How can we find the result as a 0x064C, 0x064D, 0x064E, and 0x064F? Explain detail please.

Answer

The problems that follow will help reinforce your understanding of how caches work. Assume the following: The memory is byte addressable. Memory accesses are to 1-byte words (not to 4-byte words). Addresses are 13 bits wide. The cache is two-way set associative (E = 2), with a 4-byte block size (B = 4) and eight sets (S = 8). The contents of the cache are as follows, with all numbers given in hexadecimal notation. 610 Chapter 6 The Memory Hierarchy 2-way set associative cache Line 0 Line 1 Set index Tag Valid Byte 0 Byte 1 Byte 2 Byte 3 Tag Valid Byte 0 Byte 1 Byte 2 Byte 3 0 09 1 86 30 3F 10 00 0 1 45 60 4F EO 23 38 1 00 BC OB 37 2 EB OB 0 3 06 0 32 1 12 7B AD 4 C7 1 06 78 07 C5 05 1 40 C2 3B 5 71 1 OB DE 18 4B 6E 0 6 91 1 0 B7 26 2D FO 0 7 46 0 DE 1 12 CO 88 37 0 | w-um--2 121 0 118 The following figure shows the format of an address (one bit per box). Indicate (by labeling the diagram) the fields that would be used to determine the following: CO CT The cache block offset The cache set index The cache tag 12 11 10 9 8 7 6 5 4 3 2 1 0 Practice Problem 6.17 For the cache in Problem 6.13, list all of the hex memory addresses that will hit in set 3. set 3 are Ox064C, 0x064D, Ox064E, and Ox064F