- Suppose In A Computer Storage System The Basic Addressable Unit Is One Byte I E 1 Word 1 Byte The Size Of The Ma 1 (19.55 KiB) Viewed 30 times
hexadecimal format, write the values of its tag, line, word fields
in binary format, respectively. Explain it.
(2) Assume direct mapping. Give any two memory addresses with
different tag values that would be mapped to the same Cache line.
Note, write the addresses in hexadecimal format.
(3) Assume direct mapping. Suppose for a specific line in the
cache, we found that this line current stores a byte with address
0100 0001 0011 1110 0000. What are the addresses of the other bytes
that are currently stored in this line along with that
byte?
(4) Assume associative mapping. Given a memory address A004B in
hexadecimal format, write the values of its tag and word fields in
binary format, respectively.
(5) Suppose that the processor needs to access a byte with
address F001A and assume that this byte is currently stored in the
Cache. Explain the processes of locating this byte in Cache using
direct mapping and associative mapping, respectively. What's the
major difference?
Suppose in a computer storage system, the basic addressable unit is one Byte (i.e., 1 word = 1 Byte). The size of the main memory is 1 MB (1 MB = 210 KB; 1 KB = 210 Bytes) and each block in main memory has a size of 16 Bytes. The size of the Cache is 64 KB. =