Question

Consider a cache of 8 lines of 16 bytes each. Main memory is divided into blocks of 16 bytes each. That is, block 0 has bytes with addresses 0 through 15, and so on. Now consider a program that accesses memory in the following sequence of addresses:

Loop three times: 10 through 20; 32 through 52.
Once: 20 through 35.

Suppose the cache is organized as direct mapped. Memory blocks 0, 8, 16 and so on are assigned to line 1; blocks 1, 9, and so on to line 2; and so on. Compute the hit ratio during this process. Provide details regarding how you get the answer.

Answer 1

Block size => 16 Bytes -> Block offset = log 16 = 4 bits.

Let's see how address is stored in main memory block.

Address Range	Main Memory Block #
0-15	Block 0
16 - 31	Block 1
32-47	Block 3
48 - 63	Block 4

A whole main memory block will be brought into one cache block.

Let's see how many miss we have in each reference:

10 through 20 -> These address are occupied in block 0 and 1 of memory . So these 2 block of memory will be brought in cache leading to miss. One miss per block for loading into cache.

Total 2 miss. (Block 0 and Block 1 of MM is brought in cache)

32 through 52 -> Total 2 miss (Block 1 and block 2 is brought in 1st and 2nd block of cache)

Now since there is no block replacement, all reference for next 2 loop will be hit.

Once : 20 - 35 : This is also a hit since already the required block is present in cache.

Now total # of reference =>3 *(11+21) + 16 = 112

Total # of miss = 2. So total #of hits = 112 - 2 =120

So Hit rate = 120/122 *100 = 98.36%

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