Download Network Simplex Method I

OR 215
Network Flows
Spring 1999
M. Hartmann
NETWORK SIMPLEX METHOD I
Basic (Spanning Tree) Solutions
Optimality Conditions
Simplex Multipliers
The Network Simplex Pivot
Maintaining the Tree Structure
THE NETWORK SIMPLEX ALGORITHM
A basis structure consists of a spanning tree T, a set L of
arcs, and a set U of arcs, such that T  L  U = A.
 for each (i,j)  L, xij = 0 (lower bound arcs)
 for each (i,j)  U, xij = uij (upper bound arcs)
The arc flows in T are selected so that each node satisfies
its supply/demand constraint. (This selection may violate
upper and lower bound constraints.)
minimize
cijxij
subject to
 j xij

-
 k xki
0  xij  uij

= bi
for all i  N
for all (i,j)  A
A basis structure is said to be feasible if the arc flows in
T are selected to satisfy supply/demand constraints and
they also satisfy the upper and lower bound constraints.
BASIC SOLUTIONS
If x is a feasible flow, an arc (i,j) is a free arc if 0 < xij < uij
and a restricted arc if xij = 0 or xij = uij.
Theorem: If a minimum cost flow problem has an optimal
solution, then it has an optimal solution represented by a
basis structure (T,L,U).
Proof: Let x be an optimal solution for which the set F(x)
of free arcs is minimal. If F(x) contains an undirected
cycle C, then let W be a directed cycle in G(x) corresponding to C which contains the reversal of an arc in C. Then
W must be a zero-cost cycle (why?) and augmenting along
W gives another optimal solution, which contradicts the
minimality of F(x).
Since F(x) contains no cycles, it is a forest, and T can be
any spanning tree containing F(x).
HOW TO CALCULATE FLOWS
For each non-tree arc (i,j) either xij = 0 and (i,j)  L or else
xij = uij and (i,j)  U .
e(1) = 1
e(2) = -6
e(3) = 1
e(4) = 2
4
1
2
3
7
6
5
e(7) = 3
e(6) = -4
e(5) = 3
Method 1: Calculate flows iteratively starting at leaves
and working upwards.
Method 2: Calculating flows directly (conceptually useful
but not necessarily practical.) To calculate the flow in arc
(i,j), first let S be the part of T \ {(i,j)} containing i. Then let
the flow in arc (i,j) be the net supply in S.
Example: What is the flow in arc (3,2)?
OPTIMALITY CONDITIONS II
Recall: The reduced cost of arc (i,j) is c ij = cij - i + j.
Theorem: The solution represented by a basis structure
(T,L,U) is an optimal solution if it is feasible and for some
choice of node potentials , the reduced costs satisfy:
 c ij = 0
for (i,j)  T,
 c ij  0
for (i,j)  L,
 c ij  0
for (i,j)  U.
Proof: This implies that c ij  0 for all (i,j) in G(x).
Here  is the vector of simplex multipliers associated with
the equality constraints of the minimum cost flow problem:
minimize
cijxij
subject to
 j xij

-
 k xki
0  xij  uij

= bi
for all i  N
for all (i,j)  A
SIMPLEX MULTIPLIERS
We choose multipliers so that c ij = 0 for each (i,j) in T.
Recall: There is a redundant constraint so one of the
simplex multipliers can be chosen arbitrarily (say 1 = 0).
1
-6
5
2
3
3
-2
4
-4
7
6
1
5
An iterative approach: Set 1 = 0, choose 2 so that


