With the following configuration: 8MB total memory, 256kb cache , 4kb is block size.Using direct mapping, how many different physical memory blocks can be mapped on to the cache.A. 64B. 256C. 128

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Sameer Gupta

  • Sep 28th, 2005
 

Total cache size =256kb=2^18 bytes

Block size=4kb=2^12bytes

Each block will be mapped in each cache line.

So total no. of different physical blocks that can be mapped at a time in the cache will be 2^18/2^12=64

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sameergupta

  • Sep 28th, 2005
 

Total cache size =256kb=2^18 bytesBlock size=4kb=2^12bytesEach block will be mapped in each cache line.So total no. of different physical blocks that can be mapped at a time in the cache will be 2^18/2^12=64

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hani

  • Nov 16th, 2013
 

A

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