[컴퓨터구조론] 3 4 5 장 연습문제 해답
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- 2005.05.09 / 2019.12.24
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3.1 Memory (contents in hex): 300: 3005; 301: 5940; 302: 7006
Step 1: 3005 ? IR; Step 2: 3 ? AC
Step 3: 5940 ? IR; Step 4: 3 + 2 = 5 ? AC
Step 5: 7006 ? IR; Step 6: AC ? Device 6
4.3 a. Address length: 24; number of addressable units: 224; block size: 4; number of blocks in main memory: 222; number of lines in cache: 214; size of tag: 8.
b. Address length: 24; number of addressable units: 224; block size: 4; number of blocks in main memory: 222; number of lines in cache: 4000 hex; size of tag: 22.
c. Address length: 24; number of addressable units: 224; block size: 4; number of blocks in main memory: 222; number of lines in set: 2; number of sets: 213; number of lines in cache: 214; size of tag: 9.
4.7 a. 8 leftmost bits = tag; 5 middle bits = line number; 3 rightmost bits = byte number
b. slot 3; slot 6; slot 3; slot 21
c. Bytes with addresses 0001 1010 0001 1000 through 0001 1010 0001 1111 are stored in the cache
d. 256 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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(2006.05.18 16:22:14)
