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1 1 Slide © 2008 Thomson South-Western. All Rights Reserved Slides by JOHN LOUCKS St. Edward’s University.

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Presentation on theme: "1 1 Slide © 2008 Thomson South-Western. All Rights Reserved Slides by JOHN LOUCKS St. Edward’s University."— Presentation transcript:

1 1 1 Slide © 2008 Thomson South-Western. All Rights Reserved Slides by JOHN LOUCKS St. Edward’s University

2 2 2 Slide © 2008 Thomson South-Western. All Rights Reserved Chapter 6, Part B Distribution and Network Models n Shortest-Route Problem n Maximal Flow Problem

3 3 3 Slide © 2008 Thomson South-Western. All Rights Reserved Shortest-Route Problem n The shortest-route problem is concerned with finding the shortest path in a network from one node (or set of nodes) to another node (or set of nodes). n If all arcs in the network have nonnegative values then a labeling algorithm can be used to find the shortest paths from a particular node to all other nodes in the network. n The criterion to be minimized in the shortest-route problem is not limited to distance even though the term "shortest" is used in describing the procedure. Other criteria include time and cost. (Neither time nor cost are necessarily linearly related to distance.)

4 4 4 Slide © 2008 Thomson South-Western. All Rights Reserved n Linear Programming Formulation Using the notation: Using the notation: x ij = 1 if the arc from node i to node j x ij = 1 if the arc from node i to node j is on the shortest route is on the shortest route 0 otherwise 0 otherwise c ij = distance, time, or cost associated c ij = distance, time, or cost associated with the arc from node i to node j with the arc from node i to node j continued Shortest-Route Problem

5 5 5 Slide © 2008 Thomson South-Western. All Rights Reserved n Linear Programming Formulation (continued) Shortest-Route Problem

6 6 6 Slide © 2008 Thomson South-Western. All Rights Reserved Susan Winslow has an important business meeting Susan Winslow has an important business meeting in Paducah this evening. She has a number of alternate routes by which she can travel from the company headquarters in Lewisburg to Paducah. The network of alternate routes and their respective travel time, ticket cost, and transport mode appear on the next two slides. If Susan earns a wage of $15 per hour, what route If Susan earns a wage of $15 per hour, what route should she take to minimize the total travel cost? Example: Shortest Route

7 7 7 Slide © 2008 Thomson South-Western. All Rights Reserved 6 A B C D E F G H I J K L M Example: Shortest Route Paducah Lewisburg 1 2 5 3 4 n Network Model

8 8 8 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Shortest Route Transport Time Ticket Transport Time Ticket Route Mode (hours) Cost A Train 4 $ 20 A Train 4 $ 20 B Plane 1 $115 B Plane 1 $115 C Bus 2 $ 10 C Bus 2 $ 10 D Taxi 6 $ 90 D Taxi 6 $ 90 E Train 3  $ 30 E Train 3  $ 30 F Bus 3 $ 15 F Bus 3 $ 15 G Bus 4  $ 20 G Bus 4  $ 20 H Taxi 1 $ 15 H Taxi 1 $ 15 I Train 2  $ 15 I Train 2  $ 15 J Bus 6  $ 25 J Bus 6  $ 25 K Taxi 3  $ 50 K Taxi 3  $ 50 L Train 1  $ 10 L Train 1  $ 10 M Bus 4  $ 20 M Bus 4  $ 20

9 9 9 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Shortest Route Transport Time Time Ticket Total Transport Time Time Ticket Total Route Mode (hours) Cost Cost Cost A Train 4 $60 $ 20$ 80 A Train 4 $60 $ 20$ 80 B Plane 1 $15 $115$130 B Plane 1 $15 $115$130 C Bus 2 $30 $ 10$ 40 C Bus 2 $30 $ 10$ 40 D Taxi 6 $90 $ 90$180 D Taxi 6 $90 $ 90$180 E Train 3  $50 $ 30$ 80 E Train 3  $50 $ 30$ 80 F Bus 3 $45 $ 15$ 60 F Bus 3 $45 $ 15$ 60 G Bus 4  $70 $ 20$ 90 G Bus 4  $70 $ 20$ 90 H Taxi 1 $15 $ 15$ 30 H Taxi 1 $15 $ 15$ 30 I Train 2  $35 $ 15$ 50 I Train 2  $35 $ 15$ 50 J Bus 6  $95 $ 25$120 J Bus 6  $95 $ 25$120 K Taxi 3  $50 $ 50$100 K Taxi 3  $50 $ 50$100 L Train 1  $20 $ 10$ 30 L Train 1  $20 $ 10$ 30 M Bus 4  $70 $ 20$ 90 M Bus 4  $70 $ 20$ 90

