Computer Systems: A Programmer's Perspective (3rd Edition)
Computer Systems: A Programmer's Perspective (3rd Edition)
3rd Edition
ISBN: 9780134092669
Author: Bryant, Randal E. Bryant, David R. O'Hallaron, David R., Randal E.; O'Hallaron, Bryant/O'hallaron
Publisher: PEARSON
Question
Book Icon
Chapter 4.4, Problem 4.28PP

A.

Program Plan Intro

Processing stages:

  • The processing of an instruction has number of operations.
  • The operations are organized into particular sequence of stages.
  • It attempts to follow a uniform sequence for all instructions.
  • The description of stages are shown below:
    • Fetch:
      • It uses program counter “PC” as memory address to read instruction bytes from memory.
      • The 4-bit portions “icode” and “ifun” of specifier byte is extracted from instruction.
      • It fetches “valC” that denotes an 8-byte constant.
      • It computes “valP” that denotes value of “PC” plus length of fetched instruction.
    • Decode:
      • The register file is been read with two operands.
      • It gives values “valA” and “valB” for operands.
      • It reads registers with instruction fields “rA” and “rB”.
    • Execute:
      • In this stage the ALU either performs required operation or increments and decrements stack pointer.
      • The resulting value is termed as “valE”.
      • The condition codes are evaluated and destination register is updated based on condition.
      • It determines whether branch should be taken or not in a jump instruction.
    • Memory:
      • The data is been written to memory or read from memory in this stage.
      •  The value that is read is determined as “valM”.
    • Write back:
      • The results are been written to register file.
      • It can write up to two results.
    • PC update:
      • The program counter “PC” denotes memory address to read bytes of instruction from memory.
      • It is used to set next instruction’s address.

B.

Program Plan Intro

Processing stages:

  • The processing of an instruction has number of operations.
  • The operations are organized into particular sequence of stages.
  • It attempts to follow a uniform sequence for all instructions.
  • The description of stages are shown below:
    • Fetch:
      • It uses program counter “PC” as memory address to read instruction bytes from memory.
      • The 4-bit portions “icode” and “ifun” of specifier byte is extracted from instruction.
      • It fetches “valC” that denotes an 8-byte constant.
      • It computes “valP” that denotes value of “PC” plus length of fetched instruction.
    • Decode:
      • The register file is been read with two operands.
      • It gives values “valA” and “valB” for operands.
      • It reads registers with instruction fields “rA” and “rB”.
    • Execute:
      • In this stage the ALU either performs required operation or increments and decrements stack pointer.
      • The resulting value is termed as “valE”.
      • The condition codes are evaluated and destination register is updated based on condition.
      • It determines whether branch should be taken or not in a jump instruction.
    • Memory:
      • The data is been written to memory or read from memory in this stage.
      •  The value that is read is determined as “valM”.
    • Write back:
      • The results are been written to register file.
      • It can write up to two results.
    • PC update:
      • The program counter “PC” denotes memory address to read bytes of instruction from memory.
      • It is used to set next instruction’s address.

C.

Program Plan Intro

Processing stages:

  • The processing of an instruction has number of operations.
  • The operations are organized into particular sequence of stages.
  • It attempts to follow a uniform sequence for all instructions.
  • The description of stages are shown below:
    • Fetch:
      • It uses program counter “PC” as memory address to read instruction bytes from memory.
      • The 4-bit portions “icode” and “ifun” of specifier byte is extracted from instruction.
      • It fetches “valC” that denotes an 8-byte constant.
      • It computes “valP” that denotes value of “PC” plus length of fetched instruction.
    • Decode:
      • The register file is been read with two operands.
      • It gives values “valA” and “valB” for operands.
      • It reads registers with instruction fields “rA” and “rB”.
    • Execute:
      • In this stage the ALU either performs required operation or increments and decrements stack pointer.
      • The resulting value is termed as “valE”.
      • The condition codes are evaluated and destination register is updated based on condition.
      • It determines whether branch should be taken or not in a jump instruction.
    • Memory:
      • The data is been written to memory or read from memory in this stage.
      •  The value that is read is determined as “valM”.
    • Write back:
      • The results are been written to register file.
      • It can write up to two results.
    • PC update:
      • The program counter “PC” denotes memory address to read bytes of instruction from memory.
      • It is used to set next instruction’s address.

