Flow in Network - fileGraph, oriented graph, network • A graph G =(V, E) is specified by a non...

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).

c

a b

d

V = {1, 2, 3} E = {a, b, c, d}

a = b = (1, 2) ; c = (1, 3) ; d =(2, 3)

1

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.

c c

a b

d

G g

1

2

3

1

3

• A sub graph of a graph G including all the nodes of G is a partial graph of

G.

c c

a b

d d

G g

1

2

3

1

3

2

• 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).

c

a b

d

The sequence a, c is a chain.

1

2

3

• A cycle in a graph G is a chain such that u1 = up+1

c

a b

d

The sequence c, b, d is a cycle.

1

2

3

• A graph G is connected if for all pair of nodes, there exists a chain linking

them.

c

a b

d

This graph is connected.

1

2

3

• 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

c c

a b

d d

G partial tree

1

2

3

1

3

2

• 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.

c c

a b

d d

G partial tree

1

2

3

1

3

2

• 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.

c c

a b a

d d

G fondamental cycle

1

2

3

1

3

2

• 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).

a b

c d

e 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)

2

3

1 4

• A non oriented graph obtained from an oriented graph G by eliminatingthe 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] [2, 9]

[0, 6] [3, 4]

[0, 4] [1, 6]

At each arc (i, j) is associated a pair [lij , dij].

2

3

1 4

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:

{ }{ }EjiVjP

EijVjB

i

i

∈∈=

∈∈=

),(:

),(:

Exemple

B1= Φ , B2= {1, 3}, B3= {1, 2}, B4= {2, 3}

P1= {2, 3}, P2= {3, 4}, P3= {2, 4}, P4= Φ

[0, 2] [2, 9]

[0, 6] [3, 4]

[0, 4] [1, 6]

2

3

1 4

• The minimum cost flow problem is to determine how to send a fixedquantity of flow v from a source to a destination 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:

2. flow conservation constraints (specific to network flow problems):

For each node the total flow going in the node must be equal

to the total flow going out of the node

Vs ∈ Vt ∈

( , )ij ij ijl x d i j E≤ ≤ ∈

i V∈

• For each node, the total flow going in the node must be equal to the total

flow going out of the node.

v2

3

1 4

For the source

or

s s

s s

js sj

j B j P

sj js

j P j B

s

v x x

x x v

∈ ∈

∈ ∈

+ =

− =

∑ ∑

∑ ∑



v2

3

1 4

For the destination

or

t t

t t

jt tj

j B j P

tj jt

j P j B

t

x x v

x x v

∈ ∈

∈ ∈

= +

− = −

∑ ∑

∑ ∑



2

3

1 4

For an intermediate node

or

0

i i

i i

ji ij

j B j P

ij ji

j P j B

i

x x

x x

∈ ∈

∈ ∈

=

− =

∑ ∑

∑ ∑

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 to a destination along the arcs

of the network satisfying the lower bound and the capacity constraints of

the arcs in order to minimize the total cost.

Vs ∈ Vt ∈

( , )

(MCF) min

s.t.

(flow conservation)

si

0 si ,

si

(capacity)

( , )

i i

ij iji j E

ij jij P j B

ij ij ij

c x

v i s

x x i s t

v i t

l x d i j E

∈

∈ ∈

∑

=

− = ≠∑ ∑ − =

≤ ≤ ∈

( , )

(MCF) min

s.t.

(flow conservation)

0 ,

(capacity)

( , )

s s

i i

t t

ij iji j E

sj jsj P j B

ij jij P j B

tj jtj P j B

ij ij ij

c x

x x v

x x i s t

x x v

l x d i j E

∈

∈ ∈

∈ ∈

∈ ∈

∑

− =∑ ∑

− = ≠∑ ∑

− = −∑ ∑

≤ ≤ ∈

( , )

(MCF) min

s.t.

