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Home Work Topic 3 - Networks

By:Fernando Brandao - 00146919

For:IEGR 461: Operations Research, Deterministic Models

Dr. M. SalimianFall 2013

Department of Industrial and Systems EngineeringMorgan State University

Friday, September 27, 2013

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Original Problem:

IEGR 461/440: Deterministic Models of Operations ResearchFall 2013M. Salimian

Topic 3Problem 02 Name: --FCB --Pickup Time and Date: --R/6:00pm -- Due: in 24 hours

Consider the information on arc capacities from a directed network given in the table below.(Note: if both [i,j] and [j,i] arcs capacities are defined and they are equal then the arc connecting ito j is an undirected arc.)

From\To 1 2 3 4 5 6 n S 53 44 41 1 50 27 40 2 37 38 39 3 45 21 4 37 33 58 5 40 35 38 6 39 45

Plot the network.

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.

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. Start withs-3-2-5-1-4-n flow augmenting path.

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

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Research:

1) Networks:

The term network flow program describes a type of model that is a special case of the moregeneral linear program. The class of network flow programs includes such problems as thetransportation problem, the assignment problem, the shortest path problem, the maximum flowproblem, the pure minimum cost flow problem, and the generalized minimum cost flow problem. It is animportant class because many aspects of actual situations are readily recognized as networks and therepresentation of the model is much more compact than the general linear program. When a situationcan be entirely modeled as a network, very efficient algorithms exist for the solution of the optimizationproblem, many times more efficient than linear programming in the utilization of computer time andspace resources. Network models are constructed by the Math Programming add-in and may be solvedby either the Excel Solver, LINDO or Lips. [1]

2) Maximum Flow Problem:

The maximum flow problem is structured on a network. Here the arc capacities, or upper bounds,are the only relevant parameters. The problem is to find the maximum flow possible from some givensource node to a given sink node. All arc costs are zero. Since the goal of the optimization is to minimizecost, the maximum flow possible is delivered to the sink node. Applications of this problem includefinding the maximum flow of orders through a job shop, the maximum flow of water through a storm

sewer system, and the maximum flow of product through a product distribution system, among others.[2].

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Resolution:

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

Cut N. S S-bar Max Flow1 S 1-2-3-4-5-6-n 1382 S-1 2-3-4-5-6-n 1623 S-2 1-3-4-5-6-n 2084 S-3 1-2-4-5-6-n 1635 S-1-3 2-4-5-6-n 2276 S-1-2 3-4-5-6-n 2227 S-2-3 1-4-5-6-n 1888 S-2-5 1-3-4-6-n 2839 S-1-4 2-3-5-6-n 270

10 S-3-6 1-2-4-5-n 22611 S-1-5 2-3-4-6-n 23512 S-1-2-3 4-5-6-n 20213 S-1-4-5 2-3-6-n 30314 S-1-2-4 3-5-6-n 29415 S-1-2-5 3-4-6-n 22016 S-2-3-6 1-4-5-n 21217 S-2-3-5 1-5-6-n 263

18 S-3-5-6 1-2-4-n 26519 S-1-2-3-4 5-6-n 22920 S-1-2-3-5 4-6-n 19721 S-1-2-3-6 4-5-n 20122 S-1-2-4-5 3-6-n 21123 S-2-3-5-6 1-4-n 21324 S-1-2-3-4-5 6-n 19125 S-1-2-3-4-6 5-n 21426 S-1-2-3-5-6 4-n 18627 S-1-2-3-4-5-6 n 141

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Where the cut with the maximum flow is S = {S} and S-bar = {1, 2, 3, 4, 5, n} with the flow of 138.

2. Find the maximum flow through the network from source to sink node using the labeling routine andmaximum flow algorithm and compare your result with the minimum cut capacity. Start with s-3-2-5-1-4-n flow augmenting path.

Solving the network using the labeling routine and maximum flow algorithm, we have:

Augment Patch Augment Flow Total FlowS-3-2-5-1-4-n 27 27

S-1-5-n 38 65S-1-2-4-n 15 80S-2-4-n 16 96

S-2-5-6-n 11 107S-2-6-n 17 124

S-3-6-n 14 138

The table bellow gives us the following network:

In the indirect arc between 1 and 5, we have 2 flows, the first of 27, from 5 to 1, and the second of 38,from 1 to 5, so, the resulting flow is 11, from 1 to 5.

In the indirect arc between 2 and 4, we have one flow, from 2 to 4 of 31, so, the resulting flow is thisone, since we dont have a flow in the opposite direction.

