GATE CSE 1995 | Question: 2.25

Question

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:

• A. $15, 40$
• B. $6, 4$
• C. $7, 2$
• D. $4, 6$

Comments

option D) 4,6 is the answer.

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NO. of bits required are-

block size= 64 = $2^{6}$ (6 bits)

cache size = 4k = $2^{12}$

number of lines= $\frac{cache size}{block size}$ = $\frac{4K}{64}$ = 64 = $2^{6}$

number of set = $\frac{number of lines}{block per set}$ = $\frac{64}{4}$ = 16 = $2^{4}$  (4 bits)

4

6

          Tag                SET NO      BLOCK OFFSET

Ans: D. 4,6

 

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block size = 64 word = 2^6 word

cache size = 4k = 2^12

number of lines​ = cache size / block size = 2^12 / 64 = 64

number of set = block size / blockperset = 2^6 / 4 = 16

Answer 1

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 mean WORD is $6$-bits.

So, the answer is an option (D).

Comments

determine the physical address size ?

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can not be determined with given information.

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hi, can i find out size of main memory with the available information?

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No, the given data is not enough.

Answer 2

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