Question 63

System parameters:

• TLB access time: $T_{\text{tlb}} = 20\text{ ns}$
• Main memory access time: $T_m = 100\text{ ns}$
• Disk page transfer time: $T_d = 5000\text{ ns}$
• TLB hit ratio: $h = 95\% = 0.95$
• Page fault rate: $p = 10\% = 0.10$ (on TLB miss)
• Dirty page write-back proportion: $20\% = 0.20$ of page faults

Case 1: TLB Hit (probability 0.95):

• Address translation found in TLB (20 ns), then data accessed in main memory (100 ns):
  $$T_{\text{hit}} = 20 + 100 = 120\text{ ns}$$

Case 2: TLB Miss (probability 0.05):

• TLB lookup performed first ($20\text{ ns}$).
• Page table in main memory accessed ($100\text{ ns}$).
  • Subcase 2a: No Page Fault (probability $1 - 0.10 = 0.90$):
    • Data accessed in main memory ($100\text{ ns}$).
    • Total time = $20 + 100 + 100 = 220\text{ ns}$.
  • Subcase 2b: Page Fault (probability $0.10$):
    • Disk service: 20% need write-back and read ($2 \times 5000 = 10000\text{ ns}$), 80% need only read ($5000\text{ ns}$).
    • Average disk transfer time = $0.20 \times 10000 + 0.80 \times 5000 = 2000 + 4000 = 6000\text{ ns}$.
    • Then data accessed in main memory ($100\text{ ns}$).
    • Total time = $20 + 100 + 6000 + 100 = 6220\text{ ns}$.
• Average TLB miss time:
  $$T_{\text{miss}} = 20 + 100 + [0.90 \times 100 + 0.10 \times (6000 + 100)] = 120 + [90 + 610] = 820\text{ ns}$$

Average Memory Access Time (AMAT):

(Accepted range: 154.5 to 155.5 ns).