# GROUP MEMBERS: HAMIZAH BINTI HAMZAHB050910232 NORAINA BINTI MOHD YUSOFB050910140 NUR ADILA BINTI RAMLIB050810 SHCEDULLING (PART II)

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GROUP MEMBERS: HAMIZAH BINTI HAMZAHB050910232 NORAINA BINTI MOHD YUSOFB050910140 NUR ADILA BINTI RAMLIB050810 SHCEDULLING (PART II)

1. FCFS 2. SPT 3. EDD 4. CR 5. ALGORITHM - Moore 1968 - Lawler’s SEQUENCING THEORY FOR SINGLE MACHINE

Example:FCFS FCFS (first come-first served) Jobs are processed in the sequence in which they entered the shop The simplest and nature way of sequencing as in queuing of a bank Example. 1  A machine center in a job shop for a local fabrication company has five unprocessed jobs remaining at a particular point in time. The jobs are labeled 1, 2, 3, 4, and 5 in the order that they entered the shop. The respective processing times and due dates are given in the table below.  Sequence the 5 jobs by above 4 rules and compare results based on mean flow time, average tardiness, and number of tardy jobs

Mean Flow time=268/5=53.6 Average tardiness=121/5=24.2 No. of tardy jobs=3

Example:SPT SPT (shortest processing time) Jobs are sequenced in increasing order of their processing time The job with shortest processing time is first, the one with the next shortest processing time is second, and so on Mean Flow time=135/5=27.0 Average tardiness=43/5=8.6 No. of tardy jobs=1

Example: EDD EDD (earliest due date) Jobs are sequenced in increasing order of their due dates The job with earliest due date is first, the one with the next earliest due date is second, and so on Mean Flow time=235/5=47.0 Average tardiness=33/5=6.6 No. of tardy jobs=4

Current time: t=0 Job numberProcessing TimeDue DateCritical Ratio 1234512345 11 29 31 1 2 61 45 31 33 32 61/11(5.545) 45/29(1.552) 31/31(1.000) 33/1 (33.00) 32/2 (16.00) Example: Scheduling CR CR (Critical ratio) Critical ratio is the remaining time until due date divided by processing time; Scheduling the job with the smallest CR next;

Current time: t=31 Job numberProcessing TimeDue Date-Current TimeCritical Ratio 12451245 11 29 1 2 30 14 2 1 30/11(2.727) 14/29(0.483) 2/1 (2.000) 1/2 (0.500) Current time should be reset after scheduling one job Current time=60 (31+29) Job numberProcessing TimeDue Date- Current Time Critical Ratio 145145 11 1 2 1 -27 -28 1/11(0.0909) -27/1<0 -28/2<0

Job numberProcessing TimeCompletion TimeTardiness 3245132451 31 29 1 2 11 31 60 61 63 74 0 15 28 31 13 Totals28987 Mean Flow time=289/5=57.8 Average tardiness=87/5=17.4 No. of tardy jobs=4. Both Jobs 4 and 5 are later, however Job 4 has shorter processing time and thus is scheduled first; Finally, job 1 is scheduled last.

Sequencing Rules Summary RuleMean Flow TimeAverage Tardiness Number of Tardy Jobs FCFS SPT EDD CR 53.6 27.0 47.0 57.8 24.2 8.6 6.6 17.4 31443144 Discussions  SPT results in smallest mean flow time;  EDD yields the minimum maximum tardiness (42, 43, 18, and 31 for the 4 different rules);  Always true? Yes!

Sequencing Theory for A Single Machine AN ALGORITHM (Moore 1968) Minimizing the number of Tardy Jobs that minimizes the number of tardy jobs for the single machine problem. Step1. Sequence the jobs according to the earliest due date to obtain the initial solution. That is d [1]  d [2] ,…,  d [n] ; Step2. Find the first tardy job in the current sequence, say job [i]. If none exists go to step 4. Step3. Consider jobs [1], [2], …, [i]. Reject the job with the largest processing time. Return to step2. (Why ?) Reason: It has the largest effect on the tardiness of the Job [i]. Step4. Form an optimal sequence by taking the current sequence and appending to it the rejected jobs. (Can be appended in any order?) Yes, because we only consider the number of tardiness jobs rather than tardiness.

