1
GATE CSE 2014 Set 3
Numerical
+2
-0
An operating system uses $$shortest$$ $$remaining$$ $$time$$ $$first$$ scheduling algorithm for pre-emptive scheduling of processes. Consider the following set of processes with their arrival times and $$CPU$$ burst times (in milliseconds):
Process Arrival Time Burst Time
P1 0 12
P2 2 4
P3 3 6
P4 8 5

The average waiting time (in milliseconds) of the processes is ________.

Your input ____
2
GATE CSE 2014 Set 1
Numerical
+2
-0
Consider the following set of processes that need to be scheduled on a single $$CPU.$$ All the times are given in milliseconds.
Process Name Arrival Time Execution Time
A 0 6
B 3 2
C 5 4
D 7 6
E 10 3

Using the $$shortest$$ $$remaining$$ $$time$$ $$first$$ scheduling algorithm, the average process turnaround time (in $$msec$$) is _______.

Your input ____
3
GATE CSE 2012
MCQ (Single Correct Answer)
+2
-0.6
Consider the $$3$$ processes, $$P1,$$ $$P2$$ and $$P3$$ shown in the table.
Process Arrival Time Time Units
Required
P1 0 5
P2 1 7
P3 3 4

The completion order of the $$3$$ processes under the policies $$FCFS$$ and $$RR2$$ (round robin scheduling with $$CPU$$ quantum of $$2$$ time units) are

A
$$FCFS:P1,P2,P3\,\,\,\,RR2:P1,P2,P3$$
B
$$FCFS:P1,P3,P2\,\,\,\,RR2:P1,P3,P2$$
C
$$FCFS:P1,P2,P3\,\,\,\,RR2:P1,P3,P2$$
D
$$FCFS:P1,P3,P2\,\,\,\,RR2:P1,P2,P3$$
4
GATE CSE 2011
MCQ (Single Correct Answer)
+2
-0.6
Consider the following table of arrival time and burst time for three processes $$P0,P1$$ and $$P2$$.
Process Arrival Time Burst Time
P0 0 ms 9 ms
P1 1 ms 4 ms
P2 2 ms 9 ms

The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only a arrival or completion of processes. What is the average waiting time for the three processes?

A
$$5.0$$ $$ms$$
B
$$4.33$$ $$ms$$
C
$$6.33$$ $$ms$$
D
$$7.33$$ $$ms$$
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CBSE
Class 12