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+3 votes
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In k-way set associative mapping , the tag field contains 8 bits and 64 no. of sets and propagation delay of a comparator is k/20 nsec and propagation delay of 2*1 multiplexer is k/10 nsec. then what will be the total delay? (let k=8)
asked in CO & Architecture by Loyal (2.6k points)  
edited by | 70 views
We don't know about no of blocks in one set so how many mux is required is uncertain i think .

1 Answer

+2 votes
Delay of comparator will be 8/20 or 0.4 nsec

Since it is given 2:1 comparator but we require input of size 8 so we will require

8/2= 4 4/2=2 2/2 =1

so we will require total of 7 Mux 1st level will have 4 mux 2nd will have 2 and 3rd will have 1

Delay of 1st level will be 0.8 second level will be 0.8 third level be 0.8

Total delay will be 0.4+0.8+0.8+0.8=2.8 nsec
answered by Boss (7.8k points)  
You can identify which set ?? But within that set which block ?? How you will identify??
Question is not asking about mapping it is asking about delay

Tesla!  

can you tell me h/w implementation of set associative mapping(i.e working of MUX).....please

refer this



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