Open In App
Related Articles

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

Improve Article
Improve
Save Article
Save
Like Article
Like

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)

Explanation:

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

In between two access of the same block there 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 cahe for the first 
time.  Therefore Miss ratio =(Cache miss)/(No. of request) = n/N 


Quiz of this Question
Please comment below if you find anything wrong in the above post

Last Updated : 15 Dec, 2021
Like Article
Save Article
Similar Reads