, 2}, {1, 3, 5}, {2, 4, 5}, {3}, {1}, {4, 5}}  a cost associated with each subset of S in   e.g., cost = 1 for each subset of S in   a subset of  is a cover of S if the subset contains all elements of S  {1, 2}, {1, 3, 5}, and {2, 4, 5} forms a cover of S  {1}, {3}. and {4, 5} do not form a cover of S

13 13 Set Covering Problems  given S, , and all costs of subsets in  = 1, find the minimum cost cover of S  S = {1, 2, 3, 4, 5}   = {{1, 2}, {1, 3, 5}, {2, 4, 5}, {3}, {1}, {4, 5}}  what are the decisions?  a subset is selected or not  what are the constraints?  elements of S are covered

14 14 Set Covering Problems  S = {1, 2, 3, 4, 5}   = {{1, 2}, {1, 3, 5}, {2, 4, 5}, {3}, {1}, {4, 5}}  examples   1 =  3 =  4 = 1 and  2 =  5 = 0: {1, 2}, {2, 4, 5}, {3}   2 =  5 = 1 and  1 =  3 =  4 =  6 = 0: {1, 3, 5}, {1}

15 15 Set Covering Problems  S = {1, 2, 3, 4, 5}   = {{1, 2}, {1, 3, 5}, {2, 4, 5}, {3}, {1}, {4, 5}}  i  {0, 1} element 1: element 2: element 3: element 4: element 5: set 1: set 2: set 3: set 4: set 5: set 6: Property 9.1: minimization with all  constraints Property 9.2: all RHS coefficients = 1 Property 9.3: all matrix coefficients = 0 or 1

16 16 Generalization of Set Covering Problems  weighted set covering problems: RHS coefficient  positive integers > 1  some elements covered multiple times  generalized set covering problems: a weighted set covering problem + matrix coefficients 0 or  1

17 17 Applications of Set Covering Problems  aircrew scheduling  S: the collection of flights legs to cover   : the collection of feasible rosters of air crew

18 18 Comments for Set Covering Problems  “Set covering problems have an important property that often makes them comparatively easy to solve by the branch and bound method. It can be shown that the optimal solution to a set covering problem must be a vertex solution in the same sense as for LP problems. Unfortunately, this vertex solution will not generally be (but sometimes is) the optimal vertex solution to the corresponding LP model. It is, however, often possible to move from this continuous optimum to the integer optimum in comparatively few steps.” (pp 191 of [7])

19 19 Optimal Solution at a Vertex for a Set Covering Problem  suppose there are optima not at a vertex  let {x i } and {y i } be two different optimal solutions  then {  x i +(1  )y i } are optimal solution  there must be at least one i such that x i  y i  for x i  y i,  x i +(1  )y i  {0, 1} iff  = 0 or 1  either case there is only one optimal LP: optimal at a vertex  i  {0, 1}

20 20 LP Optimum Not Set Covering Optimum Set Covering: optimal at a vertex, but not necessarily at that of LP LP: optimal at a vertex

21 21 Comments for Set Covering Problems  relatively easy to solve by Branch and Bound  optimal solution at a vertex, though not that by LP relaxation  possible to move from LP optimum to the set covering optimum in a few steps  in applications, usually many more variables than constraints  solved by column generations


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