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: Average memory access time = access time of L₁ + Miss rate of L₁ * Miss penality of L₁
Miss penality of L₁ = Access time of L₂ + Miss rate of L₂ * Miss penality of L₂
Let miss rate of L₂ = M ;
Then Let miss rate of L₁ = 2M
Miss penality of L₁ = 5 ns + M * 12 ns
= 5 ns + M * 12 ns
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
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
X – Y = 1/4
So, option (C) is correct.


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