2. Similar to problem 1, assume your computer system has a 32-bit byte-addressable architecture where addresses and data are each 32 bits. It has a 16K-byte (16,384 bytes) direct-mapped cache, but now the block size is 32 bytes. Answer the following question to observe how the design change impacts the cache size. [10pts]

Question

Engineering

Nathaly

16 August, 14:43

2. Similar to problem 1, assume your computer system has a 32-bit byte-addressable architecture where addresses and data are each 32 bits. It has a 16K-byte (16,384 bytes) direct-mapped cache, but now the block size is 32 bytes. Answer the following question to observe how the design change impacts the cache size. [10pts]

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Answers (1)

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Quinton Odonnell · Answer 1

Question:

The question is not complete. The question to answer was not added. See below the possible question and the answer.

a. How many blocks are in the cache with this new arrangement?

b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

Answer:

(a) Number of blocks = 512 blocks

(b) Tag is 18

(c) Total size of the cache = 8388608 bytes

Explanation:

a.

block size = 32 bytes

cache size = 16384 bytes

No. of blocks = 16384 / 32

No. pf blocks = 512 blocks

b.

Total address size = 32 bits

Address bits = Tag + Line index + block offset

Block Size = 32 bytes.

So block size = 25 bytes.

Hence Offset is 5

No. of Cache blocks = 512 blocks = 29 blocks

Hence line offset is 9

We know that Address bits = Tag + Line index + block offset

So, 32 = tag+9+5

tag = 32 - (9+5)

So Tag is 18

c.

Data bits = 32 bits

Tag=18 bits

Valid bit is 1 bit

so Total cache size = 25+218+20

= 223

=8388608 bytes