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A certain processor uses a fully associative cache of size 16 kB. The cache block size is 16 bytes Assume that the main memory is byte addressable and uses a 32-bit address. How many bits are required for the Tag and the Index fields respectively in the addresses generated by the processor?
  • a)
    28 bits and 4 bits
  • b)
    24 bits and 4 bits
  • c)
    24 bits and 0 bits
  • d)
    28 bits and 0 bits
Correct answer is option 'D'. Can you explain this answer?
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A certain processor uses a fully associative cache of size 16 kB. The ...
The division of fully associative cache is 
 Main memory address(32 bits)
Cache size is 16 kB
Block size is 16 Bytes , byte offset is log2 = 4bits
Number of blocks 16kb/16 = 1k = 210
Number of bits required to index is 10 bits.
Since fully associative is not having index field hence TAG field is 32 – 4 = 28 bits (D) is the Key.
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Most Upvoted Answer
A certain processor uses a fully associative cache of size 16 kB. The ...
To determine the number of bits required for the Tag and Index fields in the addresses generated by the processor, we need to consider the size of the cache, the cache block size, and the address size.

Given information:
Cache size = 16 kB
Cache block size = 16 bytes
Address size = 32 bits

1. Calculate the number of cache blocks:
Cache size = Number of cache blocks * Cache block size
16 kB = Number of cache blocks * 16 bytes
Number of cache blocks = (16 kB) / (16 bytes) = 1024 cache blocks

2. Determine the number of bits required for the Index field:
Since the cache is fully associative, there is no need for an Index field. In a fully associative cache, each cache block can be placed in any cache location. Therefore, the Index field size is 0 bits.

3. Calculate the number of bits required for the Tag field:
To calculate the number of bits required for the Tag field, we need to determine the number of bits needed to represent the cache block address.

Since there are 1024 cache blocks (as calculated in step 1), we need log2(1024) bits to represent the cache block address. This is because log2(1024) = 10, and we need 10 bits to represent the cache block address.

As the address size is 32 bits, and the Index field size is 0 bits, the remaining bits are used for the Tag field.

Number of bits for Tag field = Address size - Index field size = 32 bits - 0 bits = 32 bits

Therefore, the number of bits required for the Tag and Index fields respectively in the addresses generated by the processor is 32 bits and 0 bits, which corresponds to option 'D'.
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A certain processor uses a fully associative cache of size 16 kB. The cache block size is 16 bytes Assume that the main memory is byte addressable and uses a 32-bit address. How many bits are required for the Tag and the Index fields respectively in the addresses generated by the processor?a)28 bits and 4 bitsb)24 bits and 4 bitsc)24 bits and 0 bitsd)28 bits and 0 bitsCorrect answer is option 'D'. Can you explain this answer?
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