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)
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:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F270eecfd-706c-4133-a4f7-b35458d6bf50%2F32d4787c-9ada-478f-98c2-d012adcebb27%2F2j83i3_processed.jpeg&w=3840&q=75)
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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