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Consider the following set of processes, assumed to have arrived at time $0$. Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order$P_1, P_2, P_3, P_4$.

$\begin{array}{|l|l|l|l|l|} \hline \text{Processes} & P_1 & P_2 & P_3 & P_4 \\ \hline \text{Burst time (in ms)} &8 & 7 & 2 & 4 \\ \hline \end{array}$

If the time quantum for RR is $4$ ms, then the absolute value of the difference between the average turnaround times (in ms) of SJF and RR  (round off to $2$ decimal places is_______

### 1 comment

SJF:

$$\small \begin{array} {|c|c|c|c|} \hline \textbf{Process} & \textbf{Burst Time} & \textbf{Completion Time} & \textbf{Turn Around Time} \\\hline P_{1} & 8 & 21 & 21 \\ P_{2} & 7 & 13 & 13 \\ P_{3} & 2 & 2 & 2 \\ P_{4} & 4 & 6 & 6 \\\hline \end{array}$$Average Turn-Around Time $: \frac{21+13+2+6}{4}= 10.5$

RR:

$$\small \begin{array} {|c|c|c|c|} \hline \textbf{Process} & \textbf{Burst Time} & \textbf{Completion Time} & \textbf{Turn Around Time} \\\hline P_{1} & 8 & 18 & 18 \\ P_{2} & 7 & 21 & 21 \\ P_{3} & 2 & 10 & 10 \\ P_{4} & 4 & 14 & 14 \\\hline \end{array}$$Average Turn-Around Time $: \frac{18+21+10+14}{4}= 15.75$

Absolute Difference $= \mid 10.5 - 15.75 \mid = 5.25.$

### 1 comment

it should be 21+13+2+6 In SJF calculation. you wrote 9 in place of 6

Since arrival time of all process is same =0, So turn around time = finish time

For SJF :

Process Burst Time Completion Time Turn Around Time
P1 8 21 21
P2 7 13 13
P3 2 2 2
P4 4 6 6

Average Turn Around Time using SJF : $\frac{21+13+2+6}{4}$= 10.5

For RR(4 Time quantum):

Process Burst Time Completion Time Turn Around Time
P1 8 18 18
P2 7 21 21
P3 2 10 10
P4 4 14 14

Average Turn Around Time using RR :$\frac{18+21+10+14}{4}$= 15.75

Absolute Difference = 15.75 - 10.5= 5.25

### 1 comment

it should be 21+13+2+6 In SJF calculation. you wrote 9 in place of 6 absolute value of the difference between the average turnaround times (in ms) of SJF and RR  is 5.25