COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG
COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG
3rd Edition
ISBN: 9780134671123
Author: Bryant
Publisher: PEARSON
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Chapter 6, Problem 6.29HW

Suppose we have a system with the following properties:

• The memory is byte addressable.

• Memory accesses are to 1-byte words (not to 4-byte words).

• Addressed are 12 bits wide.

• The cache is two-way set associative (E= 2), with a 4-byte block size (B = 4) and four sets.(S = 4).

The contents of the cache are as follows, with all addresses, tags and values given in hexadecimal notation:

Chapter 6, Problem 6.29HW, Suppose we have a system with the following properties:  The memory is byte addressable.  Memory , example  1

  1. A.    The following diagram shows the format of an address (1 bit per box). Indicate (by labeling the diagram) the fields that would be used to determine the following:

    CO. The cache block offset

    CI. The cache set index

    CT. The cache tag

Chapter 6, Problem 6.29HW, Suppose we have a system with the following properties:  The memory is byte addressable.  Memory , example  2

  1. B.     For each of the following memory accesses, indicates if it will be a cache hit or miss when carried out in sequence as listed. Also give the value of a read if it can be inferred from the information in the cache.

Chapter 6, Problem 6.29HW, Suppose we have a system with the following properties:  The memory is byte addressable.  Memory , example  3

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D. S. Malik, Data Structures Using C++, 2nd Edition, 2010
Methods (Ch6) - Review 1. (The MyRoot method) Below is a manual implementation of the Math.sqrt() method in Java. There are two methods, method #1 which calculates the square root for positive integers, and method #2, which calculates the square root of positive doubles (also works for integers). public class SquareRoot { public static void main(String[] args) { } // implement a loop of your choice here // Method that calculates the square root of integer variables public static double myRoot(int number) { double root; root=number/2; double root old; do { root old root; root (root_old+number/root_old)/2; } while (Math.abs(root_old-root)>1.8E-6); return root; } // Method that calculates the square root of double variables public static double myRoot(double number) { double root; root number/2; double root_old; do { root old root; root (root_old+number/root_old)/2; while (Math.abs (root_old-root)>1.0E-6); return root; } } Program-it-Yourself: In the main method, create a program that…
I would like to know the main features about the following 3 key concepts:1. Backup Domain Controller (BDC)2. Access Control List (ACL)3. Dynamic Memory

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