1. Math-based Decision Making in Logistics and Operations

2. Where we are? The groupIntroductionDescriptionResultsConclusion

3. Math-based Decision Making in Logistics and Operations MbDMLO Grupo Inv INGOR Grupo Inv OS UD Organización de la ProducciónUD Administración EmpresasUD EconomíaUD ProyectosUD Estadística Dep. Ingeniería de Organización, Adm. de Empresas y Estadística ETSII UPM Lab IOL Math-based Decision Making in Logistic and Operations The groupIntroductionDescriptionResultsConclusion

4. The team Miguel Ortega Mier Álvaro García Sánchez Javier Diego Daniel Herrero Natalia Ibáñez Raúl Pulido Jing Shao Tamara Borreguero Others. The groupIntroductionDescriptionResultsConclusion

5. Techniques and technologies Optimization Modeling Solver SimulationOther Programming and other specific software The groupIntroductionDescriptionResultsConclusion

6. The inventory routing problem for the Mixed Car Model Assembly Line 6 The groupIntroductionDescriptionResultsConclusion

7. Relevance A high inventory level in the assembly line is a big cost contributor. Some of the car manufacturer’s objectives are keeping low stock levels, performing the replenishment of the production line, and providing the required components at the right time (Monden, 1983). 7 The groupIntroductionDescriptionResultsConclusion

8. Abstract A car assembly line usually produces hundreds of cars every day; each workstation in the assembly line needs car components to perform their task. The replenishment of the components is a critical issue for the proper operation of the assembly line. In a multi- model assembly line, this task becomes more complicated than in a single model assembly line. A lack of inventory could cause some problems in the production line; excess inventory could also create it. 8 The groupIntroductionDescriptionResultsConclusion

9. Problem description In this problem, the assembly line already exists and the production plan, for the planning period, is already known. Each model has a set of characteristics, such as types of wheels and tires, radio, sunroof, car seat, and so on. In every workstation, a kit of components is installed; these components can have different trim levels. It is assumed that each component needed for the planning horizon is available in a single warehouse from where all the routes depart. 9 The groupIntroductionDescriptionResultsConclusion

10. Problem description The early arrival of the component causes space problems with the buffers of the production lines. The late arrival causes several problems in the production line. Dispatch only with a just-in-time policy increases the transportation cost and the green impact of the production line. It is necessary to select the route and the amount of required components to get the lower cost. 10 The groupIntroductionDescriptionResultsConclusion

11. 11 Problem description The groupIntroductionDescriptionResultsConclusion

12. MIP A MIP has been developed. – Transportation Cost – Transportation Vehicle Used Cost – Holding Cos – Violation Cost There is a conflict of interest among the objectives. Right now it is being presented in the CIO 2013 by one of the authors. 12 The groupIntroductionDescriptionResultsConclusion

13. Two types of Ants 13 Total Production Rule Violations Sequence(i) Class (i) Visited Stations Total Cost Vehicle (k) Tour (n) Tour Length Time Visited (n) Sequencing Routing The groupIntroductionDescriptionResultsConclusion Initial Data Number of Cars Number of Car Classes Number of Stations Capacity of Stations Requirements per Station by each class Distance Number of Vehicles

14. Pheromones & Heuristic info structure Pheromone 1 (number of vehicles, number of vehicles) – MAX-MIN approach. – The best ant of the cycle deposit pheromone at the end. – Represent the learnt of desirability of scheduling C j after C i. Pheromone 2 (number of classes, number of classes) – Initialize in the lower bound – Each time no more cars can be scheduled without violation some pheromones are added. – Represent the difficult to sequence this class without violating rules. 14 The groupIntroductionDescriptionResultsConclusion

15. Pheromones & Heuristic info structure 2 Pheromone 3 (number of stations, number of stations) – The best ant of the cycle deposit pheromone at the end. – Represents the learnt of desirability of visiting S j after S i. Dynamic Heuristic info(number of stations, number of stations) – Different options using the stock at determinate time, Traveling time, distance, and so on. 15 The groupIntroductionDescriptionResultsConclusion

16. Simplified Algorithm Pheromones trails are initialized. For the maximum iteration allowed to do – The sequence ants construct a sequence. – Calculate the demand over the time. – Each route ant constructs a full route. – Each route ant construct a full route using one less vehicle. – Evaluate the solution using the global function. – Update the pheromone trails with their respective pheromone and rules. 16 The groupIntroductionDescriptionResultsConclusion

17. Probability 17 The groupIntroductionDescriptionResultsConclusion

18. 18 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.2 Bl.2 T1 ConsumedS1S2S3S4S5 0/0 Sj Si Sj Si

19. 19 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.2 Bl.2 T1 Sj Si Sj Si RANDOM ConsumedS1S2S3S4S5 0/10/0

