Transcript Network Simplex Algorithms

Chapter 11
Minimum Cost Flows: Network
Simplex Algorithms
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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.
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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.
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1 2 3 4 5 6 7 8 9
pred(i)
0 1 2 3 2 5 5 6 6
depth(i)
0 1 2 3 2 3 3 4 4
thread(i)
2 5 4 1 6 8 3 9 7
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i
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 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).
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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
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 1 0 0 0
 1 1 0 0

 0 0 1 0
0 0 1 1

 0 1 0  1
0
0

0
0
1
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 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)
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 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
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 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.
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(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,
is solution
otherwise
 corresponds to flow augmentation on W.
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(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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