Ip Addressing and Subnetting Workbook -Desarrollado Todo El Documento PDF (1)

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10111001010111001011000111010011011110100011010000010100101100101001010101100111111101010100010111010011010100110010100101010101010101000110010010101001011000110101100011010110101000010110010101001101001010010

IP Addressingand

Subnetting Workbook

Version 1.5

11111110

10010101

00011011

11010011

Instructor’s Edition

10000110

1 01

IP Address Classes

Class A 1 – 127 (Network 127 is reserved for loopback and internal testing) Leading bit pattern 0 00000000.00000000.00000000.00000000

Network . Host . Host . Host

Class B 128 – 191 Leading bit pattern 10 10000000.00000000.00000000.00000000Network . Network . Host . Host

Class C 192 – 223 Leading bit pattern 110 11000000.00000000.00000000.00000000Network . Network . Network . Host

Class D 224 – 239 (Reserved for multicast)

Class E 240 – 255 (Reserved for experimental, used for research)

Private Address Space

Class A 10.0.0.0 to 10.255.255.255

Class B 172.16.0.0 to 172.31.255.255

Class C 192.168.0.0 to 192.168.255.255

Default Subnet Masks

Class A 255.0.0.0

Class B 255.255.0.0

Class C 255.255.255.0

Produced by: Robb Jones

[email protected] County Career & Technology

Center Cisco Networking AcademyFrederick County Public Schools

Frederick, Maryland, USA

Special Thanks to Melvin Baker and Jim Dorsch for taking the time to check this workbook for errors,

and to everyone who has sent in suggestions to improve the series.

Workbooks included in the series:

IP Addressing and Subnetting Workbooks ACLs - Access Lists

WorkbooksVLSM Variable-Length Subnet Mask IWorkbooks

Instructors (and anyone else for that matter) please do not post the Instructors version on public websites.

When you do this you are giving everyone else worldwide the answers. Yes, students look for answers this

way.It also discourages others; myself included, from posting high quality

materials.

Inside Cover

1

00011011 27

10101010 170

01101111 111

11111000 248

00100000 32

01010101 85

00111110 62

00000011 3

11101101 237

11000000 192

Binary To Decimal Conversion

128 64 32 16 8 4 2 1 Answers Scratch Area

1 0 0 1 0 0 1 0 146 128 6416 32

0 1 1 1 0 1 1 1 119 2 16146 4

1 1 1 1 1 1 1 1 2 1

1 1 0 0 0 1 0 1 197 119

1 1 1 1 0 1 1 0 2 4 6

0 0 0 1 0 0 1 1 1 9

1 0 0 0 0 0 0 1 129

0 0 1 1 0 0 0 1 49

0 1 1 1 1 0 0 0 120

1 1 1 1 0 0 0 0 240

0 0 1 1 1 0 1 1 59

0 0 0 0 0 1 1 1 7

2

Decimal To Binary ConversionUse all 8 bits for each

problem

128 64 32 16 8 4 2 1 = 255 Scratch Area

1 1 1 0 1 1 1 0

0 0 1 0 0 0 1 0

0 1 1 1 1 0 1 1

0 0 1 1 0 0 1 0

1 1 1 1 1 1 1 1

1 1 0 0 1 0 0 0

0 0 0 0 1 0 1 0

1 0 0 0 1 0 1 0

0 0 0 0 0 0 0 1

0 0 0 0 1 1 0 1

1 1 1 1 1 0 1 0

0 1 1 0 1 0 1 1

1 1 1 0 0 0 0 0

0 1 1 1 0 0 1 0

1 1 0 0 0 0 0 0

1 0 1 0 1 1 0 0

0 1 1 0 0 1 0 0

0 1 1 1 0 1 1 1

0 0 1 1 1 0 0 1

0 1 1 0 0 0 1 0

1 0 1 1 0 0 1 1

0 0 0 0 0 0 1 0

238

34

123

50

255

200

10

138

1

13

250

107

224

114

192

172

100

119

57

98

179

2

238-128110-6446

-32 14 -8

6-42

-2 0

34-32

2-20

3

Address Class Identification

Address Class

10.250.1.1 __A___

150.10.15.0 __B___

192.14.2.0 __C___

148.17.9.1 __B___

193.42.1.1 __C___

126.8.156.0 __A___

220.200.23.1 __C___

230.230.45.58 __D___

177.100.18.4 __B___

119.18.45.0 __A___

249.240.80.78 __E___

199.155.77.56 __C___

117.89.56.45 __A___

215.45.45.0 __C___

199.200.15.0 __C___

95.0.21.90

__A___ 33.0.0.0 __A___

158.98.80.0 __B___

219.21.56.0

4

__C___

5

Network & Host Identification

Circle the network portion of these addresses:

177.100.18.4

119.18.45.0

209.240.80.78

199.155.77.56

117.89.56.45

215.45.45.0

192.200.15.0

95.0.21.90

33.0.0.0

158.98.80.0

217.21.56.0

10.250.1.1

150.10.15.0

192.14.2.0

148.17.9.1

193.42.1.1

126.8.156.0

220.200.23.1

Circle the host portion of these addresses:

10.15.123.50

171.2.199.31

198.125.87.177

223.250.200.222

17.45.222.45

126.201.54.231

191.41.35.112

155.25.169.227

192.15.155.2

123.102.45.254

148.17.9.155

100.25.1.1

195.0.21.98

25.250.135.46

171.102.77.77

55.250.5.5

218.155.230.14

10.250.1.1

6

Network Addresses

Using the IP address and subnet mask shown write out the network address:

188.10.18.2 ______1_8_8__.__1_0__._0__.__0_________255.255.0.0

10.10.48.80 ______1_0__.__1_0__._4__8__. _0_________255.255.255.0

192.149.24.191 ______1_9_2___. _1_4__9__. _2_4___.

_0______255.255.255.0

150.203.23.19 ______1_5__0__._2__0_3__.__0__. _0_______255.255.0.0

10.10.10.10 ______1_0__.__0__. _0__.__0___________255.0.0.0

186.13.23.110

______1_8_6__.__1_3__._2__3__. _0________ 255.255.255.0

223.69.230.250 ______2_2__3__._6__9__. _0__._0_________255.255.0.0

200.120.135.15 ______2_0__0__._1__2_0__.__1_3_5___.

_0____255.255.255.0

27.125.200.151 ______2_7__.__0__._0__.__0___________255.0.0.0

199.20.150.35 ______1_9_9__.__2_0___. _1_5__0__.

_0______255.255.255.0

7

191.55.165.135 ______1_9_1__.__5_5___. _1_6_5___.

