1. 1 Lecture 2 Shortest-Path Problems Assignment Problems Transportation Problems

2. 2 Agenda  shortest-path problems  assignment problems  transportation problems  transshipment problems

3. 3 Yet Another Simple Concrete Numerical Example  obvious shortest path between node 1 and node 4  But how to formulate? What is the direction of flow in the middle arc, upward or downward? Or any flow at all? 1 2 5 6 3 1 2 4 3 min length of path s.t. a set of constraints representing a path

4. 4 Yet Another Simple Concrete Numerical Example  a route from the source to the sink = a collection of arcs from the source to the sink  some restriction on the choice of arcs in to form a path  question: How to define the values of a group of x ij such that x ij s form a route from the source to the sink? 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3

5. 5 Yet Another Simple Concrete Numerical Example  properties of a route from the source to the sink:  each route represented by a collection of x ij = 1, with the other off-route variables = 0  source: one arc out  sink: one arc in  intermediate node: one arc in and one arc out 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3

6. 6 Formulation  min 3x 12 + 6x 13 + 2(x 23 +x 32 ) + 5x 24 + x 34, s.t.s.t. x 12 + x 13 = 1(node 1, source) x 12 + x 32 = x 23 + x 24 (node 2) x 13 + x 23 = x 32 + x 34 (node 3) x 24 + x 34 = 1(node 4, sink) x 12, x 13, x 32, x 24, x 34  {0, 1} 1 2 5 6 3 1 2 4 3 選最 node 1 到 node 4 最短徑，就像將一單位 （ 1 unit ）的水從 node 1 送到 node 4 ，水流 過的 arcs 就是在路徑上的 arcs.

7. 7 Formulation - Question  min 3x 12 + 6x 13 + 2(x 23 +x 32 ) + 5x 24 + x 34, s.t.s.t. x 12 + x 13 = 1(node 1, source) x 12 + x 32 = x 23 + x 24 (node 2) x 13 + x 23 = x 32 + x 34 (node 3) x 24 + x 34 = 1(node 4, sink) x 12, x 13, x 32, x 24, x 34  {0, 1}  Would both x 23 and x 32 be positive at minimum? 1 2 5 6 3 1 2 4 3

8. 8 Formulation  tricks in our formulation  only outflow from source node  only inflow to destination node  variables x 21, x 31, x 42, and x 43 set to zero in the textbook  min 3(x 12 +x 21 ) + 6(x 13 +x 31 ) + 2(x 23 +x 32 ) + 5(x 24 +x 42 ) + (x 34 +x 43 ), s.t.s.t. x 12 + x 13 = 1 (node 1, source) x 12 + x 32 + x 42 = x 21 + x 23 +x 24 (node 2) x 13 + x 23 + x 43 = x 31 + x 32 + x 34 (node 3) x 24 + x 34 = 1 (node 4, destination) x 12, x 13, x 32, x 24, x 34  {0, 1}; x 21 = x 31 = x 42 = x 43 = 0 1 2 5 6 3 1 2 4 3

9. 9 Formulation - Question  Would the following formulation be all right?  min 3(x 12 +x 21 ) + 6(x 13 +x 31 ) + 2(x 23 +x 32 ) + 5(x 24 +x 42 ) + (x 34 +x 43 ), s.t.s.t. x 12 + x 13 = 1 (node 1, source) x 12 + x 32 + x 42 = x 21 + x 23 +x 24 (node 2) x 13 + x 23 + x 43 = x 31 + x 32 + x 34 (node 3) x 24 + x 34 = 1 (node 4, destination) x 12, x 13, x 32, x 24, x 34, x 21, x 31, x 42, x 43  {0, 1} 1 2 5 6 3 1 2 4 3

10. 10 Formulation  the formulation for the shortest distance between node 1 and node 6:  Min z = 5(x 12 +x 21 )+8(x 13 +x 31 )+7(x 14 +x 41 )+6(x 23 +x 32 )+1(x 34 +x 43 )+13(x 25 +x 52 ) +7(x 35 +x 53 )+16(x 46 +x 64 )+6(x 56 +x 65 ) s.t. (i.e., subject to): x 12 +x 13 +x 14 = 1 (source node) x 12 +x 32 +x 52 = x 21 +x 23 +x 25 (node 2) x 13 +x 23 +x 43 +x 53 = x 31 +x 32 +x 34 +x 35 (node 3) x 14 +x 34 +x 64 = x 41 +x 43 +x 46 (node 4) x 25 +x 35 +x 65 = x 52 +x 53 +x 56 (node 5) x 56 +x 46 = 1 (node 6) x ij = 0 or 1 for all i, j combinations Actually it is possible to drop the red variables.

