CPU Scheduling

Question

Consider 3 processes A, B, and C to be scheduled as per SRTF scheduling. The process A is known to be scheduled first and when A has been running for 7 units of time the process C has arrived. The process C has run for 1 unit of time, then the process B has arrived and completed running for 2 units of time. Then what could be the minimum burst time of processes A and C?

1. 11, 3
2. 12,3
3. 11,4
4. 12,4

Comments

A =12

C= 4

Answer 1

Let Burst time of A, B, C are x, y, z

when A has been running for 7 units of time the process C has arrived

$x-7 > z$

The process C has run for 1 unit of time, then the process B has arrived

$z-1>y$

and completed running for 2 units of time

which means $y = 2$

Now, $z-1>y$ 

$z-1>2$

$z=4$

$x-7>z$

$x-7>4$

$x=12$

So, the minimum Burst Time of A and C is 12, 4 respectively,

Answer> $D$

Answer 2

Let BT for A is x

Let BT for C is y

As A runs for 7 units, remaining time of A is x-7

As it is SRTF and the process C preempts A therefore, x-7>y

C runs for 1 unit of time,then B preempts C

similarly,  y-1>2

y is greater than 3 ,y is 4 than x is 12

ANS: d