Operations Research_18ME735_module 5 sequencing notes.pdf
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The document discusses sequencing problems in job scheduling on machines, focusing on flow shop and job shop sequencing techniques. It outlines various rules for optimal job processing sequences, such as First Come First Serve (FCFS), Shortest Processing Time (SPT), Earliest Due Date (EDD), and others, along with their assumptions and objectives for minimizing criteria like make span, mean flow time, and tardiness. Additionally, it includes examples illustrating the application of these rules to determine optimal job sequences and performance metrics.
Operations Research_18ME735_module 5 sequencing notes.pdf
- 1. SEQUENCING It is an optimal order of arranging number of jobs (n) on available number of resources (m). Flow shop sequencing Let there are m machines and n jobs. Each job requires processing on all m machines or some of them No machine can perform more than one operation simultaneously. The order of operations on each job through the machines is specified. For each operation of each job, execution time (processing time) is specified. The problem is to find the technologically feasible sequence among (n!)m possible sequences for processing the jobs so that we optimize. For example n=2, m=3, theoretically possible no. of different sequences are = (2!)3 = 8 feasible solutions, but some may technologically infeasible (We can find solutions ------ using known algorithms) Job shop sequencing Let m no. of machines and n no. of jobs Each time a job is completed by a machine, the next job to be started is dynamic…. (to solve these types of problems -- -still lot of work is going on……) Assumptions in flow shop sequencing: 1. Only one operation is carried out on a machine at a particular time. 2. Each operation once stated must be completed. 3. An operation must be completed before its succeeding operation can start 4. Only one machine of each type is available 5. A job is processed as soon as possible, but only in the order specified. 6. Processing times are independent of order of performing the operations. 7. The transportation time i.e. time required to move job from one machine to another is negligible. 8. Jobs are completely known and are ready for processing when the period under consideration starts. 9. The cost of in process inventory for each job is same and negligible. What is optimizing here: (Problem objective may be any of the following) 1. Minimizing total elapsed time (Make Span - defined as the completion time of the last job to leave the system. The time difference between start and end time of a schedule. Is defined as the amount of time, from start to finish for completing a set of jobs, i.e. the maximum completion time of all jobs) 2. Minimizing mean flow time (a measure of the average time that a job spends in the shop) 3. Minimizing the idle time of machines
- 2. 4. Minimizing total tardiness (Tardiness is a measure of a delay in executing certain operations and earliness is a measure of finishing operations before due time. Lateness = actual completion time of job - Its due date = a positive number, then it is termed as tardiness. Total tardiness is the sum of tardiness over all the jobs in the set) 5. Minimizing number of tardy jobs 6. Minimize in process inventory Sequencing Types of sequencing problems Case 1: Processing of ‘n’ jobs through one machine Case 2: Processing of n jobs and two machines A and B, all jobs processed in the order AB. Case 3: Processing of n jobs and three machines A, B and C all jobs processed in the order ABC Case 4: Processing of n jobs and m machines. Case 5: Processing of 2 jobs m machine Case 1: Processing of ‘n’ jobs through one machine Solution method: Priority rules 1. FCFS Rule (First Come First Serve Rule) – Jobs are processed in the order in which they arrive at a machine or work centre 2. SPT Rule (Shortest Processing Time Rule) – Jobs are processed according to processing time at a machine or work centre 3. EDD Rule (Earliest Due Date Rule) – Jobs are processed according to due date, earliest due date is first 4. STR Rule (Slack time remaining rule) – Slack Time Remaining (STR) = due days – processing days. The shortest STR goes first. 5. Critical Ratio (CR) = Time remaining before due date ÷ Remaining processing time. The smallest CR goes first In case the importance of job is varying, then a weight wi is assigned to each job based on their importance. A larger value of wi indicating greater importance. The processing times are divided by the weights is computed and the jobs are arranged in order of increasing t/wi. Where, weighted mean flow time = (∑(𝑤𝑖 ∗ 𝑓𝑙𝑜𝑤 𝑡𝑖𝑚𝑒) ÷ ∑𝑤𝑖) This rule is called as WSPT rule (Weighted shortest processing time rule) The measures of Effectiveness: i. Make span i.e. total elapsed time or completion time of the jobs (is defined as the completion time of the last job to leave the system.)