10 10 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Shortest Route n LP Formulation Objective Function Objective Function Min 80 x 12 + 40 x 13 + 80 x 14 + 130 x 15 + 180 x 16 + 60 x 25 Min 80 x 12 + 40 x 13 + 80 x 14 + 130 x 15 + 180 x 16 + 60 x 25 + 100 x 26 + 30 x 34 + 90 x 35 + 120 x 36 + 30 x 43 + 50 x 45 + 100 x 26 + 30 x 34 + 90 x 35 + 120 x 36 + 30 x 43 + 50 x 45 + 90 x 46 + 60 x 52 + 90 x 53 + 50 x 54 + 30 x 56 + 90 x 46 + 60 x 52 + 90 x 53 + 50 x 54 + 30 x 56 Node Flow-Conservation Constraints Node Flow-Conservation Constraints x 12 + x 13 + x 14 + x 15 + x 16 = 1 (origin) x 12 + x 13 + x 14 + x 15 + x 16 = 1 (origin) – x 12 + x 25 + x 26 – x 52 = 0 (node 2) – x 12 + x 25 + x 26 – x 52 = 0 (node 2) – x 13 + x 34 + x 35 + x 36 – x 43 – x 53 = 0 (node 3) – x 13 + x 34 + x 35 + x 36 – x 43 – x 53 = 0 (node 3) – x 14 – x 34 + x 43 + x 45 + x 46 – x 54 = 0 (node 4) – x 14 – x 34 + x 43 + x 45 + x 46 – x 54 = 0 (node 4) – x 15 – x 25 – x 35 – x 45 + x 52 + x 53 + x 54 + x 56 = 0 (node 5) – x 15 – x 25 – x 35 – x 45 + x 52 + x 53 + x 54 + x 56 = 0 (node 5) x 16 + x 26 + x 36 + x 46 + x 56 = 1 (destination) x 16 + x 26 + x 36 + x 46 + x 56 = 1 (destination)

11 11 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Shortest Route n Solution Summary Minimum total cost = $150 x 12 = 0 x 25 = 0 x 34 = 1 x 43 = 0 x 52 = 0 x 13 = 1 x 26 = 0 x 35 = 0 x 45 = 1 x 53 = 0 x 13 = 1 x 26 = 0 x 35 = 0 x 45 = 1 x 53 = 0 x 14 = 0 x 36 = 0 x 46 = 0 x 54 = 0 x 14 = 0 x 36 = 0 x 46 = 0 x 54 = 0 x 15 = 0 x 56 = 1 x 15 = 0 x 56 = 1 x 16 = 0 x 16 = 0

12 12 Slide © 2008 Thomson South-Western. All Rights Reserved Maximal Flow Problem n The maximal flow problem is concerned with determining the maximal volume of flow from one node (called the source) to another node (called the sink). n In the maximal flow problem, each arc has a maximum arc flow capacity which limits the flow through the arc.

13 13 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow n A capacitated transshipment model can be developed for the maximal flow problem. n We will add an arc from the sink node back to the source node to represent the total flow through the network. n There is no capacity on the newly added sink-to- source arc. n We want to maximize the flow over the sink-to-source arc.

14 14 Slide © 2008 Thomson South-Western. All Rights Reserved Maximal Flow Problem n LP Formulation (as Capacitated Transshipment Problem) (as Capacitated Transshipment Problem) There is a variable for every arc. There is a variable for every arc. There is a constraint for every node; the flow out must equal the flow in. There is a constraint for every node; the flow out must equal the flow in. There is a constraint for every arc (except the added sink-to-source arc); arc capacity cannot be exceeded. There is a constraint for every arc (except the added sink-to-source arc); arc capacity cannot be exceeded. The objective is to maximize the flow over the added, sink-to-source arc. The objective is to maximize the flow over the added, sink-to-source arc.

15 15 Slide © 2008 Thomson South-Western. All Rights Reserved Maximal Flow Problem n LP Formulation (as Capacitated Transshipment Problem) (as Capacitated Transshipment Problem) Max x k 1 ( k is sink node, 1 is source node) Max x k 1 ( k is sink node, 1 is source node) s.t.  x ij -  x ji = 0 (conservation of flow) i j s.t.  x ij -  x ji = 0 (conservation of flow) i j x ij < c ij ( c ij is capacity of ij arc) x ij > 0, for all i and j (non-negativity) x ij > 0, for all i and j (non-negativity) (x ij represents the flow from node i to node j) (x ij represents the flow from node i to node j)

16 16 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow National Express operates a fleet of cargo planes and National Express operates a fleet of cargo planes and is in the package delivery business. NatEx is interested in knowing what is the maximum it could transport in one day indirectly from San Diego to Tampa (via Denver, St. Louis, Dallas, Houston and/or Atlanta) if its direct flight was out of service. NatEx's indirect routes from San Diego to Tampa, along with their respective estimated excess shipping capacities (measured in hundreds of cubic feet per day), are shown on the next slide. Is there sufficient excess capacity to indirectly ship 5000 cubic feet of packages in one day?

