CO Moemory Hierarchy

Question

Consider a  3 Level Memory System with the following specifications

• (H₁,H₂,H₃) = (0.8,0.9,1)
• (T₁,T₂,T₃) = (2,100,500) nano sec / word
• IF there is miss at L₁ and Hit at L₂ .A 2-Word block must be transfered from L₂ to L₁
• IF there is miss at both Levels then 4-Word block must be bought from L₃ to L₂ and concerned block must be moved from L₂ to L₁

The refered word must always given from L₁   only to processor .

1. Average Access Time Taken Given that the data width between L₁ to L₂ and L₂ to L₃  is 1 Word block is T_avg-1
2. Average Access Time Taken Given that the data width between L₁ to L₂ and L₂ to L₃  is 4 Word block is T_avg-2

What is the ratio of  T_avg-1 to T_avg-2

Answer 1

Lets consider for T_avg-1.

Avg access time for L2

= L2 hit ratio * (L2 access time)

   + L2 miss ratio * ( L2 search to conclude miss + Avg L3 access time + L2 access to accomodate new data) 

= 0.9 * (100) + 0.1 * (100 + 4*500 + 4*100)  

  .....(we multiply by 4 because for each word transfer, 1 mem cycle is consumed)

= 340 ns

Avg access time for L1 

= L1 hit ratio * (L1 access time)

   + L1 miss ratio * ( L1 search to conclude miss + Avg. L2 access time + L1 access to accomodate new data) 

= 0.8 * (2) + 0.2 * (2 +   2*34 + 2*2 )

   ...............(we multiply by 2 for the same reason above)

= 16.4ns

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Now, we consider T_avg-2.

Here, the data width=4 and hence, 1 mem cycle is suficient for transfer of data between any two levels of memory.

Thus, Avg L2 access time

= 0.9 * (100) + 0.1 * (100 + 1*500 + 1*100)  

= 160ns

Avg access time for L1 

= 0.8 * (2) + 0.2 * (2 +   1*34 + 1*2 )

= 9.2ns

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T_avg-1 / T_avg-2

= $\frac{16.4}{9.2}$

= 1.78

Comments

Lets consider 2 level system: cache and M.M

If the cache generates a miss, 2 options are there:

1. Bring the word to cache and then access cache ( M.M access + L2 access)

2. Transfer the word directly to processor from M.M (M.M access )

   The transfer time from M.M to CPU is considered negligible.

 

So total time= L2 access to conclude miss + (1 or 2)

Plz study from Galvin. Form your opinions and after solving lot of questions, confirm them.

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bro tavg1 =340+16.4

tavg2=160+9.2

=356.4/169.2

= 2.10

right??

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While calculating AVG access time for L1 2*34 why? Why not 2*340?