- 3. The time difference between start and end time of a schedule. Is defined as the amount of time, from start to finish for completing a set of jobs, i.e. the maximum completion time of all jobs) ii. Mean waiting time iii. Mean flow time Where, mean flow time = (∑ 𝑓𝑙𝑜𝑤 𝑡𝑖𝑚𝑒) ÷ 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑗𝑜𝑏𝑠 iv. Mean idle time v. Mean lateness vi. Number of tardy jobs vii. Total tardiness viii. Average tardiness ix. Mean number of jobs waiting as in-process inventory i.e (∑ 𝑓𝑙𝑜𝑤 𝑡𝑖𝑚𝑒) ÷ 𝑚𝑎𝑘𝑒 𝑠𝑝𝑎𝑛 Example 1: (Teaches FCFS, SPT, EDD) The 6 jobs have already arrived in the order shown below and also assume no new jobs arrive thereafter. Processing times and due dates for six jobs waiting to be processed at a work centre (machine) are given below. Determine the sequence of jobs, average flow time, average tardiness, average number of jobs at the work centre for each of these rules: FCFS, SPT, EDD and STR Jobs Processing time (in days) Due Date (in days) A 2 7 B 8 16 C 4 4 D 10 17 E 5 15 F 12 18 a) FCFS rule As per this rule it is simply, A – B – C – D – E – F. P F D Lateness = D -P W Job sequence Processing time Flow time Due date Tardiness Earliness Waiting time A 2 2 7 0 -5 0 B 8 10 16 0 -6 2 C 4 14 4 10 0 10 D 10 24 17 7 0 14 E 5 29 15 14 0 24 F 12 41 18 23 0 29 Measures: i. Make span = 41 days ii. Mean flow time = (2+10+14+24+29+41) / 6 = 120/6 = 20 days
- 4. iii. Mean idle time = NA iv. Average tardiness = 10+7+14+23 / 6 = 9 days v. Average lateness = -5-6++10+7+14+23 / 6 = 43/6 = 7.16 days vi. Number of tardy jobs = 4 vii. Mean waiting time = (0+2+10+14+24+29)/6 = 79/6 = 13.16 days viii. Mean number of jobs waiting as in-process inventory i.e (∑ 𝑓𝑙𝑜𝑤 𝑡𝑖𝑚𝑒) ÷ 𝑚𝑎𝑘𝑒 𝑠𝑝𝑎𝑛 = 120/41 = 2.93 jobs b) SPT rule As per this rule the sequence is A – C – E – B – D – F Lateness Job sequence Processing time Flow time Due date Tardiness Earliness Waiting time A 2 2 7 -5 0 C 4 6 4 2 2 E 5 11 15 -4 6 B 8 19 16 3 11 D 10 29 17 12 19 F 12 41 18 23 29 Measures: i. Make span = 41 days ii. Mean flow time = (2+6+11+19+29+41) / 6 = 108/6 = 18 days iii. Mean idle time = NA iv. Average tardiness = 2+3+12+23/ 6 = 40/6 = 6.67 days v. Average lateness = -5+2-4+3+12+23 / 6 = 5.17 vi. Number of tardy jobs = 4 vii. Mean waiting time = (0+2+6+11+19+29)/6 = 67/6 = 11.16 days viii. Mean number of jobs waiting as in-process inventory i.e (∑ 𝑓𝑙𝑜𝑤 𝑡𝑖𝑚𝑒) ÷ 𝑚𝑎𝑘𝑒 𝑠𝑝𝑎𝑛 = 108/41 = 2.63 jobs c) EDD rule As per this rule the sequence is: C A E B D F P F D Lateness = F-D Job sequence Processing time Flow time Due date Tardiness Earliness Waiting time C 4 4 4 0 0 A 2 6 7 -1 4 E 5 11 15 -4 6 B 8 19 16 3 11 D 10 29 17 12 19 F 12 41 18 23 29 i. Make span = 41 days ii. Mean flow time = (4+6+11+19+29+41) / 6 = 110/6 = 18.33 days