(flow conservation)

0 ,

(capacity

( , )

s s

i i

t t

ij iji j E

sj jsj P j B

ij jij P j B

tj jtj P j B

ij ij ij

c x

x x v

x x i s t

x x v

l x d i j E

∈

∈ ∈

∈ ∈

∈ ∈

∑

− =∑ ∑

− = ≠∑ ∑

− = −∑ ∑

≤ ≤ ∈

• The matrix associated with the flow conservation constraints are denoted

the nodes-arcs incidence matrix:

row i ↔ node i

column (i, j) ↔ arc (i, j)

( , )

(FCM) min

Sujet à

(conservation de flot)

si

0 si ,

si

(capacité)

( , )

i i

ij iji j E

ij jij P j B

ij ij ij

c x

v i s

x x i s t

v i t

l x d i j E

∈

∈ ∈

∑

=

− = ≠∑ ∑ − =

≤ ≤ ∈

−

0

0

1

0

0

1

0

0

�

�

�

ijx

ligne i

ligne j

colonne (i, j)

ij P∈

ji B∈

• Exemple of an incidence matrix

2

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1

( , )

(MCF) min

s.t.

(flow conservation)

si

0 si ,

si

(capacity)

( , )

i i

ij iji j E

ij jij P j B

ij ij ij

c x

v i s

x x i s t

v i t

l x d i j E

∈

∈ ∈

∑

=

− = ≠∑ ∑ − =

≤ ≤ ∈

• 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

s.t.

(flow conservation)

si

0 si ,

si

(capacity)

( , )

i i

ij iji j E

ij jij P j B

ij ij ij

c x

v i s

x x i s t

v i t

l x d i j E

∈

∈ ∈

∑

=

− = ≠∑ ∑ − =

≤ ≤ ∈

Properties of the incidence matrix

• Now we show that the columns of any basis correspond to arcs in a spanning tree and vice versa.

• 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 Each column of A includesonly one component equal to 1, another one equal to -1, and all the otherelements 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 1 .A m≤ −

( ) ( )rank 1 .A m≤ −

( , )

(MCF) min

s.t.

(flow conservation)

si

0 si ,

si

(capacity)

( , )

i i

ij iji j E

ij jij P j B

ij ij ij

c x

v i s

x x i s t

v i t

l x d i j E

∈

∈ ∈

∑

=

− = ≠∑ ∑ − =

≤ ≤ ∈

A ↔

• Exemple of incidence matrix

2

3

1 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

− − − −

− −

343224231312 xxxxxx

4

3

2

1

_____________________

0 0 0 0 0 0

( ) ( )Now we show that rank 1 .A m= −

1Suppose that there exist a set of rows of , ,

that are linearly dependent.

ki iK k A a a…

[ ]1

1

1

There exists scalars , , that are not all equal to 0 such that

0, ,0 .k

k

i k ia a

α α

α α+ + =

…

… …

Since each column of the matrix includes exactly one element equal

to 1 and another one equal to -1, if 0, then for all element different

from 0 in row , we must find in the set , the row of j

j

i

A

a K

α ≠

including the

corresponding complementary element different from 0.

A

• Exemple of incidence matrix

2

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1

( ) ( )Now we show that rank 1 .A m= −

1Suppose that there exist a set of rows of , ,

that are linearly dependent.

ki iK k A a a…

[ ]1

1

1

There exists scalars , , that are not all equal to 0 such that

0, ,0 .k

k

i k ia a

α α

α α+ + =

…

… …

Since each column of the matrix includes exactly one element equal

to 1 and another one equal to -1, if 0, then for all element different

from 0 in row , we must find in the set , the row of j

j

i

A

a K

α ≠

including the

corresponding complementary element different from 0.

