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FRIDAY, DECEMBER 23, 2011FRIDAY, DECEMBER 23, 2011

Explain and Solve : Priority Scheduling of OperatingSystem Concepts

If you haven't read/tried the earlier problems then click the links follow:

01. First Come First Served (FCFS). 02. Shortest Job First (SJF).

03. Round Robin (RR).

Priority Scheduling:This method is quite same as the SJF but the difference is that instead of choosing the nextprocess to work on by the shortest burst time, CPU chooses the next process by the shortest priority value. Here, all the

processes are given a priority value.The process with the shortest (The most shortest is 1) priority will be worked on

first and so on. Now consider a CPU and also consider a list in which the processes are listed as follows,

Arrival Process Burst Time Priority

0 1 3 2

1 2 2 1

2 3 1 3

Here, Arrival is the time when the process has arrived the list, Process Number is used instead of the process name,

Burst Time is the amount of time required by the process from CPU, Priority value is what we described in the first

paragraph. Well, as the unit of time you can take anything like nano-second, second, minute etc whatever. We

consider it as second.

Now for an instance, consider the above list as the ready queue for the CPU, that is the CPU will take processes from

this list and will process it.

Here in Priority Scheduling, what will happen is as follows,

AT 0s: There is only 1 job, that is Process-1 with Burst Time 3 and Priority 2. So CPU will do 1 second job of Process-

1. Thus Process-1 has 2s more job to be done.

AT 1s: Now there are 2 jobs,

Process-1 that arrived at 0s and has 2s job to be done with Priority 2.


Here, Priority of the job Process-2 is higher (Lower value of priority is higher, that is shortest priority value is

higher), So 1s job of Process-2 will be done. Thus process-2 has more 1s job.





Here, Process-2 has the highest prior ity, thus CPU will do 1s job of Process-2. So process-2 has more 0s job.




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Here, as Process-1 is with the highest priority (which is 2) so CPU will do 1s job of Process-1. So process-1

has more 1s job.

AT 4s: There are 2 jobs,



Again, CPU will do 1s job of Process-1 as it is with the highest priority . Thus Process-1 has 0s more job to be

done.

AT 5s: There is only 1 job, that is Process-3 with burst time 1 and prior ity 3. So CPU will do 1s job of Process-3. Thus

Process-3 has 0s more job to be done.

All Job Done.

We can show the above thing as the following time-line

Process-1 Process-2 Process-2 Process-1 Process-1 Process-3

0s 1s 2s 3s 4s 5s

A shortened view of the above time-line is as follows,

| Process-1 | Process-2 | Process-1 | Process-3 |0 1 3 5 6

So now came the main thing, Waiting Time. Ok, Look carefully,

See the following time-line for Process-1. Here, RED marked seconds symbolizes the seconds while Process-1 was

waiting and GREEN marked second symbolizes the seconds while Process-1 was working.


0s 1s 2s 3s 4s 5s

Thus Process-1 waited for 2 seconds.

Process-2 did not wait, it came at 1s and worked at 1s and 2s.

But Process-3 Arrived at the List of Process at 2s with a Burst Time of 1s. But CPU started processing Process-3 at 5s.

So process-3 waited for 3s.


0s 1s 2s 3s 4s 5s

So total waiting time = (Waiting Time of Process-1)+ (Waiting Time of Process-2)

+ (Waiting Time of Process-3)

= (2+0+3)s

= 5s

So the average waiting time is = (Total waiting time / Number of Processes)s

= (5/3)s

= 1.66s

Ok...I think you have understood the thing. Now lets talk about the program.

Input:You will ask the user for the number of processes. Then for each process you will take its Process Number,Arrival Time, Burst time and Priority. You dont have to worry, the number of processes wont be more than 5 or 6,

Arrival time of a process can only be equal or greater than the arrival time of its previous process and Process will be

entered as a serial number, so no problem. Priority values of 2 processes wont be same.

Output:In the output you will have to print out the shortened time-line we showed you above. For example, if the inputis as follows,


0 1 3 2

1 2 2 1

2 3 1 3

Then the shortened time-line will be,

| Process-1 | Process-2 | Process-1 | Process-3 |

0 1 3 5 6

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You can also print out the time-line vertically if you want (thats what I have done, see the output of my program)

Ok beneath this shortened time-line you will have to print a table as follows,

Process Arrival Finish Total Wait

2 1 2 2 0

1 0 4 3 2

3 2 5 1 3

Beneath this, you will have to show the Average Waiting Time.

