# Chapter 3 Subnetting - Hobe Sound Computers

Networking and Communications

Oct 24, 2013 (4 years and 6 months ago)

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Chapter 3: Subnetting, VLSMs and Troubleshooting TCP/IP

Subnetting: Allows you to take one large network and break it up into smaller networks

Subnetting Benefits

Reduced network traffic
-

routers help create smaller broadcast domain

Simplified management

and Diagnostics

IP Subnet
-
Zero

A command that allows you to use first and last subnet in your network design

Turn on by default starting the 12.x

How To Create Subnets

The power of 2

2^9= 512, 2^10= 1,024…..

every exponential increase doubles

Subnet M

Class A Subnet is 255.0.0.0

Class B Subnet is 255.255.0.0

Class C Subnet is 255.255.255.0

Classless Inter
-
Domain Routing (CIDR)

Denoted by a / followed by a number

For ex. /28 means 28 bits are turned on

The largest subnet available is /30

How to de
termine the subnet mask based on CIDR value

o

255.0.0.0 = /8 (means the first 8 bits are on)

o

255.240.0.0 = /12 (means the first 8 bits plus 4 more are on hence 128 +
64 + 32 + 16 = 240)

/8 through /15 are for the Class A network addresses

/16 through /23 ar
e for the Class A and B

/24 through /30 are for the Class A, B and C

Class A network addresses provide the maximum flexibility for subnetting

Subnetting How to:

Determine number of subnets by doing 2^x where x is defined by the number of
1s. For ex. 11
000000 means 2^2 = 4 subnets

Determine the number of hosts per subnet by doing 2^y
-
2 where y is defined by
the number of 0s. For ex. 11000000 means 2^6
-
2 = 62 hosts per subnet

Determining the valid subnet by taking 256
-
-
192 = 64.
Th
en you count up by block size starting at 0. Count up by block size always
starting at 0 and stopping at the last valid subnet.

The broadcast address always is the last number before the next subnet. For ex. If
dress would be 63, 127 and 191.

Valid host range is between the subnet and the broadcast address. For ex. 64
-
127,
valid host range is 65
-
126.

Subnetting Examples

Subnetting a Class C network address

o

255.255.255.128 ( /25 )

o

10000000

o

.168.10.0

How many subnets? 2^1 = 2 subnets

How many Host per subnet? 2^7
-
2 = 126 hosts

What are the valid subnets? 256
-
128 = 128

subnet is 127

What are the valid hosts? The subnet plus 1 and

minus 1.

First Subnet

Second Subnet

Subnet

0

128

First Host

1

128

Last Host

126

254

127

255

Another Class C subnet example

o

255.255.255.192 ( /26 )

o

o

11000000

How many subnets? 2^2 = 4 subnet
s

How many hosts per subnet? 2^6
-
2 = 62 hosts

What are the valid subnets? 256
-
192 = 64. Our subnets are 0, 64,
128, 192

What’s the broadcast number for each subnets? 63, 127, 191, 255

What are the valid hosts? The subnet plus 1 and the broadcast
minus 1.

First Subnet

Second Subnet

Third Subnet

Forth Subnet

Subnet

0

64

128

192

First Host

1

65

129

193

Last Host

62

126

190

254

63

127

191

255

Another Class C Subnet example:

o

255.255.255.224 ( /27 )

o

o

11100000

How m
any subnets? 2^3 = 8 subnets

How many hosts? 2^5
-
2 = 30 hosts

What are the valid subnets? 256
-
224 = 32. Our subnets are 0, 32,
64, 96, 128, 160, 192, and 224.

What’s the broadcast number for each subnet? 31, 63, 95, 127,
159, 191, 223, 255

What are the val
id hosts? The subnet plus 1 and the broadcast
minus 1.

First
Subnet

Second
Subnet

Third
Subnet

Forth
Subnet

Fifth
Subnet

Sixth
Subnet

Seventh
Subnet

Eighth
Subnet

Subnet

0

32

64

96

128

160

192

224

First Host

1

33

65

97

129

161

193

225

Last Host

30

62

94

126

158

190

222

254

31

63

95

127

159

191

223

255

Another class C example

255.255.255.240 (/28)

11110000

How many subnets? 2^4=16

How many hosts? 2^4
-
2=14

What are the valid subnets? 256
-
240=16. Our subnets

are 0, 16, 32, 48, 64, 80, 96,
112, 128, 144, 160, 176, 192, 208, 224, 240.

