Back to the 2023 paper

Module 5: Memory Management

20232m

If a process has 24 k bytes logical address space and the page size is 4096 bytes then the number of frames of that process is?
(i) 12
(ii) 6
(iii) 8
(iv) 16

Worked SolutionAI Assisted

Answer: (ii) 6

Logical address space = 24KB = 24*1024 = 24576 bytes. Page size = 4096 bytes. Number of pages = 24576/4096 = 6. One page per frame, so 6 frames needed for this process.

Similar questions

AI LearningWeb Platform

Coursify — Personal AI Course Generator

Struggling with a difficult subject or concept? Turn any engineering topic into a structured course with interactive lessons, quizzes, and research-backed explanations.