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A computer system has a $4 \ K$ word cache organized in block-set-associative manner with $4$ blocks per set, $64$ words per block. The number of bits in the SET and WORD fields of the main memory address format is:

  1. $15, 40$
  2. $6, 4$
  3. $7, 2$
  4. $4, 6$
asked in CO and Architecture by Veteran (52.1k points)
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1 Answer

+16 votes
Best answer

Number of sets $=\dfrac{4K}{(64\times 4)}=16$

So, we need $4-bits$ to identify a set $\Rightarrow$ SET $= 4$ bits.

$64$ words per block means WORD is $6-bits$.

So, answer is option (D)

answered by Veteran (414k points)
edited by
0
determine the physical address size ?
+4
can not be determined with given information.
0
hi, can i find out size of main memory with the available information?
Answer:

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