Homework Help is Here – Start Your Trial Now!
Engineering
Computer Engineering
f an organization is given the CIDR address 215.151.48.0/24 and decided to divide its addresses into 8 subnets. How many host addresses per ubnets are valid? (hint: this depends on the number of bits in the host ID part for each subnet but should take out the subnet address and the oroadcast address) O1. (2^5)-2 = 30 hosts O II. (2^4)-2 = 14 hosts O I. (2^3)-2 = 6 hosts O IN. (2^6)-2 = 62 hosts

f an organization is given the CIDR address 215.151.48.0/24 and decided to divide its addresses into 8 subnets. How many host addresses per ubnets are valid? (hint: this depends on the number of bits in the host ID part for each subnet but should take out the subnet address and the oroadcast address) O1. (2^5)-2 = 30 hosts O II. (2^4)-2 = 14 hosts O I. (2^3)-2 = 6 hosts O IN. (2^6)-2 = 62 hosts

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
See similar textbooks
icon
Related questions
Question

choose the correct answer

If an organization is given the CIDR address 215.151.48.0/24 and decided to divide its addresses into 8 subnets. How many host addresses per subnets are valid? (hint: this depends on the number of bits in the host ID part for each subnet but should take out the subnet address and the broadcast address) O 1. (2^5)-2 = 30 hosts O II. (2^4)-2 = 14 hosts O I. (2^3)-2 = 6 hosts O IV. (2^6)-2 = 62 hosts
Transcribed Image Text:If an organization is given the CIDR address 215.151.48.0/24 and decided to divide its addresses into 8 subnets. How many host addresses per subnets are valid? (hint: this depends on the number of bits in the host ID part for each subnet but should take out the subnet address and the broadcast address) O 1. (2^5)-2 = 30 hosts O II. (2^4)-2 = 14 hosts O I. (2^3)-2 = 6 hosts O IV. (2^6)-2 = 62 hosts
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Computer Networking: A Top-Down Approach (7th Edi…
Computer Networking: A Top-Down Approach (7th Edi…
Computer Engineering
ISBN:
9780133594140
Author:
James Kurose, Keith Ross
Publisher:
PEARSON
Computer Organization and Design MIPS Edition, Fi…
Computer Organization and Design MIPS Edition, Fi…
Computer Engineering
ISBN:
9780124077263
Author:
David A. Patterson, John L. Hennessy
Publisher:
Elsevier Science
Network+ Guide to Networks (MindTap Course List)
Network+ Guide to Networks (MindTap Course List)
Computer Engineering
ISBN:
9781337569330
Author:
Jill West, Tamara Dean, Jean Andrews
Publisher:
Cengage Learning
Concepts of Database Management
Concepts of Database Management
Computer Engineering
ISBN:
9781337093422
Author:
Joy L. Starks, Philip J. Pratt, Mary Z. Last
Publisher:
Cengage Learning
Prelude to Programming
Prelude to Programming
Computer Engineering
ISBN:
9780133750423
Author:
VENIT, Stewart
Publisher:
Pearson Education
Sc Business Data Communications and Networking, T…
Sc Business Data Communications and Networking, T…
Computer Engineering
ISBN:
9781119368830
Author:
FITZGERALD
Publisher:
WILEY
SEE MORE TEXTBOOKS