Ip Addressing And Subnetting Workbook Version 2.0 Student Answers Access

Alex later passes the CCNA – thanks to the methodical practice of subnetting by hand.

| | Hosts needed | CIDR | Subnet mask | Network ID | |----------------|--------------|------|-------------------|-----------------| | Headquarters | 2000 | /20 | 255.255.240.0 | 10.0.0.0/20 | | Branch1 | 500 | /22 | 255.255.252.0 | 10.0.16.0/22 | | Branch2 | 200 | /24 | 255.255.255.0 | 10.0.20.0/24 | | WAN1 | 2 | /30 | 255.255.255.252 | 10.0.21.0/30 | | WAN2 | 2 | /30 | 255.255.255.252 | 10.0.21.4/30 |

: Class A (1–126), B (128–191), C (192–223). D/E ignored here. Part 2 – Subnetting a Class C Address (192.168.1.0/24) Problem : Create 4 subnets with at least 25 hosts each. Alex later passes the CCNA – thanks to

| | CIDR | Subnet Mask | Network ID | Hosts usable | |-------------|----------|----------------|----------------|------------------| | LAN A | /25 | 255.255.255.128 | 192.168.10.0 | 126 | | LAN B | /26 | 255.255.255.192 | 192.168.10.128 | 62 | | LAN C | /27 | 255.255.255.224 | 192.168.10.192 | 30 | | WAN link | /30 | 255.255.255.252 | 192.168.10.224 | 2 |

: Large contiguous block from 10.0.22.0/24 onward. Epilogue Alex submits the workbook. The instructor’s note on top says: “Version 2.0 completed correctly. Next step: practice until you can do each problem in under 45 seconds.” Part 2 – Subnetting a Class C Address (192

: Perform bitwise AND with mask to find network ID; add inverse mask to find broadcast. Part 6 – Supernetting / Route Summarization Question : Summarize these Class C networks into a single route: 192.168.4.0/24, 192.168.5.0/24, 192.168.6.0/24, 192.168.7.0/24

: 192.168.10.228 – 192.168.10.255 (28 addresses left). Part 5 – Identifying Subnet for a Given IP (Tricky ones) | IP Address | Mask | Network ID | Broadcast | First host | Last host | |---------------------|-------------------|----------------|-----------------|----------------|----------------| | 10.20.30.40 | 255.255.255.240 | 10.20.30.32 | 10.20.30.47 | 10.20.30.33 | 10.20.30.46 | | 172.25.150.200 | 255.255.254.0 | 172.25.150.0 | 172.25.151.255 | 172.25.150.1 | 172.25.151.254 | | 192.168.99.199 | 255.255.255.192 | 192.168.99.192 | 192.168.99.255 | 192.168.99.193 | 192.168.99.254 | The instructor’s note on top says: “Version 2

Below is Alex’s with explanations written in the margins. Part 1 – Address Classification & Default Masks | IP Address | Class | Default Subnet Mask | Network ID | Host ID | |-----------------------|-----------|-------------------------|----------------|--------------| | 10.250.1.15 | A | 255.0.0.0 | 10.0.0.0 | 250.1.15 | | 172.31.200.99 | B | 255.255.0.0 | 172.31.0.0 | 200.99 | | 192.168.5.10 | C | 255.255.255.0 | 192.168.5.0 | 10 | | 203.0.113.88 | C | 255.255.255.0 | 203.0.113.0 | 88 | | 8.8.8.8 | A | 255.0.0.0 | 8.0.0.0 | 8.8.8 |

The story follows a fictional student, , who completes the workbook and documents the reasoning behind each answer. Story: Alex’s Subnetting Breakthrough – Workbook 2.0 Answer Log Background Alex is a networking student preparing for the CCNA exam. The instructor hands out IP Addressing and Subnetting Workbook Version 2.0 – 40 problems ranging from basic address identification to VLSM design. The catch: No multiple choice. Every answer requires calculation.