summerschool2013.ccaba.upc.edusummerschool2013.ccaba.upc.edu/documents/Lecture9_Festa.pdf · Preliminaries: notation Network Flow Problems Network Flows: Variants Special cases of

Paola Festa - DMA, University of Napoli FEDERICO II

E-mail: [email protected] - Web: http://www.dma.unina.it/∼festaNetwork Flows - p. 1/105

Operational Research aspects in Routing:Network Flows

Paola Festa

Department of Mathematics and Applications “R. Caccioppoli”

University of Napoli FEDERICO II

http://www.dma.unina.it/∼festa/

E-mail: [email protected]

Preliminaries: notation

Network Flow Problems

Network Flows: Variants

Special cases of the

Network Flow Problem

The Maximum Flow Problem

A solution method:

the Ford-Fulkerson algorithm

A simpler (not more efficient)

implementation

of Ford-Fulkerson algorithm

Ford-Fulkerson algorithm:

an exercise




Network flow problems are defined on specialdirected graphs.

Let G = (V,A) be a directed graph, where

★ V = {1, 2, . . . , n} is a set of n nodes;

★ A = {(i, j)| i, j ∈ V } is a set of m arcs;

★ for each node i ∈ V , let

FS(i) = {j ∈ V | (i, j) ∈ A}

be the forward star of node i;

★ for each node i ∈ V , let

BS(i) = {j ∈ V | (j, i) ∈ A}

be the backward star of node i.


Network Flow ProblemsNetwork Flows: Introduction,

➀

Network Flows: Introduction,

➁

A general network flow

problem, ➀


problem, ➁


problem, ➂


problem, ➃


problem, ➃

Minimum Cost Flow Problem,

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Minimum Cost Flow:

alternative form., ➀

Minimum Cost Flow:

alternative form., ➁

Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:

alternative form., ➂

Circulation









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problem, ➀


problem, ➁


problem, ➂


problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Network Flows: Introduction, ➀

Network flow problems are important combinatorialoptimization problems arising in any real-woldscenarios, whenever it is needed to organize andcoordinate distribution systems of one or morecommodities/materials

✔ gas, water;

✔ phone calls;

✔ e-mails, electronic information;

✔ ...

from one or more source/distribution locations to oneor more destinations/request locations.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Network Flows: Introduction, ➁

Network flow problems are special LinearProgramming problems.Therefore, they could be solved by any linearprogramming method, e.g. the simplex method.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Network Flows: Introduction, ➁

Network flow problems are special LinearProgramming problems.Therefore, they could be solved by any linearprogramming method, e.g. the simplex method.

Nevertheless, they have peculiar characteristics andproperties such that

☞ the methods for Linear Programming problemsbecomes “easier” when specialized to deal withthem, but also

☞ those characteristics justify the design of efficientad-hoc techniques.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






A general network flow problem, ➀

Network flow problems can be stated on specialdi-graphs, called flow networks and whose elements(nodes and arcs) have numerical info associated with.

Definition. A network flow is a di-graph N = (V,A)whose arcs (i, j) ∈ A are associated with the followingquantities:

✧ a cost cij representing the cost per unit of flow sentalong arc (i, j);

✧ a capacity kij ≥ 0 representing an upper bound onthe quantity of flow that can be sent along arc (i, j).



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






A general network flow problem, ➀

Network flow problems can be stated on specialdi-graphs, called flow networks and whose elements(nodes and arcs) have numerical info associated with.




Note:

If kij = +∞, along arc (i, j) an arbitrary quantity offlow can be sent.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






A general network flow problem, ➁




A quantity bi can be associated with each node i ∈ V :

✦ if bi > 0, bi represents the quantity of materialentering i from outside the network.bi is called supply and i is called source node;

✦ if bi < 0, |bi| represents the quantity of materialrequested by i.|bi| is called demand and i is called sink node;

✦ if bi = 0, i is called transit node.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






A general network flow problem, ➂

Example. V = {1, 2, 3, 4, 5}; |A| = 8; kij = +∞, ∀ (i, j) ∈ A:

10

b1 = 2

2

3 4

5

6

4 3

- 4

- 2

5

2

b2 = 5

b4 = - 4

b5 = - 4

b3 = 1



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problem, ➁


problem, ➂


problem, ➃


problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation








10

b1 = 2

2

3 4

5

6

4 3

- 4

- 2

5

2

b2 = 5

b4 = - 4

b5 = - 4

b3 = 1

Source nodes: 1, 2, and 3;



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problem, ➁


problem, ➂


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problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation








10

b1 = 2

2

3 4

5

6

4 3

- 4

- 2

5

2

b2 = 5

b4 = - 4

b5 = - 4

b3 = 1

Source nodes: 1, 2, and 3; Sink nodes: 4 and 5.



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problem, ➁


problem, ➂


problem, ➃


problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation








10

b1 = 2

2

3 4

5

6

4 3

- 4

- 2

5

2

b2 = 5

b4 = - 4

b5 = - 4

b3 = 1

Source nodes: 1, 2, and 3; Sink nodes: 4 and 5. No transit nodes.



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problem, ➁


problem, ➂


problem, ➃


problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation








10

b1 = 2

2

3 4

5

6

4 3

- 4

- 2

5

2

b2 = 5

b4 = - 4

b5 = - 4

b3 = 1

Source nodes: 1, 2, and 3; Sink nodes: 4 and 5. No transit nodes.

Note:∑

i∈V

bi = 0.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






A general network flow problem, ➃

Definition. Given a network flow N = (V,A), afeasible flow is an vector of flow variables

{xij}(i,j)∈A , xij ∈ R, ∀ (i, j) ∈ A

such that

① lij(= 0) ≤ xij ≤ kij , ∀ (i, j) ∈ A;

② bi +∑

(j,i)∈A

xji =∑

(i,j)∈A

xij , ∀ i ∈ V .

.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation








{xij}(i,j)∈A , xij ∈ R, ∀ (i, j) ∈ A

such that

① lij(= 0) ≤ xij ≤ kij , ∀ (i, j) ∈ A;

② bi +∑

(j,i)∈A

xji =∑

(i,j)∈A

xij , ∀ i ∈ V .

Conditions ① are easy to understand.



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problem, ➂


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation








{xij}(i,j)∈A , xij ∈ R, ∀ (i, j) ∈ A

such that

① lij(= 0) ≤ xij ≤ kij , ∀ (i, j) ∈ A;

② bi +∑

(j,i)∈A

xji =∑

(i,j)∈A

xij , ∀ i ∈ V .

Conditions ① are easy to understand.

Conditions ② are known as equation or flowconservation law.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation








{xij}(i,j)∈A , xij ∈ R, ∀ (i, j) ∈ A

such that

① lij(= 0) ≤ xij ≤ kij , ∀ (i, j) ∈ A;

② bi +∑

(j,i)∈A

xji =∑

(i,j)∈A

xij , ∀ i ∈ V .

