Now, we consider a 16-byte, four-way, fully- associative cache. Since the capacity of the cache is 16 bytes, the array "a" in our example (does/does not) fit inside the cache. We can deduce that the block size for this cache is (? ) bytes per block. So the block index size b=2 bits. For a memory trace record such as: L 1fff000116,2 the 2-bit block offset is (Ob01/0b10/0bl1/0b00). The tag bits are all the rest of the bits not part of the block offset.
Now, we consider a 16-byte, four-way, fully- associative cache. Since the capacity of the cache is 16 bytes, the array "a" in our example (does/does not) fit inside the cache. We can deduce that the block size for this cache is (? ) bytes per block. So the block index size b=2 bits. For a memory trace record such as: L 1fff000116,2 the 2-bit block offset is (Ob01/0b10/0bl1/0b00). The tag bits are all the rest of the bits not part of the block offset.
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Question
![Now, we consider a 16-byte, four-way, fully-
associative cache. Since the capacity of the
cache is 16 bytes, the array "a" in our
example (does/does not) fit inside the cache.
We can deduce that the block size for this cache
is (? ) bytes per block. So the block index size
b=2 bits. For a memory trace record such as: L
1fff000116,2 the 2-bit block offset
is (Ob01/0b10/0bl1/0b00). The tag bits are all
the rest of the bits not part of the block offset.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fffc2ae6a-0ab1-41fc-b4a7-26254b48b5e5%2F2e2e48d7-5b00-40d3-adaf-2f6133b040ac%2Fubwck4f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Now, we consider a 16-byte, four-way, fully-
associative cache. Since the capacity of the
cache is 16 bytes, the array "a" in our
example (does/does not) fit inside the cache.
We can deduce that the block size for this cache
is (? ) bytes per block. So the block index size
b=2 bits. For a memory trace record such as: L
1fff000116,2 the 2-bit block offset
is (Ob01/0b10/0bl1/0b00). The tag bits are all
the rest of the bits not part of the block offset.
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