self doubt

Question

I have a doubt regarding the no of context switches needed in any scheduling algorithm. assume we have 5 processes, which require 4, 6, 7,5 and 1 time units and arrive at times 0, 1, 2,3, and 5
respectively. How many context switches are needed if
the operating system implements shortest remaining. here if we are doing this firstly we will run |p1| p1| p4| p5| p4| p2|

                                                                                                                                                        0  1   4   5   6   10 16  23 so total no of context switches are 7 but in solution the run p1 directly 0 to 4 I wanted to ask how SRTF work internally (as i know it will check firstly for next process arrival time then it will go ahed)

Comments

yes got it , if in gantt chart p1|,p1 |p1| p1 |p3| p2 like this comes then it does'nt mean that no. of context switches are 5 , here the no of context switches are only 2

p1 to p3

p3 to p2    Am I right?

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yes correct.

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I was Very confuse on this topic thanks for these imp links

Answer 1

P5 takes minimum time but it's arrival time is 5 sec..

In case of SRTF ->

0             4                5             6              10          16         23

P1

P4

P5

P4

P2

P3

Here also first P1 execute and then P4 BUT P4 Preempted on 5th sec (Execute only one unit of time 4 unit of time left) bcs on 5th sec Process P5(1sec) active and it take CPU. And P1 first execute and then P4 execute its remaining 4 sec bcs its lesser then time taken by P2 and P3. 

Hope i make it simple. Thanks