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A computer is using a fully associative cache and has 216 bytes of memory and a cache of 64 blocks, where each block contains 32 bytes.

Required:
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 and offset fields?
c. To which cache block will the memory address F8C9 (subscript) 16 map?

User Mnickey
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1 Answer

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1) The number of blocks of main memory that are there are;

2^16/32 = 2^16/2^5 = 2^11

2) The format of a memory address as seen by the cache,that is the sizes of the tag and offset fields are;

16 bit addresses with 11 bits also in the tag field and the 5 bits in the word field.

3) The cache block that the memory address F8C9 (subscript) 16 map;

Since it is an associative cache, it can actually map anywhere.

That is, it has the ability of mapping anywhere.

User Krupesh Kotecha
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