What’s the broadcast number for each subnet? 15, 31, 47, 63, 79, 95, 111, 127,
143, 159, 175, 191, 207, 223, 229, 255.

What are the valid hosts? The subnet plus 1 and the broadca
st minus 1.

First

Subnet

Second

Subnet

Third

Subnet

Fourth

Subnet

Fifth

Subnet

Sixth

Subnet

Seventh

Subnet

Eighth

Subnet

Ninth

Subnet

Tenth

Subnet

Eleventh

Subnet

Subnet

0

16

32

48

64

80

96

112

128

144

160

First host

1

17

33

49

65

81

97

113

129

14
5

161

Last host

14

30

46

62

78

94

110

126

142

158

174

15

31

47

63

79

95

111

127

143

159

175

Another Class C Example

255.255.255.248 (/29)

11111000

How many subnets? 2^5 = 32

How
many hosts? 2^3 =6

What are the valid subnets? 256
-
248=8. Our valid subnets are 0, 8..240, 248.

What are the valid hosts? The subnet plus 1 and broadcast minus 1.

First

Subnet

Second

Subnet

Third

Subnet

Fourth

Subnet

Twenty

Ninth
Subnet

Thirtieth

Subnet

Thirty
First

Subnet

Thirty
Second
Subnet

Subnet

0

8

16

24

224

232

240

248

First Host

1

9

17

25

225

233

241

249

Last Host

6

14

22

30

230

238

246

254

7

15

23

31

231

239

247

255

Another Class C Example

255.255.255.252 (/30)

11111100

How many subnets? 2^6 = 64

How many hosts? 2^2
-
2=2

What are the valid subnets? 0, 4, 8, 12...252.

What are the broadcast address for each subnet? 3, 7, 11, 15.
..255.

What are the valid hosts? The subnet plus 1 and broadcast minus 1.

First
Subnet

Second

Subnet

Third
Subnet

Fourth
Subnet

Sixty
First
Subnet

Sixty
Second
Subnet

Sixty
Third
Subnet

Sixty
Fourth
Subnet

Subnet

0

4

8

12

240

244

248

252

First Host

1

5

9

13

241

245

249

253

Last Host

2

6

10

14

242

246

250

254

3

7

11

15

243

247

251

255

Example number one

o

o

o

First step is

to determine block size. 256
-
224=32

o

Count up in block size. Starting with 0. 0, 32, 64, 96.

o

The address of 33 falls between the 32 and 64 subnet.

o

It is a member of 192.168.10.32 subnet.

o

o

The valid host range is 19
2.188.10.33
-
62

o

Example number two

o

o

o

First step to determine block size. 256
-
240=16

o

Count up in block size. Starting with 0,16,32,48.

o

The address of 33 falls between the 32
nd

and 48
th

subnet.

o

It is a member of 192.168.10.32 subnet.

o

o

The valid host range is 192.168.10.33
-
46.

Example number three

o

o

o

First step to determine block size. 256
-
252 =
4

o

Count up in block size. Starting with 0,4, 8, 12, 16, 20...

o

The address of 17 falls between the 16
th

and 20
th

subnet.

o

It is a member of 192.168.10.16 subnet

o

o

The valid host range is 192.168.10.17
-
18

What do we know
?

CIDR

Bits

Block size

Subnets/Hosts

/25

128

10000000

128

2/126

/26

192

11000000

64

4/62

/27

224

11100000

32

8/30

/28

240

11110000

16

16/14

/29

248

11111000

8

32/6

/30

252

11111100

4

64/2

o

16 bits available for h
because 2 bits are reserved

o

You start in the third octet

o

255.255.128.0 (/17)

o

o

10000000.00000000

o

How many subnets? 2^1=2

o

How many hosts? 2^15
-
2=3
2,766

o

What are the valid subnets? 256
-
128=128.0,128

o

o

What are the valid hosts? 0.1
-
127.254 and 128.1
-
255.254

First Subnet

Second Subnet

Subnet

0.0

128.0

First Host

0.1

128.1

Last Hos
t

127.254

255.254

127.255

255.255

Another Class B Example

o

255.255.192.0 (/18)

o

o

11000000.00000000

o

How many subnets? 2^2=4

o

How many hosts? 2^14
-
2=16,382

o

What are the valid subnets? 256
-
192=64. 0, 64, 128, 192

o

What are

255.255

o

What are the valid hosts? See table

First Subnet

Second
Subnet

Third Subnet

Fourth Subnet

Subnet

0.0

64.0

128.0

192.0

First Host

0.1

64.1

128.1

192.1

Last Host

63.254

127.254

191.254

255.254

63.255

127.255

191.255

255.255

Another Class B Example

o

255.255.240.0 (/20)