- 5. iii. Mean idle time = NA iv. Average tardiness = 0+0+0+3+12+23/ 6 = 40/6 = 6.33 days v. Average lateness = 0-1-4+3+12+23/6 = 5.5 vi. Number of tardy jobs = 3 vii. Mean waiting time = (0+4+6+11+19+29)/6 = 69/6 = 11.5 days viii. Mean number of jobs waiting as in-process inventory i.e (∑ 𝑓𝑙𝑜𝑤 𝑡𝑖𝑚𝑒) ÷ 𝑚𝑎𝑘𝑒 𝑠𝑝𝑎𝑛 = 110/41 = 2.68 jobs d) STR rule Job Processing time Due date Slack = D-P A 2 7 5 B 8 16 8 C 4 4 0 D 10 17 7 E 5 15 10 F 12 18 6 Sequence is C – A – F – D – B – E P F D Lateness = F – D W Job sequence Processing time Flow time Due date Tardiness Earliness Waiting time C 4 4 4 0 0 A 2 6 7 -1 4 F 12 18 18 0 6 D 10 28 17 11 18 B 8 36 16 20 28 E 5 41 15 26 36 i. Make span = 41 days ii. Mean flow time = (4+6+18+28+36+41) / 6 = 133/6 = 22.17 days iii. Mean idle time = NA iv. Average tardiness = 11+20+26/ 6 = 57/6 = 9.5 days v. Average lateness = -1+11+20+26/6 = 56/6 = 9.33 days vi. Number of tardy jobs = 3 vii. Mean waiting time = (0+4+6+11+19+29)/6 = 69/6 = 11.5 days viii. Mean number of jobs waiting as in-process inventory i.e (∑ 𝑓𝑙𝑜𝑤 𝑡𝑖𝑚𝑒) ÷ 𝑚𝑎𝑘𝑒 𝑠𝑝𝑎𝑛 = 133/41 = 3.24 jobs Rule Mean flow time Average tardiness Average lateness Mean waiting time Mean number of jobs in- process inventory No. of tardy jobs Make span FCFS 20 9 7.16 13.16 2.93 4 41 SPT 18 6.67 5.17 11.16 2.63 4 41 EDD 18.33 6.33 5.5 11.5 2.68 3 41 STR 22.17 9.5 9.33 11.5 3.24 3 41
- 6. Example 2 (WSPT rule) Eight jobs 1, 2,3…..8 are to be processed on a single machine. The processing times, due dates and the weights of the jobs are represented. Job Processing time (Minutes) Due date (Minutes) Importance weight wi 1 5 15 1 2 8 10 2 3 6 15 3 4 3 25 1 5 10 20 2 6 14 40 3 7 7 45 1 8 3 50 2 Assuming that no new jobs arrive there after, determine using SPT and WSPT rule i) Optimal sequence ii) Completion time of jobs iii) Mean flow time iv) Weighted mean flow time v) Average in-process inventory vi) No. of jobs actually late
- 7. SPT Rule As per this rule the sequence is : 4 – 8 – 1 – 3 – 7 – 2 – 5 – 6 P F D Lateness = F – D Job sequence Processing time (Minutes) Completion time (or Flow time) (minutes) Due date (minutes) Tardiness Earliness 4 3 3 25 -22 8 3 6 50 -44 1 5 11 15 -4 3 6 17 15 2 7 7 24 45 -21 2 8 32 10 22 5 10 42 20 22 6 14 56 40 16 i) Optimal sequence: 4 – 8 – 1 – 3 – 7 – 2 – 5 – 6 ii) Completion time of jobs: 3, 6, 11,17,24,32,42,56 iii) Mean flow time : 3+6+11+17+24+32+42+56 / 8 = 191/8 = 23.87 minutes iv) Average in-process inventory: total flow time / make span = 191 / 56 = 3.41 v) No. of jobs actually late = 4 jobs WSPT Rule Job Processing time (Minutes) Due date (Minutes) Importance weight wi pi/wi 1 5 15 1 5 2 8 10 2 4 3 6 15 3 2 4 3 25 1 3 5 10 20 2 5 6 14 40 3 4.66 7 7 45 1 7 8 3 50 2 1.5 As per this rule the sequence is arranged in increasing value of pi/wi 8 – 3 – 4 – 2 – 6 - 1 – 5 – 7
