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Consider a computer system has a main memory consisting of 1M 16 bit words. It also has a 4 K-word cache organized in the block set associative manner, with 4 blocks per set and 64 words per block. What is the number of bits in each of the TAG, SET and WORD field of main memory address format

  1. 11, 4, 6
  2. 10, 5, 6
  3. 10, 4, 7
  4. 11, 4, 7
asked in CO & Architecture by Boss (34.2k points) | 196 views
0

I also need a clarification on this..they have taken word field in terms of bytes i.e. 64words*2B/word = 128 =27.

Is this what we should do since memory is not mentioned as "word addressable"?

0

MiNiPanda exactly i have so many doubts now.

0

Suppose it's Word Addressable then ? could you please solve when it's word addressable

1 Answer

+3 votes
Best answer

If nothing is mentioned in question then, by default it is byte addressable. Answer is option (3). 

Here is the explaination for the same

 

answered by Active (1.6k points)
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why here word size is taken as 2B??
why not word size=4B??
+1

 word size is given as 16 bit means 2 byte. memory   by default  taken as byte addressable ,untill it is not specify word addressable we can't take .But when we count machine cycle then we take as word addressable because in one machine cycle ,we access 1 word data. 

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