Query Regarding Direct cache mapping [closed]

Question

Thank you for looking into this, I have a problem regarding direct cache mapping, My problem really though is with the question formation itself and the problem I am about to present seems to have inconsistency and has unclear description. So the problem is as follows ( I am presenting exactly as given )

32 bit architecture
Main memory size = 8GB
cache size = 512KB
Block size = 32 words

To Find

1. Number of bits required for Byte Offset = ?
2. Number of bits required for main memory address = ?
3. Number of bits required for main memory block number = ?
4. Cache Index bits = ?
5. Tag-bits = ?
6. Size of Tag Directory = ?

Now My Interpretation of the solution is as follows

As it is 32 bit architecture which means Word size is 4 Byte ( I am confused about this part the most as the question states cache block size in Words but expects to calculate Byte offset ) but for the sake of this problem, moving on main memory size is 8GB that's 2³³Bytes, cache size ( blocks only, this does not include tag directory which is part of cache controller ) Block size of 32 Words or 128 Bytes, now If you agree with me on this much, or else this is probably where I am wrong so please correct me right here. Now assuming I am correct so far, I will first calculate main memory address bits, that will be, first let's calculate it's size then let's do binary log, that will be 2³³Bytes / 4 Bytes per word this comes out to be 2³¹ Words means 31 bits for memory address and our second answer, next I calculate number of main memory blocks that will be 2³³Bytes / 128 Bytes per block this comes out to be 2²⁶ Blocks of main memory so the number of bits required for main memory block number would be 26 bits, so this gives our third answer, moving forward as I assume a single cache/main memory block to be of 128 Bytes and not Words. So for byte offset within a block it must be 7 bits ( Point of highest concern ) and this should be our first answer according to the question. Now let's calculate number of cache blocks that will be 2¹⁹Bytes / 128 Bytes per block which is 2¹² Blocks so cache index size will be 12 bits which is our fourth answer, we're almost done ! Now Tag bits should be and since it's direct mapped it should be memory address size - ( cache index + Word or Byte offset ) that is ( 31 - (12 + 7)) = 12 bits that is our fifth answer and finally size of tag directory will be 2¹²Cache Blocks * 12bits per tag, Now not going into exact answer in KB's is this approach right or have I made a mistake or is the problem formulation incorrect ( as I think the mistake must be with cache blocks being in bytes or words ) ?

Answer 1

I am going to talk about the cache - related questions:

1. Block size is $$32 \text{words} = 2^5$$ . That, means that you need, 5 bits for offset.
2. $$\#\text{blocks}=\frac{2^{19}}{2^5}=2^{14}$$, so you need 14 bits for the index ( to indentify which block in cache is going to represent your wanted memory adresss )
3. For a 32 - bit adress, tag is the rest : $$32 - 14- 5 = 13$$ bits.