question

Question

Answer 1

Cache size = 8byte

Block size = 2byte

#f lines = Cache size/Block size

=> 8/2= 4 lines

#f sets (S)= #f lines / associativity

=> 4/2 = 2 sets

Using LRU :

mapping technique in set associativity => k mod S = i  {k=memory block no.}

only memory block "2" will hit

total no of hit =1

Comments

@Akash Just confirming this is kind of structure right?
Now,
Could you tell me the further procedure which will be hit?
 

 

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@ saxena0612

check now

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Got it !