17 17 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow n Network Model 2255 114477 3366 4 4 3 3 2 3 4 2 3 3 3 1 5 55 1 6 3 Denver SanDiego St. Louis Houston Tampa Atlanta Dallas

18 18 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow n Modified Network Model 2255 114477 3366 4 4 3 3 2 3 4 2 3 3 3 1 5 55 1 6 3 Sink Source Addedarc

19 19 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow n LP Formulation 18 variables (for 17 original arcs and 1 added arc) 18 variables (for 17 original arcs and 1 added arc) 24 constraints 24 constraints 7 node flow-conservation constraints7 node flow-conservation constraints 17 arc capacity constraints (for original arcs)17 arc capacity constraints (for original arcs)

20 20 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow n LP Formulation Objective Function Objective Function Max x 71 Max x 71 Node Flow-Conservation Constraints Node Flow-Conservation Constraints x 12 + x 13 + x 14 – x 71 = 0 (node 1) x 12 + x 13 + x 14 – x 71 = 0 (node 1) – x 12 + x 24 + x 25 – x 42 – x 52 = 0 (node 2) – x 12 + x 24 + x 25 – x 42 – x 52 = 0 (node 2) – x 13 + x 34 + x 36 – x 43 = 0 (and so on) – x 13 + x 34 + x 36 – x 43 = 0 (and so on) – x 14 – x 24 – x 34 + x 42 + x 43 + x 45 + x 46 + x 47 – x 54 – x 64 = 0 – x 14 – x 24 – x 34 + x 42 + x 43 + x 45 + x 46 + x 47 – x 54 – x 64 = 0 – x 25 – x 45 + x 52 + x 54 + x 57 = 0 – x 25 – x 45 + x 52 + x 54 + x 57 = 0 – x 36 – x 46 + x 64 + x 67 = 0 – x 36 – x 46 + x 64 + x 67 = 0 – x 47 – x 57 – x 67 + x 71 = 0 – x 47 – x 57 – x 67 + x 71 = 0

21 21 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow n LP Formulation (continued) Arc Capacity Constraints Arc Capacity Constraints x 12 < 4 x 13 < 3 x 14 < 4 x 12 < 4 x 13 < 3 x 14 < 4 x 24 < 2 x 25 < 3 x 24 < 2 x 25 < 3 x 34 < 3 x 36 < 6 x 34 < 3 x 36 < 6 x 42 < 3 x 43 < 5 x 45 < 3 x 46 < 1 x 47 < 3 x 42 < 3 x 43 < 5 x 45 < 3 x 46 < 1 x 47 < 3 x 52 < 3 x 54 < 4 x 57 < 2 x 52 < 3 x 54 < 4 x 57 < 2 x 64 < 1 x 67 < 5 x 64 < 1 x 67 < 5

22 22 Slide © 2008 Thomson South-Western. All Rights Reserved n Alternative Optimal Solution #1 Example: Maximal Flow Objective Function Value = 10.000 Variable Value x 12 3.000 x 13 3.000 x 14 4.000 x 24 1.000 x 25 2.000 x 34 0.000 x 36 5.000 x 42 0.000 x 43 2.000 Variable Value x 45 0.000 x 46 0.000 x 47 3.000 x 52 0.000 x 54 0.000 x 57 2.000 x 64 0.000 x 67 5.000 x 71 10.000

23 23 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow n Alternative Optimal Solution #1 2255 11 44 77 3366 3 4 3 2 1 2 3 2 55 5 Sink Source 10

24 24 Slide © 2008 Thomson South-Western. All Rights Reserved n Alternative Optimal Solution #2 Example: Maximal Flow Objective Function Value = 10.000 Variable Value x 12 3.000 x 13 3.000 x 14 4.000 x 24 1.000 x 25 2.000 x 34 0.000 x 36 4.000 x 42 0.000 x 43 1.000 Variable Value x 45 0.000 x 46 1.000 x 47 3.000 x 52 0.000 x 54 0.000 x 57 2.000 x 64 0.000 x 67 5.000 x 71 10.000

25 25 Slide © 2008 Thomson South-Western. All Rights Reserved Example: Maximal Flow n Alternative Optimal Solution #2 2255 11 44 77 3366 3 4 3 2 1 2 3 1 1 55 4 Sink Source 10

26 26 Slide © 2008 Thomson South-Western. All Rights Reserved End of Chapter 6, Part B


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