GATE CSE 2025 Session 1 Q38: Average waiting time with two processors and different priorities
- Operating System
- 2 marks
- MCQ
- −2/3 if wrong
Question
A computer has two processors, and . Four processes with CPU bursts of 20, 16, 25, and 10 milliseconds, respectively, arrive at the same time and these are the only processes in the system. The scheduler uses non-preemptive priority scheduling, with priorities decided as follows:
• uses priority of execution for the processes as, , i.e., and have highest and lowest priorities, respectively. • uses priority of execution for the processes as, , i.e., and have highest and lowest priorities, respectively.
A process is scheduled to a processor , if the processor is free and no other process is waiting with higher priority. At any given point of time, a process can be allocated to any one of the free processors without violating the execution priority rules. Ignore the context switch time. What will be the average waiting time of the processes in milliseconds?
Try it: multiple choice: exactly one correct option
Show answer and explanation
Answer: A (official GATE 2025 final answer key)
Concept used: Non-preemptive scheduling on two processors, each picking by its own priority order whenever it becomes free.
First step: At , takes and takes ; then see which processor frees up first ( at 16) and what it picks next.
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Source: official GATE 2025 Session 1 CS question paper and final answer key, organised by IIT Roorkee. Question paper · Answer key
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