Os doubt cpu scheduling

Question

Consider the following table showing the arrival time burst time before io time and followed by CPU time again

Process	AT	execution	time
		Cpu	io	cpu
1	0	4	3	2
2	1	1	4	1
3	4	3	1	1

Using srtf scheduling policy what is the average completion time??

Comments

WT = total TAT - ( CPU + IO ) 

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So is it 1.67?

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Tuhin Dutta  yes 5 /3   https://gateoverflow.in/75241/scheduling  check here 

Answer 1

Here we need to keep two things in mind :

a) As we know in "I/O wait state" , there are more than processes possible at the same time. Whereas in "running" state , we can have only one process . Hence if more than 1 processes are there for performing I/O and hence consume I/O time , they can do so simultaneously which means if more than one process have I/O time to execute , then both can do so simultaneously . Whereas for CPU time , we need CPU scheduling.

b) If two processes have same CPU burst time as well as same arrival time  , then the process having less process-id is given priority.

Thus keeping these points in mind , let us find the average completion time :

Comments

Thankyou...:)

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your explanation is very nice but i have one doubt why you considering cpu io cpu time saperatly. when process arrive in ready queue cpu desnt knw saperate time like for four min p1 needs cpu then io. so we should consider total time (cpu io cpu) while scheduling process p3 should be schedule.

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arch CPU also doesn't know how much time a process will take for CPU and I/O and if algorithm is predicting time then it can also predict both separately(cpu and i/o).

Explanation by Habibkhan is correct.

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at t=8 both P1 and P3 have remaining  BT=2 so P1 should be scheduled because it has lower id??

Answer 2

Here, criteria for scheduling in Gannt chart is total CPU Brust time if Process in ready state, not in I/O queue (I/o brust).

During design, Gantt chart keeps regulary update the remaining CPU burst time of process for criteria.

Process	Completion time	Turn Around Time(TAT)	Waiting time(TAT- CPU brust time)
P1	7	7	1
P2	8	7	5
P3	13	9	5

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P1

P2

P1

P1

P1

P1

P2

P3

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P3

0        1           2         4           5       6           7         8         11         12       13

                  Gantt Chart

Average waiting time = (1+5+5)/3 = 11/3

Between 11 to 12 unit time the CPU is in idle state.

Note: If you are able to make a correct Gantt chart, you solve any complex question easily.