Flow in Network
Graph, oriented graph, network
• A graph G =(V, E) is specified by a non empty set of nodes V and a set of
edges E such that each edge a is identified by a pair of nodes (u, v).
1
a
c
b
3
d
2
V = {1, 2, 3}
E = {a, b, c, d}
a = b = (1, 2) ; c = (1, 3) ; d =(2, 3)
• A graph g is a sub graph of a graph G if all the nodes and all the edges of
g are nodes and edges of G.
1
a
1
c
b
c
3
3
d
2
G
g
• A sub graph of a graph G including all the nodes of G is a partial graph of
G.
1
a
1
c
b
c
3
d
3
d
2
2
G
g
• A chain in a graph G is a sequence of distinct edges a1, a2, …, ap
such that there exist (p+1) nodes u1, u2, …, up+1 where ai= (ui, ui+1).
1
a
c
b
3
d
2
The sequence a, c is a chain.
• A cycle in a graph G is a chain such that u1 = up+1
1
a
c
b
3
d
2
The sequence c, b, d is a cycle.
• A graph G is connected if for all pair of nodes, there exists a chain linking
them.
1
a
c
b
3
d
2
This graph is connected.
• A tree is a connected graph with no cycle
Property : A tree with n nodes includes exactly (n – 1) edges
• A partial tree ( spanning tree) of a connected graph G is a partial graph of
G being a tree
1
a
1
c
b
c
3
d
d
2
2
G
partial tree
3
• A fondamental cycle with respect to a partial tree is a cycle including an
edge of the graph not included in the partial tree and edges of the tree.
1
a
1
c
b
c
3
d
d
2
2
G
partial tree
3
• A fondamental cycle with respect to a spanning tree is a cycle including an
edge of the graph not included in the partial tree and edges of the tree.
1
a
1
c
b
3
c
a
d
d
3
2
2
G
fondamental cycle
• An oriented graph G = (V, E) is specified by a non empty set of nodes V
and a set of arcs E such that each arc a is identified by an ordered pair of
nodes (u, v).
2
a
c
1
e
b
d
3
4
f
V = {1, 2, 3, 4}
E = {a, b, c, d, e, f}
a = (1, 2), b = (2, 4), c = (2, 3), d =(3, 2), e = (1, 3), f = (3, 4)
• A non oriented graph obtained from an oriented graph G by eliminating the
direction of the arc is denoted the corresponding graph.
• The notions of chain, cycle, connectedness, tree, spanning tree, and
fondamental cycle for an oriented graph refer to the corresponding graph.
• A path in an oriented graph is a sequence of distinct arcs a1, a2, …, ap
being a chain where all the arcs are oriented in the same direction.
• A directed graph is simple if the nodes identifying each arc are distinct, and
if there is no pairs of arcs specified by the same pair of nodes.
• A network is a connected oriented graph in which a flow can move over
the arcs. Each arc (i, j) est characterized as follows
a capacity dij corresponding to an upper bound on the flow in the
arc
a lower bound lij on the flow in the arc
Moreover, 0≤ lij ≤ dij
[0, 2]
[0, 6]
2
1
[2, 9]
[3, 4]
4
[0, 4]
[1, 6]
3
At each arc (i, j) is associated a pair [lij , dij].
Minimum cost flow model
• Consider a network where the following attributs are associated with each
arc (i, j) :
dij the capacity of the arc
lij the lower bound on the flow in the arc
cij flow unitary cost in the arc
xij the variable including the value of the flow in the arc
•
Furthermore, associate the following two sets of nodes with each node i:
Bi   j  V : ( j , i )  E 
Pi   j V : (i, j )  E
Exemple
B1= Φ , B2= {1, 3}, B3= {1, 2}, B4= {2, 3}
P1= {2, 3}, P2= {3, 4}, P3= {2, 4}, P4= Φ
1
[0, 2]
[0, 6]
2
[2, 9]
[3, 4]
4
[0, 4]
[1, 6]
3
• The minimum cost flow problem is to determine how to send a fixed
quantity of flow v from a source sV to a destination t V along the arcs
of the network satisfying the lower bound and the capacity constraints of
the arcs in order to minimize the total cost.
• Constraints of the problem:
1. lower bound and the capacity constraints of the arcs:
lij  xij  dij
(i, j )  E
2. flow conservation constraints (specific to network flow problems):
For each node i  V the total flow going in the node must be equal
to the total flow going out of the node
• For each node, the total flow going in the node must be equal to the total
flow going out of the node.
For the source s
v   x js   xsj
jBs
or
2
v
1
4
3
jPs
x
jPs
sj
  x js  v
jBs
• For each node, the total flow going in the node must be equal to the total
flow going out of the node.
For the destination t
x
jBt
2
1
4
3
jt
  xtj  v
jPt
v or
x  x
jPt
tj
jBt
jt
 v
• For each node, the total flow going in the node must be equal to the total
flow going out of the node.
For an intermediate node i
x
2
1
jBi
4
3
ji
  xij
jPi
or
x  x
jPi
ij
jBi
ji
0
Model for the minimum cost flow problem
• The minimum cost flow problem is to determine how to send a fixed
quantity of flow v from a source sVto a destination t V along the arcs
of the network satisfying the lower bound and the capacity constraints of
the arcs in order to minimize the total cost.
(MCF)
(MCF)
(MCF)
min
 cij xij
s.t. (flow conservation)
( i , j )E
(flow
  x v
 xconservation)
(flow conservation)
 xsj   x js  v
jPs
jBs
 xij   x ji  0 i  s, t
jPi
jBi
 xtj   x jt  v
jPt
jBt
(capacity
lij  xij  d ij
 c x
ij ij
( i ,
j )E cij xij
( i , j )E
s.t.
s.t.
jPs
min
min
(i , j )  E
sj
jBs
js
v si i  s
 xij   x ji  0 i  s, t
si i  s, t
Bi x ji  0
jPixij  j
jPi
jB
 xtj   i x jt  
vv si i  t
jPt
jBt
(capacity)
(capacity)
lijlij  xij  ddijij
(i(,ij,)j) 
EE
(FCM)
 cij xij
min
(i , j )E
Sujet à
(conservation de flot)
v si i  s

