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
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.
N ij
: a column of N corresponding to arc (i, j) = e reduced cost c ij
= c ij
– ( 
(i) -

(j)) = c ij i
– e j
-

(i) +

(j)
 c ij

.
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For nonbasic variables:
If c
If c ij
 ij


0
 x ij

0
 x ij
For basic variables:
= 0
= u ij c ij

= 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
8
2
6
5
7
3
4
9 i pred(i)
1 2 3 4 5 6 7 8 9
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
 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).
Network Theory and Applications 2010
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
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)
1
3
2
5 4
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3
1
4
2
(2, 4) (1, 2) (3, 5) (3, 1) artif






 

1
1
0
0
0

0
0
0
1
1

0
0
1
0
1

0
0
0
1
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)N ij
=
 for backward arc in W.)
(i, j)

W
(

1)(e(i) – e(j)) = 0. (Use +1 for forward arc, -1
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 = c
B for basis B)
Network Theory and Applications 2010
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 LP operations:
Computing basic feasible solution and dual solution

: x
B
: Bx
B
= b – Lx
L

Bx
B
= b’,
– Ux
U
, B: lower triangular solve by substitution
(on network, determine flows starting from leaf nodes)

:

B = c
B
, solve by substitution
(on network, determine
 starting from root node using c ij
-

(i) +

(j) = 0 )
Actually, x and
 are updated, rather than computed again from scratch.
Finding an entering nonbasic variable:
 c ij

 c ij

< 0 and x ij
> 0 and x ij
= 0 (x
= u ij ij
(x ij
< u ij
): want to increase flow on arc (i, j)
> 0): want to decrease flow on arc (i, j)
Leaving basic variable: Have x
B
= B -1 b – B -1 Lx
L
– B -1 Ux
U



If increase entering x
If decrease entering x j j
(x
(x j j
 x j
 x j
+ t): x determine largest t satisfying l
B
B
 x
B
(t) = x
 u
B
B
* - td (d: updated column of x
- t): x
B
(t) = x
B
* + td j
)
Update solution
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
Operations on the network:
Computing

:
 set

(1) = 0 (root)

Following thread(i) index, use c ij
-

(i) +

(j) = 0 i = pred(j)
(i, j)

A
 
(j) =

(i) – c ij
(j, i)

A
 
(j) =

(i) + c ij
Computing x
B
:

For (i, j)

U, set x ij
= u ij
, b’(i) = b(i) – u ij
, b’(j) = b(j) + u ij
.
Then determine tree arc flow starting from leaf node.
if j is leaf node, (i, j)

A
 x ij if j is leaf node, (j, i)

A
 x ij
= -b’(j), b’(i)
= b’(j), b’(i)

 b’(i) – x b’(i) + x ij ij
= b’(i) + b’(j)
= 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 x kl
= 0, opposite direction of (k, l) if x kl
= u ij
.
(want to increase flow along the cycle)
 ij
= u ij x ij
,
– x ij
, if (i, j)

W and forward arc 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, x
B
(t) = x
B
* - td (increase case) d = B -1 N kl or Bd = N kl
(N kl
+

W
(

1)N ij
= 0, +1 for forward, -1 for backward) choose d ij
= -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.
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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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