GATE-1995
1. Which scheduling policy is most suitable for a time-shared operating systems?
(a) Shortest Job First (b) Round Robin
(c) First Come First Server (d) Elevator
Ans: option (b)
Explanation:
Job-Id CPU-BurstTime
-----------------------
p 4
q 1
r 8
s 1
t 2
-----------------------
----------------------------
Job Arrival_Time Burst_Time
----------------------------
1 0.0 9
2 0.6 5
3 1.0 1
----------------------------
(a) {3,2,1),1 (b) (2,1,3},0
(c) {3,2,1),0 (d) {1,2,3},5
Ans: option (a)
Explanation:
Shortest Job First is an optimal solution for non-preemptive scheduling. Hence the sequence should be 3,2,1. For that sequence to occur the CPU should wait for 1 units of time.
GATE-1996
Total Execution Time = 47
CPU Idle Time = 2 + 3 = 5
Percentage of Idle Time = 5/47 = 10.6%
GATE-2007
Waiting Time = Completion Time - Arrival Time - Execution Time
1. Which scheduling policy is most suitable for a time-shared operating systems?
(a) Shortest Job First (b) Round Robin
(c) First Come First Server (d) Elevator
Ans: option (b)
Explanation:
In order to schedule processes fairly, a round-robin scheduler generally employs time-sharing, giving each job a time slot or quantum (its allowance of CPU time), and interrupting the job if it is not completed by then. It is designed especially for time-sharing systems.
GATE-2001
2. Consider a set of n tasks with known runtimes r1, r2, … rn to be run on a uniprocessor machine. Which of the following processor scheduling algorithms will result in the maximum throughput?
(a) Round-Robin (b) Shortest-Job-First
(c) Highest-Response-Ratio-Next (d) First-Come-First-Served
Ans: option (b)
Explanation:
Throughput means total number of tasks executed per unit time. Shortest Job First has maximum throughput because in this scheduling technique shortest jobs are executed first hence maximum number of tasks are completed.
Note: Highest-Response-Ratio-Next policy favors shorter jobs, but it also limits the waiting time of longer jobs.
GATE-2002
2. Consider a set of n tasks with known runtimes r1, r2, … rn to be run on a uniprocessor machine. Which of the following processor scheduling algorithms will result in the maximum throughput?
(a) Round-Robin (b) Shortest-Job-First
(c) Highest-Response-Ratio-Next (d) First-Come-First-Served
Ans: option (b)
Explanation:
Throughput means total number of tasks executed per unit time. Shortest Job First has maximum throughput because in this scheduling technique shortest jobs are executed first hence maximum number of tasks are completed.
Note: Highest-Response-Ratio-Next policy favors shorter jobs, but it also limits the waiting time of longer jobs.
GATE-2002
3. Which of the following scheduling algorithms is non-preemptive?
(a) Round-Robin (b) First In First Out
(c) Multilevel Queue Scheduling (d) Multilevel Queue Scheduling with Feedback
Ans: option (b)
GATE-2006
4. Consider three CPU-intensive processes, which require 10, 20 and 30 time units and arrive at times 0, 2 and 6, respectively. How many context switches are needed if the operating system implements a shortest remaining time first scheduling algorithm? Do not count the context switches at time zero and at the end.
(a) 1 (b) 2 (c) 3 (d) 4
Ans: option(b)
Explanation:
-------------------
PNo AT BT
-------------------
A 0 10
B 2 20
C 6 30
-------------------
Time 0: Process A arrives and its the only available process so it runs.
Time 2: Process B arrives, but A has the shortest remaining time (8), so it continues.
Time 6: Process C arrives, but A has the shortest remaining time (2), so it continues.
Time 10: Process A is completed and context switching takes place. B is scheduled as it is the shortest remaining time process.
Time 30: Process B is completed and context switching takes place. Now C is scheduled.
GATE-2007
4. Consider three CPU-intensive processes, which require 10, 20 and 30 time units and arrive at times 0, 2 and 6, respectively. How many context switches are needed if the operating system implements a shortest remaining time first scheduling algorithm? Do not count the context switches at time zero and at the end.
(a) 1 (b) 2 (c) 3 (d) 4
Ans: option(b)
Explanation:
-------------------
PNo AT BT
-------------------
A 0 10
B 2 20
C 6 30
-------------------
Time 0: Process A arrives and its the only available process so it runs.
