ccna3 mod 1.1-vlsm

8/3/2019 Ccna3 Mod 1.1-Vlsm

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CCNA v3.0 Retooling

CCNA v3.0 Retooling

Variable Length Subnet Mask

(VLSM)


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2CCNA v3.0 Retooling

What is VLSM?

A Variable Length Subnet Mask (VLSM) is a means ofallocating IP addressing resources to subnets

according to their individual need rather than some

general network-wide rule.

VLSM allows an organization to use more than one

subnet mask within the same network address space.

It is often referred to as subnetting a subnet, and can

be used to maximize addressing efficiency.

Large subnets are created for addressing LANs and

small subnets are created for WAN links (a 30 bit

mask is used to create subnets with only two host).


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Subnetting vs. VLSM

Subnetting allows you to divide big networksinto smaller, equal-sized slices.

VLSM allows you to divide big networks into

smaller, different-sized slices. This enables

you to make maximum use of your valuable

IP address space.


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Addressing a Network with

Standard Subnetting

Site A has two Ethernet networks Site B had one Ethernet network

Site C had one Ethernet network

207.21.24.0 /24

How many network addresses are needed?

How many hosts are needed for the largest LAN?

How many bits need to be borrowed to address thisnetwork?

Site A Site B Site C

25 users 25 users 10 users 8 users


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Addressing a Network with Standard

Subnetting

Borrowing 3 bits will meet the current needs of thecompany, but it leaves little room for growth.

Each network will have 30 usable addresses, including

the point-to-point WAN links (which only require two

addresses).



Subnet # Subnet AddressBits

Masked

0 . . . /27

1 . . . /27

2 . . . /27

3 . . . /27

4 . . . /27

5 . . . /27

6 . . . /27

7 . . . /27

207.21.24.0


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Subnetting in a Box

0

255

To begin, in a

class C network

there are 256addresses. When

we subnet the

address, we break

it down in to

smaller units orsubnets. 256 addresses


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Subnetting in a Box

0

255

128

127

If we were to

borrow 1 bit, it

would break the256 addresses in

to two parts

(networks) with

each part (subnet)

having 128addresses.

The subnet mask

would be

255.255.255.128.

128 addresses 128 addresses



Subnetting in a Box

0

255

128

127

64 192

63 191

If we were to

borrow 2 bits, it

would break eachof these 2

networks in half

again. This would

give us 4

networks, eachwith 64 addresses.

The subnet mask

would now be

255.255.255.192.





Subnetting in a Box

0

255

128

127

64 192

63 191

If we were to

borrow 3 bits, it


networks in half

again. This would

give us 8


The subnet mask

would now be

255.255.255.224.

32

addresses

32

addresses

31

32

32

addresses

32

addresses

95

96

32

addresses

32

addresses

159

160

32

addresses

32

addresses

223

224



Subnetting in a Box

0

255

128

127

64 192

63 191

If we were to

borrow 4 bits, it


networks in half

again. This would

give us 16


The subnet mask

would now be

255.255.255.240.

31

32

95

96

159

160

223

224

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

addresses

16

15

48

47

144

143

176

175

80

79

112

111

208

207

240

239



Addressing a Network Using VLSM

When using VLSM to subnet an address, not all of thesubnets have to be the same size.

A different subnet mask may be applied to some of the

subnets to further subnet the address.

In order to take advantage of VLSM, the proper routingprotocol must be selected.

Not all routing protocols share subnetting information in

their routing table updates.

Classful Routing Protocols

(do not share subnet info)

Classless Routing Protocols

(do share subnet info)

RIP v1 RIP v2

IGRP EIGRP

OSPF

IS-IS




To begin subnetting this network using VLSM, identify theLAN with the largest number of hosts. Subnet theaddress 207.21.24.0 /24 based on this information.

Site A has two Ethernet networks (25 hosts each)

Site B had one Ethernet network (10 hosts) Site C had one Ethernet network (8 hosts)



Subnet # Subnet AddressBits

Masked

0 207.21.24.0 /27

1 207.21.24.32 /27

2 207.21.24.64 /27

3 207.21.24.96 /27

4 207.21.24.128 /27

5 207.21.24.160 /27

6 207.21.24.192 /27

7 207.21.24.224 /27




Subnet 1 & 2 can be used to address Site A Ethernetnetworks. Subnet 5 can be subnetted to accommodate Site

