This document discusses flow shop scheduling, which involves sequencing jobs through a set of machines or processes where each job must visit the machines in the same order. It defines flow shop scheduling and provides examples of its applications in industries that require a strict production order. The document then describes two common methods for flow shop scheduling - preemptive and non-preemptive - and provides an example to illustrate the difference. Finally, it discusses algorithms for solving flow shop scheduling problems, focusing on Johnson's algorithm, which provides an optimal solution for problems with two machines.

1. Flow shop scheduling
By- Kunal Goswami
Mentor – Joy Chandra Mukherjee
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2. Content :
Definition
Application
Methods
Algorithm
Conclusion
Reference
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3. Definition
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4. Definition
• Flow shop scheduling problems, are a class
of scheduling problems with a workshop in which the
flow control shall enable an appropriate sequencing for
each job and for processing on a set of machines or with
other resources 1,2,...,m in compliance with given
processing orders.
• For an operating system – we have many task to be done.
Especially the maintaining of a continuous flow of
processing tasks is desired with a minimum of idle
time and a minimum of waiting time.
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5. Application
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6. Example
Practical application :
As we have seen strict order of all operations to be
performed on all jobs. Flow shop scheduling may apply as
well to production facilities as to computing designs.
• It is used in processing industry where a strict order of
production should be done
• In can be used in construction work.
• In medical observation
In space technology, agricultural processing basically in
every (multiprograming + strict order job).
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7. Application
It works in following case in steps :
i. we should have multiprogramming environment
ii. input and then executed
iii. Job queued for O/P
iv. O/P printed/used
• Suppose there are 1 to n job and there are m processor and we
have some more information like for task T3i for processor 3.
And it takes tj time.
Note - m >= n
• No more then 1 task for any processor at a time.
• N jobs requiring m tasks and each to be scheduled in m
processor. 7

8. Application
So in diagram view situation is like this.
Assume we have 4 jobs and each jobs can have 5 tasks.
Jobs J1 J2 J3 J4
Tasks
t1
t2
t3
t4
t5
3
1
4
2
2
2
0
3
2
2
0
1
4
2
1
1
1
3
2
1
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9. Usefuldata:
The finish time of fi(S) of job i is the time at
which all task of job i has been completed in
schedule S.
• The finish time F(S) of a schedule S is :
F(S) = max {fi(S)}
• The mean flow time MFT(S) is :
MFT(S) = (1/n). ∑ fi(S)
We try to get optimal finish time i.e. minimal
finish time. 9