Job123456 Due date1569232030 Processing time10348 6 Example Solution Job231546 Due date6915202330 Processing time3410 86 Completion time 3717273541 Longest processing time Sequencing Theory for A Single Machine

Job23546 Due date69202330 Processing time341086 Completion time37172531 Example :Solution (Cont.) Longest processing time Job2346 Due date692330 Processing time3486 Completion time371521 The optimal sequence: 2, 3, 4, 6, 5, 1 or 2, 3, 4, 6, 1, 5. In each case the number of tardy jobs is exactly 2. Sequencing Theory for A Single Machine

Sequencing Theory for A Single Machines Example Job123456 Processing time234321 Due date3697117 Precedence constraints: Lawler’s Algorithm

Example Job123456 Processing time234321 Due date3697117 Step1: find the job scheduled last (sixth) Not predecessor τ =2+3+4+3+2+1=15 356 Tardiness15-9=615-11=415-7=8 Job12346 Processing time23431 Due date36977 Step2: find the job scheduled fifth τ =15-2=13 36 Tardiness 13-9=413-7=6 Not predecessor

Example Job1246 Processing time2331 Due date3677 Step3: find the job scheduled fourth Not predecessor τ =13-4=9 26 Tardiness9-6=39-7=2 Job124 Processing time233 Due date367 Step4: find the job scheduled third τ =9-1=8 24 Tardiness8-6=28-7=1 Not predecessor Because job3 is no longer on the list, Job 2 now because a candidate. Because job6 has been scheduled, Job 4 now because a candidate along with Job 2.

Example 3 Job12 Processing time 23 Due date36 Step5: find the job scheduled second Not predecessor The optimal sequence: 1-2-4-6-3-5 JobProcessing time Flow time Due dateTardiness 124635124635 233142233142 2 5 8 9 13 15 3 6 7 9 11 001244001244 Maximum tardiness

SEQUENCING THEORY FOR MULTIPLE MACHINE

 Assume that n jobs are to be processed through m machines. The number of possible schedules is astonishing, even for moderate values of both n and m.  For each machine, there is n! different ordering of the jobs; if the jobs may be processed on the machines in any order, there are totally (n!) m possible schedules. (n=5, m=5, 25 billion possible schedules)  Let’s consider a case when each job must be processed in the following order First on machine 1, then machine 2…. Sequencing Theory for Multiple Machines 1. n jobs are to be process through m machine

 Suppose that two jobs,Y and Z, are to be scheduled on two machines, 1 and 2, the processing times are  Assume that both jobs must be processed first on machine A and then on machine B. There are four possible schedules. Sequencing Theory for Multiple Machines Machine 1Machine 2 Job Y Job Z

ScheduleTotal flow timeMean flow timeMean idle time 111(6+11)/2=8.5(5+5)/2=5 27 6.51 312 9.56 412 11.56 Sequencing Theory for Multiple Machines

Deterministic Scheduling with Multiple Machines: Johnson’s Rule Name Machine 1 = A, Machine 2 = B, then a i = processing time for job i on A and b i = processing time for job i on B Johnson’s Rule says that job i precedes job j in the optimal sequence if Algorithm: Step 1: Record the values of a i and b j in two columns Step 2: Find the smallest remaining value in two columns. If this value in column a, schedule this job in the first open position in the sequence; if this value in column b, schedule this job in the last open position in the sequence; Step 3: Cross off each job as it is scheduled

Example 1: JobMachine AMachine B 152 216 397 438 5104 Optimal sequence : 2 4 3 5 1 Sequencing Theory for Multiple Machines jobAB 152 216 397 438 5104 Johnson’s schedule: 2 –> x –> x –> x –> x 2 –> x –> x –> x –> 1 2 –> x –> x –> 5 –> 1 2 –> 4 –> x –> 5 –> 1 2 –> 4 –> 3 –> 5 –> 1

jobAB 152 216 397 438 5104 Optimal sequence : 2 4 3 5 1

Sequencing Theory for Multiple Machines 2. Extension to Three Machines  The three-machine problem can be reduced to a two-machine problem if the following condition is satisfied min A i  max B i or min C i  max B i It is only necessary that either one of these conditions be satisfied. If that is the case, then the problem is reduced to a two-machine problem  Define A i ’=A i +B i, B i ’=B i +C i  Solve the problem using the rules described above for two-machines, treating A i ’ and B i ’ as the processing times.  The resulting permutation schedule will be optimal for the three-machine problem.  If the condition are not satisfied, this method will usually give reasonable, but possibly sub-optimal results.

