1. 1 1 BA 452 Lesson B.2 Transshipment and Shortest Route ReadingsReadings Chapter 6 Distribution and Network Models

2. 2 2 BA 452 Lesson B.2 Transshipment and Shortest Route OverviewOverview

3. 3 3 Overview Transshipment Problems are Transportation Problems extended so that a shipment may move through intermediate nodes (transshipment nodes) before reaching a particular destination node. Transshipment Problems with Transshipment Origins are Transshipment Problems where goods from one origin may move through other origins before reaching a destination. Shortest Route Problems are Transshipment Problems where there is one origin, one destination, one unit supply, and one unit demand, and where that unit is indivisible, as in driving through cities to work.

4. 4 4 BA 452 Lesson B.2 Transshipment and Shortest Route Tool Summary n Define decision variable x ij = units moving from origin i to destination j. n Write origin constraints (with < or =): n Write destination constraints (with < or =): n Write transshipment constraints (with < or =): Overview

5. 5 5 BA 452 Lesson B.2 Transshipment and Shortest Route Tool Summary n Identify implicit assumptions needed to complete a formulation, such as all agents having an equal value of time. Overview

6. 6 6 BA 452 Lesson B.2 Transshipment and Shortest Route TransshipmentTransshipment

7. 7 7 TransshipmentOverview Transshipment Problems are Transportation Problems extended so that a shipment may move through intermediate nodes (transshipment nodes) before reaching a particular destination node.

8. 8 8 BA 452 Lesson B.2 Transshipment and Shortest Route 2 2 33 44 55 66 7 7 1 1 c 13 c 14 c 23 c 24 c 25 c 15 s1s1s1s1 c 36 c 37 c 46 c 47 c 56 c 57 d1d1d1d1 d2d2d2d2 Intermediate Nodes Sources Destinations s2s2s2s2 Demand SupplyTransshipment This is the network representation for a transshipment problem with two sources, three intermediate nodes, and two destinations:

9. 9 9 BA 452 Lesson B.2 Transshipment and Shortest Route Notation: Notation: x ij = number of units shipped from node i to node j x ij = number of units shipped from node i to node j c ij = cost per unit of shipping from node i to node j c ij = cost per unit of shipping from node i to node j s i = supply at origin node i s i = supply at origin node i d j = demand at destination node j d j = demand at destination node j x ij > 0 for all i and j Transshipment

10. 10 BA 452 Lesson B.2 Transshipment and Shortest Route Problem Variations Minimum shipping guarantee from i to j: Minimum shipping guarantee from i to j: x ij > L ij x ij > L ij Maximum route capacity from i to j: Maximum route capacity from i to j: x ij < L ij x ij < L ij Unacceptable route: Unacceptable route: Remove the corresponding decision variable. Remove the corresponding decision variable.Transshipment

11. 11 BA 452 Lesson B.2 Transshipment and Shortest Route Question: The Northside and Southside facilities of Zeron Industries supply three firms (Zrox, Hewes, Rockrite) with customized shelving for its offices. They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc. Currently, weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockrite. Both Arnold and Supershelf can supply up to 75 units to its customers. Because of long-standing contracts based on past orders, unit costs from the manufacturers to the suppliers are: Zeron N Zeron S Zeron N Zeron S Arnold 5 8 Arnold 5 8 Supershelf 7 4 Supershelf 7 4 The costs to install the shelving at the various locations are: Zrox Hewes Rockrite Zrox Hewes Rockrite Zeron N 1 5 8 Zeron N 1 5 8 Zeron S 3 4 4 Zeron S 3 4 4Transshipment

12. 12 BA 452 Lesson B.2 Transshipment and Shortest Route Formulate and solve a transshipment linear programming problem for Zeron Industries. Transshipment