I know you have understood, now see a screen-shot of exactly what I was talking about,

Exception:One exception is that, I said Arrival time of a process can only be equal or greater than the arrival timeof its previous process. So take a look at the following list of process,


0 1 3 2

55 2 2 1

60 3 1 3

You see, the Job of Process-1 will complete at 2s, thus from 3s to 54s the CPU will remain idle. Again the job of Process-2 will

complete at 56s and thus from 57s to 59s the CPU will again remain as idle. You have to show this in the output and also notice

that none of the processes has waited. See the following screenshot,

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Writen by Tanmay Chakrabarty

Post is About: My Works | Programs

You can download the program (.exe) I written By Clicking Here.

Our next Scheduling algorithm will be the Round Robin Scheduling Algorithm.

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20 comments:

Anonymous March 9, 2012 at 10:39 AM

nice explanation..

Reply

Tanmay Chakrabarty March 9, 2012 at 1:41 PMThank you. Have you tried to code it?

MEET0706 September 21, 2012 at 6:39 PM

thanx for sharing... :)

Reply

Tanmay Chakrabarty September 21, 2012 at 11:42 PM

Welcome....share if you have done this....

Saurabh November 23, 2012 at 11:01 AM

Great work man!!! !

Reply

Tanmay Chakrabarty November 23, 2012 at 12:20 PM

Thanks man...

sami malik January 22, 2013 at 5:07 PM

Amazing Work dude i must Say Good effort

Reply

Tanmay Chakrabarty January 22, 2013 at 5:44 PM

Thanks. Have you tried it ...

VASANTH February 18, 2013 at 9:46 PM

my doubt was cleared.keep it up tanmay

Reply

Tanmay Chakrabarty February 20, 2013 at 2:44 PM
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Nice to know that these notes are helping ...

Erin EstradaAugust 1, 2013 at 9:48 AM

i have 3 process:

Process BurstTime Arrival Time Priority

1 1 1 2

2 5 3 1

3 5 0 3

this is the result:

P1 IDLE P2 P3

1 2 3 8 13

the question is, how come that P1 was processed first when it is actually 2nd in its priority?

Reply

Tanmay Chakrabarty August 1, 2013 at 1:00 PM

Its because, P1 arrived at 1s when there was no more processes(P2 came at 3s), thus CPU process 1s of P1 at 1s,

then remained IDLE at 2s, then the P2 arrived and got processed.

Also NOTE: I stated, in my program each process will have Arrival Time greater or equal to its previous process. You

gave Arrival Time 0 to P3 where P2 is having Arrival Time 3, which is a wrong input. I thought these limitations will be

challenges for those who will code it later.

Erin EstradaAugust 2, 2013 at 10:05 AM

oh i see, thanks for the help! by the way i am done with the round robin scheduling using magic software and i really

wanna thank you and your notes for helping me :3

Tanmay Chakrabarty August 4, 2013 at 11:42 AM

You are welcome ...

Anonymous September 3, 2013 at 10:04 AM

how i wish i can output the same as what you have. We will have a program Priority scheduling for our final. I do hope this blog

can help. Thanks and more power to you sir.

Reply

Tanmay Chakrabarty September 3, 2013 at 10:54 AM

I am sure you will be able to and nice to that these notes and posts are helping other students.

Himanshu Maurya September 23, 2013 at 7:11 AM

Great work bro..this is very good site of study ,everything is in very simple way.easy to understand.i like this site very much.... :)

Reply

Tanmay Chakrabarty September 23, 2013 at 10:37 AM

Thank you for your complements.

Anonymous October 3, 2013 at 12:45 AM

A lower priority job can block a high priority one (e.g., priority inversion

problem) even a priority scheduling is used ..

Reply

Tanmay Chakrabarty October 3, 2013 at 12:54 AM

Haven't studied Priority Inversion Problem....
http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1380740082269#c4230859123982329771http://www.blogger.com/profile/08084800570982709032http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1380739552338#c1580000485136031712http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1379911039239#c3878960107555709136http://www.blogger.com/profile/08084800570982709032http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1379898693618#c7230845483566252998http://www.blogger.com/profile/06892734899042067838http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1378184048441#c5469642840519676446http://www.blogger.com/profile/08084800570982709032http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1378181055920#c1034761119001897571http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1375594947201#c5611276415333025578http://www.blogger.com/profile/08084800570982709032http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1375416344973#c8459315663172879272http://www.blogger.com/profile/06721345191331865147http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1375340402146#c2840370568968247204http://www.blogger.com/profile/08084800570982709032http://tanmayonrun.blogspot.com/2011/12/explain-and-solve-priority-scheduling.html?showComment=1375328891879#c3968552641476092410http://www.blogger.com/profile/06721345191331865147


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