A block-set associative cache memory consists of 128 blocks divided into four block sets. The main memory consists of 16,384 blocks and each block contains 256 eight bit words. a) How many bits are required for addressing the main memory? b) How many bits are needed to represent the TAG, SET and WORD fields? c) Based on the B question, illustrate the physical address

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A: For a 4-way set associative cache, the length of the TAG field is given below:

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A:

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A: I have answered this question in step 2.

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A: The solution for the above questions are:

Q: Computer Organisation Architecture

A: ANSWER:Part (a): Calculating the Total Number of Bits in a Main Memory Address Objective: Determine…

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A: Answer to the above question is in step2.

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A: From the question we are given that Main memory size = 232 bytes Cache memory size = 512 bytes…

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A: The answer for the tag and index fields respectively in the addresses generated by the processor is

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A: Given:- Size of main memory = 220 bytes           Size of blocks   = 24 bytes…

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A:

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Q: Index Tag Tag (binary) | (8 bits) 1 Valid 00 01 (8 bits) | Valid 10 86 42 69 FD 11 000 AF BC 19 6D 1…

A: The maximum size of met in kb is  =25  =64 kb

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A: ANSWER: Memory Address:

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A:

Q: A computer using a direct mapped cache has 220 Bytes of memory (byte addressable) and a cache of 32…

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A: Please find the answer to the above question below:

Q: The phrases "unified cache" and "Hadley cache" should be defined.

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