Time 2: Process B arrives, but A has the shortest remaining time (8), so it continues.
Time 6: Process C arrives, but A has the shortest remaining time (2), so it continues.
Time 10: Process A is completed and context switching takes place. B is scheduled as it is the shortest remaining time process.
Time 30: Process B is completed and context switching takes place. Now C is scheduled.
GATE-2007
5. Group 1 contains some CPU scheduling algorithms and Group 2 contains some applications. Match entries in Group 1 to entries in Group 2.
Group I Group II
(P) Gang Scheduling (1) Guaranteed Scheduling
(Q) Rate Monotonic Scheduling (2) Real-time Scheduling
(R) Fair Share Scheduling (3) Thread Scheduling
(a) P – 3 Q – 2 R – 1
(b) P – 1 Q – 2 R – 3
(c) P – 2 Q – 3 R – 1
(d) P – 1 Q – 3 R – 2
Ans: option (a)
Explanation:
Explanation:
Gang scheduling is a scheduling algorithm for parallel systems that schedules related threads or processes to run simultaneously on different processors.CLICK TO KNOW MORE
Rate monotonic scheduling is a scheduling algorithm used in real-time operating systems with a static-priority scheduling class. CLICK TO KNOW MORE
Fair Share Scheduling is a scheduling strategy in which the CPU usage is equally distributed among system users or groups, as opposed to equal distribution among processes. It is also known as Guaranteed scheduling.CLICK TO KNOW MORE
GATE-1993
6. Assume that the following jobs are to be executed on a single processor system
-----------------------Job-Id CPU-BurstTime
-----------------------
p 4
q 1
r 8
s 1
t 2
-----------------------
The jobs are assumed to have arrived at time 0 and in the order p, q, r, s, t. Calculate the departure time (completion time) for job p if scheduling is round robin with time slice 1.
(a) 4 (b) 10 (c) 11 (d) 12
Ans: option (c)
Explanation:
Execution steps are plotted below
P
|
Q
|
R
|
S
|
T
|
P
|
R
|
T
|
P
|
R
|
P
|
R
|
1
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2
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3
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4
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5
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6
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7
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8
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9
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10
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11
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16
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GATE-1995
7. The sequence …………… is an optimal non-preemptive scheduling sequence for the following jobs which leaves the CPU idle for ………………… unit(s) of time. ----------------------------
Job Arrival_Time Burst_Time
----------------------------
1 0.0 9
2 0.6 5
3 1.0 1
----------------------------
(a) {3,2,1),1 (b) (2,1,3},0
(c) {3,2,1),0 (d) {1,2,3},5
Ans: option (a)
Explanation:
Shortest Job First is an optimal solution for non-preemptive scheduling. Hence the sequence should be 3,2,1. For that sequence to occur the CPU should wait for 1 units of time.
GATE-1996
8. Four jobs to be executed on a single processor system arrive at time 0 in the order A, B, C, D. their burst CPU time requirements are 4, 1, 8, 1 time units respectively. The completion time of A under round robin scheduling with time
slice of one time unit is
(a) 10 (b) 4 (c) 8 (d) 9
Ans: option (d)
GATE-2003
9. A uni-processor computer system only has two processes, both of which alternate 10 ms CPU bursts with 90 ms I/O bursts. Both the processes were created at nearly the same time. The I/O of both processes can proceed in parallel. Which of the following scheduling strategies will result in the least CPU utilization (over a long period of time) for this system?
(a) First come first served scheduling
(b) Shortest remaining time first scheduling
(c) Static priority scheduling with different priorities for the two processes
(d) Round robin scheduling with a time quantum of 5 ms
Ans: option (a)
GATE-2004
9. A uni-processor computer system only has two processes, both of which alternate 10 ms CPU bursts with 90 ms I/O bursts. Both the processes were created at nearly the same time. The I/O of both processes can proceed in parallel. Which of the following scheduling strategies will result in the least CPU utilization (over a long period of time) for this system?
(a) First come first served scheduling
(b) Shortest remaining time first scheduling
(c) Static priority scheduling with different priorities for the two processes
(d) Round robin scheduling with a time quantum of 5 ms
Ans: option (a)
GATE-2004
10. Consider the following set of processes, with the arrival times and the CPU burst times given in milliseconds.
-------------------------------------
Process Arrival-Time Burst-Time
-------------------------------------
P1 0 5
P2 1 3
P3 2 3
P4 4 1
-------------------------------------
What is the average turnaround time for these processes with the preemptive shortest remaining processing time first (SROT) algorithm?
