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SJF(Pre-emptive) Scheduling: Operating system

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Consider 3 processes A, B and C to be scheduled as per the SJF (Pre-emptive). The process ‘A’ is known to be scheduled first and when ‘A’ has been running for 4 units of time, the process ‘C’ has arrived. The process ‘C’ has run for 2 unit of time, then the process ‘B’ has arrived and completed running in ‘3’ units of time. Then what could be the minimum burst times of the processes A and C?

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First process A arrives and runs for 4 units. Then process C arrives and runs for 2 units as BT(burst time) of C< A and B came and runs till execution(3 units) as BT of B < C. 

Process C executes as it has less remaining time than A. then Process A executes.

BT of C = 2 + (B + 1) = 2 + (3 + 1) = 6 units

BT of A = 4 + (C + 1) = 4 + (6 + 1) = 11 units

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