Sloan W.P, No. 1484-83. September 1983. -(a·RISlfU-XXIL71. --. _-. ______ .... create a strongly feasible basis (or prove that none exist) is at least t and at most ...
A POLYNOMIAL-TIME PARAMETRIC SIMPLEX ALGORITHM FOR THE MINIMUM COST NETWORK FLOW PROBLEM
James B. Orlin
Sloan W.P, No. 1484-83
-(a·RISlfU-XXIL71
--
_-
______
September 1983
A Polynomial-Time Parametric Simplex Algorithm for the Minimum Cost Network Flow Problem by James B. Orlin Sloan School of Management MIT
Abstract In this paper we consider the minimum cost network flow problem: min(cx : Ax = b, x matrix.
0), where
A is an m x n vertex-edge incidence
We show how to solve this problem as a parametric linear
program with
0(m b*) pivots, where
binary representation of
b.
b*
is the number of l's in the
The parametric formulation is non-linear and
is based on Edmonds-Karp scaling technique.
sai·igrs31--l
a
____
In this paper we consider the minimum cost network flow problem (1).
where
Minimize
cx
Subject to
Ax
= b
x
0,
A
(1)
matrix in which there is at most one
m x n
is a full row'rank
1 and at most one -1 in each column and the remaining entries are all 0. Moreover,
is an integral m-vector, and
b
c
is a real n-vector.
Zadeh (1978) showed that the simplex procedure as applied to (1) may take an exponentially large number of pivots.
Edmonds and Karp (1972)
developed a scaling procedure in conjunction with the out-of-kilter method that solves the minimum-cost network flow problem in a polynomial number The purpose of this note is to provide a parametric simplex
of steps.
procedure that solves the minimum-cost network flow problem in polynomial time. Recently, Ikura and Nemhauser (1983) have independently developed a similar parametric procedure for the transshipment problem. Strongly Feasible Bases
Preliminaries:
2.
As before, we assime that
A
is the full row rank constraint matrix of We assume without loss of
the minimum cost network flow problem (1). generality that no two columns of matrix where (2.1)
A
the directed graph
V = {0, ..., m If a column of component
(2.2)
0
are the same.
G = (V, E)
, and the edge set A
We associate with
has a
1
E
is constructed as follows.
in component
i
and a
, then we associate an edge (i, j) of
If a column of a
A
A
has a
1
-1
in
E.
[resp, -1] in component
i
and
in the remaining components then we associate an edge
(i, 0) [resp., (0, i)] of
E.
III
2.
As is well known, each basis is a tree.
downward edge of
A
induces a subgraph
G, we say that the edge
T of T
GB
of
G
e
are two vertices of PT(i, i')
e = (i, j) of
is an upward edge of and that
T
e
be the simple path in
whose terminal vertex is
i'.
T
e
whose initial edge is
and T
T.
e'
T
Suppose that
i
are two edges of
PT(e, e')
is a to j.
whose initial vertex is
Similarly
that
0.
if (i, j) is on the unique simple path from
Otherwise, we say that
let
of
We assume henceforth that the tree is rooted at vertex
For each tree
i'
B
and We
T. i
and
is the simple path in
and whose terminal edge is
e'.
We define
PT(i, e) and PT(e, i) analogously. e
We say that
e'
on
PT(O,
e'
in
T
e
if
is a descendent of
e.
We say that
(and that e' is the son of e)
if
e
also say that of
is an ancestor of
e').
spanning tree
e'
For each vertex
v
V
e
e').
We
is the father
is the second to last edge
and for each edge
ei
of the basic
T, we associate the colors red, green and yellow as follows.
(3.1)
If
x.i > 0, then we color edge
(3.2)
If
xi = 0
ei
is colored "red".
If
xi = 0
(3.3)
PT(,
E
and if
ei
and if
ei
ei "green".
is an upward edge of
T
is a downward edge of
then
T
then
ei
is colored "yellow". (3.4)
For each vertex red edge then
(3.5)
v v
For each vertex yellow edges then
We say that a red edge red and
j
is green.