 xij   x ji  0 si i  s, t
jPi
jBi
v si i  t

colonne (i, j)
x ij
(capacité)
 0
 
 
 0
 
• The matrix associated with the flow conservation constraints are
 1 denoted
ligne i
 0
the nodes-arcs incidence matrix:
 
j  Pi
row i ↔ node i

 
 0
column (i, j) ↔ arc (i, j)
 
 1
ligne j
 
i  Bj
 0
 
 
 0 
lij  xij  dij
(i, j )  E
• Exemple of an incidence matrix
(MCF)
min
 cij xij
( i , j )E
s.t.
(flow conservation)
v si i  s

 xij   x ji  0 si i  s, t
jPi
jBi
 v si i  t

(capacity)
lij  xij  d ij
2
(i , j )  E
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


• The nodes-arcs incidence matrix has the unimodularity property indicating
that the simplex algorithm generates an integer solution for problem
(MCF) when lij, dij et v are integer.
(MCF)
min
 cij xij
( i , j )E
s.t.
(flow conservation)
v si i  s

 xij   x ji  0 si i  s, t
jPi
jBi
 v si i  t

(capacity)
lij  xij  d ij
(i , j )  E
(MCF)
min
 cij xij
( i , j )E
s.t.
(flow conservation)
Properties of the incidence matrix
v
A
si i  s


 xij   x ji  0 si i  s, t
jPi
jBi
 v si i  t

• Now we show that the columns of any basis correspond to (capacity)
arcs in a
lij  xij  d ij
spanning tree and vice versa.
(i , j )  E
• Theorem: The incidence matrix A of a connected simple oriented graph
having m nodes and n arcs has a rank equal to (m–1).
Proof: First we show that rank  A   m  1 . Each column of A includes
only one component equal to 1, another one equal to -1, and all the other
elements are equal to 0.
Hence if we sum up the rows of A, then we obtain a row vector having all
its elements equal to 0.
Consequently the rows of A are linearly dependent.
Thus rank  A   m  1 .
• Exemple of incidence matrix
2
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 
 0  1  1 0 1 1