11. 11 Assignment Problems

12. 12 An Assignment Problem  4 jobs, J1 to J4, are assigned to 4 machines, M1 to M4, such that each machine can handle exactly 1 job  implicitly each job is assigned only to one machine  the costs of assigning machine i to job j are shown in the RHS matrix, where M denotes a feasible but expensive assignment  find the assignment that minimizes the total assignment cost J1J2J3J4 M19M138 M214127M M3M5M6 M4107611

13. 13 An Assignment Problem  let x ij = 1 if machine i takes up job j, and = 0, otherwise, i = 1, 2, 3, 4, j = 1, 2, 3, 4 J1J2J3J4 M19M138 M214127M M3M5M6 M4107611 (each machine has one job) (each job is assigned to one machine)

14. 14 Comments  an assignment problem is easy to solve  if there are any additional constraints other than those shown in a standard assignment model, the resulted model is no longer an assignment problem, which may be difficult to solve

15. 15 Exercise 1  Can we omit variables x ij if the corresponding c ij = M, where M is a large positive number practically representing infinity?  physical meaning: setting such x ij to zero J1J2J3J4 M19M138 M214127M M3M5M6 M4107611

16. 16 Exercise 2  Formulate a model for the 4-machine, 4- job problem such that each machine can take exact one job and jobs J1 and J2 must not be assigned to M4. J1J2J3J4 M19M138 M214127M M3M5M6 M4107611

17. 17 Exercise 2  Formulate a model for the 4-machine, 4- job problem such that each machine can take exact one job and jobs J1 and J2 must not be assigned to M4. J1J2J3J4 M19M138 M214127M M3M5M6 M4107611

18. 18 Exercise 3  Suppose that there are 5 jobs and 4 machines. Formulate an assignment model that finds the cheapest way to assign 4 out of the 5 jobs to the 4 machines, with each machine taking exactly one job. J1J2J3J4J5J5 M19413821 M21412793 M3755614 M410761117

19. 19 Exercise 4  Suppose that there are 5 jobs and 4 machines. Formulate a model that finds the cheapest way to assign all 5 jobs to the 4 machines such that only one of the 4 machines takes 2 jobs. J1J2J3J4J5J5 M19413821 M21412793 M3755614 M410761117

20. 20 Exercise 5  Suppose that there are 5 jobs and 4 machines. Formulate a model that finds the cheapest way to assign all 5 jobs to the 4 machines such that M2 takes 2 jobs and the other machines take one job. J1J2J3J4J5J5 M19413821 M21412793 M3755614 M410761117

21. 21 Exercise 6  Suppose that there are 5 jobs and 4 machines. Formulate a model that finds the cheapest way to assign all 5 jobs to the 4 machines such that each machine can take at most 2 jobs. J1J2J3J4J5J5 M19413821 M21412793 M3755614 M410761117

22. 22 Exercise 7  Formulate a model that finds the shortest paths from nodes 1 and 2 to node 6 4 5 7 8 9 11 1 2 4 3 5 6 6 3 14

23. 23 Transportation Problems

24. 24 Transportation Problem  two suppliers, A and B  60 units from A  40 units from B  three customers, C, D, and E  20 units for C  45 units for D  25 units for E  unit cost c ij for unit flow from supplier i to customer j as labeled on diagram, e.g., $5 (= c AD ) for each unit from A to D  objective: minimize the total cost to satisfy the requirement 6 1 2 5 8 3 A B D C E 60 40 25 45 20

25. 25 Transportation Problem  steps of modeling  define variables  express the objective function in terms of the variables and parameters  express the constraints in terms of the variables and parameters 6 1 2 5 8 3 A B D C E 60 40 25 45 20

26. 26 Transportation Problem  one trick to define variables: what decisions are made?  answer: the amount of goods sent from supplier i to customer j  let x ij = amount of goods sent from supplier i to customer j, i = A, B; j = C, D, F 6 1 2 5 8 3 A B D C E 60 40 25 45 20

27. 27 Transportation Problem  objective  min 3x AC + 5x AD + 2x AE + x BC + 8x AD + 6x AE  constraints  supply from A: x AC + x AD + x AE ≤ 60  supply from B: x BC + x BD + x BE ≤ 40  demand of C: x AC + x BC = 20  demand of D: x AD + x BD = 45  demand of E: x AE + x BE = 25  non-negativity: x AC, x AD, x AE, x BC, x AD, x AE  0 6 1 2 5 8 3 A B D C E 60 40 25 45 20

28. 28 Questions  is the formulation correct?  can we use x AC + x BC  20, x AD + x BD  45, x AE + x BE  25 instead?  what does it mean by can or cannot? 6 1 2 5 8 3 A B D C E 60 40 25 45 20 min 3x AC + 5x AD + 2x AE + x BC + 8x AD + 6x AE, s.t. x AC + x AD + x AE ≤ 60, x BC + x BD + x BE ≤ 40, x AC + x BC = 20, x AD + x BD = 45, x AE + x BE = 25, x AC, x AD, x AE, x BC, x AD, x AE  0.

29. 29 Questions  can we use 6 1 2 5 8 3 A B D C E 60 40 25 45 20 min 3x AC + 5x AD + 2x AE + x BC + 8x AD + 6x AE, s.t. x AC + x AD + x AE = 60, x BC + x BD + x BE = 40, x AC + x BC  20, x AD + x BD  45, x AE + x BE  25, x AC, x AD, x AE, x BC, x AD, x AE  0. min 3x AC + 5x AD + 2x AE + x BC + 8x AD + 6x AE, s.t. x AC + x AD + x AE = 60, x BC + x BD + x BE = 40, x AC + x BC = 20, x AD + x BD = 45, x AE + x BE = 25, x AC, x AD, x AE, x BC, x AD, x AE  0.

