1 1 Slide © 2009 South-Western, a part of Cengage Learning Slides by John Loucks St. Edward’s University

2 2 Slide © 2009 South-Western, a part of Cengage Learning Chapter 10, Part B Distribution and Network Models n Shortest-Route Problem n Maximal Flow Problem n A Production and Inventory Application

3 3 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning n Linear Programming Formulation (continued) Shortest-Route Problem

6 6 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 6 A B C D E F G H I J K L M Example: Shortest Route Paducah Lewisburg n Network Model

8 8 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning Example: Shortest Route n LP Formulation Objective Function Objective Function Min 80 x x x x x x 25 Min 80 x x x x x x x x x x x x x x x x x x x x x x x x x x x 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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? 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 Slide © 2009 South-Western, a part of Cengage Learning Example: Maximal Flow n Network Model Denver SanDiego St. Louis Houston Tampa Atlanta Dallas

18 Slide © 2009 South-Western, a part of Cengage Learning Example: Maximal Flow n Modified Network Model Sink Source Addedarc

19 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning 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 Slide © 2009 South-Western, a part of Cengage Learning n Alternative Optimal Solution #1 Example: Maximal Flow Objective Function Value = Variable Value x x x x x x x x x Variable Value x x x x x x x x x

23 Slide © 2009 South-Western, a part of Cengage Learning Example: Maximal Flow n Alternative Optimal Solution # Sink Source 10

24 Slide © 2009 South-Western, a part of Cengage Learning n Alternative Optimal Solution #2 Example: Maximal Flow Objective Function Value = Variable Value x x x x x x x x x Variable Value x x x x x x x x x

25 Slide © 2009 South-Western, a part of Cengage Learning Example: Maximal Flow n Alternative Optimal Solution # Sink Source 10

26 Slide © 2009 South-Western, a part of Cengage Learning A Production and Inventory Application n Transportation or transshipment models can be developed for applications that have nothing to do with the physical shipment of goods from origins to destinations. n We now show how to use a transshipment model to solve a production and inventory problem.

27 Slide © 2009 South-Western, a part of Cengage Learning Contour Carpets is a small manufacturer of carpeting for home and office installations. Production capacity, demand, production cost per square yard, and inventory holding cost per square yard for the next four quarters are shown below. Contour Carpets is a small manufacturer of carpeting for home and office installations. Production capacity, demand, production cost per square yard, and inventory holding cost per square yard for the next four quarters are shown below. Production Production Inventory Production Production Inventory Capacity Demand Cost Cost Capacity Demand Cost Cost Quarter (sq. yds.) (sq. yds.) ($/sq.yd.) ($/sq.yd.) Example: Production & Inventory Problem

28 Slide © 2009 South-Western, a part of Cengage Learning Note that production capacity, demand, and production costs vary by quarter, whereas the cost of carrying inventory from one quarter to the next is constant at $0.25 per yard. Contour wants to determine how many yards of carpeting to manufacture each quarter to minimize the total production and inventory cost for the four-quarter period. Note that production capacity, demand, and production costs vary by quarter, whereas the cost of carrying inventory from one quarter to the next is constant at $0.25 per yard. Contour wants to determine how many yards of carpeting to manufacture each quarter to minimize the total production and inventory cost for the four-quarter period. The objective is to determine a production scheduling and inventory policy that will minimize the total production and inventory cost for the four quarters. Constraints involve production capacity and demand in each quarter. The objective is to determine a production scheduling and inventory policy that will minimize the total production and inventory cost for the four quarters. Constraints involve production capacity and demand in each quarter. Example: Production & Inventory Problem

29 Slide © 2009 South-Western, a part of Cengage Learning Example: Production & Inventory Problem Let x 15 denote the number of square yards of carpet manufactured in quarter 1. The capacity of the facility is 600 square yards in quarter 1, so the production capacity constraint is: x 15 ≤ 300 x 15 ≤ 300 Using similar decision variables, we obtain the production capacities for quarters 2–4: x 26 ≤ 300 x 26 ≤ 300 x 37 ≤ 500 x 37 ≤ 500 x 48 ≤ 400 x 48 ≤ 400 n LP Formulation

30 Slide © 2009 South-Western, a part of Cengage Learning Example: Production & Inventory Problem In general, for each quarter the beginning inventory plus the production minus the ending inventory must equal demand. However, because quarter 1 has no beginning inventory, the constraint for node 5 is: x 15 – x 56 = 400 x 15 – x 56 = 400 The constraints associated with the demand nodes in quarters 2, 3, and 4 are: x 56 + x 26 – x 67 = 500 x 56 + x 26 – x 67 = 500 x 67 + x 37 – x 78 = 400 x 67 + x 37 – x 78 = 400 x 78 + x 48 = 400 x 78 + x 48 = 400 n LP Formulation

31 Slide © 2009 South-Western, a part of Cengage Learning Example: Production & Inventory Problem The objective is to minimize total production and inventory cost, so we write the objective function as: Min 2 x x x x x x x 78 n LP Formulation

32 Slide © 2009 South-Western, a part of Cengage Learning Contour Carpets should manufacture: 600 square yards of carpet in quarter square yards of carpet in quarter square yards in quarter 2, 300 square yards in quarter 2, 400 square yards in quarter 3, and 400 square yards in quarter 3, and 400 square yards in quarter square yards in quarter 4. Note also that 200 square yards will be carried over from quarter 1 to quarter 2. The total production and inventory cost is $5150. n LP Solution Example: Production & Inventory Problem

33 Slide © 2009 South-Western, a part of Cengage Learning End of Chapter 10, Part B