0
0
0

1
0

1


_____________________
0 0 0 0 0 0
Now we show that rank  A   m  1 .
Suppose that there exist a set K of k rows of A ai1 ,
, aik
that are linearly dependent.
There exists scalars 1, , k that are not all equal to 0 such that
1ai   k ai  0, ,0.
Since each column of the matrix A includes exactly one element equal
to 1 and another one equal to -1, if  j  0, then for all element different
1
k
from 0 in row ai j , we must find in the set K , the row of A including the
corresponding complementary element different from 0.
• Exemple of incidence matrix
2
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


Now we show that rank  A   m  1 .
Suppose that there exist a set K of k rows of A ai1 ,
, aik
that are linearly dependent.
There exists scalars 1, , k that are not all equal to 0 such that
1ai   k ai  0, ,0.
Since each column of the matrix A includes exactly one element equal
to 1 and another one equal to -1, if  j  0, then for all element different
1
k
from 0 in row ai j , we must find in the set K , the row of A including the
corresponding complementary element different from 0.
If we reapply the reasonning for all elements  j  0, we find that K
includes all the rows of A since the oriented graph is simple and
connected.
Hence k  m, and consequently all subsets of  rows of the matrix A
where    m  1 , are linearly independent. Then rank  A   m  1 .
It follows that rank  A   m  1 .
• Theorem: Consider an incidence matrix A associated with a simple
connected directed graph G including m nodes and n arcs. A square
(m – 1)x (m – 1) sub matrix of A is non singular if and only if the arcs
associated to its columns are those specifying a spanning tree of G.
Theorem: Consider an incidence matrix A associated with a simple
connected directed graph G including m nodes and n arcs.
A square (m  1) x (m  1) sub matrix of A is non singular if and only if
the arcs associated to its columns are those specifying a spanning tree of G.
Proof: Let T be a spanning tree of G illustrated in red.
2
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


Proof: Let T be a spanning tree of G illustrated in red. Denote by A(T) the
mx(m˗1) sub matrix A being the incidence matrix of T.
2
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


Proof: Let T be a spanning tree of G illustrated in red. Denote by A(T) the
mx(m–1) sub matrix A being the incidence matrix of T.
Since T is a simple connected graph, it follows from the preceding theorem
that rank(A(T)) = (m–1).
2
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


Proof: Let T be a spanning tree of G illustrated in red. Denote by A(T) the
mx(m–1) sub matrix A being the incidence matrix of T.
Since T is a simple connected graph, it follows from the preceding theorem
that rank(A(T)) = (m–1).
Then all (m – 1)x (m – 1) sub matrices obtained by eliminating a row of A(T)
is non singular.
2
But these sub matrices
1
4
x12 x13 x 23 x 24 x32 x34
are also sub matrices
of A.
1  1 1 0 0 0 0
3
2
3
4
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


Theorem: Consider an incidence matrix A associated with a simple
connected directed graph G including m nodes and n arcs.
A square (m  1) x (m  1) sub matrix of A is non singular if and only if
the arcs associated to its columns are those specifying a spanning tree of G
Let B be a (m – 1)x (m – 1) square non singular sub matrix of A.
2
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


Let B be a (m – 1)x (m – 1) square non singular sub matrix of A.
2
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


Let B be a (m – 1)x (m – 1) square non singular sub matrix of A.
B is obtained by eliminating a row from an incident matrix of a spanning
tree g of G.
2
1
4
3
x12 x13 x 23 x 24 x32 x34
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


Let B be a (m – 1)x (m – 1) square non singular sub matrix of A.
B is obtained by eliminating a row from an incident matrix of a spanning
tree g of G.
This sub graph g is simple, connected, including (m – 1) arcs.
2
Then g is a
spanning tree
1
4
x12 x13 x 23 x 24 x32 x34
of G.
3
1
2
3
4
 1 1 0 0 0 0
 1 0 1 1  1 0 


