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GATE | GATE CS 2010 | Question 49
  • Difficulty Level : Hard
  • Last Updated : 19 Nov, 2018

Consider the data from above question. When there is a miss in both L1 cache and L2 cache, first a block is transferred from main memory to L2 cache, and then a block is transferred from L2 cache to L1 cache. What is the total time taken for these transfers?
(A) 222 nanoseconds
(B) 888 nanoseconds
(C) 902 nanoseconds
(D) 968 nanoseconds

Answer: (C)

Explanation: Since the block size of L2 cache is 16 words and the bandwidth of mainmem->L2 cache is 4 words, it requires a transfer of 4 words 4 times and then a transfer of required 4 words from L2 cache to L1 cache.

So total time is 4*(200 + 20) + 1*(20 + 2) = 902 nanoseconds.

Option (C) is correct.

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