Note: adding all over nodes i ∈ V both sides of ②, itresults that

∑

i∈V

bi = 0. (Principle of the total divergence)

The vector {bi}i∈V is called divergence vector.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Minimum Cost Flow Problem, ➀

Definition. The general minimum cost network flowproblem consists in minimizing a linear cost functionof the form

min∑

(i,j)∈A

cijxij ,

where the vector {xij}(i,j)∈A is a feasible flow.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Minimum Cost Flow Problem, ➀

Definition. The general minimum cost network flowproblem consists in minimizing a linear cost functionof the form

min∑

(i,j)∈A

cijxij ,

where the vector {xij}(i,j)∈A is a feasible flow.

(MF) min∑

(i,j)∈A

cijxij

s.t.

(a)∑

(i,j)∈A

xij −

bi +∑

(j,i)∈A

xji

= 0, ∀ i ∈ V

(b) lij ≤ xij ≤ kij , ∀(i, j) ∈ A.



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problem, ➁


problem, ➂


problem, ➃


problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Minimum Cost Flow Problem, ➁

The general minimum cost network flow problem:

(MF) min∑

(i,j)∈A

cijxij

s.t.

(a)∑

(i,j)∈A

xij −

bi +∑

(j,i)∈A

xji

= 0, ∀ i ∈ V

(b) lij ≤ xij ≤ kij , ∀(i, j) ∈ A.

It is evident that it is a linear programming problem.



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problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation







The general minimum cost network flow problem:

(MF) min∑

(i,j)∈A

cijxij

s.t.

(a)∑

(i,j)∈A

xij −

bi +∑

(j,i)∈A

xji

= 0, ∀ i ∈ V

(b) lij ≤ xij ≤ kij , ∀(i, j) ∈ A.

It is evident that it is a linear programming problem.

If

kij = +∞, ∀ (i, j) ∈ A,

the problem is said uncapacitated and it is in standardform.



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problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation







Example: N = {1, 2, 3, 4, 5}; |A| = 8; kij = +∞, ∀ (i, j) ∈ A:

10

b1 = 2

2

3 4

5

6

4 3

- 4

- 2

5

2

b2 = 5

b4 = - 4

b5 = - 4

b3 = 1

min 5x12 − 4x23 + 6x42 − 2x13 + 0x34 + 2x15 + 4x53 + 3x45

x12 + x13 + x15 = 2, x23 − x12 − x42 = 5,

x34 − x13 − x23 − x53 = 1, x42 + x45 − x34 = −4

x53 − x15 − x45 = −4, x̄ ≥ 0.



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problem, ➃


problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Minimum Cost Flow: alternative form., ➀

Alternative and concise math formulation: by usingthe more economical matrix-vector notation of thenetwork flow N = (V,A).



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problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Minimum Cost Flow: alternative form., ➀

Alternative and concise math formulation: by usingthe more economical matrix-vector notation of thenetwork flow N = (V,A).

Let N = (V,A) be a network flow, where

✔ V = {1, . . . , n} and

✔ |A| = m,

and let D ∈ {−1, 0, 1}n×m be the associated node-arcincidence matrix s.t.

dik =

−1, if i is the tail (start node) of the k.th arc;

1, if i is the head (end node) of the k.th arc;

0, otherwise.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Minimum Cost Flow: alternative form., ➁

In matrix form:

D =

1︷︸︸︷

... · · ·

k︷︸︸︷

... · · ·

m︷︸︸︷

...

row i → 1 · · · −1 · · ·...

... · · · · · · · · · −1

... · · · 1 · · · 1

row n → −1 · · · · · · · · ·...



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problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation







In matrix form:

D =

1︷︸︸︷

... · · ·

k︷︸︸︷

... · · ·

m︷︸︸︷

...

row i → 1 · · · −1 · · ·...

... · · · · · · · · · −1

... · · · 1 · · · 1

row n → −1 · · · · · · · · ·...

Notes: for each row d′i, i = 1, . . . , n,

① the nr of “1” in d′i is |{(i, j) ∈ A}| = |FS(i)|;



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problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation







In matrix form:

D =

1︷︸︸︷

... · · ·

k︷︸︸︷

... · · ·

m︷︸︸︷

...

row i → 1 · · · −1 · · ·...

... · · · · · · · · · −1

... · · · 1 · · · 1

row n → −1 · · · · · · · · ·...


② the nr of “-1” in d′i is |{(j, i) ∈ A}| = |BS(i)|;



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problem, ➁


problem, ➂


problem, ➃


problem, ➃


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➁

Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation







In matrix form:

D =

1︷︸︸︷

... · · ·

k︷︸︸︷

... · · ·

m︷︸︸︷

...

row i → 1 · · · −1 · · ·...

... · · · · · · · · · −1

... · · · 1 · · · 1

row n → −1 · · · · · · · · ·...


③∑

(i,j)∈A

xij −∑

(j,i)∈A

xji = d′ix;



➀


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problem, ➁


problem, ➂


problem, ➃


problem, ➃


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➁


➁

Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation







In matrix form:

D =

1︷︸︸︷

... · · ·

k︷︸︸︷

... · · ·

m︷︸︸︷

...

row i → 1 · · · −1 · · ·...

... · · · · · · · · · −1

... · · · 1 · · · 1

row n → −1 · · · · · · · · ·...

Notes: for each row d′i, i = 1, . . . , n

④ the flow conservation law can be rewritten asd′ix = bi, ∀ i ∈ V or, in more compact form

Dx = b, b divergence vector.



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problem, ➁


problem, ➂


problem, ➃


problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation







In matrix form:

D =

1︷︸︸︷

... · · ·

k︷︸︸︷

... · · ·

m︷︸︸︷

...

row i → 1 · · · −1 · · ·...

... · · · · · · · · · −1

... · · · 1 · · · 1

row n → −1 · · · · · · · · ·...

Notes: for each row d′i, i = 1, . . . , n

⑤ adding all rows d′i results 0̄, i.e., the rows of D arel.d.Nevertheless, it is always possible to remove thereduntant constraints without changing thefeasible region of the problem.



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problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Minimum Cost Flow: alternative form., ➂

Notes =⇒ alternative and concise math formulation:

(MF’) min∑

(i,j)∈A

cijxij

s.t.

(a’) Dx = b

(b) lij ≤ xij ≤ kij , ∀(i, j) ∈ A.



➀


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problem, ➀


problem, ➁


problem, ➂


problem, ➃


problem, ➃


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➁

Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Minimum Cost Flow: alternative form., ➂

Notes =⇒ alternative and concise math formulation:

(MF’) min∑

(i,j)∈A

cijxij

s.t.

(a’) Dx = b

(b) lij ≤ xij ≤ kij , ∀(i, j) ∈ A.

Further notes:

❶ each network flow problem is a linearprogramming problem;

❷ the constraint matrix D of the math formulation isunimodular.

Then, it has an integer optimal solution if

lij , kij ∈ Z+ ∪ {0}, ∀ (i, j) ∈ A.



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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Circulation

Definition. Any flow vector (feasible or infeasible) thatsatisfies

Dx = 0

is called circulation (in this case, it results that b = 0).