 0  1  1 0 1 1


0
0
0

1
0

1


•
Theorem: The incidence matrix A of
a connected simple oriented graph
having m nodes and n arcs has a rank
equal to (m–1).
•
Theorem: Consider an incidence
matrix A associated with a simple
connected directed graph G including
m nodes and n arcs. A square
(m – 1)x (m – 1) sub matrix of A is
non singular if and only if the arcs
associated to its columns are those
specifying a spanning tree of G.
Any base from the incidence matrix is such that its columns
correspond to variables associated with the arcs of a spanning
tree, and vice versa.
The basic variables of a basic solution of a (MCF) problem
correspond to the arcs of a spanning tree and vice-versa.
The simplex method
for
the minimum cost flow problem
• We use the variant of the simplex method when the variables are bounded
to solve the minimum cost flow problem (MCF):
(MCF)
min

( i , j )E
cij xij
s.t.
(flow conservation)
 v si i  s

x

x

0 si i  s, t


ij
ji
jPi
jBi
  v si i  t

(capacity)
0  xij  d ij
(i , j )  E
where Bi   j V : ( j, i)  E et Pi   j V : (i, j)  E
• To analyse one iteration of the
simplex method, suppose that we
have a basic feasible solution x of
the problem.
Then the non basic variables xij
are such that
xij = 0 ou dij.
The basic variables xij correspond
to the arcs of a spanning tree T of
the network.
v4
• Entering criterion
determine the relative costs of the
variables xij :
 T  [1 ,...,  i ,...,  j ,...,  m ]
x ij
cij  cij   aij
T
that can be written as
c ij  cij   i   j
Moreover, for the basic variables
associated to the arcs (i, j )  E (T )
of the spanning tree T :
c ij  cij   i   j  0
(i, j)  E(T )
aij 
 0
 
 
 0
 
1
 0
 
 
 0
 
 1
 
 0
 
 
 0 
ligne i
ligne j
c ij  cij   i  
j
0
(i, j )  E(T )
In a tree
(# arcs) = (# nodes) –1.
The preceding system includes
(m – 1) équations (associated to
the arcs (i, j )  E (T ) ) et m unknown
πi (associated to the nodes i of T).
But since any sub matrix
(m – 1)x (m – 1) of the incidence
matrix T is non singular, we can
fix the value of one multiplier
and to evaluate the others using
the system.
The system of equations being
triangular, it is easy to show
that the multipliers are
evaluated sequentially one by
one.
c ij  cij   i   j  0
(i, j)  E(T )
In our exemple
(1) 0  c s1  c s1   s   1  4   s   1
(2) 0  c s 2  c s 2   s   2  1   s   2
(3) 0  c 23  c 23   2   3  3   2   3
(4) 0  c1t  c1t   1   t  1   1   t
Let πt = 0
πs=5
π1=1
πt=0
(4) 0  1   1   t  0  1   1   1  1
(1) 0  4   s   1  0  4   s  1   s  5
(2) 0  14   s   2  0  1  5   2   2  4
(3) 0  3
4   2  3  0  3  4  3  3 1
π2=4
π3=1
Once the values of the multipliers
are known, determine the relative
costs
c ij  cij   T aij
of the non basic variables
associated to the arcs that are not
belonging to the spanning tree T.
In our exemple
πs=5
π1=1
πt=0
c 21  c 21   2   1  2  4  1  1
c 3t  c3t   3   t  2  1  0  1
π2=4
π3=1
Once the values of the
multipliers are known, determine
the relative costs
c ij  cij   i   j
of the non basic variables
associated to the arcs that are not
belonging to the spanning tree T.
Recall the optimality criterion
If
cij  0
(i, j )  E (T ) où xij  0
cij  0
(i, j )  E (T ) où xij  d ij ,
and
then the current solution is optimal.
Otherwise select one of the variables
(admissibles) xij where the optimality
criterion is not satisfied.
If the admissible variable xij = 0, then
its value increases, and if the value
xij  d ij , then its value decreases.
Once the values of the multipliers
are known, determine the relative
costs
c ij  cij   T aij
of the non basic variables
associated to the arcs that are not
belonging to the spanning tree T.
In our exemple
πs=5
π1=1
πt=0
c 21  c 21   2   1  2  4  1  1
c 3t  c3t   3   t  2  1  0  1
x21 is admissible to increase
x3t is admissible to decrease
π2=4
π3=1
Recall: Step 1: Selecting the entering variable
The criterion to select the entering variable must be modified to account
for the non basic variables xj being equal to their upper bounds uj since
these variables can be reduced.
Hence, for an index j  JB
if x j  0 et c j  0 , it is interesting to increase xj
if x j  u j et c j  0 , it is interesting to decrease xj
c
 min  c , c 
Determine c s1  min
jJB
Let c s
s1
s2
j