V, if
PT(O,
v)
has at least one
is colored "red". v
solely of green and
V, PT(O, v) v
is colored "green".
e = (i, j) in
Thus the ancestors of
T
is maximum in e
in
T
T
if
i
is
are green.
As per Cunningham (1976), we say that a basic feasible flow is strongly
3.
feasible if no edge of the corresponding tree is red.
(Orlin (1983)
showed that a basic feasible flow is strongly feasible if and only if it is lexico-positive.) In general, the dual simplex pivot rule that we adopt will pivot out For each red (and hence upward) edge
red edges.
the vertices of vertex 0.
Let
e = (i, j), let
that are in the same connected component of
V
be the fundamental cutset of
F(T, e)
Te T - e
induced by
T
be as e
and defined as follows. F(T, e) = {e' = (i', j')
G - T
i'
Te and j'
We illustrate these terms in Figure 1 and in Tables 1 and 2. the only maximal red edge is by
e
is
e = (4, 2).
In Figure 1,
The fundamental cutset induced
6)}
F(T, e) = {(1,
Tree Edge
Edge
T - T }
Reduced Cost
Flow
Color
(0, 1)
Yes
3
green
(0, 2)
Yes
0
yellow
(1, 6)
No
5
0
(2, 1)
No
2
0
(2, 3)
Yes
4
green
(4, 2)
Yes
0
red
(4, 5)
Yes
2
green
(4, 6)
Yes
6
green
(5, 3)
No
Table 1.
1 2 3 4 5 6
_
0
The data for the edges of the graph of Figure 1.
Vertex
Table 2.
6
Supply/Demand
Color
-3 4 -4 8 -2 -6
green green green red red red
The data for the vertices of the graph in Figure 1.
slll
11
0
3
0
/0
Figure 1.
A basic feasible solution that is not strongly feasible. Edge (4,2) is red. The edge numbers refer to the edge flows.
4. 3.
Dual Pivoting to Reach Strong Feasibility In Section 4 we will develop a parametric procedure in which the
parameter decreases whenever strong feasibility is obtained.
Our parametric
algorithm relies on the efficient procedure developed below for moving from feasibility to strong feasibility using dual pivots. The dual simplex pivoting procedure is well-known; see for example Dantzig (1963).
Here we describe it using the terminology developed in the
previous section.
Moreover, we specialize it for our purposes.
Dual Simplex Pivoting (4.0).
Start with an optimal basic solution x = (xi) feasible.
(4.1).
Let
ck
denote the reduced cost of variable
Select a maximum red edge edge
e
that is not strongly
e
in the basis tree
exists, then quit because
T
T.
xk.
If no such
is strongly feasible.
Otherwise, continue. (4.2).
Select an edge If
e'
in
F(T, e)
whose reduced cost is minimum.
F(T, e) = 0, then quit as there is no strongly feasible
basis.
Otherwise, let the new basic tree consist of
T + e' - e.
Return to (4.1). REMARK.
A special case of the dual simplex pivot rule (4) is the rule of
pivoting out the lexicographically most infeasible edge.
See Orlin (1983)
for more details. We illustrate the dual simplex pivot rule in Figure 2. pivoted out corresponds to edge (2, 3), the only red edge. pivoted in corresponds to the edge (1, 6).
--1IIIII
DDil
-
-
-·^
The variable
The spanning tree obtained
subsequent to the pivot is given in Figure 2.
I'0--
The variable
.I_._.....
3
0
2
Figure 2.
A strongly feasible spanning tree solution obtained from Figure 1 by a dual pivot.
5.
We say that of T
j
and if
PT(i, j)
PT(i, j)
has no yellow edges.
induce a partial order e and e'
is a red-green chain in
ST
as follows.
are red and if
T
We write that
if edges
T.
We illustrate these concepts in Figures 3 and 4. Suppose that
problem (1) but that
B
B
PT(e, e')
T.
e
e'
in
is a red-green chain in
is not strongly feasible.
