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Question 9354 – Cache
November 12, 2023
Question 9393 – Cache
November 12, 2023
Question 9354 – Cache
November 12, 2023
Question 9393 – Cache
November 12, 2023

Question 9387 – Cache

Consider two cache organizations: The first one is 32 KB 2-way set associative with 32-byte block size. The second one is of the same size but direct mapped. The size of an address is 32 bits in both cases. A 2-to-1 multiplexer has a latency of 0.6 ns while a kbit comparator has a latency of k/10 ns. The hit latency of the set associative organization is h1 while that of the direct mapped one is h2.

The value of h1 is:

Correct Answer: A

Question 12 Explanation: 
Cache size = 32 KB = 32 × 210 B
Block size = 32 bytes
No. of blocks = 2 (2-way set associative)
No. of combinations = Cache size / (Block size×No. of blocks) = (32×210B) / (32×2) = 29
Index bits = 9
No. of set bits = 5 (∵ cache block size is 32 bytes = 25 bytes)

No. of Tag bits = 32 – 9 – 5 = 18
Hit latency = Multiplexer latency + latency
= 0.6 + (18/10)
= 0.6 + 1.8
= 2.4 ns
A
2.4 ns
B
2.3 ns
C
1.8 ns
D
1.7 ns
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