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Citrus2011 [14]
3 years ago
5

Problem 4 (25 points)Consider a byte addressing architecture with 64-bit memory addresses.(a)Which bits of the address would be

used in the tag, index and offset in a direct-mapped cache with 512 1-word blocks. 3(b)Which bits of the address would be used in the tag, index and offset in a direct-mapped cache with 64 8-word blocks.(c)What is the ratio of bits used for storing data to total bits stored in the cache in each of the above cases
Computers and Technology
1 answer:
zysi [14]3 years ago
5 0

Answer:

Following are the solution to the given points:

Explanation:

The Memory address value = 64 bit

The Size of the word = \frac{64}{8} =8 \  Byte

In point a:

The offset size = 3\ bits ( in 1-word block size)  

The Index size = 9 \  bits (as block number =512)  

Tag size = 64 - 12  = 52\  bits

In point b:

The offset size = 8 \times 8 \ bytes = 2^6 = 6 \ bits.

The Index size = 64 \ bits = 2^6 \ =6 \ bits

Tag size  = 64 - 12  = 52\  bits

In point c:

The Ratio at point a  

\to 3:64

The Ratio at point b

\to 6:64

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