Let
T
be the
be the partial order on the red
ST
edges induced by the red-green chains. S
is an ancestor
is an optimal basis for the network flow
corresponding spanning tree, and let
elements in
i
The red-green chains in
ST
THEOREM 1.
if
Let
t
be the number of minimal
Then the number of dual pivots using procedure (4) to
create a strongly feasible basis (or prove that none exist) is at least t
and at most PROOF.
tm.
Let
dual pivots, where
T,
T TO
,
...
,
Tr
be the sequence of trees obtained by
is the tree corresponding to the original basis.
In
addition, we define the following notation. vk = set of green vertices in
Tk.
Ek = set of maximum red edges in
Tk.
Sk = partial order on the red edges induced by k
t
= number of minimal red edges in
k s = dilworth number of
Tk
k
S
k Sk, i.e., the minimum number of k
red-green chains to cover the red edges of pk(i, j) = PTk(i,
) = the path on
Tk
from
i
T to
j.
To complete the proof, we will prove the following relations, all of which are valid for integers
L
I
j,
k
[O..r-l].
+l,...,k.
Here the notation [j..k] refers to the set of
0
0
0
2
Q Figure 3.
010 0
A tree with its corresponding feasible flow.
I 1 Figure 4.
I-i-----L-
The partial order induced by the red-green chains of the tree of Figure 3. Elements (2,0) and (3,1) are maximum. Elements (2,0), (7,5) and (8,6) are minimal.
_ iBrl 8. sl_
__
6. l Vk+ 1
vk
(5)
Vk + l =V
If !Ekl
1 .).
k
Vk+
[..r]. = Vk
is at most
Since the number of iterations
by (5), the total number of iterations is
m
t °m = tm.
To prove (5) - (9) we consider tree is the edge that is pivoted in at the Suppose that edge
e = (i5, i4)
Tk.
Suppose that
k-th pivot, and thus
e' = (i2, i3) e'
is the edge pivoted out of tree
Tk+l - Tk. T . Now
let us consider Figure 5, which represents the circuit created upon adding edge (i2, i3 ) to
Tk.
In Figure 5, i0
"edges" in Figure 5 represent paths in
is the source vertex T .
0, and the
Since we will often refer to
the paths in Figure 5 we will sometimes drop the superscript
k.
We do not assume that the vertices in Figure 5 are distinct. example,
i = i4
then
If, for
is the trivial path consisting of a
P(il, i4)
single vertex. The colors of the edges in
Tk + l
(10.1) Edge
is yellow in
e' = (i2, i3)
(10.2) If edge e
e
b P(i5, i3 )
may be summarized as follows.
and
is colored the same in
Tk
e
T
k+1
T - e, then and in
Tk+ l
7.
(10.3) If edge
e P(i5, i3 )
e
orientations in
Tk
then
and
e
Tk+l.
has different (Switching orientation
changes red edges to yellow and changes yellow edges to red.) We are now ready to prove (5). Tk . in
i5 i pk(o, v).
Then k+1.
T
Otherwise k+1
Since *
Let
*
P
that
k+(0, v)
V k + l = Vk .
e
e
e
E
k
is green in
and
v
is green
*
Ek+l c Ek
i2 *
pk(O, e ), and
i2
is red in
Tk.
We will now prove that
k+1 - Ek + l , this will complete the proof of (6).
be a maximum red edge in
is maximum,
k
We first claim that
yk+l _-V k , contrary to assumption.
i3
k c E.
Since
pk(O, v) =
v
Thus (5) is true.
.
Suppose now that
E
Thus
Suppose that vertex
e
P .
k+1 Tk
In fact
P
*
Let
P
*
k+l
=P
(O, e ).
n P(i5, i3) = 0. Thus k T.
is a maximal red edge in
We have thus shown
and thus (6) is true.
