Suppose we have a computer that uses a memory address word size of 8 bits. This

Question

Suppose we have a computer that uses a memory address word size of 8 bits. This computer has a 16-byte cache with 4 bytes per block. The computer accesses a number of memory locations throughout the course of running a program. Suppose this computer uses direct-mapped cache. The format of a memory address as seen by the cache is shown below: The system accesses memory addresses in this exact order: 0x6E, 0xB9, 0x17, 0xE0, 0x4E, 0x4F, 0x50, 0x91, 0xA8, 0xA9, 0xAB, 0xAD, 0x93, and 0x94. a. What is the hit ratio for the entire memory reference sequence given above, assuming we count the first four accesses as misses? b. What memory blocks will be in the cache after the last address has been accessed?

Explanation / Answer

Observation:

We can see that 2 bits are allocated for BLOCK, which indicates that the system consists of (22)4 blocks.

Before we proceed, let us break each memory addresses into TAG, BLOCK, and OFFSET

Add        TAG       BLOCK   OFFSET                 HIT/MISS

B9           1011       10           01                           MISS (compulsory miss)
17           0001       00           01                           MISS (compulsory miss)
E0           1110       00           00                           MISS (Different tag value)
4E           0100       11           10                           MISS (compulsory miss)
4F           0100       11           11                           HIT
50           0101       00           00                           MISS (Different tag value)
91           1001       00           01                           MISS (Different tag value)
A8           1010       10           00                           MISS (Different tag value)
A9           1010       10           01                           HIT
AB          1010       10           11                           HIT
AD          1010       11           01                           MISS (Different tag value)
93           1001       00           11                           HIT
94           1001       01           00                           MISS (compulsory miss)

Hit rate = 4/13 = 0.3077

b.
Cache Status

Block                     Tag                         Range (0x)
00                           1001                       90 - 93
01                           1001                       94-97
10                           1010                       A8-AB
11                           1010                       AC-AF

I tried my best to keep thing simple and add comments whenever possible. I hope you like the way I commented and my explanation. Incase you are still facing any problem with the logic and approach, please let me know in the comment section below. I shall be glad to help you with the logic.

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