: x j  0 and c s2  max c j : x j  u j
max  c
jJB
s1
, c s2


If c s  0, then the solution is optimal, and the algoithm stops.
If c s  0 and c s  c s1 , then the non basic variable xs increases, and go to Step 2.1.
If c s  0 et c s  c s1 , then the non basic variable xs decreases, and go to Step 2.2.
Once the values of the multipliers
are known, determine the relative
costs
c ij  cij   T aij
of the non basic variables
associated to the arcs that are not
belonging to the spanning tree T.
In our exemple
πs=5
π1=1
πt=0
c 21  c 21   2   1  2  4  1  1
c 3t  c3t   3   t  2  1  0  1
x21 is admissible to increase
x3t is admissible to decrease
π2=4
Entering
variable
π3=1
Recall: A fondamental cycle with respect to a spanning tree is a
cycle including an edge of the graph not included in the partial three
and edges of the three.
• Leaving criterion
Evaluate the impact of modifying
the value of the entering variable
on the values of the basic variables.
It is easy to verify that only the values
of the flow on the arcs of the
fondamental cycle associated to the
arc associated to the entering variable
are modifed.
In our exemple:
x3t is the entering variable
In cases i) or ii) where the modification
Determine the largest value of the
modification  of the entering
variable before
 of the entering variable is limited,
i) the value of a basic variable xij
from the base being replace by
reduce to 0
ii) the value of a basic variable xij
reach its capacity dij
iii) the entering variable move
between from one of its
bound to the other
then the entering variable xij exits
the entering variable. The new
basis T  where the arc  i, j   T
is remplaced by the entering arc.
In case iii), we adjust the values of
the flow on the arcs of the fondamental
cycle, and we move to the next
iteration with the same basis.
To complete the iteration, the value of
the entering variable is reduces by  .
The largest value of   2 since
the flow in arc 1, t  reach its
capacity d1t  2.
The variable x1t is replaced by the
variable x3t in the basis.
0+θ
0+θ
4-θ
4-θ
4-θ
To complete the iteration, the value of
the entering variable is reduces by  .
The largest value of   2 since
the flow in arc 1, t  reach its
capacity d1t  2.
The variable x1t is replaced by the
variable x3t in the basis.
The new solution and the spanning
tree associated are as follows
Second iteration :
Entering criterion
Determine the multipliers by
solving the system:
(1) 0  c s1  c s1   s   1  4   s   1
(2) 0  c s 2  c s 2   s   2  1   s   2
(3) 0  c 23  c 23   2   3  3   2   3
(4) 0  c 3t  c3t   3   t  2   3   t
(1) 0  c s1  c s1   s   1  4   s   1
(2) 0  c s 2  c s 2   s   2  1   s   2
(3) 0  c 23  c 23   2   3  3   2   3
(4) 0  c 3t  c3t   3   t  2   3   t
Let πt = 0
(4) 0  2   3   t  0  2   3   3  2
(3) 0 34   2   3  0  3   2  2   2  5
(2) 0 14   s   2  0  1   s  5   s  6
(1) 0  4   s   1  0  4  6   1   1  2
Determine the relavive costs of the
non basic variables
c ij  cij   i   j  0
πs= 6
4
π1= 2
πt= 0
2
1
c21  c21   2  1  2  5  2  1
c1t  c1t  1   t  1  2  0  1
π2= 5
3
π3= 2
c 21  c 21   2   1  2  5  2  1
c1t  c1t   1   t  1  2  0  1
Since x1t  2  d1t and c1t  1,
then x1t is not admissible.
The variable x21  0 and c21  1,
then x21 becomes the entering
4
variable that is increasing.
2
1
3
c 21  c 21   2   1  2  5  2  1
c1t  c1t   1   t  1  2  0  1
Since x1t  2  d1t and c1t  1,
then x1t is not admissible.
The variable x21  0 and c21  1,
then x21 becomes the entering
variable that is increasing.
The fondamental cycle associated
with x21 in the spanning tree is
• Leaving criterion
Evaluate the impact of increasing
the value of the entering variable
on the values of the basic variables.
Then θ = 2 since for this value,
xs1 = 2 – 2 = 0
x21 = 0 + 2 = 2 = d21.
We can use the same basis at the
next iteration since x21  2  d 21.
We prefer to use xs1 as the leaving
variable.
2-θ
0+θ
2+θ
• Leaving criterion
Evaluate the impact of increasing
the value of the entering variable
on the values of the basic variables.
Then θ = 2 since for this value,
xs1 = 2 – 2 = 0
x21 = 0 + 2 = 2 = d21.
We can use the same basis at the
next iteration since x21  2  d 21.
We prefer to use xs1 as the leaving
variable.
The new solution and the associated
spanning three are the following
•
Third iteration :
Entering criterion
Determine the multipliers
Determine the relative costs of
the non basic variables
c s1  c s1   s   1  4  6  3  1
c1t  c1t   1   t  1  3  0  2
πs=6
π1=3
1
πt=0
2
π2=5
2
3
π3=2
c s1  c s1   s   1  4  6  3  1
c1t  c1t   1   t  1  3  0  2
The solution is optimal
x s1  0
and
c s1  1  0
x1t  2  d1t
and
c1t  2  0
• The simplex version for the minimum cost flow problem is simplified
because of the incidence matrix structure.
• There is a lot of degeneracy in flow problem. This may induce degenerate
iteration (where the values of the variables dont change) sinply to modify
the basis. There exists procedures to guide in chosing the basic solutions in
order to reduce the number of degenerate iterations.
Multiple sources and multiple destinations
p sources si V , i  1,
, p
q destinations t j V , j  1,
where vsi denotes the flow quantity coming from si
,q
where vt j denotes the flow quantity arriving at t j
s1
s2
sp
t1
Rest
of
network
t2
tq
p sources si V , i  1,
, p
q destinations t j V , j  1,
where vsi denotes the flow quantity coming from si
,q
where vt j denotes the flow quantity arriving at t j
Generate the network G1= (V1, E1 )
V 1  V  s, t
s1
s
s2
sp
t1
Rest
of
network
t2
tq
t
p sources si V , i  1,
, p
q destinations t j V , j  1,
where vsi denotes the flow quantity coming from si
,q
where vt j denotes the flow quantity arriving at t j
Generate the network G1= (V1, E1 )
V 1  V  s, t


