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Scheduling ( Part II ). Mohd Zaidan bin Abdul Aziz B050810281 Grace Hii Leh Ung B050810211 Low Mei Ching B050810215 Sawita bt Amir B050810051. - PowerPoint PPT Presentation
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Mohd Zaidan bin Abdul Aziz B050810281Grace Hii Leh Ung B050810211Low Mei Ching B050810215Sawita bt Amir B050810051
SCHEDULING (Part II)
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1. Shortest Processing Time (SPT) Minimizes the mean flow time Minimizes waiting time Minimizes lateness for single-machine sequencing.2. Earliest Due Date Scheduling Minimizes the maximum lateness
4.4 Sequencing Theory for Single Machine
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3. Moore’s algorithm Minimizes number of tardy jobsStep 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.
4.4 Sequencing Theory for Single Machine
d[1] d[2],…, d[n]
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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: Consider number of tardiness jobs rather than tardiness
Reason: Largest effect on the tardiness of the Job[i]
4.4 Sequencing Theory for Single Machine
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Job 1 2 3 4 5 6
Due date 15 6 9 23 20 30
Processing time 10 3 4 8 10 6
Job 2 3 1 5 4 6
Due date 6 9 15 20 23 30
Processing time 3 4 10 10 8 6
Completion time 3 7 17 27 35 41
EXAMPLE 1:
Solution:
Longest processing time
4.4 Sequencing Theory for Single Machine
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Job 2 3 5 4 6
Due date 6 9 20 23 30
Processing time 3 4 10 8 6
Completion time 3 7 17 25 31
EXAMPLE 1: Solution (cont.) Longest processing time
Job 2 3 4 6
Due date 6 9 23 30
Processing time 3 4 8 6
Completion time 3 7 15 21
Optimal sequence: 2, 3, 4, 6, 5, 1 or 2, 3, 4, 6, 1, 5Number of tardy jobs is two in either case.
No lateness
4.4 Sequencing Theory for Single Machine
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Exercise 4.4.1:Seven jobs are to be processed through a single machine. The processing times and due dates are given below.
Job 1 2 3 4 5 6 7
Processing time 3 6 8 4 2 1 7
Due date 4 8 12 15 11 25 21
Determine sequence of the jobs in order to minimize a) Mean flow timeb) Number of tardy jobsc) Maximum lateness
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4. Lawler’s Algorithm Minimize maximum lateness Minimize maximum tardinessSubject to precedence constraint ( certain job must be completed before other jobs can begin)
)(maxmin1
iini
Fg
iiiii LdFFg )(
)0,max()( iiii dFFg
4.4 Sequencing Theory for Single Machine
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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:
n
i i
ivik
t
gg
1
))((min)(
Example: Job among V that has smallest tardiness, if arranged on position [n].
Processing time of current sequence.
4.4 Sequencing Theory for Single Machine
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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 tk and job scheduled next to last is now determined.Step 4:Process is continued until all jobs are scheduled.
4.4 Sequencing Theory for Single Machine
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1 2 3
4 5
6
Job 1 2 3 4 5 6
Processing time 2 3 4 3 2 1
Due date 3 6 9 7 11 7
EXAMPLE 2:
4.4 Sequencing Theory for Single Machine
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Solution:Step 1: Find the job scheduled last(sixth)
Job 1 2 3 4 5 6
Processing time 2 3 4 3 2 1
Due date 3 6 9 7 11 7
τ = Total processing time = 2+3+4+3+2+1 = 15
Job 3 5 6
Tardiness 15-9=6 15-11=4 15-7=8
Min value
Hence, Job 5 scheduled last
4.4 Sequencing Theory for Single Machine
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Solution: Cont.Step 2: Find the job scheduled fifth.
1 2 3
4 5
6
Job 1 2 3 4 6
Processing time 2 3 4 3 1
Due date 3 6 9 7 7
τ = New total processing time = 15 – 2 = 13
Job 3 6Tardiness 13-9=4 13-7=6
Min valueHence, Job 6 scheduled fifth.
4.4 Sequencing Theory for Single Machine
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Solution: Cont.Step 3: Find the job scheduled fourth.