- 8. P F D Lateness = F – D W F*W Job sequence Processing time (Minutes) Completion time (or Flow time) (minutes) Due date (minutes) Tardiness Earliness Weight Weighted flow time 8 3 3 50 -47 2 6 3 6 9 15 -6 3 27 4 3 12 25 -13 1 12 2 8 20 10 10 2 40 6 14 34 40 -6 3 102 1 5 39 15 24 1 39 5 10 49 20 29 2 98 7 7 56 45 11 1 56 i) Optimal sequence: 8 – 3 – 4 – 2 – 6 – 1 – 5 - 7 ii) Completion time of jobs: 3 9 12 20 34 39 49 56 iii) Mean flow time: 3 +9 +12 +20+ 34+ 39 +49+ 56 / 8 = 27.75 iv) Weighted Mean flow time: (6+27+12+40+102+39+98+56) / (2+3+1+2+3+1+2+1) = 380/15 = 25.33 minutes v) Average in-process inventory: total flow time / make span = 222 / 56 = 3.96 vi) No. of jobs actually late = 4 jobs vii) mean tardiness = 9.25 minutes viii) mean lateness = 0.25 minutes ix) max lateness = 29 minutes Case 2: Processing of n jobs and two machines No. of jobs = n, each job is processed in the prescribed order AB, No. of machines = 2, (A, B) Solution Method: Johnsons Rule (Johnson and Bellman algorithm): it can be shown that the shortest elapsed time occurs when all the jobs are processed on the two machines in the same order. Jobs Ma/c A M/c B 1 A1 B1 2 A2 B2 3 A3 B3 identify the smallest element in the given matrix.
- 9. If the smallest element falls under column 1 i.e under machine A then do that job first. If the smallest element falls under column 2 i.e under machine 2 then do that job last. If there is a tie one in each column, schedule the one in the first column first and the one in the second column last If there is a tie in the same column, select the one with the lowest entry in the other column Assumptions: • Only two machines are involved • If the the machine order is AB, that is to say that the job which is processed first on machine A must also be processed first on machine B. No passing is allowed. • Set up times of machines A & B are independent of the sequence in which jobs are taken up • No job is required more urgently than the other • Expected processing times of jobs are known Example (n jobs 2 machines) A machine operator has to perform two operations, turning and threading, on a number of different jobs. The time required to perform these operations in minutes for each job is given. Determine the order in which the jobs should be processed in order to minimize the total time required to turn out all the jobs and also find the idle time of machine. Jobs 1 2 3 4 5 6 Time for turning (in min) 3 12 5 2 9 11 Time for threading (in min) 8 10 9 6 3 1 Solution: Jobs 1 2 3 4 5 6 M/c A Time for turning (in min) 3 12 5 2 9 11 M/c B Time for threading (in min) 8 10 9 6 3 1 4 1 3 2 5 6 Job sequence M/c A M/c B 4 2 6 1 3 8 3 5 9 2 12 10 5 9 3 6 11 1 Job sequence Time for turning (in min) Time for threading (in min) Idle time of machine B In Out In Out 4 0 2 2 8 2 1 2 5 8 13 0
