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)
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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Question
![ONE S X
O Intrdu
6 ssd (2) x
6 Task3.
Z Zotero x
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• Identif x
ملفاتي
M Iul x
fi Fiverr x
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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
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1:11 PM
P Type here to search
85°F
12/16/2021](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafff8ad9-ffc4-4f09-b9f9-30d5e4c4679a%2F18c43918-15b8-49b0-b71c-d8c511089c9c%2Fgzhfp8_processed.png&w=3840&q=75)
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
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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
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1:39 PM
P Type here to search
85°F
12/16/2021
近](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafff8ad9-ffc4-4f09-b9f9-30d5e4c4679a%2F18c43918-15b8-49b0-b71c-d8c511089c9c%2Fp5cp4m_processed.png&w=3840&q=75)
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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