o

o

11110000.00000000

o

How many subnets? 2^4=16

o

How many hosts? 2^12
-
2= 4094

o

What are the valid subnets? 256
-
240=16. 0, 16, 32
, 48...240

o

o

What are the valid hosts? See table

First Subnet

Second
Subnet

Third Subnet

Fourth Subnet

Subnet

0.0

16.0

32..0

48.0

First Host

0.1

16.1

32.1

48.1

Last Host

15.254

31.254

47.254

63
.254

15.255

31.255

47.255

63.255

Another Class B Example

o

255.255.254.0 (/23)

o

o

11111110.00000000

o

How many subnets? 2^7=128

o

How many hosts? 2^9
-
2=510

o

What are the valid subnets? 256
-
254=2. 0,2,4,6...254

o

What are the

o

What are the valid hosts? See table

First Subnet

Second
Subnet

Third Subnet

Fourth
Subnet

Fifth Subnet

Subnet

0.0

2.0

4.0

6.0

8.0

First Host

0.1

2.1

4.1

6.1

8.1

Last Host

1.254

3.254

5.254

7.254

9.254

Br

1.255

3.255

5.255

7.255

9.255

Another Class B Example *This is a Class B subnet mask when used in a Class B

o

255.255.255.0 (/24)

o

o

11111111.00000000

o

How many subnets? 2^8=256

o

How many hosts? 2^8
-
2=254

o

What are the valid subnets? 256
-
255=1. 0,1,2,3,4,5......255.

o

o

What are the valid hosts? See table

First
Subnet

Second
Subnet

Third
Subnet

Fourth
Subnet

254
th

Subnet

255
th

Subnet

Subnet

0.0

1.0

2
.0

3.0

254.0

255.0

First Host

0.1

1.1

2.1

3.1

254.1

255.1

Last Host

0.254

1.254

2.254

3.254

254.254

255.255

0.255

1.255

2.255

3.255

254.255

255.255

Another Class B Example

o

255.255.255.128 (/25)

o

o

11111111.1000000
0

o

How many subnets? 2^9=512

o

How many hosts? 2^7
-
2=126

o

What are the valid subnets? 256
-
255=1.0, 1, 2, 3, etc

o

*borrowing 1 more bit in the fourth octet hence 2 ^ 1= 2 subnets each

o

*block size for each subnet is 256
-
128= 128

o

*Max. Block size can not be over

256

1
st

subnet

2
nd

subnet

3
rd

subnet

4
th

subnet

5
th

subnet

6
th

subnet

7
th

subnet

8
th

subnet

Subnet

0.0

0.128

1.0

1.128

2.0

2.128

3.0

3.128

First Host

0.1

0.129

1.1

1.129

2.1

2.129

3.1

3.129

Last Host

0.126

0.254

1.126

1.254

2.126

2.254

3.126

3.254

0.127

0.255

1.127

1.255

2.127

2.255

3.127

3.255

Practice example #7B

255.255.255.192 (/26)

11111111.11000000

How many subnets? 2^10= 1024

How many hosts? 2^6
-

2= 62

What are the valid subnets? 256
-
192= 64. 0, 64,
128, 192

Valid hosts? See table

*Borrowing 2 more bits in the fourth octet hence 2^2 = 4 subnets

*Block size is 256
-
192= 64

1
st

subnet

2
nd

subnet

3
rd

subnet

4
th

subnet

5
th

subnet

6
th

subnet

7
th

subnet

8
th

subnet

Subnet

0.0

0
.64

0.128

0.192

1.0

1.64

1.128

1.192

First Host

0.1

0.65

0.129

0.193

1.1

1.65

1.129

1.193

Last Host

0.62

0.126

0.190

0.254

1.62

1.126

1.190

1.254

0.63

0.127

0.191

0.255

1.63

1.127

1.191

1.255

Practice Example #8B

255.255.255.224 (/27)

