cache miss

Question

Answer 1

Assuming memory to be word addressable So we have 2 words per block means cache contains total of 4 blocks and in memory every two words present per block.

So     addresses :  0,11,4,14,9,1,8,0

Correspond Blk :    0,5, 2, 7, 4,0, 4,0                // You can simply calculate manually or block no = floor(address/block size)

So

0 = Compulsory miss

5 = Compulsory miss

2 = Compulsory miss

7 = Compulsory miss

4 = Conflict miss

0 = Conflict miss

4  = Conflict miss

0 = Conflict miss

 

So total misses = 8 , Conflict miss = 4,Compulsory miss = 4

So answer = 8+4+4 = 16

 

Am i right?

Comments

exactly my point! So those will be starting addresses of each block right ?

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Mam but according to me it's not necessary though because see in question there are two addresses 0 and 1 (Lets forget about rest of addresses for a while)

Both will be present in a single block i.e. Block 0 as I above showed.

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cache block size and main memory block size is same, that's why mapping is possible

read this http://csciwww.etsu.edu/tarnoff/labs4717/x86_sim/direct.html

http://www.ques10.com/p/11169/explain-cache-memory-and-describe-cache-mapping--1/