1 2 3
4 5
6
Job 1 2 4 6
Processing time 2 3 3 1
Due date 3 6 7 7
τ = New total processing time = 13 – 4 = 9
Hence, Job 6 scheduled fourth.
Job 2 6
Tardiness 9-6=3 9-7=2
Min value
4.4 Sequencing Theory for Single Machine
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Solution: Cont.Step 4: Find the job scheduled third.
1 2 3
4 5
6
Job 1 2 4
Processing time 2 3 3
Due date 3 6 7
2 4
Tardiness 8-6=2 8-7=1
τ = New total processing time = 9 – 1 = 8
Hence, Job 4 scheduled third.Min value
4.4 Sequencing Theory for Single Machine
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Solution: Cont.Step 5: Find the job scheduled second.
1 2 3
4 5
6
Job 1 2
Processing time 2 3
Due date 3 6
Job Processing time
Flow time Due date Tardiness
124635
233142
2589
1315
36779
11
001244
The optimal sequence: 1-2-4-6-3-5
Maximum tardiness
4.4 Sequencing Theory for Single Machine
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Exercise 4.4.2:Eight jobs are to be processed through a single machine. The processing times and due dates are given below.
Furthermore, assume the following precedence relationships must be satisfied:
Job 1 2 3 4 5 6 7 8
Processing Time 2 3 2 1 4 3 2 2
Due date 5 4 13 6 12 10 15 19
2 6 3
1 4 7 8
Determine the sequence in which jobs should be done in order to minimize maximum lateness subject to precedence restriction.
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• 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
4.5 Sequencing Theory for Multiple Machines
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• How to implement this rule:1. List value of A and B in to columns2. 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
3. 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.
4. Cross off the jobs as they are scheduled.
1. Johnson’s algorithm : Scheduling n jobs on two machines
4.5 Sequencing Theory for Multiple Machines
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EXAMPLE:• Five jobs are to be scheduled on two machines. The processing
time are:
Job Machine A Machine B1 5 22 1 63 9 74 3 85 10 4
Machine A A2 A4 A3 A5 A1
Machine B B2 B4 B3 B5 B1
0 1 4 7 13 15 22 23 27 28 30
2-4-3-5-1
4.5 Sequencing Theory for Multiple Machines
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Exercise 4.5.1: Six job are to be schedule on two machine. The processing time are
Job Machine A Machine B1 20 272 16 303 43 514 60 125 35 286 42 24
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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
4.5 Sequencing Theory for Multiple Machines
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EXAMPLE: Machine
Job A' B'1 9 132 15 163 10 84 9 105 9 15
MachineJob A B C1 4 5 82 9 6 103 8 2 64 6 3 75 5 4 11
Min A = 4 Max B = 6 Min C = 6 1-4-5-2-3
Machine A A1 A4 A5 A2 A3
Machine B B1 B4 B5 B2 B3
180 9 63 7122 27 32 42 47 52
4.5 Sequencing Theory for Multiple Machines
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Exercise 4.5.2:The following four job must be processed through a three-machine flow shop
MachineJob A B C1 4 2 62 2 3 73 6 5 64 3 4 8
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3. Two job flow shop problem• Two jobs are processed through m machines.• Present graphical procedure for solving this problem. The step are:1. 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.
2. Block out areas corresponding to each machine at the intersection of the intervals marked for the machine on the two axes.
3. 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.
4.5 Sequencing Theory for Multiple Machines
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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:
Job 1 Job 2Operation Time Operation Time
Sanding (A) 3 A 2Lacquering (B) 4 B 5Polishing ( C) 5 C 3
4.5 Sequencing Theory for Multiple Machines
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A1 A2
B1 B2
C1 C2
3
531 42 1196 7 8 10 12
12
45
7
9
6
8
10
B
A
C
2 4 6 8 10 12 14
CB
Gantt ChartSolution
A
Total time = 12 + 3 = 15
Total time = 12 + (2 + 2) = 16
4.5 Sequencing Theory for Multiple Machines
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3
531 42 1196 7 8 10 12
12
45
7
9
6
8
10
B
A
C Total time = 10 + 6 = 16
Total time = 10 + (3 + 2) = 15
2 4 6 8 10 12 14
CB
Gantt ChartSolution
A A1 A2
B1 B2
C1 C2
4.5 Sequencing Theory for Multiple Machines
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Exercise 4.5.3:Two jobs must be proceeds through four machines in same order. The processing time in the required sequence are
Job 1 Job 2Machine Time Machine Time
A 5 A 2B 4 B 4C 6 C 3D 3 D 5
Determine how the two job should be schedule in order to minimize the total makespan and draw the gantt chart indicating the optimal schedule.
