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Transportation-1 Operations Research Modeling Toolset Linear Programming Network Programming PERT/ CPM Dynamic Programming Integer Programming Nonlinear.

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Presentation on theme: "Transportation-1 Operations Research Modeling Toolset Linear Programming Network Programming PERT/ CPM Dynamic Programming Integer Programming Nonlinear."— Presentation transcript:

1 Transportation-1 Operations Research Modeling Toolset Linear Programming Network Programming PERT/ CPM Dynamic Programming Integer Programming Nonlinear Programming Game Theory Decision Analysis Markov Chains Queueing Theory Inventory Theory Forecasting Markov Decision Processes Simulation Stochastic Programming

2 Transportation-2 Network Problems Linear programming has a wide variety of applications Network problems –Special types of linear programs –Particular structure involving networks Ultimately, a network problem can be represented as a linear programming model However the resulting A matrix is very sparse, and involves only zeroes and ones This structure of the A matrix led to the development of specialized algorithms to solve network problems

3 Transportation-3 Types of Network Problems Shortest Path Special case: Project Management with PERT/CPM Minimum Spanning Tree Maximum Flow/Minimum Cut Minimum Cost Flow Special case: Transportation and Assignment Problems Set Covering/Partitioning Traveling Salesperson Facility Location and many more

4 Transportation-4 The Transportation Problem

5 Transportation-5 The Transportation Problem The problem of finding the minimum-cost distribution of a given commodity from a group of supply centers (sources) i=1,…,m to a group of receiving centers (destinations) j=1,…,n Each source has a certain supply (s i ) Each destination has a certain demand (d j ) The cost of shipping from a source to a destination is directly proportional to the number of units shipped

6 Transportation-6 Simple Network Representation 1 2 m 1 2 n SourcesDestinations …… Supply s 1 Supply s 2 Supply s m Demand d 1 Demand d 2 Demand d n x ij Costs c ij

7 Transportation-7 Example: P&T Co. Produces canned peas at three canneries Bellingham, WA, Eugene, OR, and Albert Lea, MN Ships by truck to four warehouses Sacramento, CA, Salt Lake City, UT, Rapid City, SD, and Albuquerque, NM Estimates of shipping costs, production capacities and demands for the upcoming season is given The management needs to make a plan on the least costly shipments to meet demand

8 Transportation-8 Example: P&T Co. Map 1 2 3 1 2 3 4

9 Transportation-9 Example: P&T Co. Data Warehouse Cannery1234 Supply (Truckloads) 1$ 464$ 513$ 654$ 86775 2$ 352$ 216$ 690$ 791125 3$ 995$ 682$ 388$ 685100 Demand (Truckloads)80657085 Shipping cost per truckload

10 Transportation-10 Example: P&T Co. Network representation

11 Transportation-11 Example: P&T Co. Linear programming formulation Let x ij denote… Minimize subject to

12 Transportation-12 General LP Formulation for Transportation Problems

13 Transportation-13 Feasible Solutions A transportation problem will have feasible solutions if and only if How to deal with cases when the equation doesn’t hold?

14 Transportation-14 Integer Solutions Property: Unimodularity Unimodularity relates to the properties of the A matrix (determinants of the submatrices, beyond scope) Transportation problems are unimodular, so we get the integers solutions property: For transportation problems, when every s i and d j have an integer value, every BFS is integer valued. Most network problems also have this property.

15 Transportation-15 Transportation Simplex Method Since any transportation problem can be formulated as an LP, we can use the simplex method to find an optimal solution Because of the special structure of a transportation LP, the iterations of the simplex method have a very special form The transportation simplex method is nothing but the original simplex method, but it streamlines the iterations given this special form

16 Transportation-16 Transportation Simplex Method Initialization (Find initial CPF solution) Is the current CPF solution optimal? Move to a better adjacent CPF solution Stop No Yes

17 Transportation-17 The Transportation Simplex Tableau Destination Supplyuiui Source12…n 1 c 11 c 12 … c 1n s1s1 2 c 21 c 22 … c 2n s2s2 ……………… m c m1 c m2 … c mn smsm Demandd1d1 d2d2 …dndn Z = vjvj

18 Transportation-18 Prototype Problem Holiday shipments of iPods to distribution centers Production at 3 facilities, –A, supply 200k –B, supply 350k –C, supply 150k Distribute to 4 centers, –N, demand 100k –S, demand 140k –E, demand 300k –W, demand 250k Total demand vs. total supply

19 Transportation-19 Prototype Problem Destination Supplyuiui SourceNSEW A 16132217 200 B 14131915 350 C 9202310 150 Dummy 0000 90 Demand100140300250 Z = vjvj

20 Transportation-20 Finding an Initial BFS The transportation simplex starts with an initial basic feasible solution (as does regular simplex) There are alternative ways to find an initial BFS, most common are –The Northwest corner rule –Vogel’s method –Russell’s method (beyond scope)

21 Transportation-21 The Northwest Corner Rule Begin by selecting x 11, let x 11 = min{ s 1, d 1 } Thereafter, if x ij was the last basic variable selected, –Select x i(j+1) if source i has any supply left –Otherwise, select x (i+1)j

