Class B vIP/Subnetting Problem Part 1: Given address 172.16.0.0 , create a netwo
ID: 3532593 • Letter: C
Question
Class B vIP/Subnetting Problem
Part 1: Given address 172.16.0.0, create a network with 200 subnets.
1. What class is this address? Is it a public or private address? __________
2. How many bits are network bits in this address? ______________
3. How many bits do you have to borrow to create 200 subnets? _________________
4. What is the /notation of the subnets?
5. What will the subnet mask be?
6. How many subnets does this create?
7. How many host bits are left? ______ How many usable hosts per subnet?
Fill in the chart below through the 10th subnet.
Subnet Subnet Address Hosts Range Default Gateway Broadcast
1
2
3
4
5
6
7
8
9
10
Part 2: Complete the chart below using binary but only use the 3rd and 4thoctets as shown in the example:
Subnet Address Hosts Range Broadcast
00000000.00000000 00000000.00000001-00000000.11111110 00000000.11111111
Part 3: Using the Boolean ANDING process, what network would the host 172.16.250.5 be on?
1. Convert 172.16.250.5 to binary _______________________________________
2. Convert the subnet mask to binary _______________________________________
3. AND the two address to find the network _______________________________________
4. Convert back to dotted decimal format _______________________________________
Explanation / Answer
1. It is in class B address it is public address
2. 12 bits
3. 8 bits
4. /24
5.255.255.255.0/24
6. 256
7. 20 hosts bit are left and their are 2 pow 20 usable host per subnet
subnet gateway subnet address broad cast host range default
1. 10.10.10.1 10.10.10.255 2pow8 10.10.10.10
2. 10.10.0.2 10.10.0.255 2 pow 8 10.10.10.10
3. 10.10.0.2 10.10.0.255 2 pow 8 10.10.10.10
4. 10.10.10.3 10.10.10.255 2 pow 8 10.10.10.10
5. 10.10.10.4 10.10.10.255 2 pow 8 10.10.10.10
6. 10.10.0.5 10.10.0.255 2 pow 8 10.10.10.10
7. 10.10.10.5 10.10.10.255 2 pow 8 10.10.10.10
8. 10.10.10.6 10.10.10.255 2 pow 8 10.10.10.10
9. 10.10.10.7 10.10.10.255 2 pow 8 10.10.10.10
10. 10.10.0.5 10.10.0.255 2 pow 8 10.10.10.10
2 part
00000000.00000000.00000000.11111111 00000000.00000001-00000000.111111110
00000000.00000000.00000001.11111111 00000000.00000001-00000000.111111011
00000000.00000000.00000010.11111111 00000000.00000001-00000000.111111100
00000000.00000000.00000011.11111111 00000000.00000001-00000000.111111101
00000000.00000000.00000100.11111111 00000000.00000001-00000000.111111110
00000000.00000000.00000101.11111111 00000000.00000001-00000000.111111111
00000000.00000000.00000110.11111111 00000000.00000001-00000000.111111000
3 part
172.16.50.5 si anded with 255.255.255.0
after anding 50 comes So that 50 subnet would be on after hosting
1. 10101100.00010000.11111010.00000101
2. 11111111.11111111.11111111.00000000
3. 11111010 00000101
and 11111111 00000000
=11111010.00000000
4. 172.16.250.0