Operating systems miscellaneous
- Consider a uni processor system executing three tasks T1, T2 and T3, each o 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.
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Correct Option: B
- 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 remainingtime first.
The average turnaround time of these processes is ________ milliseconds.
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Due to SRTF preemption
∵ Completion Time = Arrival Time + Process Turn Around Time
∴ Process Turn Around Time = Completion Time – Arrival TimeP1 20 P2 7 P3 1 P4 5
Average Turn Around Time (TAT) = 33 / 4 = 8.25Correct Option: A
Due to SRTF preemption
∵ Completion Time = Arrival Time + Process Turn Around Time
∴ Process Turn Around Time = Completion Time – Arrival TimeP1 20 P2 7 P3 1 P4 5
Average Turn Around Time (TAT) = 33 / 4 = 8.25
- Consider an arbitrary set of CPU-bound processes with unequal CPU burst lengths submitted at the same time to a computer system. Which one of the following process scheduling algorithms would minimize the average waiting time in the ready queue?
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SRTF is the preemptive SJF which generates less average waiting time in which jobs are schedule according to shortest remaining time.
Correct Option: A
SRTF is the preemptive SJF which generates less average waiting time in which jobs are schedule according to shortest remaining time.
- Consider the set of processes with arrival time (in milliseconds). CPU burst time (in milliseconds) and priority (0 is the highest priority) shown below. None of the processes have I/O burst time.
Process Arrival Time Burst Time Priority P1 0 11 2 P2 5 28 0 P3 12 2 3 P4 2 10 1 P5 9 16 4
The average waiting time (in milliseconds) of all the processes using preemptive priority scheduling algorithm is ________.
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Gantt chart for above problem looks like:
Waiting Time = (Completion time – Arrival time – Burst time).
∑AT = (0 + 5 + 12 + 2 + 9) = 28
∑BT = (11 + 28 + 2 + 10 + 16) = 67
∑CT = (67 + 51 + 49 + 40 + 33) = 240
Waiting time = (240 – 28 – 67) = 145Average Waiting time = 145 = 29 ms 5
Hence, 29 ms is correct answer.
Correct Option: D
Gantt chart for above problem looks like:
Waiting Time = (Completion time – Arrival time – Burst time).
∑AT = (0 + 5 + 12 + 2 + 9) = 28
∑BT = (11 + 28 + 2 + 10 + 16) = 67
∑CT = (67 + 51 + 49 + 40 + 33) = 240
Waiting time = (240 – 28 – 67) = 145Average Waiting time = 145 = 29 ms 5
Hence, 29 ms is correct answer.
- Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below:
If the pre-emptive shortest remaining time first scheduling algorithm is used to schedule the processes, then the average waiting time across all processes is _______ milliseconds.
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So, turn around time is for P1 ,P2 ,P3 ,P4 are :
P1 ⇒ (12 - 0) , P2 ⇒ (6 - 3), P3 ⇒ (17 - 5) , P4 ⇒ (8 - 6),
⇒ P1 = 12 , P2 = 3 , P3 = 12 , P4 = 2
As we know that Waiting time = Turn around time - Burst time
So, waiting time for P1 ,P2 ,P3 ,P4 are :
P1 ⇒ (12 - 7) = 5
P2 ⇒ (3 - 3) = 0
P3 ⇒ (12 - 5) = 7
P4 ⇒ (2 - 2) = 0Hence, avg. waiting time = P1 + P2 + P3 + P4 = 5 + 7 = 3.0 4 4 Correct Option: B
So, turn around time is for P1 ,P2 ,P3 ,P4 are :
P1 ⇒ (12 - 0) , P2 ⇒ (6 - 3), P3 ⇒ (17 - 5) , P4 ⇒ (8 - 6),
⇒ P1 = 12 , P2 = 3 , P3 = 12 , P4 = 2
As we know that Waiting time = Turn around time - Burst time
So, waiting time for P1 ,P2 ,P3 ,P4 are :
P1 ⇒ (12 - 7) = 5
P2 ⇒ (3 - 3) = 0
P3 ⇒ (12 - 5) = 7
P4 ⇒ (2 - 2) = 0Hence, avg. waiting time = P1 + P2 + P3 + P4 = 5 + 7 = 3.0 4 4
