Presentation Topic 3.pdf

8/14/2019 Presentation Topic 3.pdf

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Presentation Topic 3 - Networks

By:

Fernando Brandao - 00146919

For:

IEGR461: Operations Research, Deterministic ModelsDr. M. Salimian

Fall 2013

Department of Industrial and Systems Engineering

Morgan State UniversityWednesday, September 25, 2013


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Considering the network bellow respond:

1) Create a table and identify all possible cuts. For each cut provide information on nodes that arein S, nodes that are in S-bar, and cut capacity. Distinguish the minimal cut.

Cut N. S S-bar Max-Flow

1 S 1-2-3-4-5-n 80

2 S-1 2-3-4-5-n 60

3 S-2 1-3-4-5-n 75

4 S-1-2 3-4-5-n 55

5 S-1-2-3 4-5-n 70

6 S-1-2-3-4 5-n 70

7 S-1-2-3-5 4-n 60

8 S-1-2-3-4-5 n 60

Where the minimal cut is S = {S, 1, 2} and S-bar = {3, 4, 5, n} with the max flow of 55.

2) Find the maximum flow through the network from source to sink node using the labeling routineand maximum flow algorithm and compare your result with the minimum cut capacity.


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Patch Max Patch Flow Total Flow

S-1-4-n 10 10

S-1-3-4-n 10 20

S-2-3-4-n 10 30

S-2-3-5-n 5 35

S-2-5-n 15 50

Lets considerate the following patch:

S => 2 => 5 => 3 => n,

Where:

S => 2 is a forward arc (with 40 of capacity and 30 of flow, and a capacity of increase of 10);

2 => 5 is a forward arc (with 20 of capacity and 15 of flow, and a capacity of increase of 5);

5 => 3 is a backward arc (because is a forward arc from 3 to 5; a capacity of 10 and a flow of 5, and a

capacity of decrease of 5);

3 => n is a forward arc (with 10 of capacity and 0 of flow, and a capacity of increase of 10).

So, analyzing the arc above, we can augment the max flow of the network in 5 units that will give us the

network below:


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There is the optimum max flow (55).

3) Write the LP associated with this network, solve using LINDO and compare results.

The LP associated with this network is:

MAX F

Subject to:

FXS1XS2 = 0;

XS1X14X13 = 0;

XS2X23X25 = 0;

X13 + X23X34X3NX35 = 0;

X14 + X34X4N = 0;

X25 + X35X5N =0

X4N + X3N + X5NF =0;

XS1 =< 40;

XS2 =< 40;

X14 =< 10;


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X13 =< 10;

X23 =< 15;

X25 =< 20;

X34 =< 20;

X3N =< 10;

X35 =< 10;

X4N =< 30;

X5N =< 20;

Solving by LINDO we have this optimal solution:

Patch Flow

F 55

XS1 20

XS2 35

X14 10

X13 10

X23 15

X25 20

X34 15

X3N 10

X35 0

X4N 25

X5N 20

Where Xij is the patch between i and j.

The result matches with the one find in 1) and 2).

LINDO Output:

MAX F

SUBJECT TO

2) F - XS1 - XS2 = 0

3) XS1 - X14 - X13 = 0

4) XS2 - X23 - X25 = 0

5) X13 + X23 - X34 - X3N - X35 = 0


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6) X14 + X34 - X4N = 0

7) X25 + X35 - X5N = 0

8) - F + X3N + X4N + X5N = 0

9) XS1


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X13 10.000000 0.000000

X23 15.000000 0.000000

X25 20.000000 0.000000

X34 15.000000 0.000000

X3N 10.000000 0.000000

X35 0.000000 1.000000

X4N 25.000000 0.000000

X5N 20.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES

2) 0.000000 0.000000

3) 0.000000 0.000000

4) 0.000000 0.000000

5) 0.000000 -1.000000

6) 0.000000 -1.000000

7) 0.000000 0.000000

8) 0.000000 -1.000000

9) 20.000000 0.000000

10) 5.000000 0.000000

11) 0.000000 1.000000

12) 0.000000 1.000000

13) 0.000000 1.000000

14) 0.000000 0.000000

15) 5.000000 0.000000

16) 0.000000 0.000000

17) 10.000000 0.000000

18) 5.000000 0.000000

19) 0.000000 1.000000


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NO. ITERATIONS= 0

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE CURRENT ALLOWABLE ALLOWABLE

COEF INCREASE DECREASE

F 1.000000 INFINITY 1.000000

XS1 0.000000 INFINITY 1.000000

X2S 0.000000 INFINITY 1.000000

X14 0.000000 INFINITY 1.000000

X13 0.000000 INFINITY 1.000000

XS2 0.000000 INFINITY 1.000000

X23 0.000000 INFINITY 1.000000

X25 0.000000 INFINITY 1.000000

X34 0.000000 0.000000 1.000000

X3N 0.000000 INFINITY 0.000000

X35 0.000000 1.000000 INFINITY

X4N 0.000000 0.000000 1.000000

X5N 0.000000 INFINITY 1.000000

RIGHTHAND SIDE RANGES

ROW CURRENT ALLOWABLE ALLOWABLE

RHS INCREASE DECREASE

2 0.000000 35.000000 INFINITY

3 0.000000 20.000000 20.000000

4 0.000000 5.000000 35.000000


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5 0.000000 15.000000 5.000000

6 0.000000 25.000000 5.000000

7 0.000000 0.000000 20.000000

8 0.000000 35.000000 INFINITY

9 40.000000 INFINITY 20.000000

10 40.000000 INFINITY 5.000000

11 10.000000 5.000000 10.000000

12 10.000000 5.000000 10.000000

13 15.000000 5.000000 15.000000

14 20.000000 INFINITY 0.000000

15 20.000000 INFINITY 5.000000

16 10.000000 15.000000 5.000000

17 10.000000 INFINITY 10.000000

18 30.000000 INFINITY 5.000000

19 20.000000 0.000000 20.000000

References

[1]https://www.youtube.com/watch?v=J0wzih3_5Wo
https://www.youtube.com/watch?v=J0wzih3_5Wohttps://www.youtube.com/watch?v=J0wzih3_5Wohttps://www.youtube.com/watch?v=J0wzih3_5Wohttps://www.youtube.com/watch?v=J0wzih3_5Wo

Documents

Presentation Topic 3.pdf