B & C Ethernet networks. Subnet 6 can be subnetted to

accommodate the WAN links.Site A Site B Site C


Subnet # Subnet Address

0 207.21.24.0 /27

1 207.21.24.32 /27

2 207.21.24.64 /27

3 207.21.24.96 /27

4 207.21.24.128 /275 207.21.24.160 /27

6 207.21.24.192 /27

7 207.21.24.224 /27

Site A

Site B & C

WAN links

Free

Addresses Sub-subnet 0 207.21.24.192 /30

Sub-subnet 1 207.21.24.196 /30

Sub-subnet 2 207.21.24.200 /30

Sub-subnet 3 207.21.24.204 /30

Sub-subnet 4 207.21.24.208 /30

Sub-subnet 5 207.21.24.212 /30

Sub-subnet 6 207.21.24.216 /30

Sub-subnet 7 207.21.24.220 /30

Free

Addresses

WAN

1 & 2

Sub-subnet 0 207.21.24.160 /28

Sub-subnet 1 207.21.24.176 /28

Site B

Site C




Use VLSM to apply subnet addresses to the topology. Be sureto include the number of bits borrowed ( ex. /24)



_____________ _____________ _____________ _____________

______________ ______________




Exercise 1

Your company has been assigned IP network 195.39.71.0 /24.Given that headquarters (60 hosts) is connected to five branch

offices (12 hosts each) by a WAN link, and to an ISP (the ISP

owns the addresses on that link), determine an appropriate IP

addressing scheme.

Headquarters

Branch 1

60 users

12 users 12 users 12 users 12 users 12 usersBranch 2 Branch 3 Branch 4 Branch 5

ISP



Given the IP

address

195.39.71.0 /24,

subnet accordingto the largest

subnet needed.

(Headquarters 60

hosts)

0

255

128

127

64 192

63 191

You would need to

borrow 2 bits or /

26. This would

give you 4networks with 64

host addresses on

each subnet.



Playing it safe, we

will not use the first

subnet (subnet 0).

0

64

128

192We will start

addressing with

195.39.71.64 /26.

Headquarters

needs 60 hosts, so

we will assign them

.64 - .127.

Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)



The 5 Branch

offices only need

12 hosts each.

0

64

128

192

The next address

block available is

the .128 - .191

block (64

addresses). Herewe will apply

VLSM.Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)

Using a /28 mask

will give us 16hosts at each

location. This will

take care of 4 of

the Branch offices.

160

144 176

Branch 1

12 hosts

/28

(255.255.255.240)

Branch 2

12 hosts

/28

(255.255.255.240)

Branch 3

12 hosts

/28

(255.255.255.240)

Branch 4

12 hosts

/28

(255.255.255.240)



To obtain a block

for Branch 5, we

will need to subnet

the .192 - .255block using a /28

mask.

0

64

128

192

Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)

160

144 176

Branch 1

12 hosts

/28

(255.255.255.240)

Branch 2

12 hosts

/28

(255.255.255.240)

Branch 3

12 hosts

/28

(255.255.255.240)

Branch 4

12 hosts

/28

(255.255.255.240)

224

208 240

Branch 5

12 hosts

/28

(255.255.255.240)


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Now we need to

address the 5

WAN links that

connect to theBranch offices.

These are point-to-

point connections

and only require 2

addresses.

0

64

128

192

Here we will use

a /30 mask to

further subnet thesubnets.

Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)

160

144 176

Branch 1

12 hosts

/28

(255.255.255.240)

Branch 2

12 hosts

/28

(255.255.255.240)

Branch 3

12 hosts

/28

(255.255.255.240)

Branch 4

12 hosts

/28

(255.255.255.240)

224

208 240

Branch 5

12 hosts

/28

(255.255.255.240)

216

212 220

232

228 236

WAN

1

WAN

2

WAN

3

WAN

4

WAN

5


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Subnet 0 could be

used later if

needed for future

growth of HQ or formore Branch

offices.

0

64

128

192

Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)

160

144 176

Branch 1

12 hosts

/28

(255.255.255.240)

Branch 2

12 hosts

/28

(255.255.255.240)

Branch 3

12 hosts

/28

(255.255.255.240)

Branch 4

12 hosts

/28

(255.255.255.240)

224

208 240

Branch 5

12 hosts

/28

(255.255.255.240)

216

212 220

232

228 236

WAN

1

WAN

2

WAN

3

WAN

4

WAN

5


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22CCNA 3 0 Retooling

Address

provided by ISP195.39.71.64 /26

195.39.71.128 /28 195.39.71.144 /28 195.39.71.160 /28 195.39.71.176 /28 195.39.71.192 /28

195.39

.71.208

/30

195

.39

.71

.212

/30

195.

39.

71.

216

/30

195.3

9.71.220/3

0

195.39.71.224/30

Applying the Addresses to the Topology

ccna3 mod 1.1-vlsm

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