Net

11111111.11100000

How many subnets? 2^11= 2048

How many hosts? 2^5
-
2= 30

What are the valid subnets? 256
-
224= 32. 0, 32, 64, 96, 128, 160, 192, 224

Valid hosts? See table

*Borrowing 3 more bits in the

fourth octet hence 2^3= 8 subnets

Block size is 32

1
st

subnets

2
nd

subnet

3
rd

subnet

4
th

subnet

5
th

subnet

6
th

subnet

7
th

subnet

8
th

subnet

Subnets

0.0

0.32

0.64

0.96

0.128

0.160

0.192

0.224

First Host

0.1

0.33

0.65

0.97

0.129

0.161

0.193

0.225

Last

Host

0.30

0.62

0.94

0.126

0.158

0.190

0.222

0.254

0.31

0.63

0.95

0.127

0.159

0.191

0.223

0.255

Example #1

What is

256
-
224= 32. (Block Size) 0, 32, 64, 96

The IP address of 172.16.10.33 falls between 32 and 64 subnet. Our subnet is

Example #2

dress is 255.255.192.0 (/18)

Look at the 3
rd

octet and do 256
-
192=64. 0, 64, 128.

The IP address of 172.16.66.10 falls between the 64 and 128 subnet. Our subnet is
7.255.

Example #3

Look at the 3
rd

octet and do 256
-
224= 32. 0, 32, 64,

The IP Address of 172.16.50.10 falls between the 32 and 64 subnet. Our

subnet is

Example #4

Look at the 3
rd

octet and do 256
-
240= 16. 0, 16, 32, 48, 64

ess is 172.16.46.155 falls between the 32 and 48 subnet. Our subnet is

Example #5

256
-
252
= 4. 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52

IP Address is 172.16.45.14 falls between the 12 and 16 subnet. Our subnet is

Example #6

.240.0

Look in the 3
rd

octet and do 256
-
240= 16. 0, 16, 32, 48, 64, 80, 96.

IP address is 172.16.88.255 falls between the 80 and 96 subnet. Our subnet is

Example #7

residing on the 128 subnet.

24 bits are available

We have more host bits

Sta
rt in the 2
nd

octet

Practice example # 1A

255.255.0.0 (/16)

111111111.11111111.00000000.00000000

How many subnets? 2^8 = 256

How many hosts? 2^16
-
2=65,534

What are the valid subnets? 256
-
255=1. The subnets are 0, 1,2,3,4…

What

What are the valid hosts? See table

1
st

Subnet

2
nd

Subnet

255
th

Subnet

256
th

Subnet

Subnet

10.0.0.0

10.1.0.0

10.254.0.0

10.255.0.0

First Host

10.0.0.1

10.1.0.1

10.254.0.1

10.255.0.1

Last Host

10.0.255.254

10.1
.255.254

10.254.255.254

10.255.255.254

10.0.255.255

10.1.255.255

10.254.255.255

10.255.255.255

Practice example #2A

255.255.240.0 (/20)

11111111.11111111.11110000.00000000

How many subnets? 2^12 = 4096

How many hosts? 2^12
-
2= 4094

What are the valid subnets? 256
-
240=16. The valid subnets are 0,16,32,48, 64

What are the valid hosts? See table

1
st

Subnet

2
nd

Subnet

3
rd

Subnet

4096
th

Subnet

Subnet

10.0.0.0

10.0.16.0

10.0.32.0

10.255.240.0

First host

10.0.0.1

10.0.16.1

10.0.32.1

10.255.240.1

Last host

10.0.15.254

10.0.31.254

10.0.47.254

10.255.255.254

10.0.15.255

10.0.31.255

10.0.47.255

10.255.255.255

Practice example #3A

255.255.255.192 (/26)

11111111.11111111.11111111
.11000000

We are subnetting in the 2
nd
, 3
rd

and 4
th

octet.

How many subnets? 2^18
=
262,144

How many hosts? 2^6
-
2= 62

What are the valid subnets? The valid subnets for the 2
nd

octet are 256
-
255=1.
The valid subnets are 0,1,2,3… The valid

subnets for the 3
rd

octet are 256
-
255=1.
The valid subnets are 0,1,2,3. The valid subnets for the 4
th

octet are 256
-
192=64.
The valid subnets are 0,64,128,192.

The valid hosts are? See table

1
st

Subnet

2
nd

Subnet

3
rd

Subnet

4
th

Subnet

Subnet

10.0.0.0

10.0.0.64

10.0.0.128

10.0.0.192

First Host

10.0.0.1

10.0.0.65

10.0.0.129

10.0.0.193

Last Host

10.0.0.62

10.0.0.126

10.0.0.190

10.0.0.254

10.0.0.63

10.0.0.127

10.0.0.191

10.0.0.255

You start in the 2
nd

octet

Only need to be concerned with the octet that has the largest block size.

