For every job, the start time on machine j+1 is greater than or equal to the completion time on machine j.

Transcription

1 Flow Shop Scheduling - makespan A flow shop is one where all the jobs visit all the machine for processing in the given order. If we consider a flow shop with n jobs and two machines (M1 and M2), all the jobs visit M1 first and then visit M2. The processing times may be different on different machines for the jobs. We consider n jobs and m machines. Let S ij be the start time of job i on machine j. Let p ij be the known processing times. The completion times of the jobs are S im + p im. The objective is to Minimize Z where Z S im + p im For every job, the start time on machine j+1 is greater than or equal to the completion time on machine j. S i,j+1 S ij + p ij. Each machine can process only one job at a time. This gives us the usual constraints S kj - S ij - p ij Mδ ijk S ij S kj p jk M (1 -δ ijk ). δ ijk binary. Illustration 5.7 Consider five jobs that go through a 2 machine flow shop. The processing times on M1 are 10, 8, 12, 15 and 8 while the processing times on M2 are 8, 10, 12, 9 and 16 respectively. Find the sequence that minimizes makespan? Let S ij be the start time of job i on machine j. The formulation is Minimize Z Z S Z S Z S Z S Z S S 12 S S 22 S S 32 S S 42 S S 52 S S 21 - S δ S 11 - S δ S 22 - S δ S 12 - S δ S 31 - S δ S 11 - S δ S 32 - S δ

2 S 12 - S δ S 41 - S δ S 11 - S δ S 42 - S δ S 12 - S δ S 51 - S δ S 11 - S δ S 52 - S δ S 12 - S δ S 31 - S δ S 21 - S δ S 32 - S d S 22 - S d S 41 - S d S 21 - S d S 42 - S d S 22 - S d 242 = 990 S 51 - S d S 21 - S d S 52 - S d S 22 - S d S 41 - S d S 31 - S d S 42 - S d S 32 - S d S 51 - S d S 31 - S d S 52 - S d S 32 - S d S 51 - S d S 41 - S d S 52 - S d S 42 - S d δ ijk binary. The optimum solution to the binary IP is given by S 12 = 55, S 22 = 45, S 11 = 45, S 31 = 8, S 32 = 24, S 42 = 36, S 21 = 37, S 41 = 21, S 52 = 8, S 51 = 0, with makespan = 63. From the solution we observe that the jobs are sent in the sequence J5 J3 J4 J1 J2. J5 starts at M1 at time = 0 and finishes at M1 at time = 8. It starts at M2 at 8 and finishes at 24. J3 starts on M1 at 8 (when M1 is free) and finishes at 20. It starts on M2 at 24 (when M2 is free; it waits for 4 minutes to access M2) and finishes at 36. J4 starts on M1 at 20 and finishes at 35. It starts on M2 at 36 (when M2 is free) and finishes at 44. J1 starts on M1 at 35 and finishes at 45. It starts on M2 at 45 and finishes at 53. J2 starts on M1 at 45 and finishes at 53. It starts on M2 at 53 and finishes at 63. All jobs are completed by time 63, which is the makespan.

3 The optimum makespan to the two machine flowshop can be obtained by the Johnson s rule (Johnson, 1954). The algorithm is as follows: 1. Find the job not yet in the sequence that has the smallest value of the processing time. If this is on M1 schedule the job in the first available position from the left. If it is on M2 schedule it in the first available position from the right. Break ties arbitrarily. 2. Repeat step 1 till all jobs are scheduled. Compute the makespan of the complete sequence. Flow Shop Scheduling sum of completion times, tardiness etc.. We consider other objectives such as minimizing sum of completion times, minimizing total tardiness, minimizing maximum tardiness and minimizing the number of tardy jobs. The objective function to minimize sum of completion times is given by Minimize Σ S j2. To minimize total tardiness we define E j and T j as the earliness and tardiness of job j. D j is the given due date for job j. We minimize Σ T j and add constraints E j T j = D j S j2 p j2. To minimize maximum tardiness we add constraints u T j and minimize u. To minimize number of tardy jobs, we define N j = 1 if job j is tardy (N j is a binary variable) and add constraints T j M N j. Illustration 5.8 Consider five jobs that go through a 2 machine flow shop. The processing times on M1 are 10, 8, 12, 15 and 8 while the processing times on M2 are 8, 10, 12, 9 and 16 respectively. The due dates are 30, 30, 20, 30 and 60 respectively for the five jobs. Apply Johnson s algorithm to minimize makespan? Also find the optimum sequences that minimize sum of completion times, total tardiness, maximum tardiness and number of tardy jobs? We first apply Johnson s algorithm to the problem. We choose J1 on M2 first and schedule J1 as the fifth job. We choose J2 on M1 (tie) and J2 becomes the first job. J5 on M1 comes next and J5 becomes the second job. We choose J4 on M2 and J4 becomes the fourth job. J3 becomes the third job and the sequence is J2 J5 J3 J4 J1. The completion times are 18, 34, 44, 55 and 63 for jobs J2, J5, J3, J4 and J1. The makespan is 63. This sequence is different from the solution to the binary IP but the makespan is the same. The processing times are the same as in Illustration 5.7. To minimize sum of completion times we minimize S 12 + S 22 + S 32 + S 42 + S 52. The constraints are the same as in example 7. The optimum solution to the binary IP is given by S 12 = 18, S 22 = 8, S 11 = 8, S 31 = 41, S 32 = 53, S 42 = 42, S 21 = 0, S 41 = 26, S 52 = 26, S 51 = 18, with Z = 147. The optimum sequence is J2 J1 J5 J4 J3 with completion times 18, 26, 42, 51 and 65 for jobs J2, J1, J5, J4 and J3

