Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 3

Consider the problem of adding two \(n \times n\) matrices. If it takes \(t_{\alpha}\) time for one person to add two numbers, how many people do we need to minimize the total time spent to get the final answer? What will be the minimum amount of time needed to find the answer if we have enough people? Justify your answers.

Short Answer

Expert verified
The number of people needed to minimize the total time spent to get the final answer is \(n^2\) and the minimum amount of time needed to find the answer if we have enough people would be \(t_{\alpha}\)

Step by step solution

01

Determine the number of tasks

First, we need to identify how many tasks we have. Given a matrix size of \(n \times n\), when we add two matrices of this size, there will be \(n \times n = n^2\) cells or numbers to be added. So, there are \(n^2\) tasks.
02

Finding the number of people needed

Since every individual task can be done in parallel without waiting for any other task, the best way to minimize time is to do as many tasks in parallel as possible. That means we aim to get every cell added at the same time. So for adding \(n^2\) cells, if we allow one person to add one pair of numbers, we would need \(n^2\) people.
03

Calculating the minimal time needed

If each person can add a pair of numbers in \(t_{\alpha}\) time and all the people can work at the same time, the total time spent would be just the time a person takes to add a single pair of numbers, which is \(t_{\alpha}\). Therefore, if we have \(n^2\) people, the total time needed to complete the task would be minimized to \(t_{\alpha}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Write a CREW PRAM algorithm for determining for each element of an \(n\) -clement linked list if it is the middle \(\left(\int n / 2 \text { th }\right)\) element in \(\Theta(\lg n)\) time.

If we have two people to add \(n\) numbers of a list, and it takes \(t_{e}\) time for one person to add two numbers, how long will it take the two people to add all a numbers of the list considering the operation of addition as the basic operation and including \(t,\) time for passing the result of an addition from one person to the other? Justify your answer.

Write a PRAM algorithm for Quicksort using \(n\) processors to sort a list of \(n\) elements.

Consider the problem of adding \(n\) numbers of a list. If it takes \(t_{a}\) time for one person to add two numbers, and it takes no time to pass the result of an addition from one person to another, how many people do we need to minimize the total time spent to get the final answer? What will be the minimum amount of time needed to find the answer if we have enough people? Justify your answer.

Consider the problem of adding the numbers in a list of \(n\) numbers, If it takes \(t_{d}(n-1)\) time for one person to add all \(n\) numbers, is it possible for \(m\) people to compute the sum in less than \(\left[t_{e}(n-1)\right] / m\) time? Justify your answer.
See all solutions

Recommended explanations on Computer Science Textbooks

Issues in Computer Science

Read Explanation

Problem Solving Techniques

Read Explanation

Databases

Read Explanation

Computer Programming

Read Explanation

Theory of Computation

Read Explanation

Computer Organisation and Architecture

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.

Sign-up for free