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+2 votes


asked in Operating System by Active (1.2k points) | 50 views

1 Answer

+3 votes
Best answer

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


answered by Boss (5.4k points)
selected by

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



@ saxena0612

check now

Got it !

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