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problem, ➃


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Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Minimum Cost Flow:


Circulation






Circulation

Definition. Any flow vector (feasible or infeasible) thatsatisfies

Dx = 0

is called circulation (in this case, it results that b = 0).

The flow conservation law (also known as Kirchkoff’sequation) is imposed only within the network,without external supply or demand.

In other words, the flow “circulates” only inside thenetwork.













A solution method:



implementation



an exercise
















A solution method:



implementation



an exercise




There are several variants of network flow problems,all of which can be shown to be equivalent to eachother.

We will discuss now some examples.













A solution method:



implementation



an exercise




Variant ①:Every network flow problem can be reduced to onewith exactly one source s and exactly one sink node d,s, d ∈ V , s 6= d.













A solution method:



implementation



an exercise





Let us suppose that N = (V,A) has k source nodess1, . . . , sk and h sink nodes d1, . . . , dh.













A solution method:



implementation



an exercise





Let us suppose that N = (V,A) has k source nodess1, . . . , sk and h sink nodes d1, . . . , dh.

It is always possible to define

✍ a dummy source s∗ and a dummy sink d∗;

✍ k dummy arcs (s∗, sq) (q = 1, . . . , k);

✍ h dummy arcs (dl, d∗) (l = 1, . . . , h) s.t.

∀ q = 1, . . . , k,

{

ks∗sq= +∞;

cs∗sq= 0.

∀ l = 1, . . . , h,

{

kdld∗ = +∞;

cdld∗ = 0.













A solution method:



implementation



an exercise





0

s*

s1

s2

sk

0

0

+oo

+oo

+oo

d*

d1

d2

dh

0

0

+oo

+oo

+oo0













A solution method:



implementation



an exercise




Variant ②:Every network flow problem can be reduced to onewithout sources or sinks (bi = 0, ∀ i ∈ V , circulationproblems).













A solution method:



implementation



an exercise





Without loss of generality, consider a networkN = (V,A) with a single source node s and a singlesink node d.













A solution method:



implementation



an exercise





Without loss of generality, consider a networkN = (V,A) with a single source node s and a singlesink node d.

It is always possible to define a dummy arc (d, s) s.t.

kds = +∞; cds = −M, where M is a sufficiently large number.

s

= - M

d

= + oo

kdsc ds













A solution method:



implementation



an exercise




Variant ③:Every network flow problem with node capacities(upper bound on the flow that can enter in each node)can be reduced to one with only arc capacities.













A solution method:



implementation



an exercise




Variant ③:Every network flow problem with node capacities(upper bound on the flow that can enter in each node)can be reduced to one with only arc capacities.

Suppose that in N = (V,A) there is a source node i

with supply bi and capacity gi, i.e., bi +∑

(l,i)∈A

xli ≤ gi.

i

l jgi













A solution method:



implementation



an exercise




Variant ③: (cont’d)In this case, the following 3 simple operations can beperformed:

❶ split node i into 2 nodes i− and i+;

❷ substitute each arc (l, i) ∈ A with the arc (l, i−) withcapacity kli and cost cli and each arc (i, j) ∈ A withthe arc (i+, j) with capacity kij and cost cij ;

❸ define an arc (i−, i+) with capacity gi and null cost.

i -

l j

i +0

gi

kli

cli

kijcij






Special Network Flows, ➀

Special Network Flows, ➁

Special Network Flows, ➂


A solution method:



implementation



an exercise



Special cases of theNetwork Flow Problem










A solution method:



implementation



an exercise




Special cases of the network flow problem areimportant and classical optimization problems,among them

✔ the transportation problem;

✔ the assignment problem;

✔ the shortest path problems (under a certainassumption on the arc costs);

✔ the maximum flow problem;

✔ ...










A solution method:



implementation



an exercise




The shortest path problem under the assumption thatcij ≥ 0, ∀ (i, j) ∈ A, or that there are no negative cycles.










A solution method:



implementation



an exercise





(SP) min∑

(i,j)∈A

cijxij

s.t.

∀ i ∈ V,∑

j∈FS(i)

xij −∑

j∈BS(i)

xji =

1, if i = s;

−1, if i = d;

0, otherwise.

xij ∈ {0, 1}, ∀ (i, j) ∈ A.










A solution method:



implementation



an exercise





(SP) min∑

(i,j)∈A

cijxij

s.t.

∀ i ∈ V,∑

j∈FS(i)

xij −∑

j∈BS(i)

xji =

1, if i = s;

−1, if i = d;

0, otherwise.

xij ∈ {0, 1}, ∀ (i, j) ∈ A.

SP can be viewed as a minimum cost flow problem,where a single unit of flow has to be sent from a singlesource node s to a single sink node d.

All nodes i ∈ V \ {s, d} are transit nodes.










A solution method:



implementation



an exercise




The shortest path tree problem under the assumptionthat cij ≥ 0, ∀ (i, j) ∈ A, or that there are no negativecycles.

(SPT) min∑

(i,j)∈A

cijxij

s.t.

∀ i ∈ V,∑

j∈FS(i)

xij −∑

j∈BS(i)

xji =

{

n − 1, if i = s;

−1, otherwise.

xij ∈ {0, 1}, ∀ (i, j) ∈ A.










A solution method:



implementation



an exercise




The shortest path tree problem under the assumptionthat cij ≥ 0, ∀ (i, j) ∈ A, or that there are no negativecycles.

(SPT) min∑

(i,j)∈A

cijxij

s.t.

∀ i ∈ V,∑

j∈FS(i)

xij −∑

j∈BS(i)

xji =

{

n − 1, if i = s;

−1, otherwise.

xij ∈ {0, 1}, ∀ (i, j) ∈ A.

SPT can be viewed as a minimum cost flow problem,where n − 1 units of flow have to be sent from a singlesource node s to any other node i ∈ V \ {s}.

All nodes i ∈ V \ {s} are sink nodes; there are notransit nodes.







Definition

MF as MCF, ➀

MF as MCF, ➁

Math formulation, ➀

Math formulation, ➁

Math formulation, ➂

Math formulation, ➃

Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄

Max Flow / Min Cut, ➀

Max Flow / Min Cut, ➁

Satured and unloading arcs

Residual network, ➀


Residual network, ➁


Augmenting path

Optimality conditions, ➀

Optimality conditions, ➁



A solution method:










Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Definition

The Maximum Flow Problem (MF) is a special flowproblem defined on a network flow N = (V,A) s.t.

❍ ∀ (i, j) ∈ A

✧ cij represents the cost per unit of flow sent alongarc (i, j);

✧ kij ≥ 0 represents an upper bound on thequantity of flow that can be sent along arc (i, j);

❍ s, d ∈ V , s 6= d, are the source and the sink nodes inN , respectively.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Definition


❍ ∀ (i, j) ∈ A




Objective: sent the max amount of flow from thesource node s to the sink node d.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Definition


❍ ∀ (i, j) ∈ A





Note ①: we know that the hypothesis that s and d arethe only source and sink nodes is not restrictive.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Definition


❍ ∀ (i, j) ∈ A





Note ①: we know that the hypothesis that s and d arethe only source and sink nodes is not restrictive.

