Cache

Direct Mapped Cache

Cache's are directed mapped if each memory location is mapped to exactly one location in the cache. An example would be:

     location = (Block Address) MOD (Number of cache blocks in the cache)

In this way we can map a memory location to a cache location.

Example: Suppose we have a cache with 8 slots 2^3. Then a word at 45 would be found at slot 5 in the cache.

https://www.scotnpatti.com/images/directmappedcache.jpg

Tags contain the address information required to identify if the information in a cache location coresponds to the data needed. These tags contain the upper bits of the memory address.

Byte offset contains the number of bits required to specify bytes in a block. Since this example uses 4 byte ( or word) sized blocks the byte offset is 2 bits

Valid bit (or validity bit): When the computer is first started the tags contain junk. Since we don't want to recognize this junk, we start with all the valid bits set to 0 for invalid. Even as time goes on we may still have invalid information and so we have to check this bit.

https://www.scotnpatti.com/images/directmappedcache2.jpg

To determine the size of the cache above:

cache size = (number of locations=1024 = 2^10) X 
             ( Block size=32 + Validity bit=1 + (32 - TagSize=10 - Block bits=2)) 
           = 2^10 X (32 + 1 + 32 - 10 - 2) 
           = 1024 X (53)
           = 54,272 bits or 6,784 bytes

How many bits are required for a direct-mapped cache with 64 KB of data and one-word blocks, assuming a 32-bit address.

64/4 = 16 KWords = 2^16
Cache Size = 2^16 X (32 + (32-16-2) + 1) 
           = 2^16 X 49 = 802816 bits = 100,352 bytes = 98 KB.

Cache Misses

A data cache miss requires us to "freeze" the processor on the instruction that caused the miss. We then retrieve the data from memory, place it in cache and restart the instruction - this time we are guaranteed a hit.

An instruction cache miss requires us to do the following

1. Send the original PC value (Current PC - 4) to memory.
2. Instruct main memory to performa read and wait for the memory to complete it's access.
3. Write the cache entry, putting the data from maemory in the data portion of the entry, writing the upper bitsof the address (from the ALU) into the tag field, and turning the valid bit on.
4. Restart the instruction execution at the first step, which will refetch the instruction, this time finding it in the cache.

Points to Ponder

1. A 32-bit processor has 32-bit registers --> The smallest unit of data to be loaded anywhere is 32-bits = 4 bytes = 1 word. A word becomes the smallest unit of the memory hierarchy.

2. Block size is a conversion factor:

       Bytes
     C -----
       Block

Locality

The schemes of using one word blocks does help with temporal locality but it does not help with spatial locality. In order to take advantage of spatial locality we must use multi-word blocks. This will also increase the effeciency of the cache because more bits will be used for data instead of overhead (tag, valid bit)

Direct Mapped cache with multi-word blocks

Since we are now talking about multi-word blocks, we must have a way to identify not just the word, but the block!

                  |  Byte Address   |
  Block Address = | --------------- | or (Byte Address) DIV (Bytes Per Block)
                  | Bytes Per Block |
                  --               --

So here we have four different divisions of the address:

Tag is now the Block Address, we still use it to check and make sure that the block in the cache is the block that we want.

Index identifies the word within the cache, that is Block Address MOD Number of Cache Blocks (remember Cache Blocks are multi-word now). How does Blocks in general relate to Cache Blocks? I surmise that they are the same!

Block offset Identifies the word within the block to the multiplexor. This the size in bits = n where 2^n = words per block.

Byte offset Will probably always be 2 since we are dealing with 32-bit machines with 32-bit words.

https://www.scotnpatti.com/images/directmappedcache3.jpg

Cache Miss

We can handle read misses exactly as before, however write misses require a read from memory. P588 computer organization and design. Basically you can't just change the tag and go like before on a write through buffer, you must read in the correct block other wise you will have two parts of different blocks in memory with the latest tag - now that's a problem!

Increasing block size will not always result in better miss rates. You could end up with only one block in your cache... and also your miss penalty will increase as the transfer time may also increase.