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Segment Execution Activity 2 Segment Name Time (ns) Instruction Fetch & Decode (IF) Operand Fetch (OF) Instruction Execute (IE) Operand Store (OS) 52 Example 13: 40 30 • There are 4 segments as follows: · The interface delay is 3 ns. • The simple block diagram of the pipeline: 40 IE OS Draw the space time diagram and calculate the: a) cycle time of the non-pipeline and pipeline. b) execution time for 50 tasks. c) real speedup. d) Maximum speedup. 25

Segment Execution Activity 2 Segment Name Time (ns) Instruction Fetch & Decode (IF) Operand Fetch (OF) Instruction Execute (IE) Operand Store (OS) 52 Example 13: 40 30 • There are 4 segments as follows: · The interface delay is 3 ns. • The simple block diagram of the pipeline: 40 IE OS Draw the space time diagram and calculate the: a) cycle time of the non-pipeline and pipeline. b) execution time for 50 tasks. c) real speedup. d) Maximum speedup. 25

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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ONE S X O Intrdu 6 ssd (2) x 6 Task3. Z Zotero x O Differe x G Use th x • Identif x ملفاتي M Iul x fi Fiverr x + A elearning.utm.my/21221/pluginfile.php/318434/mod_resource/content/3/Module%205%20-%20Central%20Processing%20Unit%20AA24Apr2021-StudentPar. Q Intrduction 25 / 56 67% + 24 Segment Activity 2 Segment Name Execution Time (ns) Instruction Fetch & 52 Example 13: Decode (IF) Operand Fetch (OF) Instruction Execute (IE) 40 30 • There are 4 segments as follows: Operand Store (OS) 40 • The interface delay is 3 ns. • The simple block diagram of the pipeline: 国中中中中中 R OF PRP IE OS Draw the space time diagram and calculate the: a) cycle time of the non-pipeline and pipeline. b) execution time for 50 tasks. c) real speedup. d) Maximum speedup. Module 5 - Central.pdf Show all 1:11 PM P Type here to search 85°F 12/16/2021
Transcribed Image Text:ONE S X O Intrdu 6 ssd (2) x 6 Task3. Z Zotero x O Differe x G Use th x • Identif x ملفاتي M Iul x fi Fiverr x + A elearning.utm.my/21221/pluginfile.php/318434/mod_resource/content/3/Module%205%20-%20Central%20Processing%20Unit%20AA24Apr2021-StudentPar. Q Intrduction 25 / 56 67% + 24 Segment Activity 2 Segment Name Execution Time (ns) Instruction Fetch & 52 Example 13: Decode (IF) Operand Fetch (OF) Instruction Execute (IE) 40 30 • There are 4 segments as follows: Operand Store (OS) 40 • The interface delay is 3 ns. • The simple block diagram of the pipeline: 国中中中中中 R OF PRP IE OS Draw the space time diagram and calculate the: a) cycle time of the non-pipeline and pipeline. b) execution time for 50 tasks. c) real speedup. d) Maximum speedup. Module 5 - Central.pdf Show all 1:11 PM P Type here to search 85°F 12/16/2021
ONE X Intrd X O ssd ( X O Task: X Z Zote x O Diffe x G Uset x • Ident X ملفات البريد M i Fiver X C ONE X + A elearning.utm.my/21221/pluginfile.php/318427/mod_resource/content/2/Module%205%20-%20Central%20Processing%20Unit%20AA24Apr2021-StudentPar. Q Intrduction 75 / 88 75% + | 74 Memory address MAR (Memory Address Register) MBR (Memory Buffer Register) IR (Instruction Register) Memory Content Instruction/Data 39D A450 L1: XCHG CX, NUM 39E B451 SUB VAL1,CX 39F C39D JMP L1: Solution 5.2 : 450 100 NUM 451 500 VALI Clock IP/PC MAR | MBR IR CX NUM VAL1 Micro-operation to 39D 200 100 IP/PC = 39D t1 39D MAR E IP/PC MBR + [MAR] IP/PC € IP/PС + 1 t2 t3 IR E MBR t4 MAR + IR[address] t5 MBR + [MAR] Exchange: CX +→ MBR to t7 ts [MAR] + MBR Module 5 - Central.pdf Show all 1:39 PM P Type here to search 85°F 12/16/2021 近
Transcribed Image Text:ONE X Intrd X O ssd ( X O Task: X Z Zote x O Diffe x G Uset x • Ident X ملفات البريد M i Fiver X C ONE X + A elearning.utm.my/21221/pluginfile.php/318427/mod_resource/content/2/Module%205%20-%20Central%20Processing%20Unit%20AA24Apr2021-StudentPar. Q Intrduction 75 / 88 75% + | 74 Memory address MAR (Memory Address Register) MBR (Memory Buffer Register) IR (Instruction Register) Memory Content Instruction/Data 39D A450 L1: XCHG CX, NUM 39E B451 SUB VAL1,CX 39F C39D JMP L1: Solution 5.2 : 450 100 NUM 451 500 VALI Clock IP/PC MAR | MBR IR CX NUM VAL1 Micro-operation to 39D 200 100 IP/PC = 39D t1 39D MAR E IP/PC MBR + [MAR] IP/PC € IP/PС + 1 t2 t3 IR E MBR t4 MAR + IR[address] t5 MBR + [MAR] Exchange: CX +→ MBR to t7 ts [MAR] + MBR Module 5 - Central.pdf Show all 1:39 PM P Type here to search 85°F 12/16/2021 近
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