1. Suppose a computer using direct mapped cache has 2^24 bytes of byte-addressab
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Question
1. Suppose a computer using direct mapped cache has 2^24 bytes of byte-addressable main memory and a cache of 64 blocks, where each cache block contains 16 bytes.
How many blocks of main memory are there?
What are the sizes of the tag, block and offset fields?
Which cache block does the memory address (in hexadecimal) 0x0AB75 map to?
2. Suppose a computer using fully associative cache has 2^20 bytes of byte-addressable main memory and a cache of 64 blocks, where each cache block contains 32 bytes.
How many blocks of main memory are there?
What are the sizes of the tag and offset fields?
Which cache block does the memory address (in hexadecimal) 0B9E6 map to?
Explanation / Answer
1)Answer:
How many blocks of main memory are there?
2^24 / 2^4
=2^20 blocks
What are the sizes of the tag, block and offset fields?
24 bit addresses – 6 in word field and 18 in the tag field
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