Chapter 3 Subnetting - Hobe Sound Computers

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Oct 24, 2013 (3 years and 7 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
asks



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
-
subnet mask. For ex. 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
you have 0, 64, 128 and 192, the broadcast ad
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

Class C address is 192
.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



The broadcast address for each subnet? The broadcast of the 0
subnet is 127



What are the valid hosts? The subnet plus 1 and

the broadcast
minus 1.


First Subnet

Second Subnet

Subnet

0

128

First Host

1

128

Last Host

126

254

Broadcast

127

255






Another Class C subnet example

o

255.255.255.192 ( /26 )

o

Network address is 192.168.10.0

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

Broadcast

63

127

191

255




Another Class C Subnet example:

o

255.255.255.224 ( /27 )

o

Network Address is 192.168.10.0

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

Broadcast

31

63

95

127

159

191

223

255




Another class C example



255.255.255.240 (/28)



Network address is 192.188.10.0



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

Broadcast

15

31

47

63

79

95

111

127

143

159

175

*Remember the /28 network mask


Another Class C Example



255.255.255.248 (/29)



Network address is 192.188.10.0



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 broadcast address for each subnet? 7, 15,....247, 255.



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

Broadcast

7

15

23

31

231

239

247

255


Another Class C Example



255.255.255.252 (/30)



Network address is 192.168.10.0



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

Broadcast

3

7

11

15

243

247

251

255


Subnetting by example in your head (Class C addresses)



Example number one

o

Host address is 192.168.10.33

o

Subnet mask is 255.255.255.224 (/27)

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

The broadcast address is 192.168.10.63.

o

The valid host range is 19
2.188.10.33
-
62

o




Example number two

o

Host address is 192.168.10.33

o

Subnet mask is 255.255.255.240 (/28)

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

The broadcast address is 192.168.10.47

o

The valid host range is 192.168.10.33
-
46.



Example number three

o

Host address is 192.168.10.17

o

Subnet mask is 255.255.255.252 (/29)

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

The broadcast address is 192.168.10.19

o

The valid host range is 192.168.10.17
-
18


What do we know
?


CIDR

Mask


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



Subnetting Class B Addresses

o

16 bits available for h
ost addressing, 14 bits are actually used for addressing
because 2 bits are reserved

o

You start in the third octet


Class B Adressess Example

o

255.255.128.0 (/17)

o

Network address 172.16.0.0

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

What are the broadcast address for each subnet? 127.255 and 255.255

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

Broadcast

127.255

255.255


Another Class B Example

o

255.255.192.0 (/18)

o

Nework address is 172.16.0.0

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

the broadcast address for each subnet? 63.255, 127.255, 191.255,
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

Broadcast

63.255

127.255

191.255

255.255


Another Class B Example

o

255.255.240.0 (/20)

o

Network address is 172.16.0.0

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

What are the broadcast addresses for each subnet? See table

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

Broadcast

15.255

31.255

47.255

63.255


Another Class B Example

o

255.255.254.0 (/23)

o

Network address is 172.16.0.0

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
broadcast addresses for each subnet? See table

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
oadcast

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
network address.

o

255.255.255.0 (/24)

o

Network address is 172.16.0.0

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

What are the broadcast addresses for each subnet? See table

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

Broadcast

0.255

1.255

2.255

3.255

254.255

255.255


Another Class B Example

o

255.255.255.128 (/25)

o

Network address is 172.16.0.0

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

Broadcast

0.127

0.255

1.127

1.255

2.127

2.255

3.127

3.255



Practice example #7B



255.255.255.192 (/26)



Network address is 172.16.0.0



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



Broadcast address? See table



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

Broadcast

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
work address is 172.16.0.0



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



Broadcast address? See table



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

Broadcast

0.31

0.63

0.95

0.127

0.159

0.191

0.223

0.255



Subnetting in your head: Class B Addresses in your head


Example #1


IP address is 172.16.10.33

Network Address is 255.255.255.224 (/27)


What is
the subnet and broadcast address?

