Computer Science, asked by asnitha1007, 1 year ago

a computer system has a three level memory hierarchy, with access time and hit ratios as shown below:

Answers

Answered by thesmartlerner

║║║║║║hello mate║║║║║║║║

АпЗЩЁГ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓

A computer system has an L1 cache, an L2 cache, and a main memory unity 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.

cache-image

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

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

First thing that came to my mind was, how to calculate the transfer time using the given access time. During a miss, a block of data is moved from main memory to cache. Then CPU will access it. So, wouldn't be access time > transfer time ?

Then I thought, lets assume access time = transfer time & do the calculation.

Now first question. The question already states there is a miss in L1, so I will not consider L1 access time. Since there is a miss in L1 & hit in L2, a entire block from L2 has to be moved to L1. L2 block size is 16 words, but data bus size is 4 words.

So we have to move 4 words * 4 times.

To transfer 4 word it takes 20 ns. To transfer 4 words, its 80ns. Isn't it the time transferred from L2 to L1 ? The question does not say anything about accessing L1 after moving the data. But 80ns is not in the option !