In a system using segmentation with paging (paged segmentation), what is the space complexity of the combined page and segment tables relative to the number of segments S and pages per segment P?

medium🪤 Complexity Trap Q6 of Q15

Operating Systems - Paging vs Segmentation - Address Translation

AO(S * P), since each segment has its own page table with P entries

BO(S + P), because segment and page tables are stored separately

CO(P), because only page tables matter in paged segmentation

DO(S^2 * P^2), due to nested page and segment tables

Step-by-Step Solution

Solution:

Step 1: Understand paged segmentation structure
Each segment has a page table with P entries.
Step 2: Calculate total entries
Total entries = number of segments S times pages per segment P -> O(S * P).
Step 3: Evaluate options
O(S + P), because segment and page tables are stored separately underestimates by adding instead of multiplying. O(P), because only page tables matter in paged segmentation ignores segment tables. O(S^2 * P^2), due to nested page and segment tables exaggerates complexity.
Final Answer:
Option A -> Option A
Quick Check:
Segment count times pages per segment -> O(S * P) [OK]

Quick Trick: Paged segmentation tables multiply segments by pages per segment [OK]

Common Mistakes:

MISTAKES

Adding instead of multiplying segment and page table sizes
Ignoring segment table overhead

Trap Explanation:

PITFALL

Candidates often think segment and page tables are independent and add sizes, missing multiplicative effect.

Interviewer Note:

CONTEXT

Assesses understanding of space complexity in combined memory schemes

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More Operating Systems Quizzes

In a system using segmentation with paging (paged segmentation), what is the space complexity of the combined page and segment tables relative to the number of segments S and pages per segment P?

Step 1: Understand paged segmentation structure

Step 2: Calculate total entries

Step 3: Evaluate options

Final Answer:

Quick Check:

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