Note ②:Max Flow Problem ⇔ Min Cost Flow Problem (MCF).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



MF as MCF, ➀

Any instance N = (V,A) of MF, where each arc(i, j) ∈ A is associated with a capacity kij , can be

reduced to an instance N = (V ,A) of MCF s.t.

✔ V = V ; A = A ∪ {(d, s)};

✔ ∀ (i, j) ∈ A \ {(d, s)}, cij = 0 and kij is unchanged;

✔ cds = −1 and kds = +∞.

s0

d

0

0

0

0

0

k = + ooc = -1

0 0

ds ds







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



MF as MCF, ➁

s0

d

0

0

0

0

0

k = + ooc = -1

0 0

ds ds

To find a max flow from s to d in N = (V, A) is equivalent to

finding a min cost flow in N = (V , A):

max xds ⇐⇒ min∑

(i,j)∈A

cijxij







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



MF as MCF, ➁

s0

d

0

0

0

0

0

k = + ooc = -1

0 0

ds ds




(i,j)∈A

cijxij and max xds ⇐⇒ max xsd







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



MF as MCF, ➁

s0

d

0

0

0

0

0

k = + ooc = -1

0 0

ds ds




(i,j)∈A

cijxij and max xds ⇐⇒ max xsd

Nevertheless, typically the math formulation reflects the special

peculiarity of the problem.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:




For each (i, j) ∈ A let

xij be the amount of flow sent along (i, j),

(MF) max ϕ0 =∑

j∈FS(s)

xsj −∑

j∈BS(s)

xjs

s.t.

(a)∑

j∈FS(i)

xij −∑

j∈BS(i)

xji = 0, ∀ i ∈ V \ {s, d}

(b) 0 ≤ xij ≤ kij , ∀(i, j) ∈ A.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:






(MF) max ϕ0 =∑

j∈FS(s)

xsj −∑

j∈BS(s)

xjs

s.t.

(a)∑

j∈FS(i)

xij −∑

j∈BS(i)

xji = 0, ∀ i ∈ V \ {s, d}

(b) 0 ≤ xij ≤ kij , ∀(i, j) ∈ A.

Note ①: if BS(s) = ∅, the o.f. reduces to

max ϕ0 =∑

j∈FS(s)

xsj .







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:






(MF) max ϕ0 =∑

j∈FS(s)

xsj −∑

j∈BS(s)

xjs

s.t.

(a)∑

j∈FS(i)

xij −∑

j∈BS(i)

xji = 0, ∀ i ∈ V \ {s, d}

(b) 0 ≤ xij ≤ kij , ∀(i, j) ∈ A.

Note ②: each feasible solution (feasible flow) for (MF)is a circulation.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:






(MF) max ϕ0 =∑

j∈FS(s)

xsj −∑

j∈BS(s)

xjs

s.t.

(a)∑

j∈FS(i)

xij −∑

j∈BS(i)

xji = 0, ∀ i ∈ V \ {s, d}

(b) 0 ≤ xij ≤ kij , ∀(i, j) ∈ A.

Note ③: constraints (a) become

for i = s:∑

j∈FS(s)

xsj −∑

j∈BS(s)

xjs = ϕ0;

for i = d:∑

j∈FS(d)

xdj −∑

j∈BS(d)

xjd = −ϕ0.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:






(MF) max ϕ0 =∑

j∈FS(s)

xsj −∑

j∈BS(s)

xjs

s.t.

(a)∑

j∈FS(i)

xij −∑

j∈BS(i)

xji = 0, ∀ i ∈ V \ {s, d}

(b) 0 ≤ xij ≤ kij , ∀(i, j) ∈ A.

Note ④: since constraints matrix is unimodular(node-arc incidence matrix), MF has an integeroptimal solution if

kij ∈ Z+ ∪ {0},∀ (i, j) ∈ A

and it could be solved by the Simplex Method!







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➀

Definition. Given a network N = (V,A), a cut is apartition (S, V \ S) of V s.t.

s ∈ S and d ∈ V \ S.

s d

V\S

S







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➁

Let be

δ+N (S) := {(i, j) ∈ A | i ∈ S, j ∈ V \ S)};

δ−N (S) := {(i, j) ∈ A | i ∈ V \ S, j ∈ S)}.

Definition. Given a feasible flow vector x, the amountof flow crossing a cut (S, V \ S) is given by:

ϕ(S) =∑

(i,j)∈δ+N(S)

xij −∑

(j,i)∈δ−

N(S)

xji,

i.e., the difference between the total flow “leaving”from S and that “entering” S.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➁

Let be

δ+N (S) := {(i, j) ∈ A | i ∈ S, j ∈ V \ S)};

δ−N (S) := {(i, j) ∈ A | i ∈ V \ S, j ∈ S)}.

Definition. Given a feasible flow vector x, the amountof flow crossing a cut (S, V \ S) is given by:

ϕ(S) =∑

(i,j)∈δ+N(S)

xij −∑

(j,i)∈δ−

N(S)

xji,

i.e., the difference between the total flow “leaving”from S and that “entering” S.

The quantity

ϕ0 = ϕ({s})def.=

∑

(s,j)∈δ+N ({s})

xsj −∑

(j,s)∈δ−

N({s})

xjs

is called value of the flow x.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➂

Definition. Given a cut (S, V \ S) in a network flowN = (V,A), the capacity of the cut (S, V \ S) is givenby:

K(S) =∑

(i,j)∈δ+N(S)

kij .

Note: K(S) does not take into account the capacities of

the arcs “entering” S, i.e. those belonging to δ−N (S).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➃

Theorem. Given a feasible flow vector x in a networkflow N = (V,A), for every cut (S, V \ S) in N it resultsthat

ϕ(S) = ϕ0 = ϕ({s}).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➃


ϕ(S) = ϕ0 = ϕ({s}).

Note: the theorem claims that, for every cut (S, V \ S),the amount of flow that crosses it is always equal to

ϕ0, i.e. it is constant.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➃


ϕ(S) = ϕ0 = ϕ({s}).

Proof:from the flow conservation law applied to every nodei ∈ S \ {s}.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➃


ϕ(S) = ϕ0 = ϕ({s}).

Proof:

ϕ0 = ϕ({s}) =∑

(s,j)∈δ+N({s})

xsj −∑

(i,s)∈δ−

N({s})

xis

=∑

h∈S

∑

(h,j)∈δ+N ({h})

xhj −∑

(j,h)∈δ−

N ({h})

xjh

︸︷︷︸

=0,∀h6=s

=∑

h∈S

∑

(h,j)∈δ+N ({h})

xhj −∑

h∈S

∑

(j,h)∈δ−

N({h})

xjh







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➃


ϕ(S) = ϕ0 = ϕ({s}).

Proof: Note that,∑

h∈S

∑

(h,j)∈δ+N

({h})

xhj =∑

h∈S

∑

j∈FS(h)

xhj =∑

(i,j)∈A(S)

xij +∑

(i,j)∈δ+N

(S)

xij

h

j

s







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➃


ϕ(S) = ϕ0 = ϕ({s}).

