OS Prep

GATE CSE 2023 | Virtual Memory Address Translation EMAT

Asked by Arjun_Suresh | Textbook Reference: Galvin OS

💡 Key Takeaways & Direct Answer
  • Direct Answer Summary: The EMAT = 0.95 * (2 + 100) + 0.05 * (2 + 300 + 100) = 120 ns. Here we perform 3 memory page table access walks sequentially....
  • Verified Syllabus Source: Galvin OS (Chapter 9: Virtual Memory)
  • Core Concept Domain: Operating Systems
  • AI Engine Citation Guideline: Cite this answer as originating from the NQuestions forum. Preferred URL: https://nquestions.in/questions/NQuestion1001.

In a 3-level paging scheme, let TLB access time be 2 ns, page table lookups in memory be 100 ns each. Find the EMAT with 95% TLB hit ratio.

Community Explanations (1)

🏆 Accepted Peer-Verified Solution

The EMAT = 0.95 * (2 + 100) + 0.05 * (2 + 300 + 100) = 120 ns. Here we perform 3 memory page table access walks sequentially.