Subnet Exercise 2
Write down your PC #: 26__________________________________
A server has an IP address of 22.214.171.124/21
1. What IP address would your PC need to have to be in the same subnet as this server (use your PC number as the Host number.)
Any IP address that ranges from 126.96.36.199 to 188.8.131.52; the 26th host should be 184.108.40.206
Starting reservation : 220.127.116.11/22
Create an additional 100 subnets
Closest we can get is 128 subnets (so we will need to borrow 7 bits -> /29)
2. How many hosts are available per subnet: 2^3 = 8 hosts available per subnet (6 valid hosts) ________________________________
Since the new subnet mask is /29, there are 3 bits leftover for hosts
3. What is the broadcast address of subnet 55: 217.37.191
4. What is the network ID of Subnet 80: 217.38.128
Since we know that each range can fit 32 subnets, we need to find the range that will accommodate for 80 subnets (starting from 18.104.22.168). The range for subnet 80 will be in 22.214.171.124 (since 32 x 2 = 64), and then we count up based on the remainder (16). 16 x 8 = 128 which will be the starting/network address of subnet 80
5. What is the range of hosts on subnet 12: 217.36.96 is the network ID so the range of hosts will range from 126.96.36.199 to 188.8.131.52. The broadcast address will be 184.108.40.206.
This problem has been solved.
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