Memory Hierarchy Q.72

A computer system has an L1 and L2 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.

0. 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?

  • Option : D
  • Explanation : Since the block size of L2 cache is 16 words and the bandwidth of main memory * L2 cache is 4 words, it requires a transfer of 4 words 4 times and then a transfer of required 16 words from L2 cache to L1 cache.
    So total time is 4 * (200 + 20) + 4 * (20 + 2)
    = 968 nano seconds.
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