Computer Organization and Architecture, 10th Edition. William Stallings. ©2016|Pearson| Available. Share this page. Computer Organization and Architecture,. TENTH EDITION. Having been active in computer organization and architecture for many years, it is a. By William Stallings provides a very timely and comprehensive foundation. PDF files: Copies of all figures and tables from the book. William Stallings. Refers to whether memory is internal and external to the computer. Organization refers to the physical arrangement of bits to form words. Instructor Solutions Manual (Download only) for Computer Organization and Architecture, 10th Edition Download Instructor's Solutions Manual - Chapters 1-10 (application/pdf) (2.5MB) Download Instructor's Solutions Manual - Chapters 11-21 (application/pdf) (2.1MB). [PDF] Data and Computer Communications (10th Edition) (William Stallings Books on Computer and 0:21 Collection Book Data and Computer Communications (8th, Eighth Edition) - By William Stallings.
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Table of contents:
PART ONE: INTRODUCTION
Chapter 1 Basic Concepts and Computer Evolution
Chapter 2 Performance Issues
PART TWO: THE COMPUTER SYSTEM
Chapter 3 A Top-Level View of Computer Function and Interconnection
Chapter 4 Cache Memory
Chapter 5 Internal Memory
Chapter 6 External Memory
Chapter 7 Input/Output
Chapter 8 Operating System Support
PART THREE: ARITHMETIC AND LOGIC
Chapter 9 Number Systems
Chapter 10 Computer Arithmetic
Chapter 11 Digital Logic
PART FOUR: THE CENTRAL PROCESSING UNIT
Chapter 12 Instruction Sets: Characteristics and Functions
Chapter 13 Instruction Sets: Addressing Modes and Formats
Chapter 14 Processor Structure and Function
Chapter 15 Reduced Instruction Set Computers
Chapter 16 Instruction-Level Parallelism and Superscalar Processors
PART FIVE: PARALLEL ORGANIZATION
Chapter 17 Parallel Processing
Chapter 18 Multicore Computers
Chapter 19 General-Purpose Graphic Processing Units
PART SIX: THE CONTROL UNIT
Chapter 20 Control Unit Operation
Chapter 21 Microprogrammed Control
Product Detail:
Language: English
ISBN-10: 0134101618
ISBN-13: 978-0134101613
ISBN-13: 9780134101613
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Computer Organization And Architecture 9th
- Step 1 of 11
Given data:
Here, the aim is to write an IAS program to compute thesummation of n-natural numbers and save the result in a variablenamed Y.
• This notation is shown with the equation is given below:
Where,
o “”represents the summation.
o The value of X ranging from 1 to N.
o This is represented mathematically as .
• Computation does not result in arithmetic overflow.
• “X”, “Y”, and “N” values are positive integers where “N” valueis greater than or equal to 1.
- Step 2 of 11
a.
The IAS program for the equation is as shown below:
Location
Instruction/Value
Comments
0
< >
Constant (N) [initialized to some value]
1
1
Constant; Integer value = 1
2
2
Constant; Integer value = 2
3
0
Variable Y (initialized to integer zero);
Sum(Y)
4L
LOAD M(0)
N → AC
4R
ADD M(1)
AC + 1 → AC
5L
MUL M(0)
N(N+1) → AC
5R
DIV M(2)
AC/2 → AC
6L
STORE M(3)
AC → Y; saving the Sum in variable Y
6R
JUMP M(6,20:39)
Done; HALT
- Step 3 of 11
Explanation:
• An IAS instruction is a 40 bit instruction and it is dividedin to left portion with 20 bits and right portion with 20 bits.
• In the above mentioned code, initially a constant “N” is takenwith some value and it is in the location “0”.
• Later this constant is initialized with integer value, andlater with integer value “2”.
• Variable “Y” is initialized to integer value 0.
• The constant “N” is stored in an accumulator AC. That is, and this is stored in the left portion of the 4thlocation.
• Take an accumulator increment the value by “1” and store it inanother accumulator, i.e. this is stored in the right portion of the 4thlocation.
This process is as shown
Left Instruction
Right Instruction
https://commongol.netlify.app/league-of-legends-download-game-do-not-have-play-button.html. LOAD M(0)
ADD M(1)
- Step 4 of 11
• Multiply this accumulator with constant N and store it inaccumulator again, That is, this is stored in the left portion of the 5thinstruction
• Divide the accumulator by “2” and save it in accumulator, thisis stored in the right portion of the 5th location.
• This process is as shown:
Left Instruction
Right Instruction
MUL M(0) https://commongol.netlify.app/download-game-pc-nba-2k17.html.
DIV M(2)
- Step 5 of 11
• Finally, save the accumulator in the variable “Y”.
Left Instruction
Right Instruction
STORE M(3)
AC → Y
JUMP M(6,20:39)
HALT
- Step 6 of 11b.
Location
Instruction / Value
Comments
0
< >
Constant (N) [initialized to some value]
1
1
Constant (loop counter increment)
2
1
Variable i (loop index value; current)
3
1
Variable Y = Sum of X values (Initialized to One)
4L
LOAD M(0)
N → AC (the max limit)
4R
SUB M(2)
Compute N–i → AC
5L
JUMP + M(6,0:19)
Check AC > 0 ? [i < N]
5R
JUMP + M(5,20:39)
i=N; done so HALT
6L
LOAD M(2)
i
6R
ADD M(1)
i+1 in AC
7L
STOR M(2)
AC → i
7R
ADD M(3)
i + Y in AC
8L
STOR M(3)
AC → Y
8R
JUMP M(4,0:19)
Continue at instruction located at address 4L
- Step 7 of 11
Explanation:
• Take a constant “N” that indicates the maximum number.
• Make the constant as a loop counter.
• Take a variable “i” that is the loop index
• Variable “Y” is set to be the sum of all “X” values initiallyset as “1”.
• Load the “N” value to the accumulator and keep it in the leftportion of the 4th instruction.
• Subtract the “i” value from the “N” and save it in accumulatoras shown:
Left Instruction
Right Instruction
LOAD M(0)
SUB M(2)
- Step 8 of 11
• If AC value is greater than “0” and “i” value is less than “N”jump to the left portion of the 6th instruction. If “N”value is equal to “i”, then HALT and jump to right portion of the5th instruction.
Note: In an instruction 0:19 indicate the left portion 20 bitsand 20:39 indicates right portion 20 bits.
Left Instruction
Right Instruction
JUMP + M(6,0:19)
AC > 0 ? [i < N]
JUMP + M(5,20:39)
i=N
- Step 9 of 11
• If “i” value is less than “N” load in to memory save it inleft portion of the 6th instruction, increment “i” valueby “1” and save it in accumulator save it in right portion of the6th instruction as shown below:
Left Instruction
Right Instruction
LOAD M(2)
ADD M(1)
- Step 10 of 11
• Assign accumulator value to “i” and save it in left portion of7th instruction.
• Add variables “i” and “Y” and save it in accumulator.
Left Instruction
Right Instruction
STOR M(2)
AC → i
ADD M(3)
- Step 11 of 11
• Assign the value of accumulator to variable “Y”, jump to theleft portion of the 4th instruction in the same mannercontinue until the end of the loop.
Left Instruction
Right Instruction
STOR M(3)
AC → Y
JUMP M(4,0:19)
LOOPTO 4L
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