30. 30 Questions  how to model: impossible to send goods from A to E  setting c AE to   numerically c AE = M, a large positive number 6 1 M 5 8 3 A B D C E 60 40 25 45 20 min 3x AC + 5x AD + Mx AE + x BC + 8x AD + 6x AE, s.t. x AC + x AD + x AE ≤ 60, x BC + x BD + x BE ≤ 40, x AC + x BC = 20, x AD + x BD = 45, x AE + x BE = 25, x AC, x AD, x AE, x BC, x AD, x AE  0. 6 1 5 8 3 A B D C E 60 40 25 45 20

31. 31 Questions  how to model: impossible to send goods from A to E  omitting x AE (i.e., setting to x AE = 0) min 3x AC + 5x AD + x BC + 8x AD + 6x AE, s.t. x AC + x AD ≤ 60, x BC + x BD + x BE ≤ 40, x AC + x BC = 20, x AD + x BD = 45, x BE = 25, x AC, x AD, x AE, x BC, x AD, x AE  0. 6 1 5 8 3 A B D C E 60 40 25 45 20

32. 32 Balanced Transportation Problems  a balanced transportation problem: total supply = total demand  computational advantages for a balanced transportation problem  possible to transform any imbalanced transportation problem to a balanced one

33.  total supply = 100  90 = total demand  adding a dummy customer T  what is b T ?  what is c AT ?  what is c BT ? 33 To Transform into a Balanced Transportation Problem 6 1 2 5 8 3 A B D C E 60 40 25 45 20 c AT c BT 6 1 2 5 8 3 A B D C E 60 40 25 45 20 T bTbT

34.  b T = 100  90 = 10  can c AT = 0 and c BT = 0?  what does it mean by can or cannot?  can c AT = 19 and c BT = 19?  can c AT = 19 and c BT = 31?  the same trick for total supply  total demand 34 To Transform into a Balanced Transportation Problem c AT c BT 6 1 2 5 8 3 A B D C E 60 40 25 45 20 T bTbT

35. 35 Comments 0 0 6 1 2 5 8 3 A B D C E 60 40 25 45 20 T 10 min 3x AC + 5x AD + 2x AE + x BC + 8x AD + 6x AE, s.t. x AC + x AD + x AE + x AT = 60, x BC + x BD + x BE + x BT = 40, x AC + x BC = 20, x AD + x BD = 45, x AE + x BE = 25, x AT + x BT = 10, x AC, x AD, x AE, x BC, x AD, x AE, x AT, x BT  0.

36. 36 A General Transportation Problem  m suppliers  n customers  quantity supplied by supplier i, a i  quantity demanded by customer j, b j  cost of unit flow from supplier i to customer j, c ij  objective: minimize the total shipment cost

37. 37 Special Structure for the Set of Constraints  very special property: as long as all a i and b j are integral, the minimum solution is integral  no need to use integer programming  x AC  x AD  x AE  x AT =  60,  x BC  x BD  x BE  x BT =  40, x AC + x BC = 20, x AD + x BD = 45, x AE + x BE = 25, x AT + x BT = 10, x AC, x AD, x AE, x BC, x AD, x AE, x AT, x BT  0. c AT c BT 6 1 2 5 8 3 A B D C E 60 40 25 45 20 T bTbT

38. 38 Solution of a Transportation Problem  at least three approaches  linear programming, Simplex method by CPLEX, Gurobi  special transportation Simplex by Dantzig (1951) (for balanced transportation problems)  basically Simplex, but specialized steps because of the special structure of the problem  network flow algorithm, Ford and Fulkerson (1956, 1962) min 3x AC + 5x AD + 2x AE + x BC + 8x AD + 6x AE, s.t. x AC + x AD + x AE + x AT = 60, x BC + x BD + x BE + x BT = 40, x AC + x BC = 20, x AD + x BD = 45, x AE + x BE = 25, x AT + x BT = 10, x AC, x AD, x AE, x BC, x AD, x AE, x AT, x BT  0.

39. 39 Example 5.3 of [7]  transportation problems occurring in many contexts, not necessarily transportation  production capacities and demands in the next four months  unit cost of production at regular time = $1  unit cost of production at overtime = $1.5  storage cost = $0.3 per month per unit  how to produce to satisfy the demands in the four months with minimum total cost JanFebMarchApril Capacity in Regular Time (pcs)100150140160 Capacity in Overtime (pcs)50757080 Demand (pcs)80200300200

40. 40 Example 5.3 of [7]  possible to model as a transportation problem JanFebMarchApril JanRegular11.31.61.9 Overtime1.51.82.12.4 FebRegular  11.31.6 Overtime  1.51.82.1 MarchRegular  11.3 Overtime  1.51.8 AprilRegular  1 Overtime  1.5

41. 41 Capacitated Transportation Problems  lower and upper bounds for x ij  0 ≤ l ij ≤ x ij ≤ u ij  solved by extensions of the specialized algorithms