_0______255.255.255.0

28.212.250.254 ______2_8__.__2_1_2___. _0__._0_________255.255.0.0

8

Host Addresses

Using the IP address and subnet mask shown write out the host address:

188.10.18.2 ______0__._0___. _1_8__._2____________255.255.0.0

10.10.48.80 ______0__._0___. _0__._8__0___________255.255.255.0

222.49.49.11 ______0__._0___. _0__._1_1____________255.255.255.0

128.23.230.19 ______0__._0___.

_2_3__0__._1_9_________255.255.0.0

10.10.10.10 ______0__._1__0__. _1_0__.__1_0_________255.0.0.0

200.113.123.11 ______0__._0___.

_0__._1_1____________ 255.255.255.0

223.169.23.20 ______0__._0___. _2_3___. _2_0__________255.255.0.0

203.20.35.215 ______0__._0___. _0__._2__1_5__________255.255.255.0

117.15.2.51 ______0__._1__5__._2__.__5_1__________255.0.0.0

199.120.15.135 ______0__._0___. _0__._1_3__5__________255.255.255.0

191.55.165.135 ______0__._0___. _0__._1_3__5__________255.255.255.0

48.21.25.54 ______0__._0___. _2_5___.

9

_5__4_________255.255.0.0

10

Default Subnet Masks

Write the correct default subnet mask for each of the following addresses:

177.100.18.4 ______2_5__5__.__2_5__5__._0___.

_0______

119.18.45.0 _______2__5_5___. _0__.__0__. _0________

191.249.234.191

______2_5__5__._2__5__5__._0__.__0______

223.23.223.109 ____2__5_5___.

_2_5__5__.__2_5__5__._0_____

10.10.250.1 _______2__5_5___. _0__.__0__. _0________

126.123.23.1 _______2__5_5___. _0__.__0__. _0________

223.69.230.250 ____2__5_5___.

_2_5__5__.__2_5__5__._0_____

192.12.35.105 ____2__5_5___.

_2_5__5__.__2_5__5__._0_____

77.251.200.51 _______2__5_5___. _0__.__0__. _0________

189.210.50.1

______2_5__5__._2__5__5__._0__.__0______

88.45.65.35 _______2__5_5___. _0__.__0__. _0________

128.212.250.254

______2_5__5__._2__5__5__._0__.__0______

193.100.77.83 ____2__5_5___.

11

_2_5__5__.__2_5__5__._0_____

125.125.250.1 _______2__5_5___. _0__.__0__. _0________

1.1.10.50 _______2__5_5___. _0__.__0__. _0________

220.90.130.45 ____2__5_5___.

_2_5__5__.__2_5__5__._0_____

134.125.34.9

______2_5__5__._2__5__5__._0__.__0______

95.250.91.99 _______2__5_5___. _0__.__0__. _0________

12

ANDING With Default subnet masks

Every IP address must be accompanied by a subnet mask. By now you should be able to look at an IP address and tell what class it is. Unfortunately your computer doesn’t think that way. For your computer to determine the network and subnet portion of an IP address it must“AND” the IP address with the subnet mask.

Default Subnet Masks:Class A 255.0.0.0Class B 255.255.0.0Class C 255.255.255.0

ANDING Equations:1 AND 1 = 11 AND 0 = 00 AND 1 = 00 AND 0 = 0

Sample:

What you see...

IP Address: 192 . 100 . 10 . 33

What you can figure out in your head...

Address Class: CNetwork Portion: 192 . 100 . 10 . 33Host Portion: 192 . 100 . 10 . 33

In order for you computer to get the same information it must AND the IP address with the subnet mask in binary.

Network Host

IP Address: 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 1

(192 . 100 . 10 . 33)

Default Subnet Mask: 1 1 1 1 1 1 1 1 . 0 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 0 0 0 0 0 0 0 0

(255 . 255 . 255 . 0)

AND: 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 0 0 0 0 0 0

(192 . 100 . 10 . 0)

ANDING with the default subnet mask allows your computer to figure out the network portion of the address.

13

ANDING With Custom subnet masks

When you take a single network such as 192.100.10.0 and divide it into five smaller networks (192.100.10.16, 192.100.10.32, 192.100.10.48, 192.100.10.64, 192.100.10.80) the outside world still sees the network as 192.100.10.0, but the internal computers and routers see five smaller subnetworks. Each independent of the other. This can only be accomplished by using a custom subnet mask. A custom subnet mask borrows bits from the host portion of the address to create a subnetwork address between the network and host portions of an IP address. In this example each range has 14 usable addresses in it. The computer must still AND the IP address against the custom subnet mask to see what the network portion is and which subnetwork it belongs to.

IP Address: 192 . 100 . 10 . 0Custom Subnet Mask: 255.255.255.240

Address Ranges: 192.10.10.0 to 192.100.10.15 192.100.10.16 to 192.100.10.31192.100.10.32 to 192.100.10.47 (Range in the sample below)192.100.10.48 to 192.100.10.63192.100.10.64 to 192.100.10.79192.100.10.80 to 192.100.10.95192.100.10.96 to 192.100.10.111192.100.10.112 to 192.100.10.127192.100.10.128 to 192.100.10.143192.100.10.144 to 192.100.10.159192.100.10.160 to 192.100.10.175192.100.10.176 to 192.100.10.191192.100.10.192 to 192.100.10.207192.100.10.208 to 192.100.10.223192.100.10.224 to 192.100.10.239192.100.10.240 to 192.100.10.255

NetworkSub

Network Host

IP Address: Custom Subnet Mask:

AND:

1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 1 (192 . 100 . 10 .

33)

1 1 1 1 1 1 1 1 . 0 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 0 0 0 0 (255 . 255 . 255 . 240)

1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 0 (192 . 100 . 10 .

32)

Four bits borrowed from the host portion of the address for the custom subnet mask.

The ANDING process of the four borrowed bits shows which range of IP addresses this particular address will fall into.

14

In the next set of problems you will determine the necessary information to determine the correct subnet mask for a variety of IP addresses.

15

How to determine the number of subnets and the number of hosts per subnet

Two formulas can provide this basic information:

Number of subnets = 2 s (Second subnet formula: Number of subnets = 2s - 2)

Number of hosts per subnet = 2 h - 2

Both formulas calculate the number of hosts or subnets based on the number of binary bits used. For example if you borrow three bits from the host portion of the address use the number of subnets formula to determine the total number of subnets gained by borrowing the three bits. This would be 23 or 2 x 2 x 2 = 8 subnets

To determine the number of hosts per subnet you would take the number of binary bits used in the host portion and apply this to the number of hosts per subnet formula If five bits are in the host portion of the address this would be 25 or 2 x 2 x 2 x 2 x 2 = 32 hosts.

