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+12 votes
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An operating system used Shortest Remaining System 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?

  1. 5
  2. 15
  3. 40
  4. 55
asked in Operating System by Veteran (59.4k points) | 1.2k views

3 Answers

+17 votes
Best answer

The answer is B.

The Gantt chart of execution of processes
P1 P2 P3 P2 P4

0                20                 30                40                55                70

Waiting time for process P2 = Completion time – Arrival time – burst time = 55 – 15 – 25 = 15

answered by Boss (19.7k points)
selected by
–1
Ans should be A)..

@Gate Keeda- why process P2 will preempt? At 30 th time both P2 and P3 have same remaining time,  CPU will not switch to P3. it continue executing P2..right?
0

No.At time 30th both do not have same remaining time. P3 has 10 and P2 has 15. Check that again to make sure.

+1

Could anyone explain why the process is pre-empted ? IS SRT pre-emptive if not mentioned ?

EDIT  : SRTN is pre emptive form of SJF

http://www.moreprocess.com/process-2/shortest-job-first-sjf-shortest-remaining-time-next-srtn-scheduling-algorithm

0
Process is executed on the basis of their remaining time in execution  ...
+11 votes

For process $P2$ :
$\begin{align*} \text{Waiting Time} &= \text{TAT} - \text{BT}\\ &= \left ( \text{CT-AT}\right) - BT \\ &= \left( 55 - 15 \right ) - 25 \\ &= 15 \end{align*}$

 

answer = option B

answered by Boss (30.6k points)
+2 votes

we can solve using this way 

answered by Active (3.2k points)
Answer:

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