Theory-of-Computation
March 6, 2024Question 4074 – UGC NET CS 2005 Dec-Paper-2
March 6, 2024Process-Scheduling
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Question 20
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Consider the following four processes with arrival times (in milliseconds) and their length of CPU bursts (in milliseconds) as shown below:
These processes are run on a single processor using preemptive Shortest Remaining Time First scheduling algorithm. If the average waiting time of the processes is 1 millisecond, then the value of Z is _____.
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2
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7
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1
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4
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Question 20 Explanation:
This is the Gantt chart till time = 4 units
At this point P4 arrives with burst ‘Z’ & P3 is in queue with burst 3.
P1 & P2 have executed with P1 incurred delay 1unit & P2 0units.
Hence, Avg = 0+1+x/4 =1
⇒ x=3, the next delay should be 3. It would happen if assume Z=2.
It executes and completes at 6.
P3 will wait totally for 3units.
Hence, Avg=1.
Z=2
At this point P4 arrives with burst ‘Z’ & P3 is in queue with burst 3.
P1 & P2 have executed with P1 incurred delay 1unit & P2 0units.
Hence, Avg = 0+1+x/4 =1
⇒ x=3, the next delay should be 3. It would happen if assume Z=2.
It executes and completes at 6.
P3 will wait totally for 3units.
Hence, Avg=1.
Z=2
Correct Answer: A
Question 20 Explanation:
This is the Gantt chart till time = 4 units
At this point P4 arrives with burst ‘Z’ & P3 is in queue with burst 3.
P1 & P2 have executed with P1 incurred delay 1unit & P2 0units.
Hence, Avg = 0+1+x/4 =1
⇒ x=3, the next delay should be 3. It would happen if assume Z=2.
It executes and completes at 6.
P3 will wait totally for 3units.
Hence, Avg=1.
Z=2
At this point P4 arrives with burst ‘Z’ & P3 is in queue with burst 3.
P1 & P2 have executed with P1 incurred delay 1unit & P2 0units.
Hence, Avg = 0+1+x/4 =1
⇒ x=3, the next delay should be 3. It would happen if assume Z=2.
It executes and completes at 6.
P3 will wait totally for 3units.
Hence, Avg=1.
Z=2