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
172.16.10.32. Our broadcast address is 172.16.10.63


Example #2


IP address is 172.16.66.10

Network ad
dress is 255.255.192.0 (/18)


What is the subnet and broadcast address?

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
172.16.64.0. Our broadcast address is 172.16.12
7.255.


Example #3


IP Address is 172.16.50.10

Network Address is 255.255.224.0 (/19)


What is the subnet and broadcast address?

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
172.16.32.0. Our broadcast address is 172.16.63.255.



Example #4


IP Address is 172.16.46.155

Network address is 255.255.240.0 (/20)


What is the subnet and broadcast address?

Look at the 3
rd

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


IP addr
ess is 172.16.46.155 falls between the 32 and 48 subnet. Our subnet is
172.16.32.0. Our broadcast address is 172.16.47.255.


Example #5


IP Address is 172.16.45.14

Network address is 255.255.255.252 (/30)


What is the subnet and broadcast address?

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
172.16.45.12. Our broadcast address is 172.16.45.15.



Example #6


IP Address is 172.16.88.255/20

Network mask is 255.255
.240.0


What is the subnet and broadcast address?

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
172.16.80.0. Our broadcast is 172.16.95.255.


Example #7


IP Address is 172.16.46.191/26

Network mask is 255.255.255.192


The router will discard the packet because the IP address is a broadcast address
residing on the 128 subnet.



Subnetting Class A Addresses




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



The IP address is 10.0.0.0



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
are the broadcast addresses? See table.



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

Broadcast

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



IP Address 10.0.0.0.



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



The broadcast address? See table



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

Broadcast

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.



IP Address 10.0.0.0



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 broadcast address is? See table



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

Broadcast

10.0.0.63

10.0.0.127

10.0.0.191

10.0.0.255


Class A Addresses: Subet
ting in your head



You start in the 2
nd

octet



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



For example:

o

Host IP address is 10.20.80.30/20

o

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

o

What is the subnet? 10.20.80.0

o

What is the broadca
st address? 10.20.95.255

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

Host IP address is 10.1.3.65/23

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

What is the broadcast address? 10.1.3.255

o

What is the valid host range? 10.1.2.1
-
10.1.3.254








Variable Length Subnet Masks (VLSMs)



Creating many networks using different subnet masks



Classful vs. Classless

subnetting

o

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



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
ess : Support the advertisements of subnet information



RIPv2, EIGRP, OSPF



Benefits: Saving IP address space


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

What is the network addr
ess used to advertise the summary address? It is
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

The summary address is 172.16.32.0

o

Network mask is 255.255.240.0

o

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




Another example: Summary address is 192.168.144.0/20

o

The mask is 255.255.240.0

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

The IP addresses forwarded with the summary address is 172.1.4.1
-
172.1.7.255



Troubleshooting IP Addressing



Four Steps

1.

Ping 127.0.0.1



Test loopback ad
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



WAN link 96 subnet



Marketing Dept. 64
th

subnet



Valid hosts for Sale Dept. is 33
-
62, 63 is the broadcast



Valid host for the Marketing Dept. 65
-
94, 95 is the broadcast



WAN link 97
-
126, 127 is the broadcast



Lab B router is incorrect, its u
sing the broadcast address of the 64
th

subnet




Figure 3.19 pg.154



A user in sales can’t assess server B



Our mask is 255.255.255.248 (/29)



Find the valid host, subnet and broadcast addresses



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
-
30, broadcast is 31)



The WAN is the 40
th

subnet (our host 41
-
46, broadcast is 47)



Marketing LAN is in the 80
th

subnet (our host 81
-
86, broadcast is 87)



The problem is that server B has been assi
gned the broadcast address in the 80
th

subnet


Find Valid Host (figure 3.20) pg.155



Router A 192.168.10.33/27



Network Address 255.255.255.224



Block Size 256
-
224= 32



On the 32
nd

subnet



Valid host range is 33 to 62



Broadcast address is 63


Find the host addr
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?



The serial link address is 172.16.17.0/22



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



S0/0 address could be 172.16.18.
255


Figure 3.23 pg.157



What is your mask?



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