A

If we reapply the reasonning for all elements 0, we find that

includes all the rows of since the oriented graph is simple and

connected.

j K

A

α ≠

( ) ( ) ( )

Hence , and consequently all subsets of rows of the matrix

where 1 , are linearly independent. Then rank 1 .

k m A

m A m

τ

τ

=

≤ − ≥ −

( ) ( )It follows that rank 1 .A m= −

• 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

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1

Theorem: Consider an incidence matrix A associated with a simple

connected directed graph G including m nodes and n arcs.

A square ( 1) ( 1) sub matrix of A is non singular if and only if

the arcs

m x m− −

associated to its columns are those specifying a spanning tree of G.

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

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1


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

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1


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)

isis non non singularsingular.

But these sub matrices

are also sub matrices

of A.

2

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1

Let B be a (m – 1)x (m – 1) square non singular sub matrix of A.

2

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1

Theorem: Consider an incidence matrix A associated with a simple

connected directed graph G including m nodes and n arcs.

A square ( 1) ( 1) sub matrix of A is non singular if and only if

the arcs

m x m− −

associated to its columns are those specifying a spanning tree of G


2

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1


B is obtained by eliminating a row from an incident matrix of a spanning

tree g of G.

2

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1


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.

Then g is a

spanning tree

of G.

2

3

1 4

−−

−−

−−

101000

110110

011101

000011

343224231312 xxxxxx

4

3

2

1�

• Theorem: The incidence matrix A of a connected simple oriented graph having m nodes and n arcs has a rankequal 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

s.t.

(flow conservation)

0 ,

(capacity)

0 ( , )

i i

ij ij

i j E

ij ji

j P j B

ij ij

c x

v si i s

x x si i s t

v si i t

x d i j E

∈

∈ ∈

=

− = ≠− =

≤ ≤ ∈

∑

∑ ∑

{ } { }where : ( , ) et : ( , )i i

B j V j i E P 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.

4v =

• Entering criterion

determine the relative costs of the

variables xij :

that can be written as

Moreover, for the basic variables

associated to the arcs

of the spanning tree T :

Tij ij ijc c aπ •= −

−

0

0

1

0

0

1

0

0

�

�

�

ijx

T

1[ , ..., , ..., , ..., ]i j m

π π π π π=

ligne i

ligne j

=•ija

jiijij cc ππ +−=

)(),( TEji ∈

)(),(0 TEjicc jiijij ∈∀=+−= ππ

In a tree

(# arcs) = (# nodes) –1.

The preceding system includes

(m – 1) équations (associated to

the arcs ) et m unknown

πi (associated to the nodes i of T).


)(),( TEji ∈

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.

In our exemple

Let πt = 0


tttt

ssss

ssss

cc

cc

cc

cc

ππππ

ππππ

ππππ

ππππ

+−=+−==

+−=+−==

+−=+−==

+−=+−==

1111

32322323

2222

1111

10)4(

30)3(

10)2(

40)1(

143040)3(

451040)2(

514040)1(

11010)4(

3332

222

1

111

=⇒+−=⇒+−=

=⇒+−=⇒+−=

=⇒+−=⇒+−=

=⇒−=⇒+−=

ππππ

ππππ

ππππ

ππππ

s

sss

t

π3=1

πt=0π1=1πs=5

π2=41

3

Once the values of the multipliersare known, determine the relative costs

of the non basic variables associated to the arcs that are not belonging to the spanning tree T.

In our exemple

ijT

ijij acc •−= π

πt=0π1=1πs=5

π2=4 π3=11012

1142

333

122121

=+−=+−=

−=+−=+−=

ttt cc

cc

ππ

ππ

Once the values of the

multipliers are known, determine

the relative costs

of the non basic variables

associated to the arcs that are not

belonging to the spanning tree T.

jiijij cc ππ +−=

If

0 ( , ) ( ) où 0

and

0 ( , ) ( ) où ,

then the current solution is

Recall the optimality criter

opti .

i n

al

o

m

ij ij

ij ij ij

c i j E T x

c i j E T x d

≥ ∀ ∉ =

≤ ∀ ∉ =

Otherwise select one of the variables

(admissibles) where the optimality

criterion is not satisfied.