Proof: Similarly,∑

h∈S

∑

(j,h)∈δ−

N({h})

xjh =∑

h∈S

∑

j∈BS(h)

xjh =∑

(i,j)∈A(S)

xij +∑

(i,j)∈δ−

N(S)

xij

h j

s







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➃


ϕ(S) = ϕ0 = ϕ({s}).

Proof: Therefore,

ϕ0 = · · ·

=∑

h∈S

∑

(h,j)∈δ+N

({h})

xhj −∑

h∈S

∑

(j,h)∈δ−

N({h})

xjh

=

∑

(i,j)∈A(S)

xij

︸︷︷︸

=

+∑

(i,j)∈δ+N

(S)

xij

−

∑

(i,j)∈A(S)

xij

︸︷︷︸

=

+∑

(i,j)∈δ−

N(S)

xij

= ϕ(S).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➄

Theorem. For every feasible flow x in a network flowN = (V,A) and for every cut (S, V \ S) in N , it resultsthat

ϕ(S) ≤ K(S).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Properties, ➄

Theorem. For every feasible flow x in a network flowN = (V,A) and for every cut (S, V \ S) in N , it resultsthat

ϕ(S) ≤ K(S).

Proof:

ϕ(S)def.=

∑

(i,j)∈δ+N (S)

xij −∑

(j,i)∈δ−

N(S)

xji

∑

(j,i)∈δ−

N(S)

xji ≥ 0

≤∑

(i,j)∈δ+N (S)

xij

xij≤kij ∀ (i,j)∈A

≤∑

(i,j)∈δ+N (S)

kij

def.= K(S).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:




Theorem [Maximum Flow / Minimum Cut].







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:





A feasible flow x in a network flow N = (V,A) is amaximum flow iff there exists a cut (S∗, V \ S∗) in N

with minimum capacity such that

ϕ(S∗) = ϕ∗0 = K(S∗).

max = min

max { value of x } min { K(S) }x S







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:





A feasible flow x in a network flow N = (V,A) is amaximum flow iff there exists a cut (S∗, V \ S∗) in N

with minimum capacity such that

ϕ(S∗) = ϕ∗0 = K(S∗).

Proof: constructive.

It is needed to define

☞ saturated and unloading arcs and

☞ residual network N that can be built from N , onceavailable a feasible flow x.

N is an auxiliary network built to keep track of theamount of flow that can be pushed along the arcs ofthe original network N .







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:




Definition. Given a feasible flow x crossing a networkflow N = (V,A), an arc (i, j) ∈ A is said

✔ saturated, if kij − xij = 0;

✔ unloading, if xij = 0.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:




Definition. Given a feasible flow x crossing a networkflow N = (V,A), an arc (i, j) ∈ A is said

✔ saturated, if kij − xij = 0;

✔ unloading, if xij = 0.

xij

s

1

d

2

1

ikij

j 3

4

10 N=(V,A)

10

0

1

20

10

1

5

0

7

7

0

3







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:




Definition. Given a feasible flow x crossing a network

flow N = (V,A), the residual network N = (V,A)associated with N is built performing the following 2operations:







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:






① substitute each arc (i, j) ∈ A with

✧ a forward arc (i, j) with residual capacity

kij = kij − xij ≥ 0 (by def. of kij);

✧ a backward arc (j, i) with residual capacity

kji = xij ≥ 0 (by def. of feasible flow x);

i j

kij

xij

i jkij := kij - xij

forward arc

backward arc kij := xij







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:






② remove all arcs with null residual capacity ((i, j)

with kij = kij −xij = 0 and (j, i) with kji = xij = 0).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:






② remove all arcs with null residual capacity ((i, j)

with kij = kij −xij = 0 and (j, i) with kji = xij = 0).

s

1

d

2

1

21

2

2

4

1 1

3

2

s

1

d

2

2

3

1

1

2

1

N; o.f. = 3 N

2

ikij

jxij







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:




The residual network N allows to find a flow x froms ∈ V to d ∈ V higher than x,

✓ increasing the flow along non saturated arcs

(i.e., those arcs (i, j) with kij = kij − xij 6= 0) and

✓ decreasing it along non unloading arcs

(i.e., those arcs (j, i) with kji = xij 6= 0).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:









If ∃ P from s to d in N =⇒ ∃ in N x higher than x!







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:









If ∃ P from s to d in N =⇒ ∃ in N x higher than x!

P from s to d is said augmenting path and in N itcorresponds to a sequence

✍ of forward arcs along which to increase thecurrent flow and

✍ of backward arcs along which to decrease thecurrent flow.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:




s

1

d

2

1

21

2

2

4

1 1

3

2

s

1

d

2

2

3

1

2

1

1

N; f.o. = 3 N P

2

s

1

d

2

1+a

21+a

2

2

4

1-a 1

3

2

N; f.o. = 3+a, a<=1

P in N is a sequence of forward arcs along which to increase the current

flow and backward arcs along which to decrease the current flow.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Augmenting path

Definition. Given a feasible flow x in a network flowN = (V,A), an augmenting path is a path P from thesource s to the sink d in N s.t., denoting by

❒ F the set of forward arcs in P and

❒ B the set of backward arcs in P ,

it results that

❍ xij < kij , ∀ (i, j) ∈ F =⇒it is possible to increase the current flow along (i, j)!







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:



Augmenting path

Definition. Given a feasible flow x in a network flowN = (V,A), an augmenting path is a path P from thesource s to the sink d in N s.t., denoting by

❒ F the set of forward arcs in P and

❒ B the set of backward arcs in P ,

it results that

❍ xij < kij , ∀ (i, j) ∈ F =⇒it is possible to increase the current flow along (i, j)!

❍ xij > 0, ∀ (i, j) ∈ B=⇒it is possible to decrease the current flow along(i, j)!







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:




Theorem. A feasible flow x in a network flowN = (V,A) is optimal for MF iff node d can not be

reached from node s in the residual network N .







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:






Proof. Let ϕ0 be the value of flow x.

Scenario ①: d is reachable from s in N .







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:







Scenario ①: d is reachable from s in N .

Then, ∃ P augmenting from s to d in N .

Let be δ = min{kij | (i, j) ∈ P}(> 0!) and ∀ (i, j) ∈ P

xij :=

{

xij + δ, if (i, j) ∈ F ;

xij − δ, if (i, j) ∈ B,

x is a feasible flow in N with value ϕ0 + δ > ϕ0 =⇒=⇒ x is not optimal.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:







Scenario ②: d is not reachable from s in N .

Then, ∃ (S∗, V \ S∗) in N s.t. δ+N

(S∗) = ∅.







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:







Scenario ②: d is not reachable from s in N .By definition of N , in N it results that

✔ ∀ (i, j) ∈ δ+N (S∗), xij = kij (saturated);

✔ ∀ (i, j) ∈ δ−N (S∗), xij = 0 (unloading). Therefore,

ϕ0Th.= ϕ(S∗)

def=

saturated︷︸︸︷

∑

(i,j)∈δ+N (S∗)

xij −

unloading︷︸︸︷

∑

(i,j)∈δ−

N (S∗)

xij

=∑

(i,j)∈δ+N (S∗)

kijdef= K(S∗).







