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Consider a system with 24 bit addresses, 128 KB cache with 32 byte lines using direct...

Consider a system with 24 bit addresses, 128 KB cache with 32 byte lines using direct mapping. Divide the address below labeling each part of the address and specify the size of each field in bits.

Tag? Line? Offset?
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Answer #1

ANSWER:

GIVEN THAT:

GIVEN 24 BIT ADDRESSES, 128 KB CACHE WITH 32 BYTE LINES USING DIRECT MAPPING:

1. Cache memory size = 128 KB

2. Block size = Frame size = Line size = 32 bytes = 28 bits = 8 bits is the line size

3. Line Number = cache size / line size = 128 kb/ 8 bits = 214 bits = 14 bits is the Line number

4. Tag = System address bits -(Block size + Line Number) = 24-(14+8) = 2 Bits Tag

5.

Tag = 2 bits

 Line number = 14 bits offset / block size = 8 bits
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