13. 13 BA 452 Lesson B.2 Transshipment and Shortest Route Answer: Since demands by the three customer firms (Zrox, Hewes, Rockrite) are fixed, revenue for Zeron Industries is fixed, and so profit maximization is the same as cost minimization. There is data on transportation costs, but there is no data on the cost of ordering from suppliers. Nevertheless, if the unit cost from each supplier is the same, say P per unit, then the cost of ordering supply equal to the fixed demand of 50, 60, and 40 is fixed at 150P. Hence, to minimize cost, we just have to minimize transportation cost. To that end, it may help to draw a network model: Transshipment

14. 14 BA 452 Lesson B.2 Transshipment and Shortest Route ARNOLD WASH BURN HEWES 75 75 50 60 40 5 8 7 4 1 5 8 3 4 4 Arnold SuperShelf Hewes Zrox ZeronN ZeronS Rock-RiteTransshipment

15. 15 BA 452 Lesson B.2 Transshipment and Shortest Route n Next, Define decision variables: x ij = amount shipped from manufacturer i to supplier j x jk = amount shipped from supplier j to customer k x jk = amount shipped from supplier j to customer k where i = 1 (Arnold), 2 (Supershelf) where i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) Problem Features: n There will be 1 variable for each manufacturer-supplier pair and each supplier- customer pair, so 10 variables all together. n There will be 1 constraint for each manufacturer, 1 for each supplier, and 1 for each customer, so 7 constraints all together. Transshipment

16. 16 BA 452 Lesson B.2 Transshipment and Shortest Route n Define objective function: Minimize total shipping costs. Min 5x 13 + 8x 14 + 7x 23 + 4x 24 + 1x 35 + 5x 36 + 8x 37 + 3x 45 + 4x 46 + 4x 47 Min 5x 13 + 8x 14 + 7x 23 + 4x 24 + 1x 35 + 5x 36 + 8x 37 + 3x 45 + 4x 46 + 4x 47 n Constrain amount out of Arnold: x 13 + x 14 < 75 n Constrain amount out of Supershelf: x 23 + x 24 < 75 n Constrain amount through Zeron N: x 13 + x 23 - x 35 - x 36 - x 37 = 0 n Constrain amount through Zeron S: x 14 + x 24 - x 45 - x 46 - x 47 = 0 n Constrain amount into Zrox: x 35 + x 45 = 50 n Constrain amount into Hewes: x 36 + x 46 = 60 n Constrain amount into Rockrite: x 37 + x 47 = 40 Transshipment

17. 17 BA 452 Lesson B.2 Transshipment and Shortest Route Indicies: i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) Out of Arnold through Zeron N Minimized shipping costs Out of Supershelf through Zeron S Through Zeron N into Zrox Through Zeron S into Hewes Transshipment

18. 18 BA 452 Lesson B.2 Transshipment and Shortest Route ARNOLD WASH BURN ZROX HEWES 75 75 50 60 40 5 8 7 4 1 5 8 3 4 4 Arnold SuperShelf Hewes Zrox ZeronN ZeronS Rock-Rite 75 75 50 25 35 40 Indicies: i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite)Transshipment

19. 19 BA 452 Lesson B.2 Transshipment and Shortest Route Transshipment with Transshipment Origins

20. 20 BA 452 Lesson B.2 Transshipment and Shortest Route Transshipment with Transshipment Origins Overview Transshipment Problems with Transshipment Origins seek to minimize the total shipping costs of transporting goods from m origins (each with a supply s i ) to n destinations (each with a demand d j ), where goods from one origin may move through other origins (transshipment nodes) before reaching a particular destination node.

21. 21 BA 452 Lesson B.2 Transshipment and Shortest Route Transshipment with Transshipment Origins Question: Index cities i = 1 (Newbury Park), i = 2 (Thousand Oaks), i = 3 (Westlake Hills), i = 4 (Agoura Hills), i = 5 (Calabasas). Suppose you run rental car lots in each city. Newbury Park has a surplus of 3 cars (it has 3 more cars than it needs), Westlake Hills has a surplus of 2 cars (it has 2 more cars than it needs), and Calabasas has a deficit of 5 cars (it needs 5 more cars than it has).

