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GATE | GATE-CS-2014-(Set-1) | Question 65

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  • Difficulty Level : Expert
  • Last Updated : 15 Dec, 2021
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An access sequence of cache block address is of length N and contains n unique block addresses. The number of unique block addresses between two consecutive accesses to the same block address is bounded above by k. What is the miss ration is the access sequence is passed through a cache of associativity A >= k exercising least-recently used replacement policy.
(A) n/N
(B) 1/N
(C) 1/A
(D) k/n

Answer: (A)


Their are N access request for the cache blocks out this n
blocks are unique .

In between two access of the same block their are request of 
(k-1) other block block.

And if their associativity >=k and use LRU, then
there will be only one cache miss for every unique block i.e.,
n and it will be the time when the enter the cache for the first 
time.  Therefore Miss ratio =(Cache miss)/(No. of request) = n/N 

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