20Suppose a computer using set associative cache has 218 byt
20.Suppose a computer using set associative cache has 218 bytes of byte-addressable main memory and a
cache of 128 blocks, and each cache block contains 8 bytes.If this cache is 4-way set associative, what is the
size in bits of the tag field?.
Solution
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Answer:
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Given
Set associative cache has 2^18 bytes of byte-addressable main memory
Cache of 128 blocks
Each cache block contains 8 bytes
A four-way set implied 4 lines / set in the cache.
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Calculating total number of lines in the cache
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=> Cache blocks * number of bytes in each block
= 128 * 8
= 1024 lines in the cache
Therefore, The cache consists of 1024 lines/4 lines/set = 256 sets
Threfore 8 bits are required to identify the set number.
Given 2^18 bytes of byte-addressable main memory
=> 18-bit address is needed;
Main memory consists of 2^18 byte/8 bytes = 2^15 blocks
=> The length of
Tag + Set = 15
Tag = 15-8
Tag = 7
=> Hence The length of tag field is 7 bits.
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