ITPEC FE Morning October 2020 Question 18

A finishes at 6 ms, B runs 3–4 and 6–7 then starts I/O, C runs 7–9 — trace from 0 ms, letting the higher-priority task take the CPU whenever it becomes ready.

All three tasks are ready at 0 ms, and a task that returns from I/O pre-empts a lower-priority one:

• •0–2: A runs CPU(2); B and C wait
• •2–4: A does I/O(2); B, the highest ready task, runs 2 of its 3 ms of CPU
• •4–6: A is back and pre-empts B, runs CPU(2) and completes at 6; B waits with 1 ms left
• •6–7: B finishes its first CPU burst
• •7–9: B starts I/O(5), until 12; C finally gets the CPU for its CPU(2)
• •9–10: C does I/O(2)

So from 3 to 10 ms: A has I/O 3–4, CPU 4–6, complete; B has CPU 3–4, waiting 4–6, CPU 6–7, I/O 7–10; C waits 3–7, runs CPU 7–9 and has I/O 9–10.

Answer (c)

Why not others:

• •(a) — starts A's first CPU burst at 3 ms after B has used the CPU for 0–3, putting the high-priority A behind the medium-priority B
• •(b) — has B in I/O from 3 ms, so B used the CPU for all of 0–3, yet A still has 1 ms of its first burst left; that needs 4 ms of CPU in 3 ms
• •(d) — first come, first served without priority: B keeps the CPU until 5 and the low-priority C runs 5–7 while A waits from 4 to 7

Key rule: For a scheduling chart that starts part-way, trace from time 0: the state at the window's start depends on everything before it, and with pre-emptive priority a task leaving I/O takes the CPU at once.

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