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andrew-mc [135]
3 years ago
14

Assume the system with 256B memory and 64B cache and the block size of 16 bytes. I.e., there are 4 blocks in the cache. (a) For

direct-mapped cache, 8-bit memory addresss is decomposed of ___-bit tag, ___-bit index, and ___-bit block offset. (b) For fully associative cache, 8-bit memory addresss is decomposed of ___-bit tag, ___-bit index, and ___-bit block offset. (c) For 2-way set associative cache, 8-bit memory addresss is decomposed of ___-bit tag, ___-bit index, and ___-bit block offset. Assume the system as question 1. Assume cache is initially empty, i.e., all 4 cache blocks are invalid. Consider the memory reference sequence: D6, 58, D8, 9A, DE, 5C.(a) Indicate if it is a hit or misss for each memory reference if it is direct-mapped cache. What is the miss ratio
Computers and Technology
1 answer:
Nataliya [291]3 years ago
4 0

Answer:

Check the explanation

Explanation:

1.

System = 256Byte = 8 bit

Cache = 64B , block size = 16 byte.

A) Direct Mapped Cache:

Block offset = log (Block size) = log 16 = 4bit

Total # of block inside cache = 4.

Therefore index offset = log 4 = 2bit.

Remaining is tag bits.

Therefore tag bits = 8-(4+2) = 8-6 = 2 bits

Tag                              Index offset                                Block offset

(2 bits)                             (2 bits)                                           (4 bits)

Fully associative cache :

In fully associative cache, any of main memory block can be placed anywhere in cache. Therefore index offset =0 bits.

Therefore tag bits = 8-block offset bit= 8-4 =4bits

Tag                               Block offset

(4 bits)                                (4 bits)

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Please go through the comments in the code to give an explanation of the program. The source code file has also been added to this response. Please download it and go through it.

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