22. 22 BA 452 Lesson B.2 Transshipment and Shortest Route Transshipment with Transshipment Origins Suppose you calculate the following costs per car of transporting cars between the cities: transporting between 1 and 2 (that is, either 1 to 2, or 2 to 1) costs $2 transporting between 1 and 2 (that is, either 1 to 2, or 2 to 1) costs $2 transporting between 1 and 3 costs $3 transporting between 1 and 3 costs $3 transporting between 1 and 4 costs $4 transporting between 1 and 4 costs $4 transporting between 1 and 5 costs $5 transporting between 1 and 5 costs $5 transporting between 2 and 3 costs $2 transporting between 2 and 3 costs $2 transporting between 2 and 4 costs $3 transporting between 2 and 4 costs $3 transporting between 2 and 5 costs $4 transporting between 2 and 5 costs $4 transporting between 3 and 4 costs $2 transporting between 3 and 4 costs $2 transporting between 3 and 5 costs $3 transporting between 3 and 5 costs $3 transporting between 4 and 5 costs $2 transporting between 4 and 5 costs $2 How should you move cars between cities? Formulate your rental-car problem as a linear program, but you need not solve for the optimum. Tip: Your written answer should define the decision variables, and formulate the objective and constraints.

23. 23 BA 452 Lesson B.2 Transshipment and Shortest Route Transshipment with Transshipment Origins Answer: Define decision variables: x ij = amount of cars moved from City i to City j Define objective function: Minimize total costs. Min 2(x 12 +x 21 ) + 3(x 13 +x 31 ) + 4(x 14 +x 41 ) + 5(x 15 +x 51 ) + 2(x 23 +x 32 ) + 3(x 24 +x 42 ) + 4(x 25 +x 52 ) + 2(x 34 +x 43 ) + 3(x 35 +x 53 ) + 2(x 45 +x 54 ) + 3(x 24 +x 42 ) + 4(x 25 +x 52 ) + 2(x 34 +x 43 ) + 3(x 35 +x 53 ) + 2(x 45 +x 54 ) Constrain cars from City 1: x 12 + x 13 + x 14 + x 15 = 3 + x 21 + x 31 + x 41 + x 51 Constrain cars from City 2: x 21 + x 23 + x 24 + x 25 = x 12 + x 32 + x 42 + x 52 Constrain cars from City 3: x 31 + x 32 + x 34 + x 35 = 2 + x 13 + x 23 + x 43 + x 53 Constrain cars from City 4: x 41 + x 42 + x 43 + x 45 = x 14 + x 24 + x 34 + x 54 Constrain cars from City 5: x 51 + x 52 + x 53 + x 54 = -5 + x 15 + x 25 + x 35 + x 45 (Or constraints can be written with < rather than =. It does not matter since excess supply exactly cancels excess demand.)

24. 24 BA 452 Lesson B.2 Transshipment and Shortest Route Shortest Route

25. 25 BA 452 Lesson B.2 Transshipment and Shortest Route Shortest Route Overview Shortest Route Problems are Transshipment Problems where there is one origin, one destination, one unit supplied, and one unit demanded, and where that unit is indivisible. Shortest Route Problems find the shortest path in a network from one node (or set of nodes) to another node (or set of nodes). The criterion to be minimized in the shortest-route problem is not limited to distance, but can be minimum time or cost.

