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Given the IP address 165.165.8.0/21. We want to divide this networks into 7 subnetworks, such that: • Network 1: 930 hosts • Network 2: 475 hosts • Network 3: 225 hosts • Networks 4, 5, 6, and 7: 60, 57, 56, and 43 hosts respectively. Answer the following questions: 1. The IP address of Network 1 is: 2. The first IP address in Network 2 is: 3. The last IP address in Network 3 is: 4. The subnet mask of network 4 is: 5. The broadcast address of Network 5 is: 6 The maximum pumber cf possible bosts in Network 6 is:

Given the IP address 165.165.8.0/21. We want to divide this networks into 7 subnetworks, such that: • Network 1: 930 hosts • Network 2: 475 hosts • Network 3: 225 hosts • Networks 4, 5, 6, and 7: 60, 57, 56, and 43 hosts respectively. Answer the following questions: 1. The IP address of Network 1 is: 2. The first IP address in Network 2 is: 3. The last IP address in Network 3 is: 4. The subnet mask of network 4 is: 5. The broadcast address of Network 5 is: 6 The maximum pumber cf possible bosts in Network 6 is:

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...
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• Given the IP address 165.165.8.0/21. We want to divide this networks into 7 subnetworks, such that: • Network 1: 930 hosts • Network 2: 475 hosts • Network 3: 225 hosts • Networks 4, 5, 6, and 7: 60, 57, 56, and 43 hosts respectively. Answer the following questions: 1. The IP address of Network 1 is: 2. The first IP address in Network 2 is: 3. The last IP address in Network 3 is: 4. The subnet mask of network 4 is: 5. The broadcast address of Network 5 is: 6 The mavimum pumber of possible bosts in Network 6 is:
Transcribed Image Text:• Given the IP address 165.165.8.0/21. We want to divide this networks into 7 subnetworks, such that: • Network 1: 930 hosts • Network 2: 475 hosts • Network 3: 225 hosts • Networks 4, 5, 6, and 7: 60, 57, 56, and 43 hosts respectively. Answer the following questions: 1. The IP address of Network 1 is: 2. The first IP address in Network 2 is: 3. The last IP address in Network 3 is: 4. The subnet mask of network 4 is: 5. The broadcast address of Network 5 is: 6 The mavimum pumber of possible bosts in Network 6 is:
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