4 respectively. The minimum sum of completion times is 202 (= ). The objective function is the sum of start times on M2. To this we add all the processing times on M2 which is 55. The sum of completion times is = 202. To minimize total tardiness, we define T j and E j. We minimize T 1 + T 2 + T 3 + T 4 + T 5 and add five constraints E 1 - T 1 + S 12 = 22 E 2 - T 2 + S 22 = 20 E 3 - T 3 + S 32 = 8 E 4 - T 4 + S 42 = 21 E 5 - T 5 + S 52 = 44 The optimum solution to the binary IP is given by S 31 = 0, S 32 = 12, S 11 = 12, S 12 = 24, S 21 = 22, S 22 = 32, S 41 = 30, S 42 = 45, S 51 = 45, S 52 = 54, with total tardiness = 52. The optimum sequence is J3 J1 J2 J4 J5 with completion times 24, 32, 42, 54 and 70 for the jobs in the order of sequence. The corresponding due dates are 20, 30, 30, 30, 60. All jobs are tardy with tardiness 4, 2, 12, 24 and 10 for jobs J3, J1, J2, J4 and J5 respectively. The total tardiness is 52. To minimize maximum tardiness we use all the constraints that we used to minimize total tardiness. We introduce a variable u and minimize u. We add five constraints, u T 1, u T 2, u T 3, u T 4, u T 5. The optimum solution to the binary IP is given by S 11 = 35, S 12 = 45, S 21 = 12, S 22 = 24, S 31 = 0, S 32 = 12, S 41 = 20, S 42 = 35, S 51 = 59, S 52 = 67, with maximum tardiness = 23. The completion times for J1, J2, J3, J4 and J5 are 53, 34, 24, 44 and 83. The tardiness values are 23, 4, 4, 14 and 23 with maximum tardiness = 23. From the optimum solution, we can conclude that the sequence is J3 J2 J4 J1 J5 is optimum. We can compute the completion times as 24, 34, 44, 53 and 69 for the jobs in the order of sequence. The corresponding due dates are 20, 30, 30, 30, 60. All jobs are tardy with tardiness 4, 4, 14, 23 and 19 for the jobs in the order of sequence. The maximum tardiness is 23. We observe that the optimum solution delays the start of job 5 such that its tardiness is 23. Since the maximum tardiness occurs for J1, the optimum solution deliberately allows J5 to have the maximum value of 23 while it is possible to have T 5 = 19. Since the objective is to minimize maximum tardiness both solutions are optimum with maximum tardiness = 23. To minimize number of tardy jobs, we define N 1 to N 5 as binary variables indicating whether job j is tardy or not. We use the all the constraints to minimize total tardiness. We now minimize N 1 + N 2 + N 3 + N 4 + N 5 and add five constraints T N 1 T N 2 T N 3 T N 4 T N 5

5 The optimum solution to the binary IP is given by S 21 = 0, S 22 = 8, S 11 = 8, S 12 = 18, S 51 = 18, S 52 = 26, S 41 = 26, S 42 = 42, S 31 = 41, S 52 = 53, with number of tardy jobs = 2. The optimum sequence is J2 J1 J5 J4 J3. The completion times for J2, J1, J5, J4 and J3 are 18, 26, 42, 51 and 65. The due dates are 30, 30, 60, 30 and 20 in the order of the sequence. The tardiness values are 0, 0, 0, 21 and 45 for jobs J2, J1, J5, J4 and J3. Two jobs J4 and J3 are tardy. Job shop Scheduling In job shop scheduling, n jobs are processed on n machines where each job has a specified route. For example, job J1 may visit machines in the order M3, M2 and then M1 while J2 may visit M2 and then M3. Some jobs can skip some machines also. Let S ij be the start time of the processing on machine i for job j. Let p ij be the given processing time for job j on machine i. Let [i] represent the ith machine in a sequence. The completion time for job j is given by S [m]j + p [m]j. The objective is to minimize makespan given by Minimize Z where Z S [m]j + p [m]j The sequence of visit of machines has to be maintained. For job j, S [i+1],j S [i]j + p [i]j. Each machine can process only one job at a time. For a pair of jobs j and k on machine i, Either S ik S ij + p ij or S ik S jk + p jk. These are written using Mδ as usual by adding a binary variable for each pair of constraint. Illustration 5.9 Consider a job shop scheduling problem with three jobs (J1 to J3) and three machines (M1 to M3). The sequence and the processing times for the jobs are J1 = M1 (8), M3 (12), M2 (10), J2 = M1 (9), M2 (11) and J3 = M3 (10), M1 (11), M2 (12). Find the schedule that minimizes makespan? Let S ij be the start time of job j on machine i. We minimize Z such that Z S Z S Z S For each job, the sequence of visit of machines has to be maintained. This gives us the constraints S 31 S