22 Transportation-22 The Northwest Corner Rule Destination Supply SourceNSEW A 16132217 200 100 B 14131915 350 4030010 C 9202310 150 Dummy 0000 90 Demand100140300250 Z = 10770

23 Transportation-23 Vogel’s Method For each row and column, calculate its difference: = (Second smallest c ij in row/col) - (Smallest c ij in row/col) For the row/col with the largest difference, select entry with minimum c ij as basic Eliminate any row/col with no supply/demand left from further steps Repeat until BFS found

24 Transportation-24 Vogel’s Method (1): calculate differences Destination Supplydiff SourceNSEW A 16132217 2003 B 14131915 3501 C 9202310 1501 Dummy 0000 900 Demand100140300250 diff9131910

25 Transportation-25 Vogel’s Method (2): select x DummyE as basic variable Destination Supplydiff SourceNSEW A 16132217 2003 B 14131915 3501 C 9202310 1501 Dummy 0000 900 Demand100140300250 diff9131910 90

26 Transportation-26 Vogel’s Method (3): update supply, demand and differences Destination Supplydiff SourceNSEW A 16132217 2003 B 14131915 3501 C 9202310 1501 Dummy 0000 --- Demand100140210250 diff5035 90

27 Transportation-27 Vogel’s Method (4): select x CN as basic variable Destination Supplydiff SourceNSEW A 16132217 2003 B 14131915 3501 C 9202310 1501 Dummy 0000 --- Demand100140210250 diff5035 90 100

28 Transportation-28 Vogel’s Method (5): update supply, demand and differences Destination Supplydiff SourceNSEW A 16132217 2004 B 14131915 3502 C 9202310 5010 Dummy 0000 --- Demand---140210250 diff---035 90 100

29 Transportation-29 Vogel’s Method (6): select x CW as basic variable Destination Supplydiff SourceNSEW A 16132217 2004 B 14131915 3502 C 9202310 5010 Dummy 0000 --- Demand---140210250 diff---035 90 100 50

30 Transportation-30 Vogel’s Method (7): update supply, demand and differences Destination Supplydiff SourceNSEW A 16132217 2004 B 14131915 3502 C 9202310 --- Dummy 0000 --- Demand---140210200 diff---032 90 100 50

31 Transportation-31 Vogel’s Method (8): select x AS as basic variable Destination Supplydiff SourceNSEW A 16132217 2004 B 14131915 3502 C 9202310 --- Dummy 0000 --- Demand---140210200 diff---032 90 100 50 140

32 Transportation-32 Vogel’s Method (9): update supply, demand and differences Destination Supplydiff SourceNSEW A 16132217 605 B 14131915 3504 C 9202310 --- Dummy 0000 --- Demand--- 210200 diff--- 32 90 100 50 140

33 Transportation-33 Vogel’s Method (10): select x AW as basic variable Destination Supplydiff SourceNSEW A 16132217 605 B 14131915 3504 C 9202310 --- Dummy 0000 --- Demand--- 210200 diff--- 32 90 100 50 14060

34 Transportation-34 Vogel’s Method (11): update supply, demand and differences Destination Supplydiff SourceNSEW A 16132217 --- B 14131915 3504 C 9202310 --- Dummy 0000 --- Demand--- 210140 diff--- 90 100 50 14060

35 Transportation-35 Vogel’s Method (12): select x BW and x BE as basic variables Destination Supplydiff SourceNSEW A 16132217 --- B 14131915 --- C 9202310 --- Dummy 0000 --- Demand--- diff--- 90 100 50 14060 140 210 Z = 10330

36 Transportation-36 Optimality Test In the regular simplex method, we needed to check the row-0 coefficients of each nonbasic variable to check optimality and we have an optimal solution if all are  0 There is an efficient way to find these row-0 coefficients for a given BFS to a transportation problem: –Given the basic variables, calculate values of dual variables u i associated with each source v j associated with each destination using c ij – u i – v j = 0 for x ij basic, or u i + v j = c ij (let u i = 0 for row i with the largest number of basic variables) –Row-0 coefficients can be found from c ’ ij =c ij -u i -v j for x ij nonbasic

37 Transportation-37 Optimality Test (1) Destination Supplyuiui SourceNSEW A 16132217 200 B 14131915 350 C 9202310 150 Dummy 0000 90 Demand100140300250 vjvj 90 140 100 60 140210 50

38 Transportation-38 Optimality Test (2) Calculate u i, v j using c ij – u i – v j = 0 for x ij basic (let u i = 0 for row i with the largest number of basic variables) Destination Supplyuiui SourceNSEW A 16132217 2000 B 14131915 350-2 C 9202310 150-7 Dummy 0000 90-21 Demand100140300250 vjvj 16132117 90 140 100 60 140 210 50

39 Transportation-39 Optimality Test (3) Calculate c ’ ij =c ij -u i -v j for x ij nonbasic Destination Supplyuiui SourceNSEW A 16132217 2000 01 B 14131915 350-2 02 C 9202310 150-7 149 Dummy 0000 90-21 584 Demand100140300250 vjvj 16132117 90 140 100 60 140 210 50

