Gate2019 cs Q43

  • Option : B
  • Explanation :
    1 word = 4 bytes
    Page size = 8 kB = 213 B
    Number of words in 1 page = 213/22 = 211
    TLB can hold 128 valid entries so, at most 128 × 211 memory address can be addressed without TLB miss.
    128 × 211 = 256 × 210
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