Network Simplex Method Combinatorial Problem Solving (CPS) Enric Rodr´ ıguez-Carbonell April 26, 2019
Network Programs A network program is of the form ■ min c T x Ax = b ℓ ≤ x ≤ u, where c ∈ R m , b ∈ R n and A ∈ {− 1 , 0 , 1 } n × m has the following property: each column has exactly one 1 and one − 1 (and so the remaining coefficients are 0 ) Note that n is the number of constraints and m is the number of variables ■ 2 / 35
Network Programs A network program is of the form ■ min c T x Ax = b ℓ ≤ x ≤ u, where c ∈ R m , b ∈ R n and A ∈ {− 1 , 0 , 1 } n × m has the following property: each column has exactly one 1 and one − 1 (and so the remaining coefficients are 0 ) Example: min x 1 + x 2 + 3 x 3 + 10 x 4 ■ x 1 1 0 1 1 5 x 2 − 1 1 0 0 = 0 x 3 0 − 1 − 1 − 1 − 5 x 4 0 ≤ x 1 ≤ 4 0 ≤ x 3 ≤ 4 0 ≤ x 2 ≤ 2 0 ≤ x 4 ≤ 10 2 / 35
Minimum Cost Flow Problems Network programs can be seen as minimum cost flow problems in a graph ■ We associate a digraph G = ( V, E ) to the matrix of a network program: ■ Vertices V correspond to rows (constraints) ◆ Edges E correspond to columns (variables) ◆ A column with a 1 at row i and a − 1 at row k gives an edge ( i, k ) ◆ 3 / 35
Minimum Cost Flow Problems Network programs can be seen as minimum cost flow problems in a graph ■ We associate a digraph G = ( V, E ) to the matrix of a network program: ■ Vertices V correspond to rows (constraints) ◆ Edges E correspond to columns (variables) ◆ A column with a 1 at row i and a − 1 at row k gives an edge ( i, k ) ◆ Then we can reinterpret the other elements of the network program: ■ Each variable x j is the flow sent along the j -th edge ◆ The cost of sending 1 unit of flow is c j ◆ Flow cannot exceed capacity u j ◆ There must be a minimum flow ℓ j (usually, 0 ) ◆ Total production of flow at vertex i is determined by b i ◆ So solving the network program consists in ■ finding the feasible flow along the graph that minimizes the cost 3 / 35
Minimum Cost Flow Problems min x 1 + x 2 + 3 x 3 + 10 x 4 x 1 1 0 1 1 5 x 2 − 1 1 0 0 = 0 x 3 0 − 1 − 1 − 1 − 5 x 4 0 ≤ x 1 ≤ 4 0 ≤ x 3 ≤ 4 0 ≤ x 2 ≤ 2 0 ≤ x 4 ≤ 10 { 0 } 2 x 1 x 2 [1 , 0 , 2] [1 , 0 , 4] ❀ [ c j ,ℓ j ,u j ] x 3 [3 , 0 , 4] 1 x 4 3 [10 , 0 , 10] {− 5 } { 5 } ❀ b i 4 / 35
Network Simplex Method Network programs satisfy Hoffman & Gale’s conditions. ■ So simplex method is guaranteed to give integer solutions (if ℓ, u, b in Z ) Moreover we can specialize the simplex method for network programs ■ This lecture is devoted to this specialization: the network simplex method ■ In the first place we need to revisit a bit of graph theory ■ 5 / 35
Vertex-Edge Incidence Matrix The vertex-edge incidence matrix of digraph G = ( V, E ) is a matrix A s.t.: ■ Rows are labelled by vertices ◆ Columns are labelled by edges ◆ For each v ∈ V and e ∈ E , coefficient a v,e of A is ◆ • 1 if e = ( v, · ) • − 1 if e = ( · , v ) • 0 otherwise Given a network program whose matrix is A , ■ the vertex-edge incidence matrix of its associated digraph is precisely A 6 / 35
Vertex-Edge Incidence Matrix 2 2 1 3 1 3 4 1 2 3 4 1 1 0 1 1 2 − 1 1 0 0 3 0 − 1 − 1 − 1 7 / 35
Paths and Cycles A path is a finite sequence P = ( v 1 , e 1 , v 2 , . . . , v K , e K , v K +1 ) such that ■ either e k = ( v k , v k +1 ) or e k = ( v k +1 , v k ) for all 1 ≤ k ≤ K Note that paths can invert the orientation of edges ■ The orientation sequence of a path P is ( O P ( e 1 ) , . . . , O P ( e k )) , where ■ +1 if e k = ( v k , v k +1 ) − 1 O P ( e k ) if e k = ( v k +1 , v k ) 0 otherwise A cycle is a path such that the initial and the final vertices are the same ■ 8 / 35
Paths and Cycles 2 2 1 3 1 3 4 (3 , 4 , 1 , 1 , 2) is a path with orientation sequence ( − 1 , 1) 9 / 35
Paths and Cycles Prop. Let P = ( v 1 , e 1 , v 2 , . . . , v K , e K , v K +1 ) be a path. Then ■ K � O P ( e k ) · a e k = e v 1 − e v K +1 , k =1 where a e is the column of e in the vertex-edge incidence matrix A , and e v is the v -th unit vector, i.e., all zeroes except for a 1 at index v 10 / 35
