SRTF question

Question

With the use of SRTF the execution of three processes is as follows, the process A is scheduled first and A has been running for 6 units of time, the process B is arrived process B ran for 2 units of time, later process C is arrived and completed running in 4 units of time. Find the minimum possible burst time of the process A and B.

1.   9, 6
2.   10, 6
3.   2, 8
4.   11, 7

Comments

i got

A= 14

b= 7

Answer 1

Suppose burst time of A is x, its mention A has already run for 6 unit of time, so Remaining Burst time for A is x-6, now it get preempted by B
Now Suppose Burst time of B is y, its mention B has already run for 2 unit of times before getting preempted by C, so remaining burst time for B is y-2.
Now C run for 4 unit of time.

Now given B is preempted by C, so C has shortest BT as compare to B, i.e minimum value of y-2 must be 5, then only it get preempted by C

Therefore y-2 = 5, so y = 7.

Similarly x-6 = 8, and x = 14

Therefore burst time of A and B is 14 and 7, which is not matched by any option, so none of the above.

Comments

Even I got the same, answer given as d) which was wrong. Thanks !

Answer 2

i got the answer as

A= 14

b= 7