Relation (7) is immediate since the set of maximum red edges forms an Sk . (An anti-chain is a set of unrelated elements in
anti-chain in
To see that relation (8) is valid, we first note that Sk .
since the minimal red edges are an anti-chain in ei
let
Ri
denote the maximal red-green chain of
We claim that each red edge sk < tk
show that minimal in Pk(e', e)
k P .
e' of Tk is in some chain
and complete the proof.
Then
e'
Tk
ei
ei
tk
For each red edge containing ei. Ri, and this will
Suppose that
for some red edge
sk
Sk).
e'
k in Sk.
is not Then
c Ri, and thus (8) is true.
We now consider relation (9). edges of
Tk.
Mk
Let
denote the set of minimal red
Our proof first takes into consideration the set
Mk n Mk+l
We state the result as Lemma 1. LEMMA 1. Then
__s_l
e*
E
Mk
_r
Suppose that if and only if
* T
e
e*
T
u
Tk+ T1
and that
M+l.
______
*
e
P(i4 , i3 ).
III
8.
PROOF.
Suppose that
e
is a maximum red edge in
Mk
E
Tk,
and
e*
since
Tk+l
is yellow.
P(i5 , i3 )
contradicts that
Mk
e
and that
=
e*
P
e* ~ P(i0 , i4 ). kl T
(10)
P(e*, e')
e'.
e*
Mk+l
be a red-green (i2 , i3) i P
Then P
does not contain k T.
*
it follows that
P
= P(e*, e') e'.
e* I P(i3, i5 ).
and
E 1Mk+l
This
Suppose further that
is a red-green chain in Since
Thus no edge of .
(i5 , i4)
M.
E
terminates at the red edge
a path in
P
Since
We suppose that
Therefore, the chain
and by
Suppose now that e*
Let
terminating at the red edge
(i2 , i3)
any edge of
P(i4 , i3).
P(i0 , i4 ).
and we will derive a contradiction. chain in
e*
P*
(i4 , i5 )
*
e
that
is maximum, we know that
is in (i5, i3 )
Thus contradicts that
Tk
k+l E
and thus
P*
is also
, completing the proof of
the Lemma.
O
To complete the proof of (9) we will define a into (but not necessarily onto) at most one element of
Mk
Mk .
1:1
M k+ l
mapping of
However, we will show that there is
that is not in the range of the mapping, and
this will complete the proof. Let
f
M
k+l +
k Mk
be defined as follows.
let
f(e*) = e .
For
e* E P(i 3 , i5 ) n Mk+l
P(i3, i5 ) then let
For
*
e
E
By Lemma 1, we are left with edges on the path let
that is closest to
f(e*)
e .
If
P(i3,
be the unique yellow edge
Pk+l (e*, i5)
we
e'
i5 ). in
has no yellow edges,
f(e*) = (i5 , i4 ).
We first note that
f(e*)
is red in
Tk
by relation (10).
note that the mapping is
1:1.
such that
In such a case, one can show that
f(e*) = f(et).
Pk+l(e', e*) Mk+l
k k+ l Mk n Mk+,
Otherwise there are two edges
is a red-green chain and thus at most one of
We also
e P
and
e'
(e*, e') or
e*, e'
is in
i5
i2
Figure 5.
The circuit created by adding edge (i2 , i 3 ) to Tk.
ir_*___DII__·___1_111______
-
9.
in
Tk
chain in of
Pk+l(e
We claim that
P(i5 , i3 ).
be on the path
(Otherwise, P(e,
e
*
than is
that
Pk+l(e*, i')
edge
e
e".)
Tk+ l
e"
e
But then
is a
P
f(e*) k+l
e"
since
(i',
P
must
e")
is
We now claim
e".
e
than is
e',
This completes the proof of the
Pk(i, i
such that
Mk - f(Mk+l).
P(i5, i3 )
attention to the path
Let
i
'tk+l
tk - 1", we show that
By Lemma 1, we may restrict
e'
i*
e* r Mk n P(i*, i4 ), then
We claim first that if
P(i3, i5)
be the vertex on
is a red-green chain and such that
3)
To see this, we let e .
and
P(i5 , i3 ),
is closer to
f(e*).
e"
tk+lt