E1  E  (s, si ) : i  1,, p  (t j , t ) : j  1, , q
s1
s
s2
sp
t1
Reste
du
réseau
t2
tq
t
p sources si V , i  1,
, p
q destinations t j V , j  1,
where vsi denotes the flow quantity coming from si
,q
where vt j denotes the flow quantity arriving at t j
Generate the network G1= (V1, E1 )
V 1  V  s, t


E1  E  (s, si ) : i  1,, p  (t j , t ) : j  1, , q
[d ij , cij ]
[v s1 ,0]
[vs2, ,0]
s
s1
s2
[vs p ,0]
t1
Rest
of
network
t2
[vt1 ,0]
[ v t 2 ,0 ]
[vtq ,0]
sp
tq
t
p sources si V , i  1,
, p
q destinations t j V , j  1,
where vsi denotes the flow quantity coming from si
,q
where vt j denotes the flow quantity arriving at t j
Generate the network G1= (V1, E1 )
p
q
i 1
j 1
Total flow v   vsi   vt j
[v s1 ,0]
v
v
[vs2, ,0]
s
s1
s2
[vs p ,0]
t1
Rest
of
network
t2
[vt1 ,0]
[ v t 2 ,0 ]
[vtq ,0]
sp
tq
t
References
M.S. Bazaraa, J.J. Jarvis, H.D. Sherali, “ Linear Programming and Network
Flows”, 3rd edition, Wiley-Interscience (2005), Chapter9
F.S. Hillier, G.J. Lieberman, “Introduction to Operations Research”, Mc Graw
Hill (2005), Section 9.7
D. G. Luenberger, “ Linear and Nonlinear Programming ”, 2nd edition,
Addison-Wesley (1984), Chapter 5