2276 Subnetting Practice B (Answers Below… no cheating!)

1. You are the proud possessor of a class C network: 207.14.3.0/24 a. Subnet this network in order to allow for 6 subnets. b. What will the new subnet mask be? c. What are the first 3 subnet IDs in CIDR notation? d. How many new networks have you created? e. How many hosts will be in each network?

2. You are the proud possessor of a class C network: 207.14.3.0/24 a. Subnet this network in order to allow for 20 hosts per subnet. b. What will the new subnet mask be? c. What are the first 3 subnet IDs in CIDR notation? d. How many new networks have you created? e. How many hosts will be in each network?

3. You are the proud possessor of a class B network: 157.14.0.0/16 a. Subnet this network in order to allow for 12 subnets. b. What will the new subnet mask be? c. What are the first 3 subnet IDs in CIDR notation? d. How many new networks have you created? e. How many hosts will be in each network?

4. You are the proud possessor of a class B network: 157.14.0.0/16 a. Subnet this network in order to allow for 600 hosts per subnet. b. What will the new subnet mask be? c. What are the first 3 subnet IDs in CIDR notation? d. How many new networks have you created? e. How many hosts will be in each network?

5. You are the proud possessor of a class B network: 157.14.0.0/16 a. Subnet this network in order to allow for 600 networks. b. What will the new subnet mask be? c. What are the first 3 subnet IDs in CIDR notation? d. How many new networks have you created? e. How many hosts will be in each network?

Always start with the binary table:27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

1. You are the proud possessor of a class C subnet: 207.14.3.0/24 a. Subnet this subnet in order to allow for 6 subnets b. What will the new subnet mask be?

We are working for subnets, so we focus on the ones in the subnet mask2x >= 6, 23 = 8 >= 6, Therefore: 11111111.11111111.11111111.11100000, therefore 255.255.255.224

c. What are the first 3 subnet IDs in CIDR notation?

Our last binary one is in the 32s column of the fourth octet, so we will increase by 32. Checking our work, 256-224(our last positive subnet integer) = 32. Therefore207.14.3.0 /27207.14.3.32 /27207.14.3.64 /27

d. How many new subnets have you created? We had 24 binary ones, now we have 27, so we added 3 binary ones. 23 = 8 subnets

e. How many hosts will be in each subnet? There are 5 binary zeros in the subnet mask. 25-2 = 30 hosts per subnet

2. You are the proud possessor of a class C subnet: 207.14.3.0/24 a. Subnet this subnet in order to allow for 20 hosts. b. What will the new subnet mask be?

We are working for hosts, so we focus on the zeros in the subnet mask2x-2 >= 20, 25-2 = 30 >= 20, Therefore: 11111111.11111111.11111111.11100000, therefore 255.255.255.224

d. What are the first 3 subnet IDs in CIDR notation?

Our last binary one is in the 32s column of the fourth octet, so we will increase by 32. Checking our work, 256-224(our last positive subnet integer) = 32. Therefore207.14.3.0 /27207.14.3.32 /27207.14.3.64 /27

e. How many new subnets have you created? We had 24 binary ones, now we have 27, so we added 3 binary ones. 23 = 8 subnets

f. How many hosts will be in each subnet? There are 5 binary zeros in the subnet mask. 25-2 = 30 hosts per subnet

3. You are the proud possessor of a class B subnet: 157.14.0.0/16a. Subnet this subnet in order to allow for 12 subnets. b. What will the new subnet mask be?

We are working for subnets, so we focus on the ones in the subnet mask2x >= 12, 24 = 16 >= 12, Therefore: 11111111.11111111.11110000.00000000, therefore 255.255.240.0

c. What are the first 3 subnet IDs in CIDR notation?

Our last binary one is in the 16s column of the third octet, so we will increase by 16. Checking our work, 256-240(our last positive subnet integer) = 16. Therefore,157.14.0.0 /20157.14.16.0 /20157.14.32.0 /20

d. How many new subnets have you created? We had 16 binary ones, now we have 20, so we added 4 binary ones. 24 = 16 subnets

e. How many hosts will be in each subnet? There are 12 binary zeros in the subnet mask. 212-2 = 4092 hosts per subnet

3.

You are the proud possessor of a class B subnet: 157.14.0.0/16 a. Subnet this subnet in order to allow for 600 hosts. b. What will the new subnet mask be?

We are working for hosts, so we focus on the zeros in the subnet mask2x-2 >= 600, 210-2 = 1022 >= 600, Therefore: 11111111.11111111.11111100.00000000, therefore 255.255.252.0

e. What are the first 3 subnet IDs in CIDR notation?

Our last binary one is in the 4s column of the third octet, so we will increase by 4. Checking our work, 256-252(our last positive subnet integer) = 4. Therefore157.14.0.0 /22157.14.4.0 /22157.14.8.0 /22

f. How many new subnets have you created? We had 16 binary ones, now we have 22, so we added 6 binary ones. 26 = 64 subnets

g. How many hosts will be in each subnet? There are 10 binary zeros in the subnet mask. 210-2 = 1022 hosts per subnet

4. You are the proud possessor of a class B subnet: 157.14.0.0/16a. Subnet this subnet in order to allow for 600 subnets.

We are working for subnets, so we focus on the ones in the subnet mask2x >= 600, 210 = 1024 >= 600, Therefore: 11111111.11111111.11111111.11000000, therefore 255.255.255.192

b. What are the first 3 subnet IDs in CIDR notation? Our last binary one is in the 64s column of the fourth octet, so we will increase by 64. Checking our work, 256-192(our last positive subnet integer) = 64. Therefore157.14.0.0 /26157.14.0.64 /26157.14.0.128 /26

c. How many new subnets have you created? We had 16 binary ones, now we have 26, so we added 10 binary ones. 210 = 1024 subnets

d. How many hosts will be in each subnet? There are 6 binary zeros in the subnet mask. 26-2 = 62 hosts per subnet