Sequencing Theory for Multiple Machines 3. The Two-Job Flow Shop Problem: assume that two jobs are to be processed through m machines. Each job must be processed by the machines in a particular order, but the sequences for the two jobs need not be the same. Graphical procedure developed by Akers (1956): Draw a Cartesian coordinate system with the processing times corresponding to the first job on the horizontal axis and the processing times corresponding to the second job on the vertical axis. Block out areas corresponding to each machine at the intersection of the intervals marked for that machine on the two axes. Determine a path from the origin to the end of the final block that does not intersect any of the blocks and that minimizes the vertical movement. Movement is allowed only in three directions: horizontal, vertical, and 45-degree diagonal. The path with minimum vertical distance corresponds to the optimal solution.

Example 2: A regional manufacturing firm produces a variety of household products. One is a wooden desk lamp. Prior to packing, the lamps 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 Job 1Job2 OperationTimeOperationTime Sanding (A)3A2 Lacquering (B)4B5 Polishing( C )5C3 Sequencing Theory for Multiple Machines

Minimizing the flow time is the same as maximizing the time that both jobs are being processed. That is equivalent to finding the path from the origin to the end of block C that maximizes the diagonal movement and therefore minimizes either the horizontal or the vertical movement. or 10+(3+2)=15 or 10+6=16

Example 3: Job 1Job2 Order & Operation TimeOrder & Operation Time B3A2 D4D5 C2B4 A5C3 B D C A A D B C A C D B F 14+4=18 14+2+2=18 7 11 15 18 B A D C B A C D J1 J2 B A D C B A C D J1 J2

LINE BALANCING

WHAT IS LINE BALANCING??? EVERYONE IS DOING THE SAME AMOUNT OF WORK DOING THE SAME AMOUNT OF WORK TO CUSTOMER REQUIREMENT VARIATION IS “SMOOTHED” NO ONE OVERBURDENED NO ONE WAITING EVERYONE WORKING TOGETHER IN A BALANCED FASHION

Example 6 The Final assembly of NANO personal computers, a generic mail- order PC clone, requires a total of 12 tasks. The assembly is done at the Lubbock, Texas, plant using various components imported from the Far East. The network representation of this particular problem is given in the following figure.

TaskImmediate PredecessorsTime 1_12 216 326 422 522 62 73, 47 875 951 109, 64 118, 106 12117   t i =70, and the production rate is a unit /15 minutes;  The minimum number of workstations = [70/15]=5 The job times and precedence relationships for this problem are summarized in the table below.

The solution precedence requires determining the positional weight of each task. The positional weight of task i is defined as the time required to perform task i plus the times required to perform all tasks having task i as a predecessor. TaskPositional Weight 170 258 331 427 520 629 725 818 9 1017 1113 127 t 3 +t 7 +t 8 +t 11 +t 12 =31 The ranking 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12

Station123456 Tasks12, 3, 45, 6, 97, 810, 1112 Processing time12141512107 Idle time310358 PROFILE 1: C=15 TaskImmediate Predecessors Time 1_12 216 326 422 522 62 73, 47 875 951 109, 64 118, 106 12117 The ranking 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12

Station123456 Tasks12,3,45,6,97,810,1112 Processing time1214152107 Idle time310358 PROFILE 1: C=15 Cycle Time=15 T1=12 T2=6T3=6T4=2 T5=2T6=12T9=1 T8=5T7=7T10=4 T11=6T12=7 The ranking 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12 15  Evaluate the balancing results by the efficiency  t i /NC;  The efficiencies for Profiles 1 is 77.7%.

Profile 2: Increasing cycle time from 15 to 16 Alternative 1: Change cycle time to ensure 5 station balance Station12345 Tasks12,3,4,56,97,8,1011,12 Idle time40303  Increasing the cycle time from 15 to 16, the total idle time has been cut down from 20 min/units to 10; resulting in a substantial improvement in balancing rate.  However, the production rate has to be reduced from one unit/15 minutes to one unit/16minute;

Alternative 2: Staying with 6 stations, see if a six-station balance could be obtained by cycle time less that 15 minutes Profile 3: C=13 Station123456 Tasks12,364,5,7,98,1011,12 Idle time111140  13 minutes appear to be the minimum cycle time with six station balance.  Increasing the number of stations from 5 to 6 results in a great improvement in production rate;  The efficiencies for profile 1~ 3 are 77.7%, 87.5%, and 89.7%. Thus the profile 3 is the best one.

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