For example:

o

o

256
-
240=16. 0, 16, 32, 48, 64, 80, 96.

o

What is the subnet? 10.20.80.0

o

o

What is the valid host range? 10.20.80.1
-
10.20.95.254

o

In this example you’re looking at the 3
rd

octet and determining the subnet
and block size.

Another example

o

o

255.255.254.0

o

Look at the 3
rd

octet

o

2
56
-
254=2 block size

o

What is subnet? 10.1.2.0

o

o

What is the valid host range? 10.1.2.1
-
10.1.3.254

Creating many networks using different subnet masks

Classful vs. Classless

subnetting

o

Classful : RIPv1 and IGRP assume that all interfaces have the same subnet

The subnet information is not maintained during routing

If you have different subnet masks on a network that uses RIP or
IGRP the network will not function.

o

Classl

RIPv2, EIGRP, OSPF

Summarization

Allowing routing protocols to advertise many different networks as one.

Also called route aggregation

Main purpose is to reduce the
size of the routing table to save memory

o

To cut down on routing latency

To determine aggregation you must know block size in reference to subnetting
and VLSM design.

o

For example: 192.168.16.0
-
192.168.31.0

o

What is the block size? 16

o

always the first network address in the block.

192.168.16.0

o

What is the summary mask? Look at the block size. It is 16, so the mask is
240.

We are summarizing 255.255.240.0

Another example: 172.16.32.0
-
172
.16.50.0

o

What is the block size? 16

o

o

o

Networks 48 through 50 would be advertised as a single network.

Another example: Summary address is 192.168.144.0/20

o

o

Look at th
e 3
rd

octet for block size. It is 256
-
240=16

o

The network summary range is 144
-
159

o

The router that has the summary address of 192.168.144.0 will forward
any packet with destination IP address of 192.168.144.1
-
192.168.159.254

Another example: page 149
figure 3.16

o

Question is what the best summary address to R2?

o

We have five networks

172.1.4.0/25

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

172.1.4.128/25

o

Look for the interesting octet!!!!!

o

It’s the third octet

o

It’s the block size of 4

Its 4 because 0,4,8,12…
In the third octet you can’t go over 7.

o

The summary address from R1 to R2 is 172.1.4.0/30 (255.255.252.0

o

-
172.1.7.255

Four Steps

1.

Ping 127.0.0.1

dress

2.

Ping IP address of the local host

If it fails, there is a problem with the NIC

3.

Ping the default gateway

If it fails there a local physical network problem

Problem may lie between the NIC and the router

4.

Ping the remote server

If all the steps fail the
n there may be a name revolution problem (DNS)

Figure 3.18 pg.153

Problem: user in sales cannot assess server A in marketing Dept.

Lab A and B router have a mask /27 (255.255.255.224)

The block size is 256
-
224= 32

Our subnets are 0, 32, 64, 96, 128

Sale
s Dept. 32
nd

subnet

Marketing Dept. 64
th

subnet

Valid hosts for Sale Dept. is 33
-

Valid host for the Marketing Dept. 65
-

-

Lab B router is incorrect, its u
th

subnet

Figure 3.19 pg.154

A user in sales can’t assess server B

Our block size is 256
-
248= 8

Our subnet 0, 8, 16, 24, 32, 40, 48
, 56, 64, 72, 80, 88

Sale LAN is the 24
th

subnet (our host 25
-

The WAN is the 40
th

subnet (our host 41
-

Marketing LAN is in the 80
th

subnet (our host 81
-

The problem is that server B has been assi
th

subnet

Find Valid Host (figure 3.20) pg.155

Router A 192.168.10.33/27

Block Size 256
-
224= 32

On the 32
nd

subnet

Valid host range is 33 to 62

ess and subnet mask for A and B (figure 3.21) pg. 156

Router A 192.168.10.65/26 (255.255.255.192)

Router B 192.168.10.33/28 (255.255.255.240)

Router A block size is 256
-
192= 64

Router B block size is 256
-
240= 16

Host A IP address is 192.168.10.66
-
126

Fi
gure 3.22 pg. 157

What IP address can be assigned to the S0/0 interface on router A?

255.255.252.0

The block size 255
-
252= 4

We are in the 3
rd

octet

Valid host range is 16.1
-
19.254

255

Figure 3.23 pg.157

We have 5 subnets

Always look for the network that need the most hosts

Wyoming needs 16 users

Our block size is 32 or 255.255.255.224

8 subnets with 30 hosts each