The width of the physical address on a machine is 40 bits. The width of the tag field in a 512 KB 8-way set associative cache is________ bits.

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Habib Mohammad Khan @habibkhan 8 Dec 2016 07:02 pm

Physical Address = 40
=> Tag + Set + Block Offset = 40
=> T + S + B = 40 ----------- (1)
We have , Cache Size = number of sets * blocks per set * Block size
=> 512 KB = number of sets * 8 * Block size
=> number of sets * Block size = 512/8 KB = 64 KB
=> S + B = 16 ---------(2)
from (1) & (2)
T = 24 bits (Ans)
 

So the answer should be corrected..

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