Suppose a computer using a direct mapped cache has a main memory address space o

Question

Suppose a computer using a direct mapped cache has a main memory address space of 64K. This computer byte addressable. The cache in this computer contains 128 blocks of 16 words. Currently a partial cache configuration is shown with the following hex numbers below What is the format of memory address as seen by the cache? State number of blocks in the MM Determine the number of bytes per block in main memory Specify the cache size in Kbytes For the following address: 1B03 21AA 03A2 first find to what cache line, word slot is it mapped is it mapped second tell if any of the words are in cache or not

Explanation / Answer

Answer A

Because the block size is 16 bytes and the word size is 1 byte, this means there are 16 words

per block. We will need 4 bits to indicate which word we want out of a block. Each cache line/slot

matches a memory block. That means each cache line contains 16 bytes. If the cache is 64Kbytes

then 64Kbytes/16 = 4096 cache lines. To address these 4096 cache lines, we need 12 bits (212=

4096).

Answer d

Sets the aggregate storage formula cache size maximum to 2048 KB for every application and

database.

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