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In direct memory management, CPU references address of 15-bits. Main memory size is 512 * 8 and cache memory size is 128 * 8. What is the size of Tag and line in bits?

Block size=8=2^3

I have found the number of bits in the word field as follows, Block size=8=2^3 Therefore, word bits=3 bits

Now number of bits in line field is 7 (128=2^7) i guess.

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Main memory size of 512 doesn't make sense to me; it's too small for 15-bit addresses, and it also makes the cache of 128 appear huge being fully 1/4 of 512.


From the 15-bits in addresses, we subtract the 3 lowest bits because of the block size, leaving 12-bits of addresses to provide to the cache in order to lookup a byte or word.

With cache size at 2^7=128, and a block size of 2^3, the cache has 2^(7-3=4) indexes (aka elements, positions, blocks). (I take it that it is direct mapped, aka 1-way set associative).

So, out of the 12 bits provided to the cache, as 4 are index bits then remaining 8 are tag bits.

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