20. 20 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 0/10/0

21. 21 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 0/21/00/0

22. 22 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 1/21/10/10/0

23. 23 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 1/31/21/11/00/0

24. 24 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 1/31/21/11/00/0

25. 25 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.3.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 1/31/21/11/00/0

26. 26 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.3.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 2/31/32/11/10/1

27. 27 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.3.2 Bl.2 T1 StockS1S2S3S4S5 44444 Sj Si Sj Si ConsumedS1S2S3S4S5 2/41/43/11/31/1

28. 28 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2.3.2 Y Bk.3.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 3/42/44/12/31/2

29. 29 C12345678 1.8 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2.3.2 Y Bk.3.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 3/42/44/12/31/2

30. 30 C12345678 1.7 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.4.2.3.2 Y.3.2 Bk.3.4.2 Bl.2.3.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Other Ant

31. 31 C12345678 1.7 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.3.2.3.2 Y.3.2 Bk.3.2 Bl.2.3 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 We keep doing for the total number of ants

32. 32 C12345678 1.7 2 3 4 5 6 7 8 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 For the best ant Evaporate T1

33. 33 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Best Ant ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

34. Iterate 34

35. 35 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 WH T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Routing ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

36. 36 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S1 S3S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 WHS2 WH T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Routing ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

37. 37 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S1 S3 S5 S012345 0.5 1 2 3 4 5 T3 T2 WHS2 WHS4 WH T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Routing ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

38. 38 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S3 S5 S012345 0.5 1 2 3 4 5 T3 T2 WHS2 WHS4 WHS1 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Routing ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

39. 39 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S5 S012345 0.5 1 2 3 4 5 T3 T2 WHS2S3 WHS4 WHS1 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Routing ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

40. 40 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S012345 0.5 1 2 3 4 5 T3 T2 WHS2S3 WHS4S5 WHS1 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Routing ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

41. 41 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 S012345 0.5 1 2 3 4 5 T3 T2 WHS2S3WH S4S5WH S1WH T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Routing ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

42. Iterate 42

43. 43 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 WH T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 One vehicle less ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

44. 44 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S1 S3 S2 S4 S5 S012345 0.5 1 2 3 4 5 T3 T2 WHS2S3S5WH S4S1WH T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 One vehicle less ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

45. 45 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S012345 0.5 1 2 3 4 5 T3 T2 WHS2S3S5S1S4 T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 One vehicle less ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

46. 46 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S1 S3 S2 S4 S5 S012345 0.3 1 2 3 4 5 T3 T2 ClRYBkBl R.2 Y Bk.5.3.2 Bl.2 WHS2S3S5WH S4S1WH T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Update the best ant

47. 47 C12345678 1.7.9.7 2.9.7 3.9.7 4.9.7 5.9.7 6.9.7 7.9 8.7 WH S1 S3 S2 S4 S5 S012345 0.3.5.3.5.3 1.5.3 2.5.3 3.5 4.3.5.3 5.5.3 T3 T2 WHS2S3S5WH S4S1WH T1 Sj Si Sj Si ConsumedS1S2S3S4S5 4/42/54/22/31/2 Update the best ant ClRYBkBl R.2.3.2 Y Bk.3.4.2 Bl.2

48. Iterate 48

49. Repeat the entire process 49

50. We keep working Which heuristic information is better for the problem. Time to be out stock, time to arrive, distance, and so on. Type of filling, series or parallel Other updates of pheromones 50 The groupIntroductionDescriptionResultsConclusion

51. 51 Results of MIP The groupIntroductionDescriptionResultsConclusion

52. 52 Results of ACO The ACO was tuned using the instance small instance where an exact solution can be found by the MIP algorithm. The ACO algorithm finds solutions with the gap of 5% in less than 1 minute, when the MIP algorithm takes more than one hour. As bigger instance was not possible to solve by exact algorithm and no public instance for the entire problem was founded to compare with the optimal. We use the car sequencing instances reported by [Regin, 1997], then we create the replenishment data, such as capacity, vehicle number, and so on. The groupIntroductionDescriptionResultsConclusion

53. Practical conclusions In this work, the car sequencing, the inventory and the routing problem have been solved jointly. The routing model should consider more factors than just the transportation cost. The main factor in the delivery of material should not only be the decrease of the transportation costs but also the decrease of the holding cost of the components. The cost of the space is an amplifier of the savings of the model. The replenishment is made before the inventory level reaches the safety stock. Following the Lean idea, it is possible to decrease the safety stock until it reaches zero safety stock, always keeping in mind the risk of any delay with the consequence of the stoppage of the assembly line. 53 The groupIntroductionDescriptionResultsConclusion

54. Thank you! 54 The groupIntroductionDescriptionResultsConclusion