Consider the following set of processes that need to be scheduled on a single CPU. All the times are given in milliseconds. Using the shortest remaining time first scheduling algorithm, the average process turnaround (in milliseconds) is ________.Process Name Arrival Time Execution Time A 0 6 B 3 2 C 5 4 D 7 6 E 10 3
7.2
Shortest remaining time first (SRTF) is the preemptive form of shortest job first: whenever a process arrives, the scheduler compares its burst with the remaining time of the running process and runs whichever is smaller. Turnaround time = completion time − arrival time.
Step 1 — t = 0 to 3: only A is present, so A runs; at t = 3 A has 3 ms left.
Step 2 — t = 3: B arrives with 2 ms, which is less than A's 3 ms, so B preempts A and runs from 3 to 5. B completes at 5.
Step 3 — t = 5: C arrives with 4 ms; A has 3 ms left, so A resumes. At t = 7 D arrives with 6 ms while A has 1 ms left, so A continues and completes at 8.
Step 4 — t = 8: ready are C (4) and D (6), so C runs. At t = 10 E arrives with 3 ms while C has 2 ms left, so C continues and completes at 12.
Step 5 — t = 12: E (3) is shorter than D (6), so E runs 12 to 15, then D runs 15 to 21.
| Process | Arrival | Completion | Turnaround |
|---|---|---|---|
| A | 0 | 8 | 8 |
| B | 3 | 5 | 2 |
| C | 5 | 12 | 7 |
| D | 7 | 21 | 14 |
| E | 10 | 15 | 5 |
Step 6 — Average turnaround = \(\dfrac{8+2+7+14+5}{5} = \dfrac{36}{5} = 7.2\) ms.
The value 8.2 would need a total of 41 ms of turnaround, which no valid SRTF schedule of these bursts produces; 6 and 9 do not match the schedule either (non-preemptive SJF would give 7.4 and the average waiting time is 3).
7.2 option is correct.
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Important Questions from CPU Scheduling
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- $M_2$ uses priority of execution for the processes as, $P_2 > P_3 > P_4 > P_1$, i.e., $P_2$ and $P_1$ have highest and lowest priorities, respectively.
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