set associative

Question

A computer has 256KB , k way set associative write back data cache with block size of 32 B. The addresss sent to the cache controller by the processor is of 32 bits. In addition to the address tag , each cache directory contains 2 valid bis and 1 modified bit . if 16 bits are used to address tag then what is the min vallue of k

Comments

5???

Answer 1

Number of Cache entries in the cache of size $256KB$  having each block size $32B$

$=\frac{2^{8}*2^{10}} {2^{5}}=2^{13}$

Given tag size bits$=16$

As processor is of $32$ bits and cache is using set associative,

number of bit to represent tag+number of bit to represent sets+number of bit to represent block size$=32$

$16+x+5=32 \Rightarrow x=11$

number of sets$=2^{11}$

 

$\Rightarrow 2^{11}=\frac{2^{13}}{k}$

$k=2^{2}=4$

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[Edit]

To find Cache Tag Directory size-:

$\text{Cache Tag Directory size}=\text{Number of sets } \times k \times \text{tag bit+valid+modified bits}$

$=2^{11} \times 4 \times  (16+2+1) \text{bits}$

                                                        

Comments

@ A_i_$_h

@ sourav. 

My answer is correct you can see the similar question and Sourav your answer about cache tag directory size is wrong!!!!!!

https://gateoverflow.in/43311/gate2012-55

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okay thank you :)

@akb1115

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Updated.

Thanks @akb1115

Answer 2

​​​​​4 way set associative