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Memory Management in Operating Systems

Explore the intricacies of memory management in operating systems, covering topics of addressing, paging, segmentation, and access time calculations in this detailed problem set.

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Memory Management in Operating Systems

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  1. Suggested Exercise #5 Sarah Diesburg Operating Systems CS 3430

  2. Problem 1 (a) • 32-bit addressing • Page size: 4 Kbytes (212 bytes) • Pure paging • 232 bytes / 212 bytes/entry= 220 entries • Each page table entry • 20-bit physical page number • 4 status bits • Page table size = 220 entries * 3 bytes/entry = 3 Mbytes

  3. 32 – 12 = 20 bits for 32-bit machines log2(4KB) = 12 bits for 4-KB pages Virtual page number Physical page number Offset Offset Page table size > Physical page number 220 entries Physical page number Physical page number Error Paging Diagram 4 status bits

  4. Problem 1 (b) • Segmented-paging address translation • Page size: 4 Kbytes (212 bytes) • Offset: 12 bits • Three segments: code, data, stack • Virtual segment number: log23 = 2 bits • Virtual page number • 32 – 12 – 2 = 18 bits

  5. 32 - 2 - 12 = 18 bits 32 bits for 4-GB Page table base Page table bound 22 entries Page table base Page table bound Page table base Page table bound 12 bits for 4-KB pages 20 bits Seg # Virt page # Offset num of entries defined by bound; up to 218 entries Phy page # log2(3 segments) = 2 bits Phy page # Phy page # Segmented Paging

  6. 32 – 2 – 12 = 18 bits 218 Seg # Virt page # Offset Page table size > Phy page # Phy page # Phy page # Error Segmented Paging Page table base + Phy page # Offset log2(4GB) = 32 bits

  7. Problem 1 (b) • Each segment (code, data, stack) uses 16 Kbytes = 4 pages • 3 page tables: • Code-segment page table: 4 entries • Page table entry • 20 bits to address all physical page numbers • 4 status bits • 24 bits/entry * 4 entries = 96 bits • Same for data-segment page table and stack-segment page table • Total: 96 bits * 3 = 288 bits

  8. Problem 1 (b) • 1 segment table: • 4 entries • 32 bits for the page table base • 18 bits for the page table number bound • 4 status bits • Total memory: 4 * (32 + 18 + 4) = 216 bits

  9. Problem 2 • Effective access time = P(hit)*cost(hit) + P(miss)*cost(miss) = P(L1 hit)*cost(L1 hit) + P(L1 miss)*cost(L1 miss) = 98%*(1 clock cycle) + 2%*cost(L1_miss)

  10. Problem 2 • What is cost(L1_miss)? • cost(L1_miss) = P(L2_hit)*cost(L1_miss+L2_hit) + P(L2_miss)*cost(L1_miss+L2_miss+mem_hit)

  11. Problem 2 L1 98% 1 cycle 2% 1 cycle hit L2 1% 2 cycles 99% 2 cycles memory hit 100% 6 cycles hit

  12. Problem 2 • Effective access time = 98%*1 + 2%*99%(1 + 2) + 2%*1%(1 + 2 + 6) L1 98% 1 cycle 2% 1 cycle hit L2 1% 2 cycles 99% 2 cycles memory hit 100% 6 cycles hit

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