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Chapter 11 Minimum Cost Flows: Network Simplex Algorithms

Chapter 11 Minimum Cost Flows: Network Simplex Algorithms. Network Simplex Algorithm. Streamlined implementation of the bounded variable simplex method on the network. (See Application 9.6 for an application of the minimum cost flow problem.) Formulation: N is n m matrix min cx

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Chapter 11 Minimum Cost Flows: Network Simplex Algorithms

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  1. Chapter 11Minimum Cost Flows: Network Simplex Algorithms

  2. Network Simplex Algorithm • Streamlined implementation of the bounded variable simplex method on the network. (See Application 9.6 for an application of the minimum cost flow problem.) • Formulation: N is nm matrix min cx s.t. Nx = b 0  x  u Optimality conditions of the bounded variable simplex method: Let  denote the dual variable associated with the mass balance constraints. Nij: a column of N corresponding to arc (i, j) = ei – ej reduced cost cij = cij – ((i) - (j)) = cij - (i) + (j)  cij.

  3. For nonbasic variables: If cij  0  xij = 0 If cij  0  xij = uij For basic variables: cij = 0 • The set of basic variables corresponds to a spanning tree in the network. How to represent the spanning tree and perform operations on the spanning tree? Use three arrays pred(i), depth(i), and thread(i) to represent a spanning tree.

  4. thread: define a traversal of a tree. obtained by depth-first search of nodes (follows the contour of the tree ignoring revisited nodes, preorder)) can find all descendants of a node i easily (using depth). 1 2 3 5 6 4 7 8 9

  5. Relationship between basis and spanning tree • Rank of N  (n-1) (sum of rows = 0) Show than rank(N)  (n-1) by showing (n-1) linearly independent columns of N (spanning tree) • Ex) rearrange rows and columns of the submatrix for a spanning tree starting from leaf of the tree using preorder (reverse order) (2, 4) (1, 2) (3, 5) (3, 1) artif 1 4 2 2 3 5 3 1 4 5

  6. Converse) (n-1) linearly independent columns of N  spanning tree Pf) Suppose not a spanning tree, then there exists a cycle W. Take (i, j)W(1)Nij = (i, j)W(1)(e(i) – e(j)) = 0. (Use +1 for forward arc, -1 for backward arc in W.) Hence the columns are linear dependent.  • Thm) (n-1) linearly independent columns of N  spanning tree. • 2 approaches: • Drop one constraint • Add an artificial variable to a row of N (take the node as root of the tree) (can take the dual value (r) = 0 for the root node from B = cB for basis B)

  7. LP operations: Computing basic feasible solution and dual solution : xB : BxB = b – LxL – UxU, B: lower triangular  BxB = b’, solve by substitution (on network, determine flows starting from leaf nodes)  : B = cB, solve by substitution (on network, determine  starting from root node using cij - (i) + (j) = 0 ) Actually, x and  are updated, rather than computed again from scratch. Finding an entering nonbasic variable: • cij < 0 and xij = 0 (xij < uij): want to increase flow on arc (i, j) • cij > 0 and xij = uij (xij > 0): want to decrease flow on arc (i, j) Leaving basic variable: Have xB = B-1b – B-1LxL – B-1UxU • If increase entering xj (xj  xj + t): xB(t) = xB* - td (d: updated column of xj) • If decrease entering xj (xj  xj - t): xB(t) = xB* + td • determine largest t satisfying lB  xB  uB Update solution

  8. Operations on the network: Computing : • set (1) = 0 (root) • Following thread(i) index, use cij - (i) + (j) = 0 i = pred(j) (i, j)A  (j) = (i) – cij (j, i)A  (j) = (i) + cij Computing xB: • For (i, j)  U, set xij = uij, b’(i) = b(i) – uij, b’(j) = b(j) + uij. Then determine tree arc flow starting from leaf node. if j is leaf node, (i, j)A  xij = -b’(j), b’(i)  b’(i) – xij = b’(i) + b’(j) if j is leaf node, (j, i)A  xij = b’(j), b’(i)  b’(i) + xij = b’(i) + b’(j) Read thread(i), and put them into a stack. Pop operation chooses leaf nodes always.

  9. (continued) Finding initial tree: no details here Choosing an entering arc: some rules possible Determining leaving arc: • Suppose arc (k, l) entering  creates a cycle W orientation of cycle: same as (k, l) if xkl = 0, opposite direction of (k, l) if xkl = uij. (want to increase flow along the cycle) ij = uij – xij, if (i, j)W and forward arc xij, if (i, j)W and backward arc  = min{ ij: (i, j)W}. Augment  units of flow on W. A blocking arc leaves tree. • In simplex method, xB(t) = xB* - td (increase case) d = B-1Nkl or Bd = Nkl (Nkl + W(1)Nij = 0, +1 for forward, -1 for backward) choose dij = -1, if (i, j) is forward arc in W = +1, if (i, j) is backward arc in W = 0, otherwise is solution  corresponds to flow augmentation on W.

  10. (continued) Identifying the cycle: • Identify apex (join) of the cycle W. Use predecessor, depth index to identify apex. Also compute  at the same time. Updating the tree: • No details here. Refer Chapter 19, Linear Programming, Vasek Chvatal, 1983. • Can use thread to update the representation of the tree easily.

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