ijx

If the admissible variable = 0, then

its value increases, and if the value

, then its value decreases.

ij

ij ij

x

x d=



In our exemple

ijT

ijij acc •−= π

1012

1142

333

122121

=+−=+−=

−=+−=+−=

ttt cc

cc

ππ

ππ

πt=0π1=1πs=5

π2=4 π3=1x21 is admissible to increase

x3t is admissible to decrease

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

if , it is interesting to increase xj

if , it is interesting to decrease xj

JBj ∈

0et0 <= jj cx

0et >= jjj cux

{ } { }

{ } { }( )

1 2

1 2 1 2

1

then the solution is optimal

Determine min : 0 and max :

Let min , max ,

If 0, , and the algoithm stops.

If 0 an then the non basic variable increasesd ,

s j sj j jjj JB j JB

s s s s s

s

s s ss

c c x c c x u

c c c c c

c

c c c x

∈ ∈= = = =

= −

≥

< =

1then the non basic variab

, and go to Step 2.1.

If le decre 0 et , , and go to Stases ep 2.2.s s s sxc c c< <



In our exemple

ijT

ijij acc •−= π

1012

1142

333

122121

=+−=+−=

−=+−=+−=

ttt cc

cc

ππ

ππ

πt=0π1=1πs=5

π2=4 π3=1x21 is admissible to increase

x3t is admissible to decreaseEntering

variable

• Leaving criterion

In our exemple:

x3t is the entering variable

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

Reca

th

ll:

ree.

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.

Determine the largest value of the

modification of the entering

variable before

θ

i) the value of a basic variable

reduce to 0

ijx

ii) the value of a basic variable

reach its capacity

ij

ij

x

d

iii) the entering variable move

between from one of its

bound to the other

In cases i) or ii) where the modification

of the entering variable is limited,

then the entering variable exits

from the base being replace by

the entering variable. The new

basis where the arc

ijx

T i

θ

′ ( ),

is remplaced by the entering arc.

j T∈

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.

4-θ

4-θ

4-θ

0+θ 0+θ

To complete the iteration, the value of

the entering variable is reduces by .θ

( )

1

The largest value of 2 since

the flow in arc 1, reach its

capacity 2.t

t

d

θ =

=

1

3

The variable is replaced by the

variable in the basis.

t

t

x

x

To complete the iteration, the value of

the entering variable is reduces by .θ

( )

1

The largest value of 2 since

the flow in arc 1, reach its

capacity 2.t

t

d

θ =

=

1

3

The variable is replaced by the

variable in the basis.

t

t

x

x

The new solution and the spanning

tree associated are as follows

Second iteration :

Entering criterion

Determine the multipliers by

solving the system:

tttt

ssss

ssss

cc

cc

cc

cc

ππππ

ππππ

ππππ

ππππ

+−=+−==

+−=+−==

+−=+−==

+−=+−==

3333

32322323

2222

1111

20)4(

30)3(

10)2(

40)1(

Let πt = 0

tttt

ssss

ssss

cc

cc

cc

cc

ππππ

ππππ

ππππ

ππππ

+−=+−==

+−=+−==

+−=+−==

+−=+−==

3333

32322323

2222

1111

20)4(

30)3(

10)2(

40)1(

264040)1(

651040)2(

523040)3(

22020)4(

111

2

2232

333

=⇒+−=⇒+−=

=⇒+−=⇒+−=

=⇒+−=⇒+−=

=⇒−=⇒+−=

ππππ

ππππ

ππππ

ππππ

s

sss

t

3

1

Determine the relavive costs of the

non basic variables

2

3

1

4πt= 0

π3= 2π2= 5

πs= 6 π1= 2

0=+−= jiijij cc ππ

21 21 2 1 2 5 2 1c c π π= − + = − + = −

1 1 1 1 2 0 1t t tc c π π= − + = − + = −

2

3

1

4

1021

1252

111

122121

−=+−=+−=

−=+−=+−=

ttt cc

cc

ππ

ππ

1 1 1

1

Since 2 and 1,

then is not admissible.

