Last Updated :
26 Dec, 2018
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 L
2 is half of L
1. 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 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.
Quiz of this Question
Share your thoughts in the comments
Please Login to comment...