Question

g The machine in the problem has a byte-addressable memory of 216 bytes. The direct-mapped cache consists of 32 cache blocks, and the cache block size is 8 bytes. Question 1 How is the 16-bit memory address divided into byte offset, cache index, and tag? Question 2 What is the capacity of the cache in bytes? Question 3 Why tag is also stored in the cache? Question 4 For the following four memory addresses, what cache block would each be mapped to? 0001 0001 0001 1011 1100 0011 0011 0100 1101 0000 0001 1101 1010 1010 1010 1010

Answer 1

Answer / Explanation:

To properly answer this question, let us define some basic terms:

Byte: The term byte is the digital unit for information and mostly consists of eight bits. The byte used to be the number of bits used to encode a single character of text in a computer and for this reason it is the smallest unit of memory in many computer program

Cache: A cache can be refereed to as a software component that stores information or data for the purpose future recall. The data stored in a cache could be treated as a duplicate or a copy of a copy of another data stored elsewhere.

Now, with the understanding from the above definition, referring back to the narrative of the question asked, we have:

(1) 8 leftmost bits = tag; 5 middle bits = line number; 3 rightmost bits = byte number.

                                                          S

                                               Tag           Set slot           W            

Memory address = 16 bits       8                 5                  3

(2)  256 bytes.=2^8 or

      32 blocks * 8 byte = 256 byte

(3)   Because two items with two different memory addresses can be stored in the same place in the  cache. That is, The tag is used to distinguish between them.

(4) Line 3.

    Line 6.

    Line 3.

    Line 21