- 10. 3 5 10 13 22 0 2 10 22 22 32 0 5 22 31 32 35 0 6 31 42 42 43 7 Idle time of machine A = 43 – 42 = 1 hr Idle time of Machine B = 9 hrs If order is revised to BA, what difference will it make to the calculated results?. If no. of units of each job is 3, 4,2,5,2,3…. How to do … an example? Example 2 Case 3: Processing of n jobs and three machines A, B and C No. of machine – 3, A B C n = no. of jobs, processed in prescribed order ABC/BAC/CAB Jobs Ma/c A (Ai) M/c B (Bi) M/c C (Ci) 1 A1 B1 C1 2 A2 B2 C2 3 A3 B3 C3 Solution method: Check for the following conditions, i) Minimum time on machine A is >= maximum time on machine B ii) Minimum time on machine C is >= maximum time on machine B Either or both the condition holds good, then Replace the problem by fictious equivalent problem involving 2 machines and n jobs, These two fictious machines are G & H, their corresponding processing times are, Gi = Ai+Bi Hi=Bi+Ci Note: If neither of the above conditions holds good, the method fails and optimal sequence has to be found by enumerating all sequences. Example: There are five jobs, each of which are processed through the three machines A, B, C in order ABC. Processing times in hours are,
- 11. Jobs Ma/c A (hrs) M/c B (Hrs) M/c C (hrs) 1 3 4 7 2 8 5 9 3 7 1 5 4 5 2 6 5 4 3 10 Determine the optimum sequence for the five jobs and minimum elapsed time. And also find the idle time for three machines and waiting times of jobs. Solution: Min Ai = 3, Max Bi = 5 Min (Ai) >= Max (Bi) , condition fails Min Ci = 5, Max Bi = 5 Min (Ci) >= Max (Bi) – condition passes, So we can replace the given problem by Gi = Ai+Bi, Hi = Bi+Ci Jobs M/c G M/c H 1 7 11 2 13 14 3 8 6 4 7 8 5 7 13 4 1 5 2 3 When there is a ties in the same column…. select the one which is having lowest processing times in the other column. Jobs sequence Ma/c A (hrs) M/c B (Hrs) M/c C (hrs) 4 5 2 6 1 3 4 7 5 4 3 10 2 8 5 9 3 7 1 5 Jobs Sequence Ma/c A (hrs) M/c B (Hrs) M/c C (hrs) Idle time of m/c B Idle time of m/c C 4 0 5 5 7 7 13 5 7 1 5 8 8 12 13 20 1 0 5 8 12 12 15 20 30 0 0 2 12 20 20 25 30 39 5 0 3 20 27 27 28 39 44 2 0
- 12. Total elapsed time is 44 hrs. Idle time of machine A = 44 – 27 = 17 hrs Idle time of machine B = 5+1+5+2 + (44-28) = 29 hrs Idle time of machine C = 7 hrs Waiting time of jobs m/c A = 0 B = 0 C = 1+5+5+11 = 22 hrs Jobs Ma/c A (hrs) M/c B (Hrs) M/c C (hrs) 4 5 2 3 1 3 4 7 5 4 3 10 2 8 5 9 3 7 1 5
- 13. Example 2 (n jobs 3 machines) Solution: A B C Jobs M/c C M/c A M/c B 1 5 4 8 2 6 6 10 3 2 7 7 4 3 4 8 5 4 5 11 6 9 3 8 Min Ai = 2, Max Bi = 7 Min (Ai) >= Max (Bi) , condition fails Min Ci = 7, Max Bi = 7 Min (Ci) >= Max (Bi) – condition passes, So we can replace the given problem by Gi = Ai+Bi, Hi = Bi+Ci Jobs M/c G M/c H 1 9 12 2 12 16 3 9 14 4 7 12 5 9 16 6 12 11 4 1 3 5 6 2 Jobs Sequence M/c C M/c A M/c B 4 3 4 8 1 5 4 8
- 14. 3 2 7 7 5 4 5 11 6 9 3 8 2 6 6 10 Jobs Sequence Ma/c C (hrs) M/c A (Hrs) M/c B (hrs) Idle time of m/c A Idle time of m/c B In Out In Out In Out 4 0 3 3 7 7 15 3 7 1 3 8 8 12 15 23 1 0 3 8 10 12 19 23 30 0 0 5 10 14 19 24 30 41 0 0 6 14 23 24 27 41 49 0 0 2 23 29 29 35 49 59 2 0 Total elapsed time: also called as make span = 59 hrs Idle time C = 59-29 = 30 hrs Idle time A = 59-35+6 = 30 hrs Idle time B = 7 hrs Case 4: Processing of n jobs and m machines. Jobs A B C ……………… K 1 A1 K1 2 A2 K2 3 N An Kn Check Min time on m/c A > = max time on machines B, C, ……. K-1 Min time on M/c K > = max time on machines B C ………. K-1 Then apply the rule same as you applied for 3 machines problem G = A+B+…..K-1 H = B+C+……K If this both these conditions fail, then The given nXm problem is split into a number of n x 2 sub problems