26. 26 BA 452 Lesson B.2 Transshipment and Shortest Route Notation: 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 x ij > 1 Shortest Route

27. 27 BA 452 Lesson B.2 Transshipment and Shortest Route Question: 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 should she take to minimize the total travel cost? Shortest Route

28. 28 BA 452 Lesson B.2 Transshipment and Shortest Route 6 A B C D E F G H I J K L M Paducah Lewisburg 1 2 5 3 4 Shortest Route

29. 29 BA 452 Lesson B.2 Transshipment and Shortest Route Route Transport Time Ticket (Arc) Mode (hours) Cost (Arc) 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 1/3 $ 30 E Train 3 1/3 $ 30 F Bus 3 $ 15 F Bus 3 $ 15 G Bus 4 2/3 $ 20 G Bus 4 2/3 $ 20 H Taxi 1 $ 15 H Taxi 1 $ 15 I Train 2 1/3 $ 15 I Train 2 1/3 $ 15 J Bus 6 1/3 $ 25 J Bus 6 1/3 $ 25 K Taxi 3 1/3 $ 50 K Taxi 3 1/3 $ 50 L Train 1 1/3 $ 10 L Train 1 1/3 $ 10 M Bus 4 2/3 $ 20 M Bus 4 2/3 $ 20 Shortest Route

30. 30 BA 452 Lesson B.2 Transshipment and 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 1/3 $50 $ 30$ 80 E Train 3 1/3 $50 $ 30$ 80 F Bus 3 $45 $ 15$ 60 F Bus 3 $45 $ 15$ 60 G Bus 4 2/3 $70 $ 20$ 90 G Bus 4 2/3 $70 $ 20$ 90 H Taxi 1 $15 $ 15$ 30 H Taxi 1 $15 $ 15$ 30 I Train 2 1/3 $35 $ 15$ 50 I Train 2 1/3 $35 $ 15$ 50 J Bus 6 1/3 $95 $ 25$120 J Bus 6 1/3 $95 $ 25$120 K Taxi 3 1/3 $50 $ 50$100 K Taxi 3 1/3 $50 $ 50$100 L Train 1 1/3 $20 $ 10$ 30 L Train 1 1/3 $20 $ 10$ 30 M Bus 4 2/3 $70 $ 20$ 90 M Bus 4 2/3 $70 $ 20$ 90 Using the wage of $15 per hour, compute total cost. Shortest Route

31. 31 BA 452 Lesson B.2 Transshipment and Shortest Route n Define indices: Nodes 1 (origin), 2, …, 6 (destination) n Define decision variables: x ij = 1 if the route from node i to node j is on the shortest route n Define objective function: Minimize total transportation costs. Min 80x 12 + 40x 13 + 80x 14 + 130x 15 + 180x 16 + 60x 25 + 100x 26 + 30x 34 + 90x 35 + 120x 36 + 30x 43 + 50x 45 + 90x 46 + 60x 52 + 90x 53 + 50x 54 + 30x 56 Problem Features: n There is 1 decision variable for each possible route. n There is 1 constraint for each node. Cost of route from 1 to 4 Shortest Route

32. 32 BA 452 Lesson B.2 Transshipment and Shortest Route n 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) Shortest Route

33. 33 BA 452 Lesson B.2 Transshipment and Shortest Route Variables: Variables: 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 Possible arcs: x 12, x 13, x 14, x 15, x 16, x 25, x 26, x 34, Possible arcs: x 12, x 13, x 14, x 15, x 16, x 25, x 26, x 34, x 35, x 36, x 43, x 45, x 46, x 52, x 53, x 54, x 56 x 35, x 36, x 43, x 45, x 46, x 52, x 53, x 54, x 56 From Origin to Node 3 From Node 3 to Node 4 Minimized transportation cost From Node 4 to Node 5 From Node 5 to Destination Shortest Route

34. 34 BA 452 Lesson B.2 Transshipment and Shortest Route Shortest Route

35. 35 BA 452 Lesson B.2 Transshipment and Shortest Route Cost of Arc from Node 1 to Node 3 Input window Output window Minimized transportation costs Shortest Route

36. 36 BA 452 Lesson B.2 Transshipment and Shortest Route End of Lesson B.2 BA 452 Quantitative Analysis