cpu scheduling -SRTF with I/O example

Question

Pno.      AT      BT1    I/O    BT2

1           0         3        2       2

2           0         2        4       1

3           2         1        3       2

4           5         2        2       1

 

We are using SRTF  scheduling algorithm with I/O. Please give me the gantt chart.

Comments

@arjun sir please provide me gantt chart sir

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I have constructed the gantt chart . check whether the sequence is correct or not.

P2 ->(0-2)

P3->(2-3)

P1->(3-6)

P2->(6-7)

P4->(7-9)

P3->(9-11)

P4->(11-12)

P1->(12-14)

At time =7 units,P3 and P4 are available in the ready queue. Recent  AT of P3=6 and arrival time of P4=5. Choose P4 since burst time of P3= burst time of P4.

At time =9 units, similar case arises between P1 and P3.

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at time=7, p3 and p4 are available

AT of p3=2

AT of p4=5 as given in question

so, select p3.

do not get confused btw GIVEN AT  and THE I/O COMPLETION TIME, which is not equal to AT.

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I am getting confused because of the following two lectures . I solved this problem based on these lectures.

They are  keep on updating the arrival times and making decisions based on the updated arrival times.I also did in the same way what they have done.

Link 1:https://www.youtube.com/watch?v=rWEOJR5ZXFc

Link 2:https://www.youtube.com/watch?v=FYcc9D8llF0&index=18&list=PLgre7dUq8DGKbtnlMuJPvPYlvLdXOC9uh

I want to know whether these 2 lectures are wrong.

please help me.

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this is wrong!

Answer 1

Hope this helps,

Comments

According process scheduling, it is based on complete burst time.

correct me if Iam wrong.

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There is no other option. Because this is not mentioned in any of the text books. The only hope is @Arjun suresh sir. we have to ask him regarding this issue.

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@AnilGoudar , here you assumed that i/o resource will we available at required time to all processes?

I'm trying for I/O waiting queue (considering only 1 i/o device is available).

Answer 2

P3 WILL BE SELECTED AT TIME=7 because it has lower process id than p4.

then p1 will be selected as it has lesser BT2 then the BT1 OF P4

IDLE- P4 is gone for i/o for 2 units of time.

Comments

At time t=7 units, we have to consider the arrival time before we consider the process id. The arrival time of p3=6 units and the arrival time of p4=5 units. so only i have decided to choose p4 instead of p3.