Definition

MF as MCF, ➀

MF as MCF, ➁





Properties, ➀

Properties, ➁

Properties, ➂

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➃

Properties, ➄








Augmenting path





A solution method:







Scenario ②: d is not reachable from s in N .

Summarizing, ∃ (S∗, V \ S∗) in N s.t. δ+N

(S∗) = ∅ and

ϕ0Th.= ϕ(S∗) = K(S∗).

Remember: ϕ(S) ≤ K(S), ∀ (S, V \ S).







A solution method:


Ford-Fulkerson algorithm, ➀

Ford-Fulkerson algorithm, ➁

Ford-Fulkerson algorithm, ➂




Ford-Fulkerson algorithm, ➃



Ford-Fulkerson: complexity, ➀

Ford-Fulkerson: complexity, ➁

Ford-Fulkerson: complexity, ➂


implementation



an exercise



A solution method:the Ford-Fulkerson algorithm







A solution method:















implementation



an exercise




Given a network flow N = (V,A) and a feasible flow x

from s to d, it iteratively searches in N an augmentingpath P .







A solution method:















implementation



an exercise






If such a path P exists, then it is possible to increasethe current flow along the arcs in F and to decreasethat sent along those in B.

In order to not violate the capacities constraints, thecurrent flow along the arcs must be changed of thesame quantity δ(P ) properly chosen...







A solution method:















implementation



an exercise







In order to not violate the capacities constraints, thecurrent flow along the arcs must be changed of thesame quantity δ(P ) properly chosen... but

δ(P ) = ?







A solution method:















implementation



an exercise







In order to not violate the capacities constraints, thecurrent flow along the arcs must be changed of thesame quantity δ(P ) properly chosen... but

δ(P ) = ?

δ(P ) = min

{

min(i,j)∈F

{kij − xij}, min(i,j)∈B

xij

}

.

Note: P ≡ F (B = ∅), kij = +∞, ∀ (i, j) ∈ P =⇒δ(P ) = +∞ e ϕ0 = +∞.







A solution method:















implementation



an exercise




algorithm ford-fulkerson(V ,A,x,s,d)

1 (P, δ(P )) := search-augmenting-path(V ,A,x,s,d);

2 if (P = ∅) then return(x,ϕ({s}));

3 do

if (δ(P ) < +∞) then

for each (i, j) ∈ F do /* F: forward arcs in P */

xij := xij + δ(P );

kij := kij − δ(P );

endfor

for each (i, j) ∈ B do /* B: backward arcs in P */

xij := xij − δ(P );

kij := kij + δ(P );

endfor

else return(x,+∞); /* case δ(P ) = +∞:unlim.max flow */

endif

(P, δ(P )) := search-augmenting-path(V ,A,x,s,d);

4 while (P 6= ∅)

5 return (x,ϕ({s}));

end ford-fulkerson







A solution method:















implementation



an exercise




Function search-augmenting-path.

Input: V , A, x, s, and d.

Output: P , δ(P ).







A solution method:















implementation



an exercise






Output: P , δ(P ).

P = ∅ =⇒ x optimal;







A solution method:















implementation



an exercise






Output: P , δ(P ).


P 6= ∅, δ(P ) = +∞ =⇒ ϕ0 = +∞;







A solution method:















implementation



an exercise






Output: P , δ(P ).


P 6= ∅, δ(P ) = +∞ =⇒ ϕ0 = +∞;

P 6= ∅, δ(P ) < +∞ =⇒ increase of the current flow.







A solution method:















implementation



an exercise






Output: P , δ(P ).


P 6= ∅, δ(P ) = +∞ =⇒ ϕ0 = +∞;


But how to look for an augmenting path P ?







A solution method:















implementation



an exercise






Output: P , δ(P ).


P 6= ∅, δ(P ) = +∞ =⇒ ϕ0 = +∞;


But how to look for an augmenting path P ?

It is used a labeling algorithm, that is a variant of a visitof the flow network N starting from the source s.







A solution method:















implementation



an exercise




Let x be a feasible flow and let P be a partial path P

from s to a node l ∈ V such that

✔ p(·) predecessor array in P (p(s) = s);

✔ ∀ (i, j) ∈ F (forward arc in P ), xij < kij andp(j) = i;

✔ ∀ (i, j) ∈ B (backward arc in B), xij > 0 andp(i) = j.

In this case, P is said partial augmenting path from s

to l and visiting labeled nodes.







A solution method:















implementation



an exercise











Starting from the labeled node l, 2 cases can occur:







A solution method:















implementation



an exercise












① ∃ an arc (l, j) ∈ A s.t. xlj < klj :it is possible to extend path P to include theforward arc (l, j), so obtaining a partialaugmenting path from s to node j, that becomeslabeled and p(j) := l;







A solution method:















implementation



an exercise












② ∃ an arc (i, l) ∈ A s.t. xil > 0:it is possible to extend path P to include thebackward arc (i, l), so obtaining a partialaugmenting path from s to node i, that becomeslabeled and p(i) := l.







A solution method:















implementation



an exercise




search-augmenting-path stops when

✗ node d is labeled:

in this case, an augmenting path P from the sources to the sink d has been found and it can be builtby means of the predecessor array p;

or

✗ all labeled nodes are processed and the sink node d

results not labeled:

in this case, ∄ an augmenting path P from thesource s to the sink d.







A solution method:















implementation



an exercise




function search-augmenting-path (V ,A,x,s,d)

1 P := ∅; δ(P ) := +∞;

2 for each i ∈ V \ {s} do p(i) := 0; label[i] := 0; endfor

3 p(s) := s; label[s] := 1; Q := {s};

4 while (Q 6= ∅) do

l :=remove(Q); /* breadth, depth, l := minh∈Q

h, ...*/

for each j ∈ FS(l) do

if (label[j] = 0 and xlj < klj) then

label[j] := 1; p(j) := l; Q := Q ∪ {j};

endif

endfor

for each i ∈ BS(l) do

if (label[i] = 0 and xil > 0) then

label[i] := 1; p(i) := l; Q := Q ∪ {i};

endif

endfor

if (label[d] = 1) then break;

5 endwhile

6...

......;







A solution method:















implementation



an exercise




function search-augmenting-path (V ,A,x,s,d)

1...

......;

· · · · · · · · ·

6 if (label[d] = 0) then return(∅,+∞);

7 j := d;

8 while (j 6= s) do

i := p(j);

if ((i, j) ∈ A) then

P := P ∪ {(i, j)};

if (δ(P ) > kij − xij) then δ(P ) := kij − xij;

else /* case (j, i) ∈ A */

P := P ∪ {(j, i)};

if (δ(P ) > xji) then δ(P ) := xji;

endif

j := i;

9 endwhile

10 return(P,δ(P ));

end search-augmenting-path







A solution method:















implementation



an exercise




subroutine search-augmenting-path (V ,A,x,s,d)

1...

