Consider a two-level cache memory L₁, L₂:

Cache	Hit rate	access time
L1	X	2 ns
L2	Y	5 ns

Main memory access time is 12 ns. The average memory access time of this architecture is 6 ns. Miss rate of L₂ is half of L₁. What will be the value of (X – Y)?

(A) 0.75
(B) 0.5
(C) 0.25
(D) None of these.


Answer: (C)

Explanation: Let, miss rate of L₂ = M,
So, miss rate of L₁ = 2M

Now, average memory access time
= access time of L₁ + Miss rate of L₁ * Miss penality of L₁

And, miss penality of L₁
= Access time of L₂ + Miss rate of L₂ * Miss penality of L₂

Therefore, miss penality of L₁

= 5 ns +  M * 12 ns 

So, average memory access time

= access time of L1 + Miss rate of L1 * Miss penality of L1 
6 ns = 2 ns + 2M * (5 ns +  M * 12 ns) 

On solving this equation:

M = 1 / 4, -2 / 3 

Therefore,

M(Miss rate of L2) = 1 / 4 (rejecting negative rate)
2M(Miss rate of L1) = 1 / 2 

Since,

1 - miss rate = hit rate 
Y = 1 - 1/4 = 3/ 4
Similarly, 
X = 1 - 1/2 =  1/2 

Hence,

X - Y = 1/4 

So, option (C) is correct.


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