Day 5.5 subnetting

INTRODUCTION TO CLASSLESS ROUTING

CCNA v3.0 Semester 3

CLASSFUL ADDRESSING

The original IPv4 address architecture used an 8 bit network number for Class A addresses, a 16 bit network number for Class B addresses, and a 24 bit network number for Class C addresses.

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1 - 126

128 - 191

192 - 223

Class B

Network Host

1 0

Class C

Network Host

1 1 0

Class A

Network Host

0


CLASSFUL ADDRESSING

Classful addressing (A, B, C, etc.) is basically obsolete.

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Address Class

ApplicationNumber of

Network BitsNumber of Host Bits

Decimal Address Range

Number of Addresses

Number of Possible Host

Class ALarge

Networks8 bits 24 bits 1 - 126 126 16,777,214

Class BMedium-sized

Networks16 bits 16 bits 128 - 191 65,534 65,534

Class C Small Networks 24 bits 8 bits 192 - 223 2,097,152 254

The Class System


WHAT IS VLSM?

A Variable Length Subnet Mask (VLSM) is a means of allocating 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 networks into 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.

• So basically, you are now utilizing subnet masks in the same IP address space.

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ROUTING PROTOCOLS SUPPORTING VLSM

•RIP v2•EIGRP•OSPF

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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 this network?

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Site A Site B Site C

25 users 25 users 10 users 8 users



• Site A has two Ethernet networks

• Site B had one Ethernet network

• Site C had one Ethernet network

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If we borrow 3 bits from a class C address, that will give us eight networks, but we can only use six of them. Each network will have 30 usable addresses.

It will take four network addresses to accommodate the Ethernet networks at each site. That leaves us with two extra networks.

There is also a point-to-point WAN connection between each site. These two connections will take up the remaining two networks.



Borrowing 3 bits will meet the current needs of the company, 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).

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Subnet # Subnet AddressBits Masked

0 207.21.24.0 /271 207.21.24.32 /272 207.21.24.64 /273 207.21.24.96 /274 207.21.24.128 /275 207.21.24.160 /276 207.21.24.192 /277 207.21.24.224 /27

207.21.24.0


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We can use subnet 0

To enable subnet 0 on a Cisco router (if not already

enabled), it is necessary to use the global configuration

command ip subnet-zero.

Router# configure terminal (config t)

Router(config)# ip subnet-zero

To disable subnet 0, use the no form of this command.

Router# configure terminal

Router(config)# no ip subnet-zero


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Subnetting in a Box

0

255

To begin, in a class C network there are 256 addresses. When we subnet the address, we break it down in to smaller units or subnets. 256 addresses


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Subnetting in a Box

0

255

128

127

If we were to borrow 1 bit, it would break the 256 addresses in to two parts (networks) with each part (subnet) having 128 addresses.

The subnet mask would be 255.255.255.128.

128 addresses 128 addresses


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Subnetting in a Box

0

255

128

127

64

63

If we were to borrow 2 bits, it would break each of these 2 networks in half again. This would give us 4 networks, each with 64 addresses.

The subnet mask would now be 255.255.255.192.

64 addresses

64 addresses

64 addresses

64 addresses

192

191


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Subnetting in a Box

0

255

128

127

64 192

63 191



32addresses

32addresses

31

32

32addresses

32addresses

95

96

32addresses

32addresses

159

160

32addresses

32addresses

223

224


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Subnetting in a Box

0

255

128

127

64 192

63 191



31

32

95

96

159

160

223

224

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

16addresses

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 the subnets 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 routing protocol must be selected.

• Not all routing protocols share subnetting information in their routing table updates.

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Classful Routing Protocols(do not share subnet info)

Classless Routing Protocols(do share subnet info)

RIP v1 RIP v2IGRP EIGRP

OSPFIS-IS


ADDRESSING A NETWORK USING VLSMTo begin subnetting this network using VLSM,

identify the LAN with the largest number of hosts. Subnet the address 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)

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Subnet # Subnet AddressBits Masked

0 207.21.24.0 /271 207.21.24.32 /272 207.21.24.64 /273 207.21.24.96 /274 207.21.24.128 /275 207.21.24.160 /276 207.21.24.192 /277 207.21.24.224 /27



Subnet 1 & 2 can be used to address Site A Ethernet networks. Subnet 5 can be subnetted to accommodate Site B & C Ethernet networks. Subnet 6 can be subnetted to accommodate the WAN links.

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Free Addresses

Site A

Subnet # Subnet Address

0 207.21.24.0 /271 207.21.24.32 /272 207.21.24.64 /273 207.21.24.96 /274 207.21.24.128 /275 207.21.24.160 /276 207.21.24.192 /277 207.21.24.224 /27

Site B & C

Sub-subnet 0 207.21.24.160 /28Sub-subnet 1 207.21.24.176 /28

Site BSite C


0 207.21.24.0 /271 207.21.24.32 /272 207.21.24.64 /273 207.21.24.96 /274 207.21.24.128 /275 207.21.24.160 /276 207.21.24.192 /277 207.21.24.224 /27

WAN links

Sub-subnet 0 207.21.24.192 /30Sub-subnet 1 207.21.24.196 /30Sub-subnet 2 207.21.24.200 /30Sub-subnet 3 207.21.24.204 /30Sub-subnet 4 207.21.24.208 /30Sub-subnet 5 207.21.24.212 /30Sub-subnet 6 207.21.24.216 /30Sub-subnet 7 207.21.24.220 /30

Free Addresses

WAN 1 & 2


0 207.21.24.0 /271 207.21.24.32 /272 207.21.24.64 /273 207.21.24.96 /274 207.21.24.128 /275 207.21.24.160 /276 207.21.24.192 /277 207.21.24.224 /27



Through applying VLSM, the topology was able to be addressed and still have two complete subnets available for future growth.