t t t

t

x d c

x

= = = −

21 21

21

The variable 0 and 1,

then becomes the entering

variable that is increasing.

x c

x

= = −

1021

1252

111

122121

−=+−=+−=

−=+−=+−=

ttt cc

cc

ππ

ππ

1 1 1

1

Since 2 and 1,

then is not admissible.

t t t

t

x d c

x

= = = −

21 21

21

The variable 0 and 1,

then becomes the entering

variable that is increasing.

x c

x

= = −

21

The fondamental cycle associated

with in the spanning tree isx


Then θ = 2 since for this value,

xs1 = 2 – 2 = 0

x21 = 0 + 2 = 2 = d21.

0+θ

2+θ

2-θ

Evaluate the impact of increasing



21 21

We can use the same basis at the

next iteration since 2 .x d= =

1We prefer to use as the leaving

variable.

sx


Then θ = 2 since for this value,

xs1 = 2 – 2 = 0

x21 = 0 + 2 = 2 = d21.

•

Evaluate the impact of increasing



21 21

We can use the same basis at the

next iteration since 2 .x d= =

1We prefer to use as the leaving

variable.

sx

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

2

3

21

πt=0πs=6

π2=5 π3=2

π1=3

2031

1364

111

111

−=+−=+−=

=+−=+−=

ttt

sss

cc

cc

ππ

ππ

The solution is optimal

2031

1364

111

111

−=+−=+−=

=+−=+−=

ttt

sss

cc

cc

ππ

ππ

1 1

1 1 1

0 and 1 0

2 and 2 0

s s

t t t

x c

x d c

= = >

= = = − <

• 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

s1

s2

sp

t1

t2

tq

Rest

of

network

sources , 1, , where denotes the flow quantitycoming from

destinations , 1, , where denotes the flow quantity arriving at

i

j

i s i

j t j

p s V i p v s

q t V j q v t

∈ =

∈ =

�

�

Generate the network G1= (V1, E1 )

s1

s2

sp

t1

t2

tq

Rest

of

network

s t

{ }tsVV ,1∪=



i

j

i s i

j t j

p s V i p v s

q t V j q v t

∈ =

∈ =

�

�


s1

s2

sp

t1

t2

tq

Reste

du

réseau

s t

{ }tsVV ,1∪=

{ } { }qjttpissEE ji ,,1:),(,,1:),(1�∪�∪ ===



i

j

i s i

j t j

p s V i p v s

q t V j q v t

∈ =

∈ =

�

�


s1

s2

sp

t1

t2

tq

Rest

of

network

s t

{ }tsVV ,1∪=

{ } { }qjttpissEE ji ,,1:),(,,1:),(1�∪�∪ ===

]0,[1s

v

]0,[,2sv

]0,[psv

]0,[1t

v

]0,[2t

v

]0,[qt

v

],[ ijij cd



i

j

i s i

j t j

p s V i p v s

q t V j q v t

∈ =

∈ =

�

�


v

v

s1

s2

sp

t1

t2

tq

Rest

of

network

s t

]0,[1s

v

]0,[,2sv

]0,[psv

]0,[1t

v

]0,[2t

v

]0,[qt

v

1 1

Total flowi j

p q

s t

i j

v v v= =

= =∑ ∑



i

j

i s i

j t j

p s V i p v s

q t V j q v 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 GrawHill (2005), Section 9.7

D. G. Luenberger, “ Linear and Nonlinear Programming ”, 2nd edition, Addison-Wesley (1984), Chapter 5

Flow in Network - fileGraph, oriented graph, network • A graph G =(V, E) is specified by a non...

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Transcript of Flow in Network - fileGraph, oriented graph, network • A graph G =(V, E) is specified by a non...