A two-way set-associative cache has lines of 16 bytes and a total size of 8 Kbytes. The 64-Mbyte main memory is byte addressable. Show the format of main memory addresses.
Answers
Answer:
There are a total of 8 kbytes/16 bytes = 512 lines in the cache. Thus the cache consists of 256 sets of 2 lines each. Therefore 8 bits are needed to identify the set number. For the 64-Mbyte main memory, a 26-bit address is needed.
Answer:
The size of the physical address space of a processor is 2p bytes. The word length is 2w bytes.
Explanation:
There are a total of 8 kbytes/16 bytes = 512 lines in the cache. Thus the cache consists of 256 sets of 2 lines each. Therefore 8 bits are needed to identify the set number. For the 64-Mbyte main memory, a 26-bit address is needed.
Number of sets = 64/4 = 16
The cache is divided into 16( = 2⁴) sets
Therefore, 4 bits are needed to identify the set number.
Main memory consists of 4K = 2¹² blocks.
Therefore, the set + tag lengths must be 12 bits
tag length = 12 - 4 = 8
Each block contains 128 words = 2⁷ words
Therefore, 7 bits are needed to specify the word field
So the main memory can be represented as
TAG - 8
SET - 4
WORD - 7
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