1 exercise: ipv4 subnetting. 2 task 1 given is an ip network with address 194.141.0.0: divide this...

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1 Exercise: IPv4 subnetting Exercise: IPv4 subnetting

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Page 1: 1 Exercise: IPv4 subnetting. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets

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Exercise: IPv4 subnettingExercise: IPv4 subnetting

Page 2: 1 Exercise: IPv4 subnetting. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets

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Task 1Task 1

Given is an IP network with address 194.141.0.0:

Divide this network into 8 subnets.

Page 3: 1 Exercise: IPv4 subnetting. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets

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SolutionSolution This network is This network is class Cclass C because the because the

starting bits are starting bits are 110110.. In binary form:In binary form:

11011000010. 00010. 10001101.00000000.0000000010001101.00000000.00000000

The first 3 octets represent the network The first 3 octets represent the network address. The fourth octet is for subnet and address. The fourth octet is for subnet and host address.host address.

We need a subnet mask with 3 bits ones, We need a subnet mask with 3 bits ones, because because 8 = 28 = 23 3 ::11111111.11111111.11111111.11111111.11111111.11111111.1111110000000000

In decimal form: 255.255.255.224In decimal form: 255.255.255.224

Page 4: 1 Exercise: IPv4 subnetting. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets

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CommentsComments We use this mask in the corresponding We use this mask in the corresponding

field of the IP configuration.field of the IP configuration. The available subnets have the following The available subnets have the following

addresses:addresses:194.141.0.0194.141.0.0194.141.0.1194.141.0.1194.141.0.2194.141.0.2194.141.0.3194.141.0.3194.141.0.4194.141.0.4194.141.0.5194.141.0.5194.141.0.6194.141.0.6194.141.0.7194.141.0.7

Can not be used. In fact we have6 real subnets.

Hosts with address 0 or 255 can not be used.

Therefor each subnet contains up to 256/8 - 2 = 30 hosts.

Page 5: 1 Exercise: IPv4 subnetting. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets

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Task 2Task 2

Given is an IP network with address 162.251.0.0:

Divide this network into 32 subnets.

Page 6: 1 Exercise: IPv4 subnetting. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets

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SolutionSolution

The subnet mask isThe subnet mask is11111111.11111111.11111111.11111111.1111111111000.0000000000.000000000

in decimal: in decimal: 255.255.248.0255.255.248.0 3030 subnets with up to subnets with up to 20462046 hosts each. hosts each.

Page 7: 1 Exercise: IPv4 subnetting. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets

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Task 3Task 3

Given is a subnet mask: 255.254.0.0Given is a subnet mask: 255.254.0.0

Which class is the network?Which class is the network? How many real subnets?How many real subnets? How many real hosts may contain each How many real hosts may contain each

subnet?subnet?

Page 8: 1 Exercise: IPv4 subnetting. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets

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SolutionSolution

This is a subnet mask for network This is a subnet mask for network class class AA..

There are 2There are 277 – 2 = – 2 = 126126 real subnets real subnets..

Each subnet may has up to Each subnet may has up to 221717 – 2 = – 2 = 131070131070 real hosts. real hosts.