D.

Program Plan Intro

Processing stages:

  • The processing of an instruction has number of operations.
  • The operations are organized into particular sequence of stages.
  • It attempts to follow a uniform sequence for all instructions.
  • The description of stages are shown below:
    • Fetch:
      • It uses program counter “PC” as memory address to read instruction bytes from memory.
      • The 4-bit portions “icode” and “ifun” of specifier byte is extracted from instruction.
      • It fetches “valC” that denotes an 8-byte constant.
      • It computes “valP” that denotes value of “PC” plus length of fetched instruction.
    • Decode:
      • The register file is been read with two operands.
      • It gives values “valA” and “valB” for operands.
      • It reads registers with instruction fields “rA” and “rB”.
    • Execute:
      • In this stage the ALU either performs required operation or increments and decrements stack pointer.
      • The resulting value is termed as “valE”.
      • The condition codes are evaluated and destination register is updated based on condition.
      • It determines whether branch should be taken or not in a jump instruction.
    • Memory:
      • The data is been written to memory or read from memory in this stage.
      •  The value that is read is determined as “valM”.
    • Write back:
      • The results are been written to register file.
      • It can write up to two results.
    • PC update:
      • The program counter “PC” denotes memory address to read bytes of instruction from memory.
      • It is used to set next instruction’s address.

Blurred answer
Students have asked these similar questions
In completing this report, you may be required to rely heavily on principles relevant, for example, the Work System, Managerial and Functional Levels, Information and International Systems, and Security. apply Problem Solving Techniques (Think Outside The Box) when completing. should reflect relevance, clarity, and organisation based on research. Your research must be demonstrated by Details from the scenario to support your analysis, Theories from your readings, Three or more scholarly references are required from books, UWIlinc, etc, in-text or narrated citations of at least four (4) references. “Implementation of an Integrated Inventory Management System at Green Fields Manufacturing” Green Fields Manufacturing is a mid-sized company specialising in eco-friendly home and garden products. In recent years, growing demand has exposed the limitations of their fragmented processes and outdated systems. Different departments manage production schedules, raw material requirements, and…
1. Create a Book record that implements the Comparable interface, comparing the Book objects by year - title: String > - author: String - year: int Book + compareTo(other Book: Book): int + toString(): String Submit your source code on Canvas (Copy your code to text box or upload.java file) > Comparable 2. Create a main method in Book record. 1) In the main method, create an array of 2 objects of Book with your choice of title, author, and year. 2) Sort the array by year 3) Print the object. Override the toString in Book to match the example output: @Javadoc Declaration Console X Properties Book [Java Application] /Users/kuan/.p2/pool/plugins/org.eclipse.justj.openjdk.hotspo [Book: year=1901, Book: year=2010]
Q5-The efficiency of a 200 KVA, single phase transformer is 98% when operating at full load 0.8 lagging p.f. the iron losses in the transformer is 2000 watt. Calculate the i) Full load copper losses ii) half load copper losses and efficiency at half load. Ans: 1265.306 watt, 97.186%

Chapter 4 Solutions

Computer Systems: A Programmer's Perspective (3rd Edition)

Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Operations Research : Applications and Algorithms
Computer Science
ISBN:9780534380588
Author:Wayne L. Winston
Publisher:Brooks Cole
Text book image
C++ for Engineers and Scientists
Computer Science
ISBN:9781133187844
Author:Bronson, Gary J.
Publisher:Course Technology Ptr
Text book image
Systems Architecture
Computer Science
ISBN:9781305080195
Author:Stephen D. Burd
Publisher:Cengage Learning
Text book image
COMPREHENSIVE MICROSOFT OFFICE 365 EXCE
Computer Science
ISBN:9780357392676
Author:FREUND, Steven
Publisher:CENGAGE L
Text book image
Principles of Information Systems (MindTap Course...
Computer Science
ISBN:9781285867168
Author:Ralph Stair, George Reynolds
Publisher:Cengage Learning
Text book image
Principles of Information Systems (MindTap Course...
Computer Science
ISBN:9781305971776
Author:Ralph Stair, George Reynolds
Publisher:Cengage Learning