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207.21.24.32 /27 207.21.24.64 /27 207.21.24.160 /28

207.21.24.176 /28

207.21.24.192 /30

207.21.24.196 /30


ADDRESSING A NETWORK USING VLSMEXERCISE 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.

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Headquarters

Branch 1

60 users

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

ISP


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Given the IP address 195.39.71.0 /24, subnet according to 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 4 networks with 64 host addresses on each subnet.


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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)


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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). Here we will apply VLSM.

Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)

Using a /28 mask will give us 16 hosts at each location. This will take care of 4 of the Branch offices.

160

144 176

Branch 112 hosts

/28(255.255.255.240)

Branch 212 hosts

/28(255.255.255.240)

Branch 312 hosts

/28(255.255.255.240)

Branch 412 hosts

/28(255.255.255.240)


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To obtain a block for Branch 5, we will need to subnet the .192 - .255 block using a /28 mask.

0

64

128

192

Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)

160

144 176

Branch 112 hosts

/28(255.255.255.240)

Branch 212 hosts

/28(255.255.255.240)

Branch 312 hosts

/28(255.255.255.240)

Branch 412 hosts

/28(255.255.255.240)

224

208

Branch 512 hosts

/28(255.255.255.240)


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Now we need to address the 5 WAN links that connect to the Branch 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 the subnets.

Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)

160

144 176

Branch 112 hosts

/28(255.255.255.240)

Branch 212 hosts

/28(255.255.255.240)

Branch 312 hosts

/28(255.255.255.240)

Branch 412 hosts

/28(255.255.255.240)

224

208

Branch 512 hosts

/28(255.255.255.240)

232

228 236

WAN 1

WAN 2

WAN 3

WAN 4

248

244

WAN 5

240


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The remaining networks could be used for future growth of either LANs or WANs.

Subnet 0 could also be further subnetted according to the needs of the network.

0

64

128

192

Headquarters

60 hosts

26 bit mask or /26

(255.255.255.192)

160

144 176

Branch 112 hosts

/28(255.255.255.240)

Branch 212 hosts

/28(255.255.255.240)

Branch 312 hosts

/28(255.255.255.240)

Branch 412 hosts

/28(255.255.255.240)

224

208

Branch 512 hosts

/28(255.255.255.240)

232

228 236

WAN 1

WAN 2

WAN 3

WAN 4

248

244

WAN 5

240


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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 /3019

5.39

.71.

212

/30

195.

39.7

1.21

6 /3

0

195.39.71.220 /30

195.39.71.224 /30

Applying the Addresses to the Topology


CLASSLESS INTERDOMAIN ROUTING

CIDR (pronounced “cider”) ignores address class.

With CIDR, a router use a bit mask to determine the network and host portions of an address.

CIDR replaced the categories (A, B, C, etc.) with a more generalized network prefix. This prefix could be of any length rather than just 8, 16, or 24 bits. This allows CIDR to craft network address spaces according to the size of a network instead of force-fitting networks into presized network address spaces.

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CIDR sounds a lot like VLSM

CIDR is usually discussed in general Internet context (ISPs)

Uses custom length prefixes to reduce workload in key Internet routers

VLSM is usually discussed in enterprise context

Uses custom length prefixes to have better usage of enterprise address space

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Routers use the network-prefix, rather than the first 3 bits of the IP address, to determine the dividing point between the network number and the host number.

In the CIDR model, each piece of routing information is advertised with a bit mask or prefix-length ( /x ). The prefix-length is a way of specifying the number of leftmost contiguous bits in the network-portion of each routing table entry.

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For example, a network with 20 bits of network-number and 12 bits of host-number would be advertised with a 20 bit prefix (/20).

The clever thing is that the IP address advertised with the /20 prefix could be a former Class A, Class B, or Class C.

All addresses with a /20 prefix represent the same amount of address space (212 or 4,096 host addresses).

20 bits network + 12 bits host

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Instead of handing out an entire A, B, or C network to an organization, address space can be assigned in “chunks” that fit the need.

If an organization needs 254 host addresses, what difference does it make whether they are given:

a Class C (200.23.76.0 /24) 1/256th of a Class B (145.38.20.0 /24) 1/65,536th of a Class A (91.187.7.0 /24)

Using a /24 prefix, each of these specifies eight host bits which will support 254 hosts.

Note: Each former Class C can be referred to as a /24.

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Network Prefix Equivalent Number of Class Addresses Number of Hosts

/27 1/8th of a Class C 32

/26 1/4th of a Class C 64

/25 1/2 of a Class C 128

/24 1 Class C or 1 /24 256

/23 2 Class C or 2 /24s 512

/22 4 Class C or 4 /24s 1,024

/21 8 Class C or 8 /24s 2,048

/20 16 Class C or 16 /24s 4,096

/19 32 Class C or 32 /24s 8,192

/18 64 Class C or 64 /24s 16,384

/17 128 Class C or 128 /24s 32,768

/16 256 Class C or 1 Class B 65,536

/15 512 Class C or 2 Class B 131,072

/14 1,024 Class C or 4 Class B 262,144

/13 2048 Class C or 8 Class B 524,288

/12 4096 Class C or 16 Class B 1,048,576

/11 8192 Class C or 32 Class B 2,097,152

/10 16384 Class C or 64 Class B 4,194,304

/9 32768 Class C or 128 Class B 8,388,608

/8 65,536 Class C or 256 Class B or 1 Class A 16,777,216

Prefix Equivalents

Day 5.5 subnetting

Education

Transcript of Day 5.5 subnetting