61 the memory mountain the rate that a program reads

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Unformatted text preview: e Cache Line 0 Line 1 Valid Byte 0 Byte 1 Byte 2 Byte 3 Tag Valid Byte 0 Byte 1 Byte 2 Byte 3 1 86 30 3F 10 00 0 – – – – 1 60 4F E0 23 38 1 00 BC 0B 37 0 – – – – 0B 0 – – – – 0 – – – – 32 1 12 08 7B AD 1 06 78 07 C5 05 1 40 67 C2 3B 1 0B DE 18 4B 6E 0 – – – – 1 A0 B7 26 2D F0 0 – – – – 0 – – – – DE 1 12 C0 88 37 Set Index 0 1 2 3 4 5 6 7 Tag 09 45 EB 06 C7 71 91 46 The box below 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 CI 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 6.10: Suppose a program running on the machine in Problem 6.9 references the 1-byte word at address 0x0E34. Indicate the cache entry accessed and the cache byte value returned in hex. Indicate whether a cache miss occurs. If there is a cache miss, enter “–” for “Cache byte returned”. A. Address format (one bit per box): 12 11 10 9 8 7 6 5 4 3 2 1 0 B. Memory reference: Parameter Cache block offset (CO) Cache set index (CI) Cache tag (CT) Cache hit? (Y/N) Cache byte returned Value 0x 0x 0x 0x 318 Practice Problem 6.11: Repeat Problem 6.10 for memory address 0x0DD5. A. Address format (one bit per box): 12 11...
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