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Line balancing: the process of assigning tasks to workstations in such a way that the workstations have approximately equal time requirements.
Line Balancing
Cycle Time
Cycle time: maximum time allowed at each workstation tocomplete 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.
Maximum cycle time = summation of the task times.Minimum cycle time = the longest task times.
4.6 ASSEMBLY LINE BALANCING
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Introduction (cont.)Factors that contribute to the
difficulty of the problem:- There are precedence
constraints- Some tasks cannot be
performed at the same workstation
DOT = timecycle = CT
CTOT = D rate,Output
rateoutput desired = Ddayper timeoperating OT
4.6 ASSEMBLY LINE BALANCING
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• 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.
Some Heuristic (intuitive) Rules:Line Balancing Rules
4.6 ASSEMBLY LINE BALANCING
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Precedence diagram: Tool used in line balancing to display elemental tasks and sequence requirements
A Simple Precedence Diagrama b
c d e
0.1 min.
0.7 min.
1.0 min.
0.5 min. 0.2 min.
Figure 4.6.1
Precedence Diagram
4.6 ASSEMBLY LINE BALANCING
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task timeof sum = tCT
t)( = N
Determine the Minimum Number of Workstations Required
What is the minimum number of workstations for the previous precedence diagram? (assume minimum cycle time)
mins 2.5 = t
35.2min 1
mins 2.5 = N
4.6 ASSEMBLY LINE BALANCING
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WorkstationTimeRemaining Eligible
AssignTask
RevisedTime Remaining
StationIdle Time
1 1.00.90.2
a, ccnone
ac–
0.90.2
0.2
2 1.0 b b 0.0 0.0
3 1.00.50.3
de–
de–
0.50.3 0.3
0.5
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
4.6 ASSEMBLY LINE BALANCING
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%100*(N)(CT)cycleper timeIdle
= timeidle %
Efficiency = 100% – Percent idle time
Calculate Percent Idle Time
What’s the % idle time and efficiency for the above example?
%7.16%100*(3)(1.0)
0.5 = timeidlePercent
Efficiency = 100% - 16.7% = 83.3%
4.6 ASSEMBLY LINE BALANCING
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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
EXAMPLE:
4.6 ASSEMBLY LINE BALANCING
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Assign the tasks above to workstations in the order of greatest positional weight.
Steps:1) Arrange the task in the decreasing order of positional weights.2) Find out the number of workstations
Number of workstations = = = 3.2 =4
timecycle s task timeof sum
1445
Task Task time
F 5
D 7
G 6
A 3
B 2
C 4
E 4
H 9
I 5
Arrange tasks in decreasing order of positional weight
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%64.19
%100*(4)(14)
11 = timeidlepercent
D=7
Cycle Time=14 min
F=5 G=6 A=3 B=2
C=4 E=4I=5H=9
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Stati
on
I
II
III
IV
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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:
a) In the order of most following task.b) In the order of greatest positional weight.
EXERCISE 4.6.1
4.6 ASSEMBLY LINE BALANCING
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EXAMPLE:
4.6 ASSEMBLY LINE BALANCING
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The job times & precedence relationship for this problem:
Task ImmediatePredecessor
Time
1 - 122 1 63 2 64 2 25 2 26 2 127 3,4 78 7 59 5 1
10 9,6 411 8,10 612 11 7
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Suppose that the company is willing to hire enough workers to produce one assembled machine every 15 minutes.
Sum of task time = 70 minutesMinimum 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.
