1 / 74

Mohd Zaidan bin Abdul Aziz B050810281

Scheduling ( Part II ). Mohd Zaidan bin Abdul Aziz B050810281 Grace Hii Leh Ung B050810211 Low Mei Ching B050810215 Sawita bt Amir B050810051.

shandi
Download Presentation

Mohd Zaidan bin Abdul Aziz B050810281

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Scheduling (Part II) MohdZaidan bin Abdul Aziz B050810281 Grace HiiLehUng B050810211 Low Mei Ching B050810215 Sawitabt Amir B050810051

  2. 4.4 Sequencing Theory for Single Machine • Shortest Processing Time (SPT) • Minimizes the mean flow time • Minimizes waiting time • Minimizes lateness for single-machine sequencing. • Earliest Due Date Scheduling • Minimizes the maximum lateness

  3. 4.4 Sequencing Theory for Single Machine • Moore’s algorithm • Minimizes number of tardy jobs Step 1: Sequence the jobs according to the earliest due date to obtain the initial solution Step 2: Find the first tardy job in the current sequence, say job [i]. If none exists go to step 4. d[1] d[2],…, d[n]

  4. 4.4 Sequencing Theory for Single Machine Step 3: Consider jobs [1], [2], …, [i]. Reject the job with the largest processing time. Return to step2. Step 4: Form an optimal sequence by taking the current sequence and appending to it the rejected jobs. The job appended to the current sequence scheduled in any order. Reason: Largest effect on the tardiness of the Job[i] • Reason: Consider number of tardiness jobs rather than tardiness

  5. 4.4 Sequencing Theory for Single Machine EXAMPLE 1: Solution: Longest processing time

  6. 4.4 Sequencing Theory for Single Machine EXAMPLE 1: Solution (cont.) Longest processing time No lateness Optimal sequence: 2, 3, 4, 6, 5, 1 or2, 3, 4, 6, 1, 5 Number of tardy jobs is two in either case.

  7. Exercise 4.4.1: Seven jobs are to be processed through a single machine. The processing times and due dates are given below. Determine sequence of the jobs in order to minimize • Mean flow time • Number of tardy jobs • Maximum lateness

  8. 4.4 Sequencing Theory for Single Machine 4. Lawler’s Algorithm • Minimize maximum lateness • Minimize maximum tardiness Subject to precedence constraint ( certain job must be completed before other jobs can begin)

  9. 4.4 Sequencing Theory for Single Machine Rule: Schedule job in reverse order. Step 1: At each stage, determine the set of jobs (named V) not require to precedes any other. Among set V, choose job k that satisfies: Example: Job among V that has smallest tardiness, if arranged on position [n]. Processing time of current sequence.

  10. 4.4 Sequencing Theory for Single Machine Step 2: Now, Job k scheduled last. Consider remaining jobs and again determine set of jobs that not require to precede any other remaining job. Step 3: After scheduling Job k, τ reduced by tkand job scheduled next to last is now determined. Step 4: Process is continued until all jobs are scheduled.

  11. 4.4 Sequencing Theory for Single Machine EXAMPLE 2: 1 2 3 4 5 6

  12. 4.4 Sequencing Theory for Single Machine Solution: Step 1: Find the job scheduled last(sixth) τ= Total processing time = 2+3+4+3+2+1 = 15 Min value • Hence, Job 5 scheduled last

  13. 4.4 Sequencing Theory for Single Machine Solution: Cont. 1 2 3 Step 2: Find the job scheduled fifth. 4 5 6 τ= New total processing time = 15 – 2 = 13 Min value • Hence, Job 6 scheduled fifth.

  14. 4.4 Sequencing Theory for Single Machine Solution: Cont. 1 2 3 Step 3: Find the job scheduled fourth. 4 5 6 τ= New total processing time = 13 – 4 = 9 Min value • Hence, Job 6 scheduled fourth.

  15. 4.4 Sequencing Theory for Single Machine Solution: Cont. 1 2 3 Step 4: Find the job scheduled third. 4 5 6 τ= New total processing time = 9 – 1 = 8 Min value • Hence, Job 4 scheduled third.

  16. 4.4 Sequencing Theory for Single Machine Solution: Cont. 1 2 3 Step 5: Find the job scheduled second. 4 5 6 Maximum tardiness The optimal sequence: 1-2-4-6-3-5

  17. Exercise 4.4.2: Eight jobs are to be processed through a single machine. The processing times and due dates are given below. 1 4 Furthermore, assume the following precedence relationships must be satisfied: 2 6 3 7 8 Determine the sequence in which jobs should be done in order to minimize maximum lateness subject to precedence restriction.

