Question 44

Given processes:

• A: arrival = 0, burst = 3
• B: arrival = 1, burst = 6
• C: arrival = 4, burst = 4
• D: arrival = 6, burst = 2

1. Shortest Remaining Time First (SRTF):

• $t = 0$: Only A arrives. A runs until $t = 1$.
• $t = 1$: B arrives (burst 6). A remaining burst is 2. Since $2 < 6$, A continues. A finishes at $t = 3$.
  Completion time of A $= 3$.
• $t = 3$: Only B is present (remaining 6). B runs from $t = 3$ to $4$.
• $t = 4$: C arrives (burst 4). B has remaining burst 5. Since $4 < 5$, C preempts B! C runs from $t = 4$ to $6$.
• $t = 6$: D arrives (burst 2). C has remaining burst 2. D has burst 2. C continues (or D runs). C finishes at $t = 8$.
  Completion time of C $= 8$.
• $t = 8$: Ready queue has D (burst 2) and B (remaining 5). D runs from $t = 8$ to $10$.
  Completion time of D $= 10$.
• $t = 10$: B runs from $t = 10$ to $15$.
  Completion time of B $= 15$.

Turnaround time ($TAT = \text{Completion} - \text{Arrival}$):

• A: $3 - 0 = 3$
• B: $15 - 1 = 14$
• C: $8 - 4 = 4$
• D: $10 - 6 = 4$
  Average TAT for SRTF $= \frac{3 + 14 + 4 + 4}{4} = \frac{25}{4} = 6.25$.

By optimality theorem of scheduling, Shortest Remaining Time First (SRTF) is provably optimal for minimizing average turnaround / waiting time among all scheduling algorithms. Hence, SRTF gives the lowest average turnaround time (Option C).