GATE CSE 1993 | Question: 7.10

Question

Assume that  the following jobs are to be executed on a single processor system$$\begin{array}{|c|c|} \hline  \textbf{Job Id} & \textbf{CPU Burst Time} \\\hline \text{p} & 4 \\\hline  \text{q} & 1 \\\hline \text{r} & 8 \\\hline \text{s} & 1 \\\hline \text{t} & 2 \\\hline \end{array}$$The jobs are assumed to have arrived at time $0^+$ and in the order $p,q,r,s,t$. Calculate the departure time (completion time) for job $p$ if scheduling is round robin with time slice $1$

• A. $4$
• B. $10$
• C. $11$
• D. $12$
• E. None of the above

Comments

@G.K.T  @ankit3009 0+ means that the all the processes comes immediately after 0 i,e every process comes after 0  together  . 

---

does the 0+ mean anything here?
 

---

It only means that all the process will arrives at time 0.

Answer 1

Answer: (C)

Execution order: $p\;q\;r\;s\;t\;p\;r\;t\;p\;r\;\mathbf{p}\;r\;r\;r\;r\;r$

Comments

when all process arrives at 0 then it will work like FCFS with TQ=1???

---

Yes...for 1st iteration since nothing is mentioned so we give priority based on Job ID.

Even when the time quantum becomes too large then also Round Robin behaves like FCFS

Answer 2

Round Robin

TQ=1

Arrival time of all the process is same

gantt chart

p	q	r	s	t	p	r	t	p	r	p
0	1	2	3	4	5	6	7	8	9	10     11

Answer - option c

Answer 3

p will depart at t = 11

Answer 4

Best Answer