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GATE | GATE CS 2010 | Question 48

  • Difficulty Level : Basic
  • Last Updated : 30 Sep, 2021

A computer system has an L1 cache, an L2 cache, and a main memory unit connected as shown below. The block size in L1 cache is 4 words. The block size in L2 cache is 16 words. The memory access times are 2 nanoseconds. 20 nanoseconds and 200 nanoseconds for L1 cache, L2 cache and main memory unit respectively.


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When there is a miss in L1 cache and a hit in L2 cache, a block is transferred from L2 cache to L1 cache. What is the time taken for this transfer?
(A) 2 nanoseconds
(B) 20 nanoseconds
(C) 22 nanoseconds
(D) 88 nanoseconds

Answer: (C)

Explanation: A block to access in L2 cache requires 20 nanoseconds, and 2 seconds to place in L1-cache.

The block size in L1 cache is 4 words, so total time is =time to access L2 + time to place in L1 = 20+2 = 22 ns.

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