GATE CSE 2017 Set 2 | Question: 51

Question

Consider the set of process with arrival time (in milliseonds), CPU burst time (in millisecods) and priority ($0$ is the highest priority)  shown below. None of the process have I/O burst time$$\small \begin{array}{|c|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time} & \textbf{Burst Time} & \textbf{Priority} \\\hline \text{$P_1$} & 0 & 11 & 2 \\\hline \text{$P_2$} & 5 & 28 & 0 \\\hline \text{$P_3$} & 12 & 2 & 3 \\\hline \text{$P_4$} & 2 & 10 & 1\\\hline \text{$P_5$} & 9 & 16 & 4  \\\hline \end{array}$$The average waiting time (in milli seconds) of all the process using premtive priority scheduling algorithm is ______

Comments

getting 29.6

148/5

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Avg waiting time=29msec

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You didn't mention anything about priority @mcjoshi ???

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Didn't see the arrival time for p4 there. Why they give jumbled arrival time aah.

Answer 1

Gantt Chart for above problem looks like :

Waiting Time $=$ Completion time $-$ Arrival time $-$ Burst Time

$\sum AT = 0 + 5 + 12 + 2 + 9 = 28$

$\sum BT = 11 + 28+ 2 + 10 + 16 = 67$

$\sum CT = 67 + 51 + 49 + 40 + 33 = 240$

Waiting time $ = 240 - 28 - 67 = 145$

Average Waiting Time $ = \frac{145}{5} = 29$ msec.

Comments

I'm getting like this, for

P1: WT= 0+(40-2=38)=38

P2: WT= 5

P3: WT= 49

P4: WT= 33-3=30

P5: WT= 51

38+5+30+49+51 = 173 173/5 = 34.6 is it correct?

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For $P_1$ you are right. But, Waiting time for $P_2 = 0$ because it arrived at $t = 5$ and got scheduled at $t = 5$.

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When i Solved it first time i still remember the mistake i committed then , but then on again revision of OS (It was marked as tricky at that time ) i committed the same mistake. 

No matter how obvious it might sound but Paper setters can create illusion of representing the data in a way that we can make mistake or lots of time might be consumed .

My Mistake:- Arrival time  is not given in ascending order and i tend to forget that and  always focusing on priority and Burst time.  -> which in result my time got wasted in redoing this part.

(DUMB ME)

Answer 2

Hope it might help.....

Answer 3

First try to draw the gantt chart. then from than chart try to find the completion time ,turn around time and finally waiting time. So the average wt is 29msec