Problem #2 in comuter organization and architecture Suppose a computer using dir

Question

Problem #2 in comuter organization and architecture
Suppose a computer using direct mapped cache has 232 bytes of main memory and a cache of 1024 blocks, where each block contains 32 bytes. (a) How many blocks of main memory are there?
(b) What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, block and byte fields?
(c) To which cache block will the memory reference 000063FAH map?
(d) What is the total size of the cache (bits) including tag, valid and dirty bits?
Problem #2 in comuter organization and architecture
Suppose a computer using direct mapped cache has 232 bytes of main memory and a cache of 1024 blocks, where each block contains 32 bytes. (a) How many blocks of main memory are there?
(b) What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, block and byte fields?
(c) To which cache block will the memory reference 000063FAH map?
(d) What is the total size of the cache (bits) including tag, valid and dirty bits?
Problem #2 in comuter organization and architecture
Suppose a computer using direct mapped cache has 232 bytes of main memory and a cache of 1024 blocks, where each block contains 32 bytes. (a) How many blocks of main memory are there?
(b) What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, block and byte fields?
(c) To which cache block will the memory reference 000063FAH map?
(d) What is the total size of the cache (bits) including tag, valid and dirty bits?

Explanation / Answer

1K = 102410 = 210 which is 40016 1M = 1,048,57610 = 220 which is 10000016 lines of cache m = 512 = 29 blocks in memory M = 524,288 = 219 memory is 16M = 224 so addresses are 24 bits each block contains 32 bytes; 32 = 25 so 5 bits are needed 9 bits are used for the block field the high order 10 bits are the tag field

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