Paths and Cycles Prop. Let P = ( v 1 , e 1 , v 2 , . . . , v K , e K , v K +1 ) be a path. Then ■ K � O P ( e k ) · a e k = e v 1 − e v K +1 , k =1 where a e is the column of e in the vertex-edge incidence matrix A , and e v is the v -th unit vector, i.e., all zeroes except for a 1 at index v Proof. Let k be s.t. 1 ≤ k ≤ K . There are two cases: 1. If e k = ( v k , v k +1 ) then a e k = e v k − e v k +1 and O P ( e k ) = 1 2. If e k = ( v k +1 , v k ) then a e k = e v k +1 − e v k and O P ( e k ) = − 1 In any case O P ( e k ) · a e k = e v k − e v k +1 . So K � O P ( e k ) · a e k = ( e v 1 − e v 2 )+( e v 2 − e v 3 )+ . . . +( e v K − e v K +1 ) = e v 1 − e v K +1 k =1 10 / 35
Paths and Cycles Prop. Let P = ( v 1 , e 1 , v 2 , . . . , v K , e K , v K +1 ) be a path. Then ■ K � O P ( e k ) · a e k = e v 1 − e v K +1 , k =1 where a e is the column of e in the vertex-edge incidence matrix A , and e v is the v -th unit vector, i.e., all zeroes except for a 1 at index v Cor. If C = ( v 1 , e 1 , v 2 , . . . , v K , e K , v K +1 ) is a cycle, ■ the columns a e 1 , a e 2 , . . . , a e K of A are linearly dependent. 10 / 35
Paths and Cycles Prop. Let P = ( v 1 , e 1 , v 2 , . . . , v K , e K , v K +1 ) be a path. Then ■ K � O P ( e k ) · a e k = e v 1 − e v K +1 , k =1 where a e is the column of e in the vertex-edge incidence matrix A , and e v is the v -th unit vector, i.e., all zeroes except for a 1 at index v Cor. If C = ( v 1 , e 1 , v 2 , . . . , v K , e K , v K +1 ) is a cycle, ■ the columns a e 1 , a e 2 , . . . , a e K of A are linearly dependent. Proof. If v 1 = v K +1 then K � O P ( e k ) · a e k = e v 1 − e v K +1 = 0 k =1 10 / 35
Paths and Cycles 2 1 2 3 4 1 1 0 1 1 2 1 2 − 1 1 0 0 3 − 1 − 1 − 1 0 3 1 3 4 Path P = (3 , 4 , 1 , 1 , 2) has orientation sequence ( − 1 , 1) 1 1 0 K � + (1) = = e 3 − e 2 O P ( e k ) · a e k = ( − 1) − 1 − 1 0 − 1 0 1 k =1 11 / 35
Trees A graph is ■ acyclic if it has no cycles ◆ connected if for any pair of vertices u, v there is a path from u to v ◆ a tree if it is acyclic and connected ◆ Thm. For a graph T with at least one vertex the following are equivalent: ■ T is a tree ◆ For any pair of vertices u, v there is a unique path from u to v ◆ T has one less edge than vertices and is connected ◆ T has one less edge than vertices and is acyclic ◆ A subgraph S of G is spanning if it covers all vertices in G ■ Thm. Every connected graph has a subgraph that is a spanning tree. ■ 12 / 35
Trees Thm. For any T subgraph of G that is a tree with at least two vertices, ■ the columns { a e | e ∈ T } of A are linearly independent. 13 / 35
Trees Thm. For any T subgraph of G that is a tree with at least two vertices, ■ the columns { a e | e ∈ T } of A are linearly independent. Proof. By contradiction. Let T be a tree with the minimum number of vertices N such that { a e | e ∈ T } are linearly dependent, i.e., there are λ e not all null s.t. � λ e a e = 0 e ∈ T If N = 2 then T would have one edge, say e , and a e � = 0 So N > 2 . Let v be a leaf of T and let e v be the only edge in T that has v as an endpoint. Let T ′ be the tree obtained from T by removing e v . From � λ e v a e v + λ e a e = 0 e ∈ T,e � = e v by projecting onto the row of v we have λ e v = 0 . Hence the tree T ′ is a subgraph of G with N − 1 ≥ 2 vertices whose columns are linearly dependent. Contradiction! 13 / 35
Paths and Cycles 2 1 2 3 4 1 1 0 1 1 2 1 2 − 1 1 0 0 3 − 1 − 1 − 1 0 3 1 3 4 Edges { 4 , 1 } induce a subgraph that is a tree, and 1 1 = 2 − 1 rank 0 − 1 0 14 / 35
Reformulating Network Programs Thm. If G is a connected graph with n > 0 nodes then rank( A ) = n − 1 ■ 15 / 35
Reformulating Network Programs Thm. If G is a connected graph with n > 0 nodes then rank( A ) = n − 1 ■ Proof. G has a spanning tree T , which has n − 1 edges. Its columns are linearly independent, so rank( A ) ≥ n − 1 . But since adding all rows of A we get 0 , finally rank( A ) = n − 1 . 15 / 35
Reformulating Network Programs Thm. If G is a connected graph with n > 0 nodes then rank( A ) = n − 1 ■ Let us assume graphs of network programs are connected, so m ≥ n − 1 ■ (otherwise, work independently on the connected components) So the matrix of a network program has rank n − 1 . ■ But the simplex method requires to have a full-rank matrix! We add an extra variable w with a unit column e r , where r is taken ■ arbitrarily from { 1 , . . . , n } , and such that it is forced to have value 0 : min c T x Ax + e r w = b ℓ ≤ x ≤ u, 0 ≤ w ≤ 0 15 / 35
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