  18. 4.5 Sequencing Theory for Multiple Machines Analysis of the previous section which several jobs must be proceed on more than one machine. The optimal solution for scheduling n jobs on two machines is always a permutation schedule (that is, jobs are done in the same order on both machines). Permutation schedules provide better performances in term of both total and average flow time Minimization of the mean idle time in the system

  19. 4.5 Sequencing Theory for Multiple Machines 1. Johnson’s algorithm : Scheduling n jobs on two machines • How to implement this rule: • List value of A and B in to columns • Find the smallest element in the 2 columns. If it appears in column A, then schedule that at front of the sequence. If appears in column B, schedule that at the back of sequence • Find the remaining element in the two columns. If it appears on column A, then schedule that next job. If appears in column B, then schedule that job last. • Cross off the jobs as they are scheduled.

  20. 4.5 Sequencing Theory for Multiple Machines • Five jobs are to be scheduled on two machines. The processing time are: EXAMPLE: 2-4-3-5-1 1 4 7 13 15 22 23 27 28 30 0

  21. Exercise 4.5.1: Six job are to be schedule on two machine. The processing time are

  22. 4.5 Sequencing Theory for Multiple Machines 2. Extension to three machine • The 3 machine problem can be reduced to a 2 machine problem if the satisfied the following condition: min A ≥ max B or min C≥ max B • only either one of these conditions be satisfied. • Then reduced to 2 machine problem in the following way: A’ = A + B and B’ = B + C

  23. 4.5 Sequencing Theory for Multiple Machines EXAMPLE: Min A = 4 Max B = 6 Min C = 6 1-4-5-2-3 0 9 18 22 27 32 42 47 52 63 71

  24. Exercise 4.5.2: The following four job must be processed through a three-machine flow shop

  25. 4.5 Sequencing Theory for Multiple Machines 3. Two job flow shop problem • Two jobs are processed through m machines. • Present graphical procedure for solving this problem. The step are: • Draw a cartesian coordinate system with the processing times corresponding the first job on the horizontal axis and the processing time corresponding to the second job on the vertical axis. • Block out areas corresponding to each machine at the intersection of the intervals marked for the machine on the two axes. • Determine the part from the origin to the end of the final block that does not intersect any of the blocks and that minimize the vertical movement.

  26. 4.5 Sequencing Theory for Multiple Machines EXAMPLE: • A regional manufacturing firm produces a variety of household products. One is a wooden desk lamp. Prior to packing, the lamp must be sanded, lacquered and polished. Each operation requires a different machine. There are currently shipments of two models awaiting processing. The times required for the three operations for each of the two shipments are:

  27. 4.5 Sequencing Theory for Multiple Machines C Total time = 12 + (2 + 2) = 16 10 9 8 Total time = 12 + 3 = 15 B 7 6 5 4 2 3 1 A 11 5 3 9 2 4 8 12 7 10 1 6 A Gantt Chart Solution B C 8 2 6 10 4 12 14

  28. 4.5 Sequencing Theory for Multiple Machines C Total time = 10 + 6 = 16 10 9 8 Total time = 10 + (3 + 2) = 15 B 7 6 5 4 2 3 1 A 11 5 3 9 2 4 8 12 7 10 1 6 A Gantt Chart Solution B C 8 2 6 10 4 12 14

  29. Exercise 4.5.3: Two jobs must be proceeds through four machines in same order. The processing time in the required sequence are Determine how the two job should be schedule in order to minimize the total makespan and draw the gantt chart indicating the optimal schedule.

  30. Cycle time:maximum time allowed at each workstation to complete its set of tasks on a unit.  amount of time allotted to each workstation determined in advance, based on the desired rate of production of assembly line. Line balancing: the process of assigning tasks to workstations in such a way that the workstations have approximately equal time requirements. 4.6 ASSEMBLY LINE BALANCING Line Balancing Cycle Time Maximum cycle time = summation of the task times. Minimum cycle time = the longest task times.

  31. 4.6 ASSEMBLY LINE BALANCING Introduction (cont.) Factors that contribute to the difficulty of the problem: • There are precedence constraints • Some tasks cannot be performed at the same workstation

  32. 4.6 ASSEMBLY LINE BALANCING Line Balancing Rules Some Heuristic (intuitive) Rules: • Assign tasks in order of most following tasks. • Count the number of tasks that follow • Assign tasks in order of greatest positional weight. • Positional weight is the sum of each task’s time and the times of all following tasks.

