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Question 116

Non-preemptive priority scheduling (lower number = higher priority), all at time 0: P1 (burst 10, pri 3), P2 (1, 1), P3 (2, 4), P4 (1, 5), P5 (5, 2).…

Non-preemptive priority scheduling (lower number = higher priority), all at time 0: P1 (burst 10, pri 3), P2 (1, 1), P3 (2, 4), P4 (1, 5), P5 (5, 2). Average waiting time is:
  1. A8.2
  2. B9.6
  3. C6.4
  4. D7.2
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Correct answer: (A) 8.2

Explanation

Order P2, P5, P1, P3, P4 → waits 0, 1, 6, 16, 18 → 41/5 = 8.2.

From DSSSB TGT Computer Science Full Mock Test 4

This is question 116 of DSSSB TGT Computer Science Full Mock Test 4. See all questions of this paper with answers, or attempt the paper as a timed online test to get your score and all-India rank.