OS Gateforum Section test

Question

Consider four processes with a burst time of $10, 20, 30, 40$ all process arrived at time $0$. Each process spends first $10$% of its execution time doing $i/0$, next $40$% time doing CPU operations, next $20$% time doing $I/0$ and the last $30$% time doing CPU operations. The system uses shortest remaining time next algorithm for scheduling. Calculate the completion time of $P3$?

Comments

Is it 43??

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No, I will tell the answer soon :), but please share your respective gantt charts.

Answer 1

p3 completes at 49

 

	Arrival	Burst	10% IO	40%CPU	20%IO	30%CPU
P1	0	10	1	4	2	3
P2	0	20	2	8	4	6
P3	0	30	3	12	6	9
P4	0	40	4	16	8	12

Gant chart

0--- no process--- -1--P1---5---P2---7---P1---10---P2----16----P3----20----P2----26-----P3----34----P4----40----P3----49----P4---59-----no process----67----P4---79

Comments

got it!! Thanks. Missed the part where it is mentioned shortest remaining time next algo.

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hey how u made the gant chart,  u are overlapping the i/o with other job computation . spooling technique, why u just not overlap the job own computation with i/o . pre fetching technique.

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Hi Pooja Palod,

I am sorry , but could not understand the gnatt chart. 

At time t=0 all processes arrive . Then P1 does its first IO till 1 , so nothing need to be done. next P1 does first CPU till 5. Then P1 goes for second IO till 8 , so no P1 now. 

Now , P1 has only 3 remaining. So , why you are scheduling P2 ? 

Can you please explain a little bit ?

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Process p1 does io from 5 to 7  so at that time we can schedule p2 now p1 needs 3 unit cpu burst so we preempt p2 and schedule p1