Introduction to Networking ( Yarnfield )
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Transcript of Introduction to Networking ( Yarnfield )
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Introduction to Networking (Yarnfield)
Variable Length Subnet Masking (VLSM)
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Objectives Define VLSM Describe the difference between classful
subnetting Describe the advantages of VLSM Be able to perform VLSM operations on give IP
addresses
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Classful subnetting exercise 172.80.0.0 255.255.248.0 Find
The first five subnet addresses First host, last host and broadcast of each subnet Default gateway How many subnets can be made? How many hosts per subnet?
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VLSM defined More than one subnet mask Using classful subnetting wastes IP addresses Why?
172.80.8.0172.80.8.1 – 15.254
172.80.40.0172.80.40.1 – 47.254
172.80.32.0172.80.32.1 – 39.254
172.80.24.0172.80.24.1 – 31.254
172.80.16.0172.80.16.1 – 23.254
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We need An IP address to perform VLSM on The number of hosts involved in each part of
the network
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We will... Create a number of subnet masks that suit
our needs more efficiently than a classful subnetting scheme could
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Example using a Class C network address
120 hosts
60 hosts
30 hosts
192.168.1.0
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Process1. Find the segment with the largest number of
hosts connected to it2. Find an appropriate subnet mask for the largest
segment3. Write down the subnet addresses to fit the
subnet mask4. Take one of the newly created subnet addresses
and apply a new subnet mask to it that is more appropriate
5. Write down the subnet addresses to fit the new subnet mask
6. Repeat from step 4 for smaller segments
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Example continued1. Find the segment with the largest number of
hosts connected to it In the example the largest segment has 120
hosts connected so we must start with this segment
To accomodate120 hosts we need to use 7 bits from the host portion of the address (27 - 2 = 126)
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Example continued2. Find an appropriate subnet mask for the
largest segment If we have borrowed 7 bits for our hosts the
subnet mask (in binary) will be 11111111.1111111.1111111.1000000
Convert this to decimal and we get 255.255.255.128
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Example continued3. Write down the subnet addresses to fit the
subnet mask Now we need to find the subnet addresses that
this subnet mask will create 256 – 128 = 128 Therefore the subnets would be 192.168.1.0 and
192.168.1.128 (remember we can now use subnet zero!)
We can now assign 192.168.1.0/25 to accommodate the 120 segment and have 192.168.1.128 to use for the other two segments
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120 hosts (126 in total)192.168.1.0/25
60 hosts (62 in total)
30 hosts (30 in total)
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Example continued4. Take one of the newly created subnet addresses
and apply a new subnet mask to it that is more appropriate
We still have two segments to deal with and we have a new subnet address to work with of 192.168.1.128
We must start with the larger segment, which has 60 hosts
To accommodate 60 hosts we need to borrow 6 bits from the host portion of the given IP address
26 – 2 = 62 hosts This will give us a subnet mask of
1111111.1111111.1111111.11000000 which is the same as 255.255.255.192
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Example continued5. Write down the subnet addresses to fit the
new subnet mask Now we need to find the subnet addresses
that this subnet mask will create 256 – 192 = 64 Therefore the new subnet addresses would
be 192.168.1.128 and 192.168.1.192 We can now use 192.168.1.128/26 for the
segment with 60 hosts
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120 hosts (126 in total)192.168.1.0/25
60 hosts (62 in total)192.168.1.128/26
30 hosts (30 in total)
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Example continued4. Take one of the newly created subnet addresses
and apply a new subnet mask to it that is more appropriate
We still have the segment with 30 hosts to deal with
We work this out in the same way as before To accommodate 30 hosts we need to borrow 5
bits from the host portion of the IP address 25 – 2 = 30 hosts This will give us a subnet mask of
1111111.1111111.1111111.11100000 which is 255.255.255.224
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Example continued5. Write down the subnet addresses to fit the
new subnet mask Now we need to find the subnet addresses
that this subnet mask will create 256 – 224 = 32 Therefore the new subnet addresses would
be 192.168.1.192 and 192.168.1.224 We can now use 192.168.1.192/27 for the
segment with 30 hosts We still have the new 192.168.1.224 subnet
which could be used for future growth
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Result
120 hosts (126 in total)192.168.1.0/25
60 hosts (62 in total)192.168.1.128/26
30 hosts (30 in total)192.168.1.192/27
192.168.1.0
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Summary To determine the number of hosts a subnet
can support use the formula 2n – 2 Always start the process with the segment
with the largest amount of hosts to accommodate
Classless subnetting deals with the hosts as opposed to classful subnetting which deals more with subnets
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Exercise 192.168.2.0/24 7 remote sites, 30 hosts
each P to P links
between routers
Remote A 30 hosts
Remote B 30 hosts
Remote C 30 hosts
Remote D 30 hosts
Remote E 30 hosts
Remote F 30 hosts
Remote G 30 hosts
Central
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Exercise 192.168.3.0
Backbone126 hosts
6 hosts
6 hosts
6 hosts
30 hosts
30 hosts
30 hosts
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Questions... ...are there any?