1. Flow in Network

2. 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 3 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 1 3 3 2

4. 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 1 3 3 2 2

5. 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 3 2

6. A cycle in a graph G is a chain such that u1 = up+1
a b d
The sequence c, b, d is a cycle. 1 3 2

7. This graph is connected.
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 3 2

8. A tree is a connected graph with no cycle
Property : A tree with n nodes includes exactly (n – 1) edges

9. 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 1 3 3 2 2

10. 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 1 3 3 2 2

11. 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 1 3 3 2 2

12. a = (1, 2), b = (2, 4), c = (2, 3), d =(3, 2), e = (1, 3), f = (3, 4)
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 1 4 3

13. 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.

14. a capacity dij corresponding to an upper bound on the flow in the arc
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 1 4 3

15. 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:

16. 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 1 4 3

17. 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.
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

18. For each node, the total flow going in the node must be equal to the total flow going out of the node. v 2 1 4 3

19. For each node, the total flow going in the node must be equal to the total flow going out of the node. v 2 1 4 3

20. For each node, the total flow going in the node must be equal to the total flow going out of the node. 2 1 4 3

21. 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.

22. 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)
colonne (i, j)
ligne i
ligne j

23. Exemple of an incidence matrix2 1 4 3

24. 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.

25. 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 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

26. Exemple of incidence matrix2 1 4 3

28. Exemple of incidence matrix2 1 4 3

30. 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.

31. Proof: Let T be a spanning tree of G illustrated in red.2 1 4 3

32. mx(m˗1) sub matrix A being the incidence matrix of T.
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

33. mx(m–1) sub matrix A being the incidence matrix of T.
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

34. mx(m–1) sub matrix A being the incidence matrix of T.
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.
But these sub matrices
are also sub matrices
of A. 2 1 4 3

35. Let B be a (m – 1)x (m – 1) square non singular sub matrix of A.2 1 4 3

36. Let B be a (m – 1)x (m – 1) square non singular sub matrix of A.2 1 4 3

37. 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

38. 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.
Then g is a
spanning tree
of G. 2 1 4 3

39. Any base from the incidence matrix is such that its columns
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: The incidence matrix A of a connected simple oriented graph having m nodes and n arcs has a rank equal to (m–1).
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.

40. 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):

41. 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.

42. determine the relative costs of the variables xij :
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 :
ligne i
ligne j

43. 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).
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.

44. In our exemple
Let πt = 0
π1=1
πs=5
πt=0 1
π3=1
π2=4 3

45. 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.
In our exemple
πs=5
π1=1
πt=0
π3=1
π2=4

46. 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.

47. x21 is admissible to increase
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.
In our exemple
πs=5
π1=1
πt=0
x21 is admissible to increase
π3=1
π2=4
x3t is admissible to decrease

48. 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

49. x21 is admissible to increase
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.
In our exemple
πs=5
π1=1
πt=0
x21 is admissible to increase
π3=1
π2=4
x3t is admissible to decrease
Entering
variable

50. Leaving criterion
In our exemple:
x3t is the entering variable

52. 0+θ
0+θ
4-θ
4-θ
4-θ

54. Determine the multipliers by solving the system:
Second iteration :
Entering criterion
Determine the multipliers by
solving the system:

55. Let πt = 0 3 1

56. Determine the relavive costs of the non basic variables
π1= 2
πs= 6
πt= 0 4 2 1 3
π2= 5
π3= 2

57. 4 2 1 3

59. Then θ = 2 since for this value, xs1 = 2 – 2 = 0
Leaving criterion
Then θ = 2 since for this value,
xs1 = 2 – 2 = 0
x21 = = 2 = d21.
2-θ
0+θ
2+θ

60. Leaving criterion
Then θ = 2 since for this value,
xs1 = 2 – 2 = 0
x21 = = 2 = d21.

61. Determine the multipliers Determine the relative costs of
Third iteration :
Entering criterion
Determine the multipliers
Determine the relative costs of
the non basic variables
πs=6
π1=3
πt=0 2 2 1 3
π2=5
π3=2

62. The solution is optimal

63. 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.

64. Multiple sources and multiple destinations
Rest of
network t1 s2 t2 sp tq

65. s1 Rest of network t1 s2 t2 s t sp tq
Generate the network G1= (V1, E1 ) s1
Rest of
network t1 s2 t2 s t sp tq

66. s1 Reste du réseau t1 s2 t2 s t sp tq
Generate the network G1= (V1, E1 ) s1
Reste du
réseau t1 s2 t2 s t sp tq

67. s1 Rest of network t1 s2 t2 s t sp tq
Generate the network G1= (V1, E1 ) s1
Rest of
network t1 s2 t2 s t sp tq

68. s1 Rest of network t1 s2 t2 s t sp tq
Generate the network G1= (V1, E1 )
v v s1
Rest of
network t1 s2 t2 s t sp tq

69. 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