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Dafna11 [192]
3 years ago
6

Put the steps in order to produce the output shown below. Assume the indenting will be correct in the program.

Computers and Technology
1 answer:
V125BC [204]3 years ago
7 0

Answer:

I took a screenshot of the test withe that question

Explanation:

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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
Nataliya [291]

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)

4 0
3 years ago
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