......;

· · · · · · · · ·

6 if (label[d] = 0) then return(∅,+∞);

7 j := d;

8 while (j 6= s) do

i := p(j);

if ((i, j) ∈ A) then

P := P ∪ {(i, j)};

if (δ(P ) > kij − xij) then δ(P ) := kij − xij;

else /* case (j, i) ∈ A */

P := P ∪ {(j, i)};

if (δ(P ) > xji) then δ(P ) := xji;

endif

j := i;

9 endwhile /* O(|A|) */

10 return(P,δ(P ));

end search-augmenting-path







A solution method:















implementation



an exercise




algorithm ford-fulkerson(V ,A,x,s,d)

1 (P, δ(P )) := search-augmenting-path(V,A, x, s, d); /*O(|A|)*/

2 if (P = ∅) then return(x,ϕ({s}));

3 do

if (δ(P ) < +∞) then

for each (i, j) ∈ F do /* F archi forward di P */

xij := xij + δ(P );

kij := kij − δ(P );

endfor

for each (i, j) ∈ B do /* B archi backward di P */

xij := xij − δ(P );

kij := kij + δ(P );

endfor

else return(x,+∞); /* caso δ(P ) = +∞: max flusso ill. */

endif

(P, δ(P )) := search-augmenting-path(V,A, x, s, d); /*O(|A|)*/

4 while (P 6= ∅) /*How many times?*/

5 return (x,ϕ({s}));

end ford-fulkerson







A solution method:















implementation



an exercise




ford-fulkerson: nr of iterations in the worst case.







A solution method:















implementation



an exercise





At each iteration, ϕ0 = ϕ({s}) is augmented in a strictlymonotone manner and the increment is δ(P ) > 0.







A solution method:















implementation



an exercise






W.l.g., we suppose that

∀ (i, j) ∈ A, kij ∈ Z+ ∪ {0} =⇒ δ(P ) ∈ Z+.

Therefore, the max nr of iterations is given by ϕ∗0.







A solution method:















implementation



an exercise






W.l.g., we suppose that

∀ (i, j) ∈ A, kij ∈ Z+ ∪ {0} =⇒ δ(P ) ∈ Z+.

Therefore, the max nr of iterations is given by ϕ∗0.

Moreover,

ϕ∗0 = ϕ∗({s})

Th≤ K({s}) = O(|A| · kmax),

where kmax = max(i,j)∈A

kij .

Summarizing, the worst case computational

complexity of the algorithm is O(|A|2 · kmax).







A solution method:















implementation



an exercise




Instance where ford-fulkerson is not efficient:

s

1

d

2

1

M

ikij

j 3

4

MM

M

M

M N=(V,A)

s

1

d

2

3

4

s

1

d

2

3

4

P odd it.delta=1

+1

+1

+1

+1 +1

+1

+1

-1

P even it.delta=1







A solution method:















implementation



an exercise




Instance where ford-fulkerson is not efficient:

s

1

d

2

1

M

ikij

j 3

4

MM

M

M

M N=(V,A)

s

1

d

2

3

4

s

1

d

2

3

4

P odd it.delta=1

+1

+1

+1

+1 +1

+1

+1

-1

P even it.delta=1

Nr of iterations: 2M .







A solution method:



implementation


Ford-Fulkerson 2, ➀

Ford-Fulkerson 2, ➁



an exercise



A simpler (not more efficient)implementation








A solution method:



implementation






an exercise




At each main iteration (do while loop lines 3-4), eachnode j ∈ V is associated with a label with 2 entries

[±p(j), ǫj ],

where

➜ p(j) is the predecessor node of j along theaugmenting path P from s to j;

➜ ǫj is the increment of flow at node j along P .







A solution method:



implementation






an exercise





[±p(j), ǫj ],

where



✍ “+”=⇒ j is reached traversing the forward arc(p(j), j) ∈ A;

✍ “−”=⇒ j is reached traversing the backward arc(j, p(j)) ∈ A.







A solution method:



implementation






an exercise





[±p(j), ǫj ],

where



✍ “+”=⇒ j is reached traversing the forward arc(p(j), j) ∈ A;

✍ “−”=⇒ j is reached traversing the backward arc(j, p(j)) ∈ A.

Note: initially, also a trivial flow vector xij = 0,∀ (i, j) ∈ A can be used.







A solution method:



implementation






an exercise




algorithm ford-fulkerson2(V ,A,s,d)

1 for each (i, j) ∈ A do xij := 0;

2 ϕ0 := 0;

3 do /* main loop */

for each j ∈ V do p(j) := 0;

[p(s), ǫs] := [+s,+∞]; Q := {s};

4 while (Q 6= ∅ and p(d) = 0) do

h :=remove(Q);

for each j ∈ FS(h), xhj < khj do /* non saturated forw arcs */

if (p(j) = 0) then

[p(j), ǫj ] := [+h, min{ǫh, khj − xhj}]; Q := Q ∪ {j};

endif

endfor

for each i ∈ BS(h), xih > 0 do /* non unloading backw arcs */

if (p(i) = 0) then

[p(i), ǫi] := [−h, min{ǫh, xih}]; Q := Q ∪ {i};

endif

endfor

5 endwhile

· · · · · · · · · · · · · · ·







A solution method:



implementation






an exercise




algorithm ford-fulkerson2(V ,A,s,d)

· · · · · · · · · · · · · · ·

if (p(d) 6= 0) then /* augmenting path P from s to d found */

δ(P ) := ǫd; ϕ0 := ϕ0 + δ(P ); j := d;

while (j 6= s) do /* building P */

i := p(j);

if (i > 0) then xij := xij + δ(P );

else xj|i| := xj|i| − δ(P );

j := |i|;

endwhile

endif

6 until (p(d) = 0)

7 print ‘‘Current flow x is optimal and the cut

(S∗, V \ S∗) has minimum capacity,

where S∗ = {j ∈ V | pj 6= 0}’’;

8 return(x,ϕ0,S∗);

end ford-fulkerson2







A solution method:



implementation



an exercise

Exercise, ➀

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Ford-Fulkerson algorithm:an exercise







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➀

Exercise. Find the maximum flow from the sourcenode s = 1 to the sink node d = 7 in the followingnetwork flow N = (V,A).

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4







A solution method:



implementation



an exercise

Exercise, ➀

Exercise, ➁

Exercise, ➂

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Exercise, ➁

Inizialization:

p(i) = 0, ∀ i ∈ V \ {s}; xij = 0, ∀ (i, j) ∈ A.

ϕ0 = 0.

[+s, +∞] = [+1,+∞].

