Subnet Exercise 2

Subnet Exercise 2

Write down your PC #: 26__________________________________

A server has an IP address of 185.27.67.213/21

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 185.27.64.0 to 185.27.71.255; the 26^th host should be 185.27.64.26

Starting reservation : 167.217.36.0/22

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 167.217.36.0). The range for subnet 80 will be in 167.217.38.0 (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 167.217.36.97 to 167.217.36.102. The broadcast address will be 167.217.36.103.