Consider four processes P, Q, R and S scheduled on a CPU as per round robin algorithm with a time quantum of 4 units. The processes arrive in the order P, Q, R, S, all at time t = 0. There is exactly one context switch from S to Q, exactly one context switch from R to Q, and exactly two context switches from Q to R. There is no context switch from S to P. Switching to a ready process after the termination of another process is also considered a context switch. Which one of the following is NOT possible as CPU burst time (in time units) of these processes?
A
P = 4, Q = 10, R = 6, S = 2
B
P = 2, Q = 9, R = 5, S = 1
C
P = 4, Q = 12, R = 5, S = 4
D
P = 3, Q = 7, R = 7, S = 3
2
GATE CSE 2020
Numerical
+2
-0
Consider the following set of processes, assumed to have arrived at time 0. Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order P1, P2, P3, P4.
If the time quantum for RR is 4 ms, then the absolute value of the difference between the average turnaround times (in ms) of SJF and RR (round off to 2 decimal places) is _____.
Your input ____
3
GATE CSE 2019
Numerical
+2
-0
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 _____.
Your input ____
4
GATE CSE 2016 Set 2
Numerical
+2
-0
Consider the following processes, with the arrival time and the length of the CPU burst given in milliseconds. The scheduling algorithm used is preemptive shortest remaining-time first.
Process
Arrival Time
Burst Time
P1 P2 P3 P4
0 3 7 8
10 6 1 3
The average turn around time of these processes is milliseconds.