TASK POSITIONAL WEIGHT
1 70
2 58
3 31
4 27
5 20
6 29
7 25
8 18
9 18
10 17
11 13
12 7
TASK POSITIONAL WEIGHT
1 70
2 58
3 31
6 29
4 27
7 25
5 20
8 18
9 18
10 17
11 13
12 7
Arrange tasks in decreasing order of positional weight
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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 stationsUse C = 16 min
No. of stations decreases 16 %, cycle time increases 7%Assume production day = 7 hrsC = 15 daily production level = 28 units/assembly operation
Station 1 2 3 4 5 6 Total idle timeTasks 1 2,3,4 5,6,9 7,8 10,11 12
Idle time 3 1 0 3 5 8 20
Station 1 2 3 4 5 Total idle timeTasks 1 2,3,4,5 6,9 7,8,10 11,12
Idle time 4 0 3 0 3 10
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C = 16 daily production level = 26.25 units/assembly operationManagement 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/operationIncreasing No. of stations from 5 to 6 substantial improvement in the throughput rate.
Station 1 2 3 4 5 6 Total idle timeTasks 1 2,3 6 4,5,7,9 8,10 11,12
Idle time 1 1 1 1 4 0 8
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EXERCISE 4.6.2
Consider the assembly line balancing problem represented by the figure above. Determine a balance for a. C = 20
4.6 ASSEMBLY LINE BALANCING
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1. Single machine Uncertainty of processing times Exact completion time of one or more jobs may not be
predictableObjective: Minimize expected average weighted flow
time. Job i precedes job i+1 if
4.7 Stochastic Scheduling 4.7.1 Static Analysis
where job times are t1, t2, …, tn and weights are u1, u2, …, ui.
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Jobi 1 2 3 4 5 6 7 8 9 10
Processing Time, ti 1 10 5 2 8 7 8 4 3 6
Importance weight, ui 3 2 1 1 4 2 3 3 2 4
Due date, di 10 20 15 10 10 25 15 25 10 20
4.7.1 Stochastic Scheduling: Static Analysis
EXAMPLE:
Find the optimal sequence that minimize expected average weighted flow time.
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Jobi ti ui ti/ui di Ci Ti
1 1 3 0.3 10 1 08 4 3 1.3 25 5 09 3 2 1.5 10 8 0
10 6 4 1.5 20 14 04 2 1 2 10 16 65 8 4 2 10 24 147 8 3 2.7 15 32 176 7 2 3.5 25 39 143 5 1 5 15 44 292 10 2 5 20 54 34
∑Ci = 237
∑Ti= 114
4.7.1 Stochastic Scheduling: Static Analysis
Solution:
Expected average weighted flow time = 237/10= 23.7
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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
4.7.1 Stochastic Scheduling: Static Analysis
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Dynamic: Jobs arrive randomly over time, and decisions must be made on an ongoing basis as how to schedule those jobs.When jobs arrive shop dynamically over time, queuing theory provides a means of analyzing the results. Standard M/M/1 queue applies to case of purely random arrivals to single machine with random processing times.
4.7.2 Stochastic Scheduling: Dynamic Analysis
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Selection independent of job processing times Same mean flow times, but differ variance
flow timesSelection dependent of job processing times Job times realized when job joins the queue
rather than when job enters service SPT results lowest expected flow time.
4.7.2 Stochastic Scheduling: Dynamic Analysis
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4.8.1 FINITE CAPACITY PLANNING (FCS)FCS - jobs are being scheduled through a number of work centers, each with one or more machines.• jobs can pass each other or change their order as they are processed, depending on their priority.• a job can be split into two or more parts if this will facilitate scheduling
Advantage:-The addition of capacity improves completion dates of jobs & reduces waiting times.
Bottleneck: a work centre whose capacity is less than the demand placed on it & less than the capacities of all other resources.
4.8 ADVANCED TOPIC FOR OPERATIONS SCHEDULING
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Batch operation:Job shop “shop”, “job”, “work center”Job “customer”, “patient”, “client”, “paperwork”Work centre “room”, “office”, “facility”, “skill specialty”
In a job shop, batch corresponds to what customer orders – can include one or several parts or items.
Each part or job is scheduled through the various machines & work centres according to the equipment & labour needed to process the job.
4.8.2 BATCH SCHEDULING
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Batch scheduling:• each batch flowing through a batch process typically moves along with many starts & stops, not smoothly – due to layout of batch process jobs or customers wait in line as each batch is transferred from one work centre to the next.
• batch scheduling problem: network or queues.
•Jobs or customers spend most of their time waiting in line – amount of time waiting varies with the load of the process
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Thank You ^^