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Consider the following four processes with their corresponding arrival time and burst time:

$$\begin{array} \text{Process}&\text{Arrival time}&\text{Burst time(in ms)}\\ \text{P1}&0.0&8\\ \text{P2}&0.6&6\\ \text{P3}&3.8&4\\ \text{P4}&4.4&2\end{array}$$

What is the average turn around time (in ms) for these processes using FCFS scheduling algorithm?

  1. $15$
  2. $12.8$
  3. $13$
  4. none of the options
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4 Answers

6 votes
6 votes

Gantt Chart:- 

P1 P2 P3 P4

0.0                            8                                14                              18                               20

Table:-

Process Arrival Time Burst Time Completion Time Turn Around Time
P1 0.0 8 8 8
P2 0.6 6 14 13.4
P3 3.8 4 18 14.2
P4 4.4 2 20 15.6

Avg. Turn Around Time =  (8+13.4+14.2+15.6)/4

                                      =  (51.2) / 4

                                      =  12.8

So, the ans. is option B)

1 votes
1 votes

Turnaround time=completion time - arrival time

Average T.T=(8+13.4+14.2+15.6)/4=12.8

option B)

0 votes
0 votes
$P_{1}$ $P_{2}$ $P_{3}$ $P_{4}$

$0$                               $8$                              $14$                             $18$                             $20$

 

The Gantt Chart is drawn above.

Process Arrival Time(AT) Burst Time Completion Time(CT) turnaround time(CT $-$ AT)
$P_{1}$ $0.0$ $8$ $8$ $8$
$P_{2}$ $0.6$ $6$ $14$ $13.4$
$P_{3}$ $3.8$ $4$ $18$ $14.2$
$P_{4}$ $4.4$ $2$ $20$ $15.6$

     

So the average Turn-Around time is :  $\frac{8+13.4+14.2+15.6}{4}=12.8$ ms

Option B is correct

Answer:

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