  33. 0.1 min. 1.0 min. A Simple Precedence Diagram a b c d e 0.7 min. 0.5 min. 0.2 min. 4.6 ASSEMBLY LINE BALANCING Precedence Diagram Precedence diagram: Tool used in line balancing to display elemental tasks and sequence requirements Figure 4.6.1

  34. 4.6 ASSEMBLY LINE BALANCING Determine the Minimum Number of Workstations Required What is the minimum number of workstations for the previous precedence diagram? (assume minimum cycle time)

  35. 4.6 ASSEMBLY LINE BALANCING • EXAMPLE: • Arrange tasks shown in Figure 4.6.1 into 3 workstations. • Use a cycle time of 1.0 minute • Assign tasks in order of the most number of followers Solution Idle time per cycle

  36. 4.6 ASSEMBLY LINE BALANCING Calculate Percent Idle Time Efficiency = 100% – Percent idle time What’s the % idle time and efficiency for the above example? Efficiency = 100% - 16.7% = 83.3%

  37. 4.6 ASSEMBLY LINE BALANCING EXAMPLE: A manager wants to assign workstations in such a manner that hourly output is 4 units. Working time is 56 minutes per hour. What is the cycle time? Cycle time = operating time/output rate = 14 min

  38. Arrange tasks in decreasing order of positional weight • Assign the tasks above to workstations in the order of greatest positional weight. • Steps: • Arrange the task in the decreasing order of positional weights. • Find out the number of workstations • Number of workstations = = = 3.2 =4

  39. 14 Station Cycle Time=14 min I II III IV F=5 D=7 G=6 A=3 B=2 C=4 E=4 H=9 I=5

  40. 4.6 ASSEMBLY LINE BALANCING EXERCISE 4.6.1 A shop wants an hourly output of 33.5 units per hour. The working time is 60 minutes per hour. Assign the tasks using the rules: • In the order of most following task. • In the order of greatest positional weight.

  41. 4.6 ASSEMBLY LINE BALANCING EXAMPLE:

  42. The job times & precedence relationship for this problem:

  43. Suppose that the company is willing to hire enough workers to produce one assembled machine every 15 minutes. Sum of task time = 70 minutes Minimum no. Workstation = 70/15 = 4.67  5 Positional weight of task i: the time required to perform task i plus the times required to perform all task having task i ask a predecessor. Arrange tasks in decreasing order of positional weight

  44. Rank the step in the order of decreasing positional weight: Assume C = 15 min But, minimum possible No. of workstation = 5!!! This method is heuristic  possible that there is a solution with 5 stations Use C = 16 min No. of stations decreases  16 %, cycle time increases  7% Assume production day = 7 hrs C = 15  daily production level = 28 units/assembly operation

  45. C = 16  daily production level = 26.25 units/assembly operation Management have to determine whether the decline in the production rate of 1.75 units/day/operation is justified by the savings realised with 5 rather than 6 stations. Alternative choice: Stay with 6 stations, but use C = 13 min 13 min: minimum cycle time with 6 stations. Why not 12 min???!! Production rate = 32.3 units/day/operation Increasing No. of stations from 5 to 6  substantial improvement in the throughput rate.

  46. 4.6 ASSEMBLY LINE BALANCING EXERCISE 4.6.2 Consider the assembly line balancing problem represented by the figure above. Determine a balance for C = 20

  47. 4.7 Stochastic Scheduling 4.7.1 Static Analysis • Single machine • Uncertainty of processing times • Exact completion time of one or more jobs may not be predictable Objective: Minimize expected average weighted flow time. • Job i precedes job i+1 if where job times are t1, t2, …, tn and weights are u1, u2, …, ui.

  48. 4.7.1 Stochastic Scheduling: Static Analysis EXAMPLE: Find the optimal sequence that minimize expected average weighted flow time.

  49. 4.7.1 Stochastic Scheduling: Static Analysis Solution: Expected average weighted flow time = 237/10 = 23.7

  50. 4.7.1 Stochastic Scheduling: Static Analysis 2. Multiple Machine • n jobs are to processed through two identical parallel machines. Each jobs needs to be processed only once on either machine. • The objective is to minimize the expected makespan. • Parallel is different from flow shop problem. • In flow shop, jobs are processed first one machine 1 then on machine 2. • Then the optimal sequence is to schedule the jobs according to LEPT (longest expected processing time first) opposite with the SPT rule

More Related