Consider a system with a single CPU core and three processes A, B, C. Process A arrives at t = 0, and runs on the CPU for 10 time units before it finishes. Process B arrives at t = 6, and requires an initial CPU time of 3 units after which it blocks to perform 
for 3 time units. After returning from wait, it executes for a further 5 units before terminating. Process C arrives at t = 8, and runs for 2 units for time on the CPU before terminating. For shortest remaining time first (preemptive) scheduling policy, calculate the time of completion of each of the three processes. Recall that only the size of the current CPU burst (excluding the time spent for waiting on ) is considered as the “job size” in these schedulers.
A. A = 10, B = 21, C = 15
B. A = 10, B = 23, C = 12
C. A = 15, B = 20, C = 11
D. A = 15, B = 21, C = 12
It is mentioned in the question
only the size of the current CPU burst (excluding the time spent for waiting on ) is considered as the “job size” in these schedulers.
Suppose it was not given then we should consider the whole CPU time (before and after the I/O) when choosing the process with the shortest remaining time. Right ?
If the method given in the question is followed then C would be the answer otherwise if the default method is followed then B would be the answer. Am I right ?
THE ANSWER KEY for this question is C.