c12
= 0, choose 3 so that c 32
= 0, etc.
Rule: Always select an arc (i,j) for which either i or j is
known. Setting c ij = 0 will fix the other value of .
An alternate approach: Let (i) denote the cost of disposing one unit of flow at node i (sending it to the root node.)
 (i) is the cost of the path from i to the root.
 (i) = cij + (j) for each arc (i,j) in the tree.
Then let j = (j) so that cij i + j = 0 for (i,j) in the tree.
1
-6
5
2
3
3
-2
4
-4
6
1
5
7
THE NETWORK SIMPLEX PIVOT
Suppose that T is a rooted spanning tree and let x be the
basic (feasible) solution associated with T. Let  be the
vector of simplex multipliers and let c ij = cij i + j.
A non-tree arc is a violating arc for entering if
i.
c ij < 0 and xij = 0 or
ii.
c ij > 0 and xij = uij.
1. Choose a violating non-tree arc. If no such arc exists,
then the solution x is optimal.
2. Add (i,j) to T creating a unique “directed” circuit C.
i. Orient C in the appropriate direction. [ If xij = 0,
then orient C in the direction of (i,j). Otherwise
orient C in the direction of (j,i).]
xij = 0
i
j
ii. Let  = min { min { ukl-xkl : (k,l) is forward on C },
min { xkl : (k,l) is backward on C } }
Send  units of flow in cycle C. Pivot out an arc
(p,q) whose flow increases to upq or decreases
to 0, obtaining a new tree T';
iii. T \ (p,q) partitions into two subtrees, T1 and T2
with the root node in T1. Let  = | c ij |. [Update
 so that c ij = 0 for (i,j)  T.]
 If i  T1, then add  to each node v  T2.
 If i  T2, then subtract  from each node in T2.
MAINTAINING THE TREE STRUCTURE
Comment: The data structures that you will see are not
the only ones that are appropriate. They are data
structures that have worked well in practice.
1. Hang the tree from a root node.
2. For each node, store pred(i), the parent of i in the tree.
3. For each node i store the depth(i), the number of arcs
on the path from i to the root.
4. For each node i, store a pointer to the next node on the
depth first search ordering of T. This data structure is
called the thread.
1
2
3
7
6
4
5
Note: These three data structures used in conjunction are
excellent for finding sub-trees and for locating circuits.
Note: Updating these data structures requires a lot of
bookkeeping, but it's fast.
node i
pred(i)
depth(i)
thread(i)
1
2
3
4
5
6
7
FINDING THE CYCLE
Suppose that arc (k,l) enters the basis.
procedure IDENTIFY CYCLE;
begin
set i := k and j := l;
while i  j do
begin
if depth(i) > depth(j) then
i := pred(i)
else if depth(j) > depth(i) then
j := pred(j)
else
i := pred(i) and j := pred(j);
end;
set w := i;
end
This subroutine locates the least common ancestor w
of k and l. We could simulataneously locate the arc that
drops out as well as the change in the flows.
UPDATING THE MULTIPLIERS
Suppose that (i,j) is pivoted in and (p,q) is pivoted out.
(Assume for now that xij = 0 before the pivot.) Let T1
be the subtree of T \ (p,q) containing the root node 1
and T2 = [T \ (p,q)] \ T1 be the other subtree.
procedure UPDATE POTENTIALS;
begin
if i  T2 then  = -| c ij | else if j  T2 then  = | c ij |;
for each i  T2, add  to i;
end
Note: Use the thread and depth indices to identify T2.
1
2
7
T2
3
4
6
5
EXAMPLE OF A MIN-COST FLOW PROBLEM
b(1) = 2
1
1
4, $1
b(4) = -4
2, $4
4
4, $2
3, $5
2
3, $4
3
1
4, $2
1, $4
5
5, $5
b(2) = 5
b(5) = -3
The minimum cost flow problem
b(1) = 2
1
1
0
1
4
1
2
2
b(2) = 5
b(4) = -4
3
3
1
3
0
0
5
b(5) = -3
The initial spanning tree solution
(1) = 0
1
1
(4) = -4
$4
4
$2
$5
2
3
1
(3) = -2
5
$5
(2) = 3
The initial
simplex
multipliers
(5) = -3
(1) = 0
(4) = -4
1
1
4
$2
-$2
3
1
Reduced costs
for the initial
spanning tree
(3) = -2
$4
2
5
(2) = 3
(5) = -3
1
1
4
0, 4
2
1, 4
2, 3
The fundamental cycle
3
1
5
1
1
1
2
4
3
0
2
3
1
0
0
The second
spanning tree
solution.
5
3
(1) = 0
1
1
3
(4) = -4
$4
4
$2
$1
3
1
2
(3) = -2
5
$5
(2) = 1
The second
simplex
multipliers
(5) = -4
(1) = 0
(4) = -4
1
1
4
$2
$2
3
1
Reduced costs
for the second
spanning tree
(3) = -2
$4
2
5
(2) = 1
(5) = -4
1, 2
1
1
4
3, 4
3, 3
The fundamental cycle
3
1
2
5
2
1
1
2
4
2
0
2
2
3
1
0
5
3
(1) = 0
1
1
$1
2
(2) = 1
0
The third
spanning tree
solution.
(4) = -6
4
$4
$2
3
1
$5
The third
simplex
multipliers
(3) = -2
5
(5) = -4
SOME REMAINING ISSUES
1. How can we avoid cycling in the simplex method?
2. What is the worst case performance of the simplex
method?
3. What are some good heuristics to speed up
performance in practice?
4. Is there a good way of doing sensitivity analysis?