10. Methods
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11. Above scheduling problem is done by two method
Two possible solution : i) pre-emptive
ii) non preemptive.
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12. i) pre-emptive Job
In it we do permits preemption (halting) of
tasks, from a cooperative multitasking system
wherein processes/tasks must be explicitly
programmed to yield when they do not need
system resources.
Rules :
a) tji cannot start unit tj-1,i finishes.
b) Task ti of any job will go to processor i only.
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13. Example
J1 J2
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14. Example
First task T1 is coming from two jobs we have.
t11 = 2 and t12 = 0
Both will be processed sequentially ;
all task-1 will complete in 2 sec.
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15. Example
Now Task-2 comes ;
T21 = 3 and T22 = 3
Task-2 of Job-1 cannot start as it’s first Task-1 needed to
be completed. So Task-2 of Job-2 starts.
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16. Example
Now Task-3 comes ;
T31 = 5 and T32 = 2
Task-3 of Job-1 cannot start as it’s first Task-2 needed
to be completed.
So is Task-3 of Job-2 which require Task-2 to complete.
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17. Example
Finish time = Max (F1 , F2) = max (5,11) = 11
5 and 11 are finish time of respective jobs.
Mean flow time = ½ .(5+11) = 8
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18. Example
ii) pre-emptive Job
Preemptive Scheduling is a CPU scheduling technique
that works by dividing time slots of CPU to a given
process.
When the burst time of the process is greater than CPU
cycle, it is placed back into the ready queue and will
execute in the next chance. This scheduling is used when
the process switch to ready state.
Rules :
a) tji cannot start unit tj-1,i finishes.
b) Task ti of any job will go to processor i only.
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19. Example
J1 J2
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20. Example
First task T1 is coming from two jobs we have.
t11 = 2 and t12 = 0,
It donot have any conflict job in machine. So execute
normally.
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21. Example
Task T2 comes.
t21 = 3 and t22 = 3,
• t21 , t22 comes. As t11 not complete. So t21 goes into
queue. And t22 starts.
• After t22 finishes t21 again complete it’s remaining cycle.
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22. Example
Task-3 comes.
t31 = 5 and t32 = 2,
• t31 , t32 comes. As t21 not complete. So t31 goes into
queue. Also same with t32 .
• t31 starts after 5 unit time. Then t32 completes.
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23. Example
Finish time = Max (F1 , F2) = max (10,12) = 12
10 and 12 are finish time of respective jobs.
Mean flow time = ½ .(10+12) = 11
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24. Algorithm
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25. Algorithms
• The machine sequence of all jobs is the same. The
problem is to find the job sequences on the machines
which minimize the makespan , i.e. the maximum of the
completion times of all tasks.
• It is similar to 2-machine problem with 2 machines to
solve different burst jobs. It is well known that in case of
real time situations - the problem is NP-hard.
• There are some optimized algorithms which uses
Dynamic algorithm, Branch and Bound and Heuristic
algorithm such as genetic algorithm.
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26. Algorithms
 Popular algorithms are :
• Johnson algorithm – nlogn
• GS algorithm (Gonzalez and sahni)
• Genetic algorithm
 n.Logn is most optimized solution for finite no. of tasks.
 We will see Johnson algorithm.
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27. Algorithms
Johnson's Algorithm for 2 machine
Step 1 : Form set1 containing all the jobs with p1j < p2j -- (nlogn)
Step 2 : Form set2 containing all the jobs with p1j > p2j, the jobs with
p1j=p2j may be put in either set. -- (n)
Step 3 :
Form the sequence as follows:
(i) The job in set1 go first in the sequence and they go in increasing
order of p1j(SPT) – shortest process time -- O.(c)
(ii) The jobs in set2 follow in decreasing order of p2j (LPT). Ties are
broken arbitrarily. – longest process time -- O.(c) 27

28. Example
Let’s try Johnson's algorithm in our example :
We remember it had took us 11 unit for non-preemptive
and 12 unit for pre-emptive scheduling using naive
process.
Our job sequence was :
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J1 J2

29. Example
t1 t2 t3
J1 2 3 5
J2 0 3 2
step 1 – find least among all
• 0 is ans.
• cut that column and put in array from left or right
according to job1 or job2.
• 0 task is of J2. So choose from right
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J1 J2

30. Example
t2 t3 (task-1 removed)
J1 3 5
J2 3 2
step 1 – find least among all now
• 2 is ans.
• cut that column and put in array from left or right
according to job1 or job2.
• 0 task is of J2. So choose from right
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J1 J2

31. Example
t2 (task-3 removed)
J1 3
J2 3
step 1 – find least among all now
• 3 is ans.
• cut that column and put in array from left or right
according to job1 or job2.
• 0 task is of J2. So choose from right
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J1 J2

32. Example
Above is job sequence for algorithm
Now make table including 2 machines
So, 10 is execution time of give sequence of task using
johnson algorithm which is better from earlier.
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J1 J2

33. Conclusion &
future work
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34. Conclusion
• We got to know about popular scheduling algorithm of
tasks in multiprogramming environment.
• A lot of work is being done in GS and genetic algorithm.
Also it is tried with many other field then scheduling.
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35. Reference
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36. Reference
• https://en.wikipedia.org/wiki/Flow_shop_scheduling
• https://www.youtube.com/watch?v=R08ql752oL0
• https://www.sciencedirect.com/science/article/abs/pii/S03608
35299000236
• https://www.sciencedirect.com/science/article/pii/S14746670
15357499#:~:text=Johnson'%20algorithm%20(JA)%20is,algorit
hms%20for%20more%20general%20cases.
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37. THANKU
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