When dealing with the number of hosts per subnet you have to subtract two addresses from the range. The first address in every range is the subnet number. The last address in every range is the broadcast address. These two addresses cannot be assigned to any device in the network which is why you have to subtract two addresses to find the number of usable addresses in each range.

For example if two bits are borrowed for the network portion of the address you can easily determine the number of subnets and hosts per subnets using the two formulas.

195. 223 . 50 . 0 0 0 0 0 0 0 0

The number of subnets created by borrowing 2 bits is 22 or 2 x 2 = 4 subnets.

The number of hosts created by leaving 6 bits is 26 - 2 or2 x 2 x 2 x 2 x 2 x 2 = 64 - 2 = 62usable hosts per subnet.

What about that second subnet formula:

Number of subnets = 2 s

- 2

In some instances the first and last subnet range of addresses are reserved. This is similar to the first and last host addresses in each range of addreses.

The first range of addresses is the zero subnet. The subnet number for the zero subnet is also the subnet number for the classful subnet address.

The last range of addresses is the broadcast subnet. The broadcast address for the last subnet in the broadcast subnet is the same as the classful broadcast address.

16

Class C Address unsubnetted:

195. 223 . 50 . 0

195.223.50.0 to 195.223.50.255

Class C Address subnetted (2 bits borrowed):

Notice that the subnet and broadcast addresses match.

195. 223 . 50 . 0 0 0 0 0 0 0 0

(Invalid range)

(Invalid range)

(0)(1)(2)(3)

195.223.50.0 to195.223.50.64 to195.223.50.128 to195.223.50.192 to

195.223.50.63195.223.50.127195.223.50.191195.223.50.255

The primary reason the the zero and broadcast subnets were not used had to do pirmarily with the broadcast addresses. If you send a broadcast to 195.223.255 are you sending it to all 255 addresses in the classful C address or just the 62 usable addresses in the broadcast range?

The CCNA and CCENT certification exams may have questions which will require you to determine which formula to use, and whehter or not you can use the first and last subnets. Use the chart below to help decide.

When to use which formula to determine the number of subnets

Use the 2s

- 2 formula and don’t use the zero and broadcast ranges if...

Use the 2s

formula and use the zero and broadcast ranges if...

Classful routing is used Classless routing or VLSM is used

RIP version 1 is used RIP version 2, EIGRP, or OSPF is used

The no ip subnet zero command is configured on your router

The ip subnet zero command is configured on your router (default setting)No other clues are given

Bottom line for the CCNA exams; if a question does not give you any clues as to whether or not to allow these two subnets, assume you can use them.

This workbook has you use the number of subnets = 2s

formula.

17

Problem 1

Custom Subnet Masks

Number of needed subnets 14Number of needed usable hosts 14

Network Address 192.10.10.0

Address class ____C______

Default subnet mask ____2__5__5__._2__5__5__._2__5_5___. _0______

Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._2__4_0___

Total number of subnets _________1_6_________

Total number of host addresses _________1_6_________

Number of usable addresses _________1_4_________

Number of bits borrowed _________4__________

Show your work for Problem 1 in the space below.

Number of256 128 64 32 16 8 4 2 - Hosts

Number ofSubnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

192 . 10 . 10 . 0 0 0 0 0 0 0 0

Add the binary value numbers to the left of the line to create the custom subnet mask.

1286432

+16

240

16

18

Observe the total number of hosts.

-2 Subtract 2 for the

number of

14 usable hosts.

19

Problem 2

Custom Subnet Masks



Address class ____B______

Default subnet mask ____2__5__5__._2__5__5__._0__.__0_________

Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._1_9__2___

Total number of subnets _______1_,0__2_4________



Number of bits borrowed _________1_0_________


Number of Hosts - . 256 128 64 32 16 8 4 2

Number ofSubnets - 2 4 8 16 32 64 128 256.

Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

Add the binary value numbers to the left of the line to create the custom subnet mask.

128643216

842

+1 128+64 192

64

20

Observe thetotal

number

of hosts.

-2 Subtract 2 for the number of

62 usable hosts.

255

1286432

Add the binary value 16numbers to the left of the line to 8create the custom subnet mask. 4

2+1

64

21

Problem 3

Custom Subnet Masks

Network Address 148.75.0.0 /26

Address class _____B_____

/26 indicates the total number of bits used for the network and subnetwork portion of the address. All bits remaining belong to the host portion of the address.

Default subnet mask _____2_5__5__.__2_5__5__._0___.

_0_________

Custom subnet mask _____2_5__5__.__2_5__5__.__2_5__5__._1__9_2___






Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

128+64

192 Observe the total number ofhosts.

-2 Subtract 2 for the number of

62 usable hosts.

22

255 1024-2

1,022

Subtract 2 for the total number of subnets to get the usable number of subnets.

23

Problem 4

Custom Subnet Masks



Address class ____C___

Default subnet mask ____2__5_5___. _2__5_5___.

_2_5__5___. _0______ Custom subnet mask

____2__5_5___. _2__5_5___. _2_5__5___.

_2_2__4___

Total number of subnets

________8___________ Total number of host

addresses ________3__2_________ Number of

usable addresses ________3__0_________

Number of bits borrowed ________3___________


Number of256 128 64 32 16 8 4 2 - Hosts


128 64 32 16 8 4 2 1 - Binary values

210 . 100 . 56 . 0 0 0 0 0 0 0 0

12864 8

+3 2 -2224 6

32 -2 30

24

Problem 5

Custom Subnet Masks



Address class ____C___

Default subnet mask ____2__5_5___. _2__5_5___.

_2_5__5___. _0______ Custom subnet mask

____2__5_5___. _2__5_5___. _2_5__5___.

_2_2__4___



addresses ________3__2_________ Number of

usable addresses ________3__0_________



Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

195 . 85 . 8 . 0 0 0 0 0 0 0 0

25

12864

+32224

32 -2 30

8 -2

6

26

Problem 6

Custom Subnet Masks

Number of needed subnets 126Number of needed usable hosts 131,070


Address class ___A____

Default subnet mask ____2_5__5__._0___.

_0__.__0_____________ Custom subnet mask

____2_5__5__._2__5__4_._0___. _0___________


________1_2__8________ Total number of host

addresses ________1_3_1__,0__7_2____ Number of

usable addresses ________1_3_1__,0__7_0____



Number of

Hosts -. 256 128 64 32 16 8 4 2

Number ofSubnets - 2 4 8 16 32 64 128 256 . .

Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

1286432

27

1684

+2 254

128 -2 126

131,072 - 2 131,070

18 255

Problem 7

Custom Subnet Masks



Address class _____B_____

Default subnet mask

_____2_5__5__.__2_5__5__._0___. _0_________

Custom subnet mask

___

__2_5__5__.__2_5__5__._2__5__5__._2__2_4___


_________2_,_0_4__8_____ Total number of host

addresses _________3_2_________ Number of

usable addresses _________3_0_________ Number

of bits borrowed _________1_1_________


Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

29

128643216

842

+1

2,048 - 2 2,046

32 -2 30

18 255

Problem 8

Custom Subnet Masks



Address class ___C____

Default subnet mask

____2_5__5__._2__5__5__._2__5__5__._0_______

Custom subnet mask

__

__2_5__5__._2__5__5__._2__5__5__._1__9_2___

_



addresses ________6_4__________ Number of

usable addresses ________6_2__________ Number

of bits borrowed ________2___________


Number of256 128 64 32 16 8 4 2 - Hosts


128 64 32 16 8 4 2 1 - Binary values

200 . 175 . 14 . 0 0 0 0 0 0 0 0

31

128 4 +6 4 -2240 2

64 -2 62

20

Problem 9

Custom Subnet Masks

Number of needed subnets 60Number of needed usable hosts 1,000


Address class ____B___

Default subnet mask

____2__5__5__._2__5__5__._0__.__0_________

Custom subnet mask

____2__5__5__._2__5__5__._2__5_2___. _0______


_________6_4_________ Total number of host

addresses _________1_,0__2_4______ Number of

usable addresses _________1_,0__2_2______



Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

128 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

192 62 221

128643216

8 +4 252

64 1,024- 2 - 2 62 1,022

22

Problem 10

Custom Subnet Masks

Number of needed usable hosts 60Network Address 198.100.10.0

Address class ___C____

Default subnet mask

____2_5__5__._2__5__5__._2__5__5__._0_______

Custom subnet mask

__

__2_5__5__._2__5__5__._2__5__5__._1__9_2____





of bits borrowed ________2___________

Show your work for Problem 1 0 in the space below.

Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

198 . 100 . 10 . 0 0 0 0 0 0 0 0

192 62 223

128 64 4+6 4 - 2 -2

24

Problem 1 1

Custom Subnet Masks

Number of needed subnets 250Network Address 101.0.0.0

Address class ___A____

Default subnet mask

___2__5_5___._0__.__0__._0______________ Custom

subnet mask ___2__5_5___._2__5_5___.

_0__.__0__________


________2_5__6________ Total number of host

addresses ________6_5__,5__3_6_____ Number of

usable addresses ________6_5__,5__3_4_____



Number of

Hosts -. 256 128 64 32 16 8 4 2

Number of

Subnets - 2 4 8 16 32 64 128 256 . .Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

1286432

1684

2 +1 255

192 62 225

256 -2 254

65,536 - 65,534

224 62 223

Problem 12

Custom Subnet Masks


Address class

___C____

Default subnet mask

____2_5__5__._2__5__5__._2__5__5__._0_______

Custom subnet mask

__

__2_5__5__._2__5__5__._2__5__5__._2__2_4___

_





of bits borrowed ________3___________


Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

218 . 35 . 50 . 0 0 0 0 0 0 0 0

24

12864

+3264 4 - 2 -2

224 62 225

Problem 13

Custom Subnet Masks


Address class

___C____

Default subnet mask ___2__5_5___._2__5_5___.

_2__5_5___. _0_______ Custom subnet mask

___2__5_5___._2__5_5___. _2__5_5___.

_2__2_4____





of bits borrowed ________3___________


Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

218 . 35 . 50 . 0 0 0 0 0 0 0 0

26

12864 8

+3 2 -2224 6

32 -2 30

25

Problem 14

Custom Subnet Masks


Address class

___B____

Default subnet mask

____2_5__5__._2__5__5__._0___. _0__________

Custom subnet mask

____2_5__5__._2__5__5__._2__4_0___. _0_______


________1_6__________ Total number of host

addresses ________4_,_0_9__6______ Number of

usable addresses ________4_,_0_9__4______



Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

26

1286432

+16 240

16 -2 14

4,096 -2 4,094

27

Problem 15

Custom Subnet Masks


Address class

___B____

Default subnet mask ___2__5_5___._2__5_5___.

_0__.__0__________ Custom subnet mask

___2__5_5___._2__5_5___. _2__5_5___.

__1_9_2____


________1_,0__2_4_______ Total number of host



of bits borrowed ________1_0__________


Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

128 6432

168

28

42

+1 255

128 64

+6 4 -2192 62

1,024 - 2 1,022

29

Problem 16

Custom Subnet Masks


Address class

___A____

Default subnet mask ____2_5__5__._0___.

_0__.__0_____________ Custom subnet mask

__

__2_5__5__._2__5__5__._2__5__5__._2__2_4___

_


________5_2__4__,2__8_8___ Total number of host



of bits borrowed ________1_9__________


Number of

Hosts -. 256 128 64 32 16 8 4 2

Number of

Subnets - 2 4 8 16 32 64 128 256 . .

Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

23 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

30

12864

+32224

32 524,288- 2 - 2 30 524,286

31

Problem 1

Subnetting




Default subnet mask ____2__5__5__._2__5__5__._2__5_5___. _0______

Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._2__4_0___

Total number of subnets _________1_6_________




What is the 4thsubnet range? 192.10.10.48 to 192.10.10.63

What is the subnet number for the 8th subnet?

What is the subnet broadcast address for

the 13th subnet?

___1_9_2___. _1_0__.__1_0__._1_1__2___

___1_9_2___. _1_0__.__1_0__._2__0_7___

32

What are the assignable addresses for the 9th

subnet? _1_9__2_._1_0_._1_0__.1_2__9___t_o___1_9_2__.1_0__.1_0__.1__4_2___

33


Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

192. 10 . 10 . 0 0 0 0 0 0 0 0

(0) 0 0 0 0 192.10.10.0 to 192.10.10.15(1) 0 0 0 1 192.10.10.16 to 192.10.10.31(2) 0 0 1 0 192.10.10.32 to 192.10.10.47(3) 0 0 1 1 192.10.10.48 to 192.10.10.63(4) 0 1 0 0 192.10.10.64 to 192.10.10.79(5) 0 1 0 1 192.10.10.80 to 192.10.10.95(6) 0 1 1 0 192.10.10.96 to 192.10.10.111(7) 0 1 1 1 192.10.10.112 to 192.10.10.127(8) 1 0 0 0 192.10.10.128 to 192.10.10.143(9) 1 0 0 1 192.10.10.144 to 192.10.10.159(10)

1 0 1 0 192.10.10.160 to 192.10.10.175(11) 1 0 1 1 192.10.10.176 to 192.10.10.191(12)

1 1 0 0 192.10.10.192 to 192.10.10.207(13)

1 1 0 1 192.10.10.208 to 192.10.10.223(14)

1 1 1 0 192.10.10.224 to 192.10.10.239(15)

1 1 1 1 192.10.10.240 to 192.10.10.255

Custom subnet mask

1286432

+16240

16

Usable subnets -2

14

16

Usable hosts -2

14

The binary value of the last bit borrowed is the range. In this problem the range is 16.

