/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Suppose a computer contained \(5... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 14

Suppose a computer contained \(512 \mathrm{MB}\) (MiB) of main memory, and an operating system needed to create a virtual memory of twice that size using pages of \(2 \mathrm{~KB}\) (KiB). How many pages would be required?

Short Answer

Expert verified
524288 pages are required.

Step by step solution

01

Determine Virtual Memory Size

The physical memory is given as \(512 \text{ MiB}\). The virtual memory needed is twice that size. Calculate the virtual memory size: \(512 \text{ MiB} \times 2 = 1024 \text{ MiB}\).
02

Convert Virtual Memory Size to Bytes

Convert the virtual memory size from MiB to bytes. Since 1 MiB = \(2^{20}\) bytes, 1024 MiB equals \(1024 \times 2^{20} = 2^{30}\) bytes.
03

Convert Page Size to Bytes

The page size is given as 2 KiB. Convert this page size to bytes: \(2 \text{ KiB} = 2 \times 2^{10} = 2^{11}\) bytes.
04

Calculate Number of Pages

Determine the number of pages by dividing the virtual memory size (in bytes) by the page size (in bytes). Using the previous conversions: \[\frac{2^{30}}{2^{11}} = 2^{19} = 524288 \text{ pages}.\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Memory Management
Memory management is a crucial component of computer systems. It involves the handling and organization of memory in a way that allows an operating system to efficiently allocate resources to processes and applications. Effective memory management ensures that applications have access to adequate memory resources.
There are different approaches to memory management:
  • Contiguous Memory Allocation: Allocates a single contiguous section of memory to a process.
  • Segmentation: Memory is divided into varying-sized segments often aligned to different needs.
  • Paging: Divides memory into fixed-sized pages to optimize space usage and manage virtual memory.
Memory management impacts system performance, ensuring programs run smoothly and efficiently without unnecessary delays.
Paging
Paging is a memory management scheme that eliminates the need for contiguous blocks of memory and thus reduces fragmentation. In paging, both physical memory and virtual memory are divided into fixed-sized pages.
The advantages of paging are numerous:
  • Reduces the complexity of memory allocation by using pages of fixed size.
  • Utilizes the virtual memory technique to give the illusion of infinite memory.
  • Improves the management of memory by enabling non-contiguous storage.
With paging, when a process requests data that is not in physical memory, a page fault occurs. The system then retrieves the required page from secondary storage, ensuring efficient memory use across multiple processes.
Operating Systems
The operating system (OS) is the backbone of a computer system, managing hardware and software resources. It acts as an intermediary between users and the computer hardware. One key function of the OS is memory management, where it employs techniques like paging to manage virtual memory.
Key functions of an operating system include:
  • Process Management: Handling active processes on the system by allocating CPU time.
  • Memory Management: Utilizing techniques like paging for efficient memory use.
  • File System Management: Organizing, storing, and tracking data in the file system.
  • Device Management: Controlling the operation of hardware devices through drivers.
The OS ensures seamless operation of applications, optimal resource allocation, and user-friendly interaction with the computer system.
Binary Conversions
Binary conversions are essential in computer systems as computers operate using binary (base-2) numeral system. Understanding binary conversions helps in memory size calculations, such as converting between units used in memory (bytes, kilobytes, megabytes).
Here's how some binary conversions work:
  • Kibibytes (KiB) to Bytes: 1 KiB = \(2^{10}\) bytes.
  • Megibytes (MiB) to Bytes: 1 MiB = \(2^{20}\) bytes.
  • Gibibytes (GiB) to Bytes: 1 GiB = \(2^{30}\) bytes.
By understanding these conversions, you can perform various calculations required in memory management, such as determining the number of pages in a virtual memory setup. These conversions ensure precision in calculations necessary in operating system functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Students who want to enroll in Model Railroading II at the local university are required to obtain permission from the instructor and pay a laboratory fee. The two requirements are fulfilled independently in either order and at different locations on campus. Enrollment is limited to twenty students; this limit is maintained by both the instructor, who will grant permission to only twenty students, and the financial office, which will allow only twenty students to pay the laboratory fee. Suppose that this registration system has resulted in nineteen students having successfully registered for the course, but with the final space being claimed by two students-one who has only obtained permission from the instructor and another who has only paid the fee. Which requirement for deadlock is removed by each of the following solutions to the problem? a. Both students are allowed in the course. b. The class size is reduced to nineteen, so neither of the two students is allowed to register for the course. c. The competing students are both denied entry to the class and a third student is given the twentieth space. d. It is decided that the only requirement for entry into the course is the payment of the fee. Thus the student who has paid the fee gets into the course, and entry is denied to the other student.

Suppose a password consisted of a string of nine characters from the English alphabet (twenty-six characters). If each possible password could be tested in a millisecond, how long would it take to test all possible passwords?

Identify two activities that can be performed by an operating system's administrator but not by a typical user.

What is a context switch?

What is meant by an interrupt handler in multiprogramming systems and what is its significance?
See all solutions

Recommended explanations on Computer Science Textbooks

Cybersecurity in Computer Science

Read Explanation

Computer Network

Read Explanation

Databases

Read Explanation

Theory of Computation

Read Explanation

Fintech

Read Explanation

Computer Systems

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.