Multiple choice

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.

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?

  1. 2 nanoseconds

  2. 20 nanoseconds

  3. 22 nanoseconds

  4. 88 nanoseconds
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Memory access time for L1 = 2 ns. Memory access time for L2 = 20 ns. Memory access time for L3 = 200 ns. If there is a miss in L1, then transfer will go to L2. So, total time in this transfer = 20 + 2 = 22 ns. The block size in L1 = 4 words. So, the time taken in transfer from L2 to L1 = 22 * 4 = 88 ns. So, this option is correct.

Keep practicing — related questions