Consider a machine with a byte addressable main memory of 28 bytes and block siz

Question

Consider a machine with a byte addressable main memory of 28 bytes and block size of 4 bytes. Assume that a direct mapped cache consisting of 8 lines is used with this machine.

a. How is a 8-bit memory address divided into tag, line number and byte number?

b. Into what line would bytes with each of the following addresses be stored?

0001 1011

00110100

11010000

1010 1010

c. Suppose the byte with address 1010 0001 is stored in the cache. What are the addresses of the other bytes stored along with it?

d. How many total bytes of memory can be stored in the cache memory?

e. why is the tag stored in cache.

Explanation / Answer

a.4 leftmost bits = tag; 3 middle bits = line number; 1 rightmost bits = byte number.

                                       TAG                          LINE             WORD

4 3 1

b.Line 1. Line 3.

Line 1.

Line 6.

c.Bytes with addresses 1010 0001 through 1010 1110 are stored in the cache.

d.16 bytes.

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

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