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If T1 and T2 are average access time of upper level memory M1 and lower level memory M2 in a 2- level memory hierarchy and H is the hit rate in M1, then the overall average access time is given by _______, assuming that in case of a miss in M1, a block is first copied from M2 to M1 and then accessed from M1.
  • a)
    T1 + (1 - H) × T2
  • b)
    (T1 + T2) (1 - H)
  • c)
    HT1 + (1 - H) × (T1 + T2)
  • d)
    (1 - H) × T1 + T2
Correct answer is option 'A,B'. Can you explain this answer?
Most Upvoted Answer
If T1 and T2 are average access time of upper level memory M1 and lowe...
We know that,
First, a page is accessed in upper-level memory, if it is present there then it is called a hit. But if the page is not found in upper-level memory, then it is called a miss, for this page is first copied from lower memory to upper-level memory. During miss, it involves first access time for upper-level memory then access time for lower-level memory.
Now,
T1 is the memory access time for memory M1.
T2 is the memory access time for memory M2.
H is the hit rate in M1.
So, 1 − H is the hit rate in M2.
In serial manner:
Average access time will be = H1 T + (1 − H) × (T1 + T2) = T1 + (1 − H)T2
Because the first page is accessed in memory M1 and it takes T1 time. If page not found there, then it will be:
Accessed in M2 and takes T2 time with miss rate (1 − H).
Hence, the correct options are (A) and (B).
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Community Answer
If T1 and T2 are average access time of upper level memory M1 and lowe...
B) T1 + (1 - H) * T2
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If T1 and T2 are average access time of upper level memory M1 and lower level memory M2 in a 2- level memory hierarchy and H is the hit rate in M1, then the overall average access time is given by _______, assuming that in case of a miss in M1, a block is first copied from M2 to M1 and then accessed from M1.a)T1 + (1 - H)× T2b)(T1 + T2) (1 - H)c)HT1 + (1 - H)× (T1 + T2)d)(1 - H)× T1 + T2Correct answer is option 'A,B'. Can you explain this answer?
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