Direction: The computer system has an L1 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 ns, 20 ns and 200 ns, for L1 cache, L2 cache and main memory unit respectively.
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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 transfer ?
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- 222 ns
- 888 ns
- 902 ns
- 968 ns
- 222 ns
Correct Option: A
As given memory access time for main memory = 200 ns
Memory access time L2 = 2 ns
Memory access time L2 = 20 ns
Total access time = Block transfer time from main memory to L2 cache + Access time of L2 + Acces time of L1
Now, the required time of transfer = 200 + 20 + 2 = 222 ns