The maximum flow found was 138, the same flow from Ex. 1

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

The problem above can be write as:

MAX F

Subject to:

F XS1 XS2 XS3 = 0

XS1 + X51 X12 X15 X14 = 0

XS2 + X42 + X12 + X32 + X62 X24 X25 X26 = 0

XS3 X32 X36 = 0

X14 + X24 + X64 X42 X45 X4N = 0

X15 + X25 + X45 X51 X56 X5N = 0

X36 + X26 + X56 X64 X62 X6N = 0

X4N + X5N + X6N F = 0

XS1

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X4N

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5) XS3 - X32 - X36 = 0

6) X14 - X42 + X24 + X64 - X45 - X4N = 0

7) - X51 + X15 + X25 + X45 - X56 - X5N = 0

8) - X62 + X26 + X36 - X64 + X56 - X6N = 0

9) - F + X4N + X5N + X6N = 0

10) XS1

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VARIABLE VALUE REDUCED COST

F 138.000000 0.000000

XS1 53.000000 0.000000

XS2 44.000000 0.000000

XS3 41.000000 0.000000

X51 0.000000 0.000000

X12 0.000000 0.000000

X15 32.000000 0.000000

X14 21.000000 0.000000

X42 0.000000 0.000000

X32 20.000000 0.000000

X62 0.000000 0.000000

X24 37.000000 0.000000

X25 6.000000 0.000000

X26 21.000000 0.000000

X36 21.000000 0.000000

X64 0.000000 0.000000

X45 0.000000 0.000000

X4N 58.000000 0.000000

X56 0.000000 0.000000

X5N 38.000000 0.000000

X6N 42.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES

2) 0.000000 1.000000

3) 0.000000 0.000000

4) 0.000000 0.000000

5) 0.000000 0.000000

6) 0.000000 0.000000

7) 0.000000 0.000000

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8) 0.000000 0.000000

9) 0.000000 0.000000

10) 0.000000 1.000000

11) 0.000000 1.000000

12) 0.000000 1.000000

13) 50.000000 0.000000

14) 8.000000 0.000000

15) 40.000000 0.000000

16) 6.000000 0.000000

17) 0.000000 0.000000

18) 37.000000 0.000000

19) 32.000000 0.000000

20) 18.000000 0.000000

21) 39.000000 0.000000

22) 25.000000 0.000000

23) 0.000000 0.000000

24) 33.000000 0.000000

25) 39.000000 0.000000

26) 35.000000 0.000000

27) 0.000000 0.000000

28) 0.000000 0.000000

29) 3.000000 0.000000

NO. ITERATIONS= 12

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE CURRENT ALLOWABLE ALLOWABLE

COEF INCREASE DECREASE

F 1.000000 INFINITY 1.000000

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XS1 0.000000 INFINITY 1.000000

XS2 0.000000 INFINITY 1.000000

XS3 0.000000 INFINITY 1.000000

X51 0.000000 0.000000 INFINITY

X12 0.000000 0.000000 INFINITY

X15 0.000000 0.000000 0.000000

X14 0.000000 0.000000 0.000000

X42 0.000000 0.000000 INFINITY

X32 0.000000 0.000000 1.000000

X62 0.000000 0.000000 INFINITY

X24 0.000000 INFINITY 0.000000

X25 0.000000 0.000000 0.000000

X26 0.000000 0.000000 0.000000

X36 0.000000 INFINITY 0.000000

X64 0.000000 0.000000 INFINITY

X45 0.000000 0.000000 INFINITY

X4N 0.000000 INFINITY 0.000000

X56 0.000000 0.000000 INFINITY

X5N 0.000000 INFINITY 0.000000

X6N 0.000000 0.000000 1.000000

RIGHTHAND SIDE RANGES

ROW CURRENT ALLOWABLE ALLOWABLE

RHS INCREASE DECREASE

2 0.000000 0.000000 0.000000

3 0.000000 0.000000 0.000000

4 0.000000 0.000000 0.000000

5 0.000000 0.000000 0.000000

6 0.000000 0.000000 0.000000

7 0.000000 0.000000 0.000000

8 0.000000 0.000000 0.000000

9 0.000000 0.000000 0.000000

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10 53.000000 3.000000 21.000000

11 44.000000 3.000000 21.000000

12 41.000000 3.000000 20.000000

13 50.000000 INFINITY 50.000000

14 40.000000 INFINITY 8.000000

15 40.000000 INFINITY 40.000000

16 27.000000 INFINITY 6.000000

17 37.000000 6.000000 6.000000

18 37.000000 INFINITY 37.000000

19 38.000000 INFINITY 32.000000

20 39.000000 INFINITY 18.000000

21 39.000000 INFINITY 39.000000

22 45.000000 INFINITY 25.000000

23 21.000000 20.000000 18.000000

24 33.000000 INFINITY 33.000000

25 39.000000 INFINITY 39.000000

26 35.000000 INFINITY 35.000000

27 58.000000 6.000000 3.000000

28 38.000000 21.000000 3.000000

29 45.000000 INFINITY 3.000000

References:

[1] http://www.me.utexas.edu/~jensen/ORMM/models/unit/network/

[2] http://lpsolve.sourceforge.net/5.5/DIMACS_maxf.htm

http://www.me.utexas.edu/~jensen/ORMM/models/unit/network/http://www.me.utexas.edu/~jensen/ORMM/models/unit/network/http://www.me.utexas.edu/~jensen/ORMM/models/unit/network/http://lpsolve.sourceforge.net/5.5/DIMACS_maxf.htmhttp://lpsolve.sourceforge.net/5.5/DIMACS_maxf.htmhttp://lpsolve.sourceforge.net/5.5/DIMACS_maxf.htmhttp://lpsolve.sourceforge.net/5.5/DIMACS_maxf.htmhttp://www.me.utexas.edu/~jensen/ORMM/models/unit/network/