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THE size of memory required at cache controller to store the metadata is 2kbyte The metadata include tag bits 1 modified bit and 1 valid bit The cache contain 1KBlocks of 32 bytes each organized as directed mapped The size of Main memory is __ Mbytes?
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Answer :  $512 MB$


The size of memory required at cache controller to store the metadata = $2KB = 2^{11}B$

Tag entry size in metadata = $x$ tag bits +$1$ modified bit +$1$ valid bit

#Cache Lines given = $1K = 2^{10}$

Block Size = $32 = 2^5B$

Given, Total size of metadata = $2^{11} * 8bits  = 2^{14}b$

Total size of metadata = Tag entry size ×Number of cache blocks = $(x+1+1)* 2^{10}$

Hence, $2^{14} = (Tag+1+1)*2^{10} \implies Tag = 14 bits$

Thus Size of Main Memory address = Tag + Line Offset + Block Size = $14 + 10+5 = 29bits$

Therefore, the size of MM = $2^{29} = 512 MB$ (ANS)

 

 

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