Confused on how they figured out bytes, book example:

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Discussion Overview

The discussion revolves around understanding the calculations related to cache memory, specifically the Intrinsity caches, including the size of data, tag fields, and total bits. Participants are examining a book example that presents these calculations, seeking clarification on specific values and the reasoning behind them.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant notes that the Intrinsity caches are 16 KB with 256 blocks and questions how the data size of 64 bytes relates to 512 bytes.
  • Another participant confirms that 64 bytes equals 512 bits, suggesting this may clarify the confusion.
  • A third participant corrects the interpretation of 16 KB, stating it should be understood as 16*1024 rather than 16E3, and emphasizes that 16*1024/16 equals 2^10, not approximately.
  • Further discussion reveals that the 8 in the tag field calculation is understood as the number of bits in a byte, but the reasoning for the 6 remains unclear to some participants.
  • Participants are trying to reconcile the total bits calculation, which involves multiplying the number of blocks by the sum of data, tag, and valid bits.

Areas of Agreement / Disagreement

Participants generally agree on the calculations related to bits and blocks, but there remains uncertainty regarding the specific values of 64 bytes, 512 bytes, and the components of the tag field calculation.

Contextual Notes

Some participants express confusion over the derivation of certain values and the assumptions made in the book example, particularly regarding the tag field calculation and the interpretation of cache sizes.

Who May Find This Useful

Readers interested in computer architecture, cache memory design, and those studying related technical materials may find this discussion relevant.

mr_coffee
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The Intrinsity caches are 16 KB caches and have 256 blocks with 16 words per block.

Thus data is 64 bytes = 512 bytes.
The tag field is 18 bits (32 - (8 + 6)).
Total bits = 256 x (Data + Tag + Valid)
= 256 x (512 bits + 18 bits + 1 bit)
= 135,936 bitsI'm confused on how they got 64 bytes then said = 512 bytes.

Also where did they get (32 - (8+6))?
I think i know where they got 32, because its a 32 bit address.Thank you.

I took 16E3/265 = 62.5 not 64.

EDIT:

Okay I got alittle farther...

16E3/16 = 1000 which is ~ 2^10 words, and with a block size of 4 words (2^2), but still not sure how they got 64 and 512, nor the (8+6).
 
Last edited:
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64 bytes = 512 bits

Isn't that just what they meant?
 
16KB is not 16E3 but 16*1024.
16*1024/16 is not ~ but = 2^10.
 
Ahhh thanks guys!

that makes sense... it was a typo in the book.
okay so I understand the 512 bits.

for the:
The tag field is 18 bits (32 - (8 + 6)).
Total bits = 256 x (Data + Tag + Valid)

32 bits address, is the 8 because 8 bits in a byte? still not seeing the 6 though.

Also is the 256 x (data + tag + valid) because they said there is 256 blocks?
 

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