Consider a computer with a byte-addressable memory. A 40-bit memory address is d

Question

Consider a computer with a byte-addressable memory. A 40-bit memory address is divided as follows for cache processing. First, the 8 low-order bits are chopped off to expose the cache-line number. Second, the next 17 low-order bits are inspected to get the cache-container index. Third, the remaining 15 bits are used as the cache tag. a) What is the cache size in bytes? b) What is the cache-mapping scheme? c) For a given byte in the cache, how many different bytes in the main memory could possibly be mapped to it?

Explanation / Answer

Given information is:

Main memory in address format = 40-bit addressable.

block-off = 8
Then block size will be = 2^8

Index= 17

tag= 15

Given that it is a direct mapped.

a) Cache size?
.............

The formula for no. of lines = cache size/block size.

In direct cache map no. of lines is equal to index.

Therefore, 2^17 = cachesize/(2^8)

=>(2^17)*(2^8) = cache size
  
therefore, cache size = 2^25.
b) cache mapping scheme is in the form of,

Tag Index Blockoff 15 17 8

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