The first address in each subnet range is the subnet number.

The last address in each subnet range is the subnet broadcast address.

34

Problem 2

Subnetting




Default subnet mask ____2__5__5__._2__5__5__._0__.__0_________

Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._1_9__2___








the 6th subnet?

___1_6_5___. _1_0__0__._1__._6__4____

___1_6_5___. _1_0__0__._1__._1__2_7___


35

subnet? _1_6__5_._1_0__0_._2_._1__t_o___1_6__5_._1_0_0__.0__.6__2______

Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

(0) . 0 165.100.0.0 to 165.100.0.63

64 128-2 6462 32

16

1 01

01

1

1 . 01 . 01 . 1 01 . 1 1

1 1 1 1 1 1 1 . 1 0 165.100.255.128 to 165.100.255.1911 1 1 1 1 1 1 . 1 1 165.100.255.192 to 165.100.255.255

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Number of Hosts - . 256 128 64 32 16 8 4 2


Usable hosts

Custom subnet mask

128 8

+644

192+1

(1) 1

(2)(3)(4)(5)(6)(7)(8) 1 0 . 0 0

165.100.0.64 to 165.100.0.127165.100.0.128 to 165.100.0.191165.100.0.192 to 165.100.0.255

165.100.1.0 to 165.100.1.63165.100.1.64 to 165.100.1.127165.100.1.128 to 165.100.1.191165.100.1.192 to 165.100.1.255

165.100.2.0 to 165.100.0.63The binary value of the last bit borrowed is the range. In this problem the range is 64.

The first address in each subnet range is the subnet number.

The last address in each subnet range is the subnet broadcast address.

(1022) 1(1023) 1

255(9) 1 0 . 0 1(10) 1 0 . 1 0(11) 1 0 . 1 1(12) 1 1 . 0 0(13) 1 1 . 0 1(14) 1 1 . 1 0(15) 1 1 . 1 1

165.100.2.64 to 165.100.0.127165.100.2.128 to 165.100.0.191165.100.2.192 to 165.100.0.255

165.100.3.0 to 165.100.3.63165.100.3.64 to 165.100.3.127165.100.3.128 to 165.100.3.191165.100.3.192 to 165.100.3.255

Down to

32

Problem 3

Subnetting



Default subnet mask

____2__5__5__._2__5__5__._2__5_5___.

_0______ Custom subnet mask

__

__2__5__5__._2__5__5__._2__5_5___._1_9__2__

_


_________4__________ Total number of host



of bits borrowed _________2__________

What is the 3rdsubnet range? 195.223.50.128 - 195.223.50.191

What is the subnet number for the 2nd subnet?


the 1st subnet?

_____1_9_5__._2_2__3_._5_0__.6__4____

___ __1_9_5__._2_2__3_._5_0__.6__3____

What are the assignable addresses for the 3rd

subnet? _1_9__5_._2_2__3_._5_0__.1_2__9__-___1_9_5__.2__2_3__.5__0_._1_9_0_

34


Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

195. 223 . 50 . 0 0 0 0 0 0 0 0

(0) 0 195.223.50.0 to 195.223.50.63(1) 1 195.223.50.64 to 195.223.50.127(2) 1

0 195.223.50.128 to 195.223.50.191(3) 1 1 195.223.50.192 to 195.223.50.255

128 +64 192 64

-2 62

33

34

Problem 4

Subnetting



Default subnet mask

____2__5__5__._2__5__5__._0__.__0_________

Custom subnet mask

__

__2__5__5__._2__5__5__._2__5_5___._1_9__2__

_


_________1_,0__2_4______ Total number of host



of bits borrowed _________1_0_________




the 10th subnet?

What are the assignable addresses for the 6th subnet?

______1_9__0_._3_5__.3__.0________

______1_9__0_._3_5__.2__.1_2__7_____

__1__9_0_._3_5__.1_._6

_5____t_o___1_9__0_.3__5_._1_._1_2_6_____

(0) . 0 190.35.0.0 to 190.35.0.63(1) 1 190.35.0.64 to 190.35.0.127(2) 1 0 190.35.0.128 to 190.35.0.191(3) 1 1 190.35.0.192 to 190.35.0.255(4) 1 . 0 0 190.35.1.0 to 190.35.1.63(5) 1 . 0 1 190.35.1.64 to 190.35.1.127(6) 1 . 1 0 190.35.1.128 to 190.35.1.191(7) 1 . 1 1 190.35.1.192 to 190.35.1.255(8) 1 0 . 0 0 190.35.2.0 to 190.35.2.63(9) 1 0 . 0 1 190.35.2.64 to 190.35.2.127(10)

1 0 . 1 0 190.35.2.128 to 190.35.2.191(11) 1 0 . 1 1 190.35.2.192 to 190.35.2.255(12)

1 1 . 0 0 190.35.3.0 to 190.35.3.63(13)

1 1 . 0 1 190.35.3.64 to 190.35.3.127(14)

1 1 . 1 0 190.35.3.128 to 190.35.3.191(15)

1 1 . 1 1 190.35.3.192 to 190.35.3.255

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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

190 . 35 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

128643216

842

+1252

64 128

-2 +6462 252

36

Problem 5

Subnetting


Address class ____A______

Default subnet mask ____2__5__5__._0___.

_0__._0_____________ Custom subnet mask

____2__5__5__._2__5__5__._2__5_5___.

_2__4_8___


_________2_,_0_9__7_,1_5__2_ Total number of

host addresses _________8_______

Number of usable addresses

_________6__________ Number of bits

borrowed _________2_1_________

What is the 2ndsubnet range? 126.0.0.8 to 126.0.0.15



the 7th subnet?

_____1_2__6_._0_._0_._3_2_________

_____1_2__6_._0_._0_._5_5_________


subnet? ______1_2_6__.0__.0__.7_3____t_o___1_2__6_.0__.0__.7__8_____

(0) 0 126.0.0.0 to 126.0.0.7(1) 1 126.0.0.8 to 126.0.0.15(2) 1 0 126.0.0.16 to 126.0.0.23(3) 1 1 126.0.0.24 to 126.0.0.31(4) 1 0 0 126.0.0.32 to 126.0.0.39(5) 1 0 1 126.0.0.40 to 126.0.0.47(6) 1 1 0 126.0.0.48 to 126.0.0.55(7) 1 1 1 126.0.0.56 to 126.0.0.63(8) 1 0 0 0 126.0.0.64 to 126.0.0.71(9) 1 0 0 1 126.0.0.72 to 126.0.0.79(10)

1 0 1 0 126.0.0.80 to 126.0.0.87(11) 1 0 1 1 126.0.0.88 to 126.0.0.95(12)

1 1 0 0 126.0.0.96 to 126.0.0.103(13)

1 1 0 1 126.0.0.104 to 126.0.0.111(14)

1 1 1 0 126.0.0.112 to 126.0.0.119(15)

1 1 1 1 126.0.0.120 to 126.0.0.127

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Number of

Hosts -. 256 128 64 32 16 8 4 2

Number of

Subnets - 2 4 8 16 32 64 128 256 . .

Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

126. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

128643216

842

+1 255

8-26

12864

3216+8

248

38

Problem 6

Subnetting


Address class

____C______

Default subnet mask

____2__5__5__._2__5__5__._2__5_5___.


__

__2__5__5__._2__5__5__._2__5_5___._2__4_0__

_



addresses _________1_6_________ Number of usable

addresses _________1_4_________


What is the 9th subnet range?



the 12th subnet?

192.70.10.128 to 192.70.10.143

____1_9_2__.7__0_._1_0_._4__8_______

____1_9_2__.7__0_._1_0__.1_9__1______


subnet? _1_9__2_._7_0_._1_0_._1_4_5_____t_o__1__9_2_._7_0__.1_0__.1_5__8__

40

Show your work for Problem 6 in the space below.Number of

256 128 64 32 16 8 4 2 - HostsNumber of

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

192 . 70 . 10 . 0 0 0 0 0 0 0 0


1 0 1 0 192.70.10.160 to 192.70.10.175(11) 1 0 1 1 192.70.10.176 to 192.70.10.191(12)

1 1 0 0 192.70.10.192 to 192.70.10.0207(13)

1 1 0 1 192.70.10.208 to 192.70.10.223(14)

1 1 1 0 192.70.10.224 to 192.70.10.239(15)

1 1 1 1 192.70.10.240 to 192.70.10.255

128 +64 240

16 -2 14

39

40

Problem 7

Subnetting



Default subnet mask ____2__5__5__._0___. _0__._0_____________

Custom subnet mask ____2__5__5__._2__5__5__._0__.__0_________

Total number of subnets _________2_5__6_______

Total number of host addresses _________6_5__,5__3_6____

Number of usable addresses _________6_5__,5__3_4____





the 2nd subnet?

_______1_0_._5__.0__.0__________

_______1_0_._1_._2_5__5_._2_5__5____


subnet? ___1_0__.8_._0_._1___t_o___1_0__.8_._2_5__5__.2__5_4________

Subnets - 2 4 8 16 32 64 128 256 . .

Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

10. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

(0) 0 10.0.0.0 to 10.0.255.255(1) 1 10.1.0.0 to 10.1.255.255(2) 1 0 10.2.0.0 to 10.2.255.255(3) 1 1 10.3.0.0 to 10.3.255.255(4) 1 0 0 10.4.0.0 to 10.4.255.255

128 (5) 1 0 1 10.5.0.0 to 10.5.255.255

64 (6) 1 1 0 10.6.0.0 to 10.6.255.255

32 (7) 1 1 1 10.7.0.0 to 10.7.255.255

16 (8) 1 0 0 0 10.8.0.0 to 10.8.255.255

8 (9) 1 0 0 1 10.9.0.0 to 10.9.255.255

4 1 0 1 0 10.10.0.0 to 10.10.255.255

2 (11) 1 0 1 1 10.11.0.0 to 10.11.255.255

+1 1 1 0 0 10.12.0.0 to 10.12.255.255

255 1 1 0 1 10.13.0.0 to 10.13.255.255(14) 1 1 1 0 10.14.0.0 to 10.14.255.255

36 (15) 1 1 1 1 10.15.0.0 to

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Number of

Hosts - . 256 128 64 32 16 8 4 2

Number of

65,5 - 2 65,534

42

Problem 8

Subnetting



Default subnet mask

____2__5__5__._2__5__5__._0__.__0_________

Custom subnet mask

____2__5__5__._2__5_5___._2__2_4___. _0______









the 6th subnet?

____1_7__2_._5_0__.1__2_8_._0_______

____1_7__2_._5_0__.1__9_1_._2_5__5____


subnet? _1_7__2_._5_0__.6__4_._1___t_o___1_7_2__.5__0_._9_5__.2__5_4____

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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

172 . 50 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

(0) 0 172.50.0.0 to 172.50.31.255(1) 1 172.50.32.0 to 172.50.63.255(2) 1 0 172.50.64.0 to 172.50.95.255(3) 1 1 172.50.96.0 to 172.50.127.255(4) 1 0 0 172.50.128.0 to 172.50.159.255(5) 1 0 1 172.50.160.0 to 172.50.191.255(6) 1 1 0 172.50.192.0 to 172.50.223.255(7) 1 1 1 172.50.224.0 to 172.50.255.255

8,192 - 2 8,190

12864

+32224

44

Problem 9

Subnetting



Default subnet mask

____2__5__5__._2__5__5__._0__.__0_________

Custom subnet mask

____2__5__5__._2__5_5___._2__5_5___.

_2__2_4___





of bits borrowed _________1_1_________

What is the 2ndsubnet range? 172.50.0.32 to 172.50.0.63



the 4th subnet?

____1_7__2_._5_0__.1_._3_2_________

____1_7__2_._5_0__.0__.1__2_7_______


subnet? _1__7_2_._5_0__.0__.1_6__1___t_o__1__7_2_._5__0_.0__.1__9_0_____

(0) . 0 172.50.0.0 to 172.50.0.31(1) 1 172.50.0.32 to 172.50.0.63(2) 1 0 172.50.0.64 to 172.50.0.95(3) 1 1 172.50.0.96 to 172.50.0.127(4) 1 0 0 172.50.0.128 to 172.50.0.159(5) 1 0 1 172.50.0.160 to 172.50.0.191(6) 1 1 0 172.50.0.192 to 172.50.0.223(7) 1 1 1 172.50.0.224 to 172.50.0.255(8) 1 . 0 0 0 172.50.1.0 to 172.50.1.31(9) 1 . 0 0 1 172.50.1.32 to 172.50.1.63(10)

1 . 0 1 0 172.50.1.64 to 172.50.1.95(11) 1 . 0 1 1 172.50.1.96 to 172.50.1.127(12) 1 .

0 0 172.50.1.128 to 172.50.1.159(13) 1 . 1 0 1 172.50.1.160 to 172.50.1.191(14) 1 .