(a) 5.50 (b) 5.75 (c) 6.00 (d) 6.25
Ans: option (a)
Explanation:
Execution chart is shown below:
P1
|
P2
|
P4
|
P3
|
P1
|
1
|
4
|
5
|
8
|
12
|
Calculate the Turn Around Time (TAT) for each process as shown in the table below.
TAT = Completion Time - Arrival Time
-------------------------------------
Pro AT BT CT TAT(CT-AT)
-------------------------------------
P1 0 5 12 12
P2 1 3 4 3
P3 2 3 8 6
P4 4 1 5 1
-------------------------------------
Avg TAT = (12+3+6+1)/4 = 5.50
GATE-2006
11. Consider three processes (process id 0, 1, 2 respectively) with compute time bursts 2, 4 and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF ties are broken by giving priority to the process with the lowest process id. The average turn around time is:
(a) 13 units (b) 14 units (c) 15 units (d) 16 units
Ans: option (a)
Explanation:
Execution chart is shown below:
11. Consider three processes (process id 0, 1, 2 respectively) with compute time bursts 2, 4 and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF ties are broken by giving priority to the process with the lowest process id. The average turn around time is:
(a) 13 units (b) 14 units (c) 15 units (d) 16 units
Ans: option (a)
Explanation:
Execution chart is shown below:
P2
|
P1
|
P2
|
P1
|
P2
|
P0
|
P1
|
P2
|
P0
|
P1
|
P2
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
13
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14
|
Calculate the Turn Around Time (TAT) for each process as shown in the table below.
TAT = Completion Time - Arrival Time
-------------------------------------
Pro AT BT CT TAT(CT-AT)
-------------------------------------
P1 0 2 12 12
P2 0 4 13 13
P3 0 8 14 14
-------------------------------------
Avg TAT = (12+13+14)/3 = 13
GATE-2006
GATE-2006
12. Consider three processes, all arriving at time zero, with total execution time of 10, 20 and 30 units, respectively. Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again. The operating system uses a shortest remaining compute time first scheduling algorithm and schedules a new process either when the running process gets blocked on I/O or when the running process finishes its compute burst. Assume that all I/O operations can be overlapped as much as possible. For what percentage of time does the CPU remain idle?
(a) 0% (b) 10.6% (c) 30.0% (d) 89.4%
Ans: option (b)
Explanation:
--------------------------
Pro AT IO BT IO
--------------------------
P1 0 2 7 1
P2 0 4 14 2
P3 0 6 21 3
--------------------------
Execution chart is shown below:
P1
|
P2
|
P3
| ||
2
|
9
|
23
|
44
|
47
|
Total Execution Time = 47
CPU Idle Time = 2 + 3 = 5
Percentage of Idle Time = 5/47 = 10.6%
GATE-2007
13. An operating system uses Shortest Remaining Time first (SRT) process scheduling algorithm. Consider the arrival times and execution times for the following processes:
-------------------------------------
Process Execution time Arrival time
-------------------------------------
P1 20 0
P2 25 15
P3 10 30
P4 15 45
-------------------------------------
What is the total waiting time for process P2?
(a) 5 (b) 15 (c) 40 (d) 55
Ans: option (b)
Explanation:
Execution chart is shown below:
P1
|
P2
|
P3
|
P2
|
P4
|
20
|
30
|
40
|
55
|
70
|
Waiting Time = 55 - 15 - 25 = 15
GATE-2011
14. 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 at 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
Ans: option (a)
Explanation:
Execution chart is shown below:
Waiting Time = Completion Time - Arrival Time - Execution Time
-----------------------------------
Pro AT BT CT WT
-----------------------------------
P0 0 9 13 4
P1 1 4 5 0
P2 2 9 22 11
-----------------------------------
14. 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 at 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
Ans: option (a)
Explanation:
Execution chart is shown below:
P0
|
P1
|
P0
|
P2
|
1
|
5
|
13
|
22
|
-----------------------------------
Pro AT BT CT WT
-----------------------------------
P0 0 9 13 4
P1 1 4 5 0
P2 2 9 22 11
-----------------------------------
Average Waiting Time = (4+0+11)/3 = 5ms
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