Process Scheduling Algorithms

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 arrives runs for 2 units of time, later process C  arrives and completes its execution in 4 units of time. Find the minimum possible burst time of the process A and B.

Answer 1

A=14  units and B=7 units.

A=6

B=2

C=4

As in SRTF the new process is scheduled when its Remaining time is minimum.

B.T. of A= 6 + value greater than 2 + value greater than 4 (as B replaced it so its remaining time is lesser A than remaining time of B and after that C replaces B which means remaining time of B is greater than remaining time of C ) 

B.T. of A =6+3+5=14

B.T. of B=2+5=7