GATE IT 2006 | Question: 54

Question

The arrival time, priority, and duration of the CPU and I/O bursts for each of three processes $P_1, P_2 $ and $P_3$ are given in the table below. Each process has a CPU burst followed by an I/O burst followed by another CPU burst. Assume that each process has its own I/O resource.$$\small \begin{array}{|c|c|c|c|c|c|} \hline \textbf{Process} & \textbf{Arrival} & \textbf{Priority}& \textbf{Burst duration} & \textbf{Burst duration} & \textbf{Burst duration) }\\& \textbf{Time} & & \textbf{(CPU)} & \textbf{(I/O)} & \textbf{(CPU) }
\\\hline P_1 & 0 & 2 & 1 & 5 &3 \\\hline P_2 & 2 & \text{3 (lowest)}& 3 & 3 & 1 \\\hline  P_3 & 3 & \text{1 (highest)}& 2 & 3 & 1 \\\hline  \end{array}$$

The multi-programmed operating system uses preemptive priority scheduling. What are the finish times of the processes $P_1, P_2 $ and $P_3$?

• A. $\text{11, 15, 9}$
• B. $\text{10, 15, 9}$
• C. $\text{11, 16, 10}$
• D. $\text{12, 17, 11}$

Comments

please how i apply this as burst time , i am not getting to find actual burst time ?

please help

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i/o bursts are characterized by the time a process waits for data to be retrieved or stored, so make the gantt chart accordingly

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I think this is the best way to do it

Answer 1

GIVEN : assuming that each process has its own i/o resource.

(GANTT CHART FOR I/O OF PROCESSOR $\mathbf{P_1, P_2 ,P_3}$)

EXPLANATION :

Here, $P_2$ has the least priority and $P_1$ has the highest.

$P_1$ enters CPU at $0$ and utilizes it for $1$ time unit. Then it performs i/o for $5$ time units.

Then $P_2$ enters at time unit $2$ and requires $3$ time units of CPU. But $P_3$ whose priority is greater than $P_2$ arrives at time unit $3$.

So, $P_2$ IS PREEMPTED (only $1$ unit of $P_2$ is done out of $3$ units. Therefore $2$ units of $P_2$ are left out) AND $P_3$ ACQUIRES THE CPU. Once $P_3$ finishes, $P_2$ enters the CPU to complete its pending $2$ units job at time unit $5$. AGAIN BY THEN $P_1$ finishes its i/o and arrives with a higher priority. Therefore of $2$ units $P_2$ performs only one unit and the CPU is given to $P_1$.Then when $P_1$ is performing in CPU, $P_3$ completes its i/o and arrives with a higher priority.Thus the CPU is given to $P_3$ ($1$ UNIT IS USED). $\mathbf{P_3}$ FINISHES AT TIME UNIT 9. NOW PRIORITY OF $P_1$ IS MORE THAN $P_2$, SO, CPU IS USED BY $P_1$. $\mathbf{P_1}$ FINISHES BY TIME UNIT 10. THEN CPU IS  ALLOCATED FOR PROCESS $P_2$. $\mathbf{P_2}$ PERFORMS REST OF ITS WORK AND FINISHES AT TIME UNIT 15.

THEREFORE,

 FINISH TIME OF $P_1,P_2,P_3$ ARE $10,15$ AND $9$ RESPECTIVELY. 😊

Correct Answer: $B$

Comments

Gantt Chart : 

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@rohith1001

Best answer Rohit

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GATE smashed with this question

Answer 2

0---p1----1--- no process---2---p2---3----p3----5----p2---6----p1----8----p3---9----p1----10----p2----11---no process----14--p2---15

From time  1 to 6 p1 is doing i/o

From time 5 to 8 p3 is doing i/o

P2 is doing i/o from11 to 14

So finishing time of p1 p2 p3 are 10 15 9

Ans is b

Comments

6---p1---8   6 to 8 should be p1..

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please explain how we get actual burst time to solve the problem

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Nice and Good Explanation.

Answer 3

People having trouble understanding refer this pipeline way-

Comments

Most useful answer

Answer 4

Grant chat:-

Comments

After 5 sec P1 and P3 are in i/o .Why we r allowing P1 to execute? Why we r not applying priority and preempt P1 and allow P3?

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Its GANTT CHART

not GRANT CHAT!

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naam me kya rakha hai!!😅

Answer 5

BEST ANSWER

Answer 6

HERE IS THE DETAILS EXPLANATION