11 0 172.50.1.192 to 172.50.1.223

(15) 1 . 1

1 1 172.50.1.224 to 172.50.1.255

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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

172 . 50 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

128643216

842

+1252

32-230

12864+32224

46

Problem 10

Subnetting



Default subnet mask

____2__5__5__._2__5__5__._2__5_5___.


__

__2__5__5__._2__5__5__._2__5_5___._2__5_2__

_



addresses _________4__________ Number of usable

addresses _________2__________ Number of bits

borrowed _________6__________




the 13th subnet?

_____2_2_0__.1_0__0_.1__0_0_._1_2_____

_____2_2_0__.1_0__0_.1__0_0_._5_1_____


subnet? __2_2__0_.1_0__0_.1_0__0_._4_5___t_o__2__2_0_._1_0_0_._1_0_0_._4_6__


1 0 1 0 220.100.100.40 to 220.100.100.43(11) 1 0 1 1 220.100.100.44 to 220.100.100.47(12)

1 1 0 0 220.100.100.48 to 220.100.100.51(13)

1 1 0 1 220.100.100.52 to 220.100.100.55(14)

1 1 1 0 220.100.100.56 to 220.100.100.59(15)

1 1 1 1 220.100.100.60 to 220.100.100.63

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Number of256 128 64 32 16 8 4 2 - Hosts


128 64 32 16 8 4 2 1 - Binary values

220 . 100 . 100 . 0 0 0 0 0 0 0 0

128643216

8+4

252

4-22

48

Problem 1 1

Subnetting

Number of needed usable hosts 8,000Network Address 135.70.0.0


Default subnet mask

____2__5__5__._2__5__5__._0__.__0_________

Custom subnet mask

____2__5__5__._2__5_5___._2__2_4___. _0______









the 3rd subnet?

_____1_3_5__.7__0_._1_9_2__.0_______

_____1_3_5__.7__0_._9_5__.2__5_5_____


subnet? __1_3_5__._7_0_._1_2_8__.1___t_o___1_3_5__.7__0_.1__5_9__.2__5_4__

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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

135 . 70 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0


8,192 -2 8,190

12864

+32224

50

Problem 12

Subnetting



Default subnet mask

____2__5__5__._2__5__5__._2__5_5___.


__

__2__5__5__._2__5__5__._2__5_5___._1_9__2__

_




usable addresses _________6_2_________


What is the 2nd subnet range?

What is the subnet number for the 2nd subnet?


the 4th subnet?

198.125.50.64 to 98.125.50.127

__

__1_9_8__.1_2__5__.5__0_._6_4______

____1_9_8__.1_2__5__.5__0_._2_5__5____


subnet? _1_9__8_.1_2__5__.5__0_._1_2_9___t_o___1_9__8_.1__2_5__.5__0_._1_9_0

52


Number of

Number of256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

198 . 125 . 50 . 0 0 0 0 0 0 0 0

(0) 0 198.125.50.0 to 198.125.50.63(1) 1 198.125.50.64 to 198.125.50.127(2) 1 0 198.125.50.128 to 198.125.50.191(3) 1 1 198.125.50.192 to 198.125.50.255

128+64192

64 -2 62

51

52

Problem 13

Subnetting


Address class

____B______

Default subnet mask

____2__5__5__._2__5__5__._0__.__0_________

Custom subnet mask

__

__2__5__5__._2__5__5__._2__5_5___._1_9__2__

_


_________1_,0__2_4______ Total number of host



of bits borrowed _________1_0_________



What is the subnet broadcast address for the 1023rd subnet?

_____1_6_5__.2_0__0_.2__.1_2__8______

_____1_6_5_._2_0__0_.2__5_5__.1_9_1____

What are the assignable addresses for the 1022nd

subnet? _1_6_5__.2__0_0_._2_5_5__.6__5___to___1_6_5__.2_0__0_._2_5_5__.1_2__6

(0) . 0 165.200.0.0 to 165.200.0.63(1) 1 165.200.0.64 to 165.200.0.127(2) 1 0 165.200.0.128 to 165.200.0.191(3) 1 1 165.200.0.192 to 165.200.0.255(4) 1 0 0 165.200.1.0 to 165.200.1.63(5) 1 0 1 165.200.1.64 to 165.200.1.127(6) 1 1 0 165.200.1.128 to 165.200.1.191(7) 1 1 1 165.200.1.192 to 165.200.1.255(8) 1 0 . 0 0 165.200.2.0 to 165.200.2.63(9) 1 0 . 0 1 165.200.2.64 to 165.200.2.127(10)

1 0 . 1 0 165.200.2.128 to 165.200.2.191(11)1 0 . 1 1 165.200.2.192 to 165.200.2.255

0 0 165.200.3.0 to 165.200.3.630 1 165.200.3.64 to 165.200.3.1271 0 165.200.3.128 to 165.200.3.1911 1 165.200.3.192 to 165.200.3.255

(12)

1 1(13)

1 1(14)

1 1(15)

1 1

165.200.255.64 to 165.200.255.127165.200.155.128 to 165.200.255.191165.200.255.192 to 165.200.255.255

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.

Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

165 . 200 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

12864

3216

842

+1 252

64 128

-2 +64 62 252

.

..

(1021) 1 1 1 1 1 1 1 1 . 0 1(1022) 1 1 1 1 1 1 1 1 . 1 0(1023) 1 1 1 1 1 1 1 1 . 1 1

54

Problem 14

Subnetting



Default subnet mask

____2__5__5__._2__5__5__._2__5_5___.


____2__5__5__._2__5_5___._2__5_5___.

_2__2_4___





of bits borrowed _________3__________




the 4th subnet?

____2__0__0_._1_0_._1_0__.1_2__8_____

__

__2__0__0_._1_0_._1_ 0__.1_2__7_____


subnet? __2_0__0_._1_0_._1_0_._1_6_1___t_o___2_0__0_.1__0_._1_0_._1_9_0___

56


Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

200 . 10 . 10 . 0 0 0 0 0 0 0 0


12864 32

+3 2 -2224 30

55

56

Problem 15

Subnetting

Network Address 93.0.0.0 \19


Default subnet mask ____2__5__5__._0___.

_0__._0_____________ Custom subnet mask

____2__5__5__._2__5_5___._2__2_4___. _0______









the 7th subnet?