Q = {1}.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

0

0

0

0

0

0

0

00

0

0

xij

[+1, +oo]







A solution method:



implementation



an exercise

Exercise, ➀

Exercise, ➁

Exercise, ➂

Exercise, ➂

Exercise, ➂

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Exercise, ➂

Iteration I: (breadth search)h :=remove(Q);

for each j ∈ FS(h), xhj < khj do /* non saturated forward arcs */

if (p(j) = 0) then {

[p(j), ǫj ] := [+h, min{ǫh, khj − xhj}]; Q := Q ∪ {j};}

for each i ∈ BS(h), xih > 0 do /* non unloading backward arcs */

if (p(i) = 0) then { [p(i), ǫi] := [−h, min{ǫh, xih}]; Q := Q ∪ {i}; }

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

0

0

0

0

0

0

0

00

0

0

xij

[+1, +oo]

[+1, 2]

[+1, 2]

[+1, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➂







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

0

0

0

0

0

0

0

00

0

0

xij

[+1, +oo]

[+1, 2]

[+1, 2]

[+1, 2]

[+2, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➂







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

0

0

0

0

0

0

0

00

0

0

xij

[+1, +oo]

[+1, 2]

[+1, 2]

[+1, 2] [+2, 2]

[+2, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➂







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

0

0

0

0

0

0

0

00

0

0

xij

[+1, +oo]

[+1, 2]

[+1, 2]

[+1, 2] [+2, 2]

[+2, 2]

[+3, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➂

Iteration I:

Augmenting path P = {(1, 2), (2, 3), (3, 7)}.

δ(P ) = ǫd = 2.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

0

0

0

0

0

0

0

00

0

0

xij

[+1, +oo]

[+1, 2]

[+1, 2]

[+1, 2] [+2, 2]

[+2, 2]

[+3, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➂

Iteration I:

Increasing/decreasing of δ(P ) = ǫd = 2 along the augmenting path

P = {(1, 2), (2, 3), (3, 7)}.

s=1

5

d=7

4

1

ikij

j 6

3

2

22

21

2

2

2

1

1

2

4

0

0

0

0

0

0

0

0

00

0

0

xij

[+1, +oo]

[+1, 2]

[+1, 2]

[+1, 2] [+2, 2]

[+2, 2]

[+3, 2]

2

22







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➂

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Exercise, ➂

End of Iteration I:

ϕ0 = 2.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

2

0

0

0

0

0

0

02

0

2

xij

[+1, +oo]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➃

Iteration II: (breadth search)h :=remove(Q);






s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

2

0

0

0

0

0

0

02

0

2

xij

[+1, +oo]

[+1, 2]

[+1, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➃







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

2

0

0

0

0

0

0

02

0

2

xij

[+1, +oo]

[+1, 2]

[+5, 1]

[+1, 2]

[+5, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➃







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

2

0

0

0

0

0

0

02

0

2

xij

[+1, +oo]

[+1, 2]

[+5, 1]

[+1, 2]

[+5, 2]

[+6, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➃

Iteration II:

Augmenting path P = {(1, 5), (5, 6), (6, 7)}.

δ(P ) = ǫd = 2.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

2

0

0

0

0

0

0

02

0

2

xij

[+1, +oo]

[+1, 2]

[+5, 1]

[+1, 2]

[+5, 2]

[+6, 2]







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➃

Iteration II:


P = {(1, 5), (5, 6), (6, 7)}.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

2

0

0

0

0

0

0

02

0

2

xij

[+1, +oo]

[+1, 2]

[+5, 1]

[+1, 2]

[+5, 2]

[+6, 2]

2

2

2







A solution method:



implementation



an exercise

Exercise, ➀

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Exercise, ➃

End of Iteration II:

ϕ0 = 4.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

0

0

2

0

0

2

02

0

2

xij

[+1, +oo]







A solution method:



implementation



an exercise

Exercise, ➀

Exercise, ➁

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Exercise, ➃

Iteration III: (breadth search)h :=remove(Q);






s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

0

0

0

0

0

2

02

0

2

xij

[+1, +oo]

[+1, 2]







A solution method:



implementation



an exercise

Exercise, ➀

Exercise, ➁

Exercise, ➂

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Exercise, ➃







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

0

0

0

0

0

2

02

0

2

xij

[+1, +oo]

[+1, 2] [+4, 1]







A solution method:



implementation



an exercise

Exercise, ➀

Exercise, ➁

Exercise, ➂

Exercise, ➂

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Exercise, ➃







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

0

0

0

0

0

2

02

0

2

xij

[+1, +oo]

[+1, 2] [+4, 1]

[-3, 1]







A solution method:



implementation



an exercise

Exercise, ➀

Exercise, ➁

Exercise, ➂

Exercise, ➂

Exercise, ➂

Exercise, ➂

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Exercise, ➃







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

0

0

0

0

0

2

02

0

2

xij

[+1, +oo]

[+1, 2] [+4, 1]

[-3, 1]

[+2, 1]







A solution method:



implementation



an exercise

Exercise, ➀

Exercise, ➁

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Exercise, ➂

Exercise, ➂

Exercise, ➂

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Exercise, ➃







s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

0

0

0

0

0

2

02

0

2

xij

[+1, +oo][-3, 1]

[+1, 2]

[+2, 1]

[+6, 1]

[+4, 1]







A solution method:



implementation



an exercise

Exercise, ➀

Exercise, ➁

Exercise, ➂

Exercise, ➂

Exercise, ➂

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Exercise, ➄

Iteration III:

Augmenting path P = {(1, 4), (4, 3), (2, 3), (2, 6), (6, 7)}.

δ(P ) = ǫd = 1.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

0

0

0

0

0

2

02

0

2

xij

[+1, +oo][-3, 1]

[+1, 2]

[+2, 1]

[+6, 1]

[+4, 1]







A solution method:



implementation



an exercise

Exercise, ➀

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Iteration III:


P = {(1, 4), (4, 3), (2, 3), (2, 6), (6, 7)}.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

0

2

0

0

0

0 0

2

0 2

0

2

xij

[+1, +oo][-3, 1]

[+1, 2]

[+2, 1]

[+6, 1]

[+4, 1]

1

1

3

1

1







A solution method:



implementation



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End of Iteration III:

ϕ0 = 5.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

1

0

2

1

0

3

01

1

2

xij

[+1, +oo]







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Exercise, ➅

Iteration IV:

d can not be reached from s.

(S∗, V \ S∗) = ({1, 4}, {2, 3, 5, 6, 7}) is the minimum cut.

ϕ∗0 = 5.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

1

0

2

1

0

3

01

1

2

xij

[+1, +oo]

[+1, 1]

S*={1,4}







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Exercise, ➆

(S∗, V \ S∗) = ({1, 4}, {2, 3, 5, 6, 7}); ϕ∗0 = ϕ(S∗) = K(S∗) = 5.

s=1

5

d=7

4

1

ikij

j 6

3

2

N=(V,A)

22

21

2

2

2

1

1

2

4

2

2

1

0

2

1

0

3

01

1

2

xij

[+1, +oo]

[+1, 1]

S*={1,4}

Notes:

✔ every arc (i, j) ∈ δ+N (S∗) is saturated;

✔ every arc (i, j) ∈ δ−N (S∗) is unloading.

Documents

summerschool2013.ccaba.upc.edusummerschool2013.ccaba.upc.edu/documents/Lecture9_Festa.pdf · Preliminaries: notation Network Flow Problems Network Flows: Variants Special cases of