GATE Papers >> CSE >> 2019 >> Question No 51

Question No. 51 CSE | GATE 2019

Consider the following four processes with arrival times (in milliseconds) and their length of CPU bursts (in milliseconds) as shown below:

Process P1 P2 P3 P4
Arrival time 0 1 3 4
CPU burst time 3 1 3 Z

These processes are run on a single processor using preemptive Shortest Remaining Time First scheduling algorithm. If the average waiting time of the processes is 1 millisecond, then the value of Z is________.

Answer : 2 to 2

Solution of Question No 51 of GATE 2019 CSE Paper


Now, we have P3 and P4 process in ready queue based on 'z' value one she process being scheduled

$ \mathrm{Case}\;\boxed1\;\mathrm z>3\;\mathrm{if}\;\left(\mathrm z=4\right) $

We have so schedule process 'P4' Now

wt = TAT - BT

TAT=completion -A           

$ \begin{array}{l}\mathrm{average}\;\mathrm{wt}=\frac{1+0+1+3}4\Rightarrow\frac54\neq1\;(\mathrm{not}\;\mathrm{matching})\\\mathrm{Case}\;\boxed2\;\mathrm z<3\;\mathrm{if}\;\left(\mathrm z=2\right)\end{array} $

Now, schedule P4                    

$ \mathrm{average}\;\mathrm{wt}=\frac{1+0+3+0}4\Rightarrow\frac44=\boxed1 $

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Posted on  18/10/2020 19:12:11  by  dobsonz
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