- 15. The no. of such problems will be m-1 problems Thus, a 3 machine problem will involve 3-1 = 2 subproblems.. Each 2 machine sub problem is solved as per johnsons rule All the solutions are examined and the sequence that yields the least processing times is optimal Case 5: Processing of 2 jobs m machine EXAMPLE: Four jobs 1,2,3,4 are to be processed on each of the five machines A B C D E in the order ABCDE. Find the total minimum elapsed time is no passing of jobs is permitted. Also determine the idle time for each machine. Jobs A B C D E 1 7 5 2 3 9 2 6 6 4 5 10 3 5 4 5 6 8 4 8 3 3 2 6 Min A = 5, Max B C D = 6 Min E = 6, Max B C D = 6 G = A+B+C+D H = B+C+D+E Jobs G H 1 17 19 2 21 25 3 20 23 4 16 14 1 3 2 4
- 16. Example 2 Jobs A B C D 1 7 15 14 21 2 11 18 18 6 3 2 13 11 16 4 14 4 27 14 5 18 11 32 16 Min A = 2 Max B C = 32 Min D = 6 3 sub problem – A D Case 5: Processing of 2 Jobs on ‘m’ Machines Solution method : Graphical Method Example: A machine shop has five jobs A,B,C,D & E. Two jobs must be processed through each of these machines. The time (in hours) taken on each of these machines and necessary sequence of jobs through the shops are given below: Find total elapsed time
- 17. Job 1 Sequence A B C D E Time 2 4 3 6 6 Job 2 Sequence C A D E B Time 4 6 3 3 6
- 18. We now have to find an optimal sequence for processing of the jobs 1 and 2, i.e., the shortest line consisting of combinations of horizontal, vertical and diagonal (45) lines from the origin. For this, we start from the origin and move diagonally first up to the point (9, 9) since both the jobs 1 and 2 are being processed simultaneously. After being processed through machines A, B and C, job 1 becomes idle since job 2 is being processed in machine A and it is then being processed in machine D. Therefore, we move vertically which means Job 2 is under process and Job 1 is idle. The idle time for job 1 is 4 hours in this case. Again we move diagonally up to the point (18, 22) since both the jobs 1 and 2 are being processed simultaneously. At this stage, job 2 is completed but job 1 is being processed through machine E. Therefore, we move horizontally, which means Job 1 is being processed and Job 2 is idle. The idle time for job 2 is 3 hours since job 1 is processed through machine E between 15 to 21 hours. In this case, the idle time for job 1 is found to be 4 hours. Therefore, the total elapsed time for Job 1 is 21 + 4 = 25 hours. Similarly, the idle time for job 2 is seen to be 3 hours when job 1 was under processing between 18 to 21 hrs and the processing for job 2 had been completed Therefore, the total elapsed time for job 2 is 22 + 3 = 25 hours,