6 S 21 S S 22 S S 13 S S 23 S A machine can process only one job at a time. Machines M1 and M2 have all three jobs being processed while M2 processes only two jobs. There are seven pairs of constraints. These are S 11 - S δ S 12 - S δ S 11 - S δ S 13 - S δ S 12 - S δ S 13 - S δ S 21 - S δ S 22 - S δ S 21 - S δ S 23 - S δ S 22 - S δ S 23 - S δ S 31 - S δ S 33 - S δ δ ijk 0,1. (In the above formulation we have used M = 1000). The optimum solution is given by S 33 = S 11 = 0, S 21 = 28, S 22 = 17, S 23 = 38, S 31 = 10, S 21 = 8, S 13 = 17 with makespan = 50. From the optimum solution we observe that M1 processes the three jobs in the order J1, J2 and J3 starting at 0, 8 and 17. M2 processes the jobs in the order J2, J1 and J3 starting at 17, 28 and 38 while M3 processes them in the order J3 and J1 starting at 0 and 10. The completion times of the jobs J2, J1 and J3 are 28, 38 and 50 and the makespan is 50. Job shop Scheduling Minimizing sum of completion times, tardiness etc.. We consider other objectives such as minimizing sum of completion times, minimizing total tardiness, minimizing maximum tardiness and minimizing the number of tardy jobs. The objective function to minimize sum of completion times is given by Minimize Σ S [jn] To minimize total tardiness we define E j and T j as the earliness and tardiness of job j. D j is the given due date for job j. We minimize Σ T j and add constraints E j T j = D j S [jn] p [jn]. To minimize maximum tardiness we add constraints u T j and minimize u.

7 To minimize number of tardy jobs, we define N j = 1 if job j is tardy (N j is a binary variable) and add constraints T j M N j. Illustration 5.10 Consider a job shop scheduling problem with three jobs (J1 to J3) and three machines (M1 to M3). The sequence and the processing times for the jobs are J1 = M1 (8), M3 (12), M2 (10), J2 = M1 (9), M2 (11) and J3 = M3 (10), M1 (11), M2 (12). Find the schedule that minimizes sum of completion times. If the due dates for the jobs are 30, 30, 30 find the minimum values of total tardiness, maximum tardiness and number of tardy jobs? The processing times are the same as in Illustration 5.9. We use the same decision variables. To minimize sum of completion times we define an objective function Minimize S 21 + S 22 + S 23. The optimum solution is given by S 33 = S 12 = 0, S 21 = 41, S 22 = 9, S 23 = 21, S 31 = 29, S 11 = 21, S 13 = 10 with makespan = 50. From the optimum solution we observe that M1 processes the three jobs in the order J2, J3 and J1 starting at 0, 10 and 21. M2 processes the jobs in the order J2, J3 and J1 starting at 9, 21 and 41 while M3 processes them in the order J3 and J1 starting at 0 and 29. The completion times of the jobs J2, J3 and J1 are 20, 33 and 51 and the minimum sum of completion times is 104. To minimize total tardiness, we define E j and T j and minimize ΣT j. We also add constraints E 1 - T 1 + S 21 = E 2 - T 2 + S 22 = E 3 - T 3 + S 23 = The optimum solution is given by S 33 = S 11 = 0, S 21 = 41, S 22 = 9, S 23 = 21, S 31 = 29, S 11 = 21, S 13 = 10 with makespan = 50. From the optimum solution we observe that M1 processes the three jobs in the order J2, J3 and J1 starting at 0, 10 and 21. M2 processes the jobs in the order J2, J3 and J1 starting at 9, 21 and 41 while M3 processes them in the order J3 and J1 starting at 0 and 29. The completion times of the jobs J2, J3 and J1 are 20, 33 and 51. Two jobs J3 and J1 are tardy and the minimum total tardiness is = 24. To minimize maximum tardiness, we Minimize U and add constraints U T 1, U T 2, U T 3. The optimum solution gives us makespan = 50 and minimum value of the maximum tardiness as 20. The solution that minimizes makespan is optimal for minimizing maximum tardiness. To minimize number of tardy jobs, we define N j = 1 if job j is tardy and add constraints T N 1, T N 2, T N 3. The optimum solution gives us 2 tardy jobs. The solution that minimizes total tardiness also minimizes number of tardy jobs in this case.