40 Transportation-40 Optimal Solution A B C N S W SourcesDestinations Supply = 200 Supply = 350 Supply = 150 Demand = 100 Demand = 140 Demand = 250 E Demand = 300 (shortage of 90) 60 140 210 140 50 100 Cost Z = 10330

41 Transportation-41 An Iteration Find the entering basic variable –Select the variable with the largest negative c ’ ij Find the leaving basic variable –Determine the chain reaction that would result from increasing the value of the entering variable from zero –The leaving variable will be the first variable to reach zero because of this chain reaction

42 Transportation-42 Destination Supplyuiui SourceNSEW A 16132217 2000 32 B 14131915 3500 - 2 C 9202310 150-5 - 2129 Dummy 0000 90-15 - 12- 4 Demand100140300250 vjvj 16131915 Initial Solution Obtained by the Northwest Corner Rule 100 150 10 300 40 100 90

43 Transportation-43 Destination Supplyuiui SourceNSEW A 16132217 2000 B 14131915 3500 C 9202310 150-5 Dummy 0000 90-15 Demand100140300250 vjvj 16131915 Iteration 1 100 150 10 300 40 100 90 ? + - + -

44 Transportation-44 End of Iteration 1 Destination Supplyuiui SourceNSEW A 16132217 200 B 14131915 350 C 9202310 150 Dummy 0000 90 Demand100140300250 vjvj 100 150 100 210 40 100 90

45 Transportation-45 Optimality Test Destination Supplyuiui SourceNSEW A 16132217 2000 32 B 14131915 3500 - 2 C 9202310 150-5 - 2129 Dummy 0000 90-19 364 Demand100140300250 vjvj 16131915 100 150 100 210 40 100 90

46 Transportation-46 Iteration 2 Destination Supplyuiui SourceNSEW A 16132217 2000 B 14131915 3500 C 9202310 150-5 Dummy 0000 90-19 Demand100140300250 vjvj 16131915 100 150 100 210 40 100 90 ? + - + -

47 Transportation-47 End of Iteration 2 Destination Supplyuiui SourceNSEW A 16132217 200 B 14131915 350 C 9202310 150 Dummy 0000 90 Demand100140300250 vjvj 60 150 100 210 140 90 40

48 Transportation-48 Optimality Test Destination Supplyuiui SourceNSEW A 16132217 2002 10 B 14131915 3500 2 C 9202310 150-5 0149 Dummy 0000 90-19 584 Demand100140300250 vjvj 14111915 60 150 100 210 140 90 40 Z = 10330

49 Transportation-49 Optimal Solution A B C N S W SourcesDestinations Supply = 200 Supply = 350 Supply = 150 E 60 140 40 100 150 210 Demand = 100 Demand = 140 Demand = 250 Demand = 300 (shortage of 90) Cost Z = 10330

50 Transportation-50 The Assignment Problem The problem of finding the minimum-costly assignment of a set of tasks (i=1,…,m) to a set of agents (j=1,…,n) Each task should be performed by one agent Each agent should perform one task A cost c ij associated with each assignment We should have m=n (if not…?) A special type of linear programming problem, and A special type of transportation problem, with s i =d j = ?

51 Transportation-51 Prototype Problem Assign students to mentors Each assignment has a ‘mismatch’ index Minimize mismatches Mentor Supply StudentSnapeMcGonagallLupin Harry 523 1 Draco 145 1 Goyle 244 1 Demand111

52 Transportation-52 Prototype Problem Linear programming formulation Let x ij denote… Minimize subject to

53 Transportation-53 General LP Formulation for Assignment Problems

54 Transportation-54 Solving the Assignment Problem It is a linear programming problem, so we could use regular simplex method It is a transportation problem, so we could use transportation simplex method However, it has a very special structure, such that it can be solved in polynomial time Many such algorithms exist, but the best known (and one of the oldest) is the Hungarian Method

55 Transportation-55 The Hungarian Method 1.Subtract row minimums from each element in the row 2.Subtract column minimums from each element in the column 3.Cover the zeroes with as few lines as possible 4.If the number of lines = n, then optimal solution is hidden in zeroes 5.Otherwise, find the minimum cost that is not covered by any lines 1.Subtract it from all uncovered elements 2.Add it to all elements at intersections (covered by two lines) 6.Back to step 3

56 Transportation-56 The Hungarian Method – Optimal Solution How to identify the optimal solution: Make the assignments one at a time in positions that have zero elements. Begin with rows or columns that have only one zero. Cross out both the row and the column involved after each assignment is made. Move on to the rows and columns that are not yet crossed out to select the next assignment, with preference given to any such row or column that has only one zero that is not crossed out. Continue until every row and every column has exactly one assignment and so has been crossed out.

57 Transportation-57 Hungarian Method Mentor StudentSnapeMcGLupin Harry 523 Draco 145 Goyle 244 Mentor StudentSnapeMcGLupin Harry Draco Goyle Mentor StudentSnapeMcGLupin Harry Draco Goyle Mentor StudentSnapeMcGLupin Harry Draco Goyle


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