GATE CSE 1995 | Question: 2.6

Question

The sequence __________ is an optimal non-preemptive scheduling sequence for the following jobs which leaves the CPU idle for ________ unit(s) of time.$$\small \begin{array}{|c|c|c|} \hline  \textbf{Job} & \textbf{Arrival Time} & \textbf{Burst Time} \\\hline 1 & 0.0 & 9 \\\hline  2 & 0.6 & 5 \\\hline 3 & 1.0 & 1 \\\hline \end{array}$$

• A. $\{3, 2, 1\}, 1$
• B. $\{2, 1, 3\}, 0$
• C. $\{3, 2, 1\}, 0$
• D. $\{1, 2, 3\}, 5$

Answer 1

Answer is (A).
Here, in option B and C they have given CPU idle time is $0$ which is not possible as per schedule (B) and (C).
So, (B) and (C) are eliminated.

Now, lets see (A) and (D):

For (A),

So, idle time is between $0$ and $1$ which is $1$ in case of option (A).

For option (D),

We can see that there is no idle time at all, but in option given idle time is $5$, which is not matching with our chart so option (D) is eliminated.

Therefore, the correct sequence is option (A).

Comments

So basically, since the question doesn’t mention any specific criteria, we go with average waiting time by default. And that means we need to calculate it for all the options and then choose the one with the least value, right ?

 

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@manya03
Yes, you could calculate all options.
But, generally SJF gives the least "Average Waiting Time" when all the jobs arrive at the same time. Just by looking, we can choose option A

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Okay okay...I get it now !!

 

Answer 2

(D) CPU idle time will be 0.
(C) CPU idle time will be 1.
(B) CPU Idle Time will be 0.6
(A) CPU idle time will be 1.

Hence, (A) is correct answer!

Answer 3

Assuming optimal is with respect to the total waiting time (a) is correct.

Total waiting times are

(a) 0 + 1.4 + 7 = 8.4, idle time = 1

(b) 0 + 5.6 + 14.6 = 20.1, idle time = 0.6

(c) 0 + 1.4 + 7 = 8.4, idle time = 1

(d) 0 + 9 + 14 = 23, idle time = 0

http://www.cs.uic.edu/~jbell/CourseNotes/OperatingSystems/5_CPU_Scheduling.html

Comments

Idle time is also specified in option and in c choice it is given as 0.

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Sir can u explain this question in detail

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Optimal Non-Premptive Scheduling is Shortest job-first (SJF)scheduling..

Answer 4

Answar is A, CPU will be idle for first 1 unit

Comments

can u explain ur answer !

Answer 5

BEST ANSWER

Answer 6

SJF is best among all non-preemptive methods with its disadvantages. SJF gives less average waiting time and less Turn around time than others and more

in all scheduling algorithms SRJF(i.e SJF with preemption )gives optimal av. waiting time and av. Turn around time

So we will use SJF here to get optimal

3,2,1 will be the sequence but from 0 to 1 CPU will be idle

Comments

But if in sjf we don't have any idle time as here the first process arrives immediately at 0s so if we use sfj algo we should be processing job S1 instead we are creating an optimal sequence of our own