1
GATE CSE 2019
Numerical
+2
-0.67
Consider the following four processes with arrival times (in milliseconds) and their length of CPU bursts (in milliseconds) as shown below :
GATE CSE 2019 Operating Systems - Process Concepts and Cpu Scheduling Question 9 English
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 ____
2
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.

Your input ____
3
GATE CSE 2015 Set 3
MCQ (Single Correct Answer)
+2
-0.6
For the processes listed in the following table, which of the following scheduling schemes will give the lowest average turnaround time?

Process Arrival Time Processing Time
A 0 3
B 1 6
D 4 4
E 6 2

A
First Come First Serve
B
Non-preemptive Shortest Job First
C
Shortest Remaining Time
D
Round Robin with Quantum value two
4
GATE CSE 2015 Set 1
Numerical
+2
-0
Consider a uniprocessor system executing three tasks T1, T2 and T3, each of which is composed of an infinite sequence of jobs (or instances) which arrive periodically at intervals of 3, 7 and 20 milliseconds, respectively. The priority of each task is the inverse of its period, and the available tasks are scheduled in order of priority, with the highest priority task scheduled first. Each instance of T1, T2 and T3 requires an execution time of 1, 2 and 4 milliseconds, respectively. Given that all tasks initially arrive at the beginning of the 1st millisecond and task preemptions are allowed, the first instance of T3 completes its execution at the end of _____________ milliseconds.
Your input ____
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