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Direct Mapping Example: CPU is searching an Instruction stored at RAM address 1110011010 in cache and doesn't find it. 0000000000 ?? 0000000001 ?? Bloc 0000000010 Instruction-1 0000000011 Instruction-2 What happens then? 0000000100 Instruction-3 0000000101 Instruction-4 Bloc Solution: Line 1 = 1 0000000110 Instruction-5 0000000111 Instruction-6 0000001100 Instruction-7 0000001101 Instruction-8 Bloc 0000001110 Data-1 0000001111 Data-2 Line Oi =0 Line 2 (= 2),

Direct Mapping Example: CPU is searching an Instruction stored at RAM address 1110011010 in cache and doesn't find it. 0000000000 ?? 0000000001 ?? Bloc 0000000010 Instruction-1 0000000011 Instruction-2 What happens then? 0000000100 Instruction-3 0000000101 Instruction-4 Bloc Solution: Line 1 = 1 0000000110 Instruction-5 0000000111 Instruction-6 0000001100 Instruction-7 0000001101 Instruction-8 Bloc 0000001110 Data-1 0000001111 Data-2 Line Oi =0 Line 2 (= 2),

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Direct Mapping Example: CPU is searching an Instruction stored at RAM address 1110011010 in cache and doesn't find it. What happens then? Tag 0000000000 ?? 0000000001 ?? Block:0, (j=0) 0000000010 Instruction-1 Line1 = 1 Tag Tag 0000000011 Instruction-2 0000000100 Instruction-3 Instruction-4 Block:1, G=1) Solution: 0000000101 Line 2 (= 2) Tag 0000000110 Instruction-5 0000000111 Instruction-6 0000001100 Instruction-7 0000001101 Instruction-8 Block:2 0000001110 Data-1 (j=2) 0000001111 Data-2 Line.3 (3) 0000011100 Data-3 0000011101 Data-4 Tag 0000011110 Data-5 0000011111 Data-6 Cache Size = 64B Size of a line = 4B Total number of Lines in Cache: m 16 RAM Size = 1KB %3D 1111111100 %3D Size of a Block= 4B Total number of Blocks, M = 256 j = 0, 1, 2, ..255 (M-1) Block 255. G =255) 1111111101 1111111110 i = 0, 1, 2, ... 15 (m-1) 1111111111
Transcribed Image Text:Direct Mapping Example: CPU is searching an Instruction stored at RAM address 1110011010 in cache and doesn't find it. What happens then? Tag 0000000000 ?? 0000000001 ?? Block:0, (j=0) 0000000010 Instruction-1 Line1 = 1 Tag Tag 0000000011 Instruction-2 0000000100 Instruction-3 Instruction-4 Block:1, G=1) Solution: 0000000101 Line 2 (= 2) Tag 0000000110 Instruction-5 0000000111 Instruction-6 0000001100 Instruction-7 0000001101 Instruction-8 Block:2 0000001110 Data-1 (j=2) 0000001111 Data-2 Line.3 (3) 0000011100 Data-3 0000011101 Data-4 Tag 0000011110 Data-5 0000011111 Data-6 Cache Size = 64B Size of a line = 4B Total number of Lines in Cache: m 16 RAM Size = 1KB %3D 1111111100 %3D Size of a Block= 4B Total number of Blocks, M = 256 j = 0, 1, 2, ..255 (M-1) Block 255. G =255) 1111111101 1111111110 i = 0, 1, 2, ... 15 (m-1) 1111111111
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