Below is a list of 32-bit memory address references, given as word addresses.

3, 180, 43, 2, 191, 88, 190, 14, 181, 44, 186, 253 5.2.1

a) For each of these references, identify the binary address, the tag, and the index given a direct-mapped cache with 16 one-word blocks. Also list if each reference is a hit or a miss, assuming the cache is initially empty.

b) For each of these references, identify the binary address, the tag, and the index given a direct-mapped cache with two-word blocks and a total size of 8 blocks. Also list if each reference is a hit or a miss, assuming the cache is initially empty.

Question

Damian Lane

Computers & Technology

25 April, 11:52

Below is a list of 32-bit memory address references, given as word addresses.

3, 180, 43, 2, 191, 88, 190, 14, 181, 44, 186, 253 5.2.1

a) For each of these references, identify the binary address, the tag, and the index given a direct-mapped cache with 16 one-word blocks. Also list if each reference is a hit or a miss, assuming the cache is initially empty.

b) For each of these references, identify the binary address, the tag, and the index given a direct-mapped cache with two-word blocks and a total size of 8 blocks. Also list if each reference is a hit or a miss, assuming the cache is initially empty.

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Answers (1)

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Jenny Arias · Answer 1

For a direct mapped cache the general rule is: first figure out the bits of the offset (the right-most bits of the address), then figure out the bits of the index (the next-to right-most address bits), and then the tag is everything left over (on the left side).

One way to think of a direct mapped cache is as a table with rows and columns. The index tells you what row to look at, then you compare the tag for that row, and if it matches, the offsettells you which column to use. (Note that the order you use the parts: index/tag/offset, is different than the order in which you figure out which bits are which: offset/index/tag.)

So in part (a) The block size is 1 word, so you need 0 offset bits (because 20 = 1 20 = 1). You have 16 blocks, so you need 4 index bits to give 16 different indices (because 24 = 16 24 = 16). That leaves you with the remaining 28 bits for the tag. You seem to have gotten this mostly right (except for the rows for "180" and "43" where you seem to have missed a few bits, and the row for "181" where you interchanged some bits when converting to binary, I think). You are correct that everything is a miss.

For part (b) The block size is 2 words, so you need 1 offset bit (because 21 = 2 21 = 2). You have 8 blocks, so you need 3 index bits to give 8 different row indices (because 23 = 8 23 = 8). That leaves you with the remaining 28 bits for the tag. Again you got it mostly right except for the rows for "180" and "43" and "181". (Which then will change some of the hits and misses.)