______9_3__.1_._0_._0___________

______9_3__.0__.2__2_3__.2__5_5_____


subnet? ___9_3__.1_._9_6_._1___t_o__9__3_.1__.1_2__7_.2__5__4________

(0) . 0 93.0.0.0 to 93.0.31.255(1) 1 93.0.32.0 to 93.0.63.255(2) 1 0 93.0.64.0 to 93.0.95.255(3) 1 1 93.0.96.0 to 93.0.127.255(4) 1 0 0 93.0.128.0 to 93.0.159.255(5) 1 0 1 93.0.160.0 to 93.0.191.255(6) 1 1 0 93.0.192.0 to 93.0.223.255(7) 1 1 1 93.0.224.0 to 93.0.255.255(8) 1 . 0 0 0 93.1.0.0 to 93.1.31.255

1 93.1.32.0 to 93.1.63.2550 93.1.64.0 to 93.1.95.2551 93.1.96.0 to 93.1.127.2550 93.1.128.0 to 93.1.159.2551 93.1.160.0 to 93.1.191.2550 93.1.192.0 to 93.1.223.255

(9) 1 . 0 0(10) 1 . 0 1(11) 1 . 0 1(12) 1 .

1 0(13) 1 . 1 0(14) 1 . 1 1

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Number of

Hosts -. 256 128 64 32 16 8 4 2

Number of

Subnets - 2 4 8 16 32 64 128 256 . .

Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

93. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

12864

3216

842

+1255

12864

+32224

8,192 - 2 8,190

58

Practical Subnetting 1Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnet s , and allow enough extra subnets and hosts for 100% growth in both areas. Circle each subnet on the graphic and answer the questions below.

Marketing 24 Hosts

F0/0Router A

IP Address 172.16.0.0

S0/0/0 S0/0/1

F0/0

Reasearch 60 Hosts

Router B

F0/1

Management 15 Hosts

Address class

Custom subnet mask

Minimum number of subnets needed

Extra subnets required for 100% growth(Round up to the next whole number)

Total number of subnets needed

Number of host addressesin the largest subnet group

Number of addresses needed for 100% growth in the largest subnet

(Round up to the next whole number)

Total number of address needed for the largest subnet

____________

__B____________________

__2__5_5_._2_5__5_._2_2_4__.

0________

____4_____

_+___4_____

_=___8_____

____6_0____

_+___6_0____

_=__1_2_0____

Start with the first subnet and arrange your sub-networks from the largest group to the smallest.

IP address range for Research

IP address range for Marketing

IP address range for Management

IP address range for Router A to Router B serial connection _

__1_7_2__.1_6_._0_._0__t_o_1_7_2__.3_1__.2_5__5___

__1__7_2_.1_6__.3_2__.0__t_o__1_7_2_._6_3_._2_5__5__

__1_7__2_.1_6__.6_4__.0__t_o__1_7_2_._9_5_._2_5__5__

__1_7_2__.1_6_._9_6_._0__t_o_1_7_2__.1_2__7_.2_5__5__


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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

172 . 16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

4x1.0

4

60 x1.0 60

60

Practical Subnetting 2Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subn e t , and allow enough extra subnets andhosts for 30% growth in all areas. Circle each subnet on the graphic and answer the questions below.

F0/0Router A

IP Address 135.126.0.0S0/0/0

S0/0/1

S0/0/1 F0/0 Router B F0/1

Science Lab 10 Hosts

S0/0/0

F0/1Router C

Tech Ed Lab 20 Hosts

Address class

Custom subnet mask








English Department 15 Hosts

______________B_______________

______2_5__5_._2_5_5__.2__5_5__.2_2__4______

_____5____

_+____2____

_=____7____

____2_0____

_+___6_____

_=___2_6____


IP address range for Tech Ed

IP address range for English

IP address range for Science

IP address range for Router A to Router B serial connection

IP address range for Router A

_1_3__5_._1_2_6_._0_.0___to__1_3_5__.1_2__6_.0__.3_1__

_1_3_5_._1_2_6_._0_._3_2__t_o__1_3_5_._1_2_6__.0_._6_3_

_1_3_5_._1_2_6_._0_._6_4__t_o__1_3_5_._1_2_6_._0_._9_5_

1_3__5_.1_2__6_.0__.9_6__t_o__1_3_5__.1_2__6_.0__.1_2_7_

to Router B serial connection1_3_5_._1_2_6_._0_._1_2_8__t_o__1_3_5_._1_2_6_._0_.1__5_9

0 135.126.0.0 to 135.126.0.311 135.126.0.32 to 135.126.0.63

1 0 135.126.0.64 to 135.126.0.951 1 135.126.0.96 to 135.126.0.127

1 0 0 135.126.0.128 to 135.126.0.1591 0 1 135.126.0.160 to 135.126.0.1911 1 0 135.126.0.192 to 135.126.0.2231 1 1 135.126.0.224 to 135.126.0.2550 0 0 135.126.1.0 to 135.126.1.310 0 1 135.126.1.32 to 135.126.1.630 1 0 135.126.1.64 to 135.126.1.950 1 1 135.126.1.96 to 135.126.1.1271 0 0 135.126.1.128 to 135.126.1.1591 0 1 135.126.1.160 to 135.126.1.1911 1 0 135.126.1.192 to 135.126.1.2231 1 1 135.1261.224 to 135.126.1.255

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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

135. 126 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

5 x.3 1.5

(Round up to 2)

20 x.3

6

(0) .(1)(2)

(3)(4)(5)(6)(7)(8) 1 .(9) 1 .(10) 1 .(11) 1 .

(12) 1 .(13) 1 .(14) 1 .(15) 1 .

62

Practical Subnetting 3Based on the information in the graphic shown, design a classfull network addressing scheme that will supply the minimum number of hosts per subn e t , and allow enough extra subnets and hosts for 25% growth in all areas. Circle each subnet on the graphic and answer the questions below.


Administrative

F0/0

Router A

S0/0/1F0/0 Sales

185 Hosts30 Hosts F0/1 S0/0/0

Router B

Marketing 50 Hosts

Address class

Custom subnet mask








______________B_______________

_______2__5_5__.2_5__5_._2_5__5_.

_0_______

____4_____

_+___1_____

_=___5_____

___1_8_5____

_+___4_7____

_=__2_3__2___


IP address range for Sales


IP address range for Administrative


__1_7_2__.1_6__.0_._0__t_o_1_7__2_.1_6__.0_._2_5__5__

__1_7_2__.1_6_._1_.0___t_o__1_7

_2_._1_6_._1_.2_5__5__

__1_7_2_._1_6_.2__.0___t_o__1_7

_2_._1_6_._2_.2__5_5__

__1_7_2_._1_6_.3__.0___t_o__1_7

_2__.1_6_._3_.2__5_5__

(0)(1)(2)(3)(4)(5)(6)(7)(8) 1(9) 1(10)

1(11) 1(12)

1(13)

1(14)

1(15)

1

0 172.16.0.0 to 1172.16.0.2551 172.16.1.0 to 1172.16.1.255

1 0 172.16.2.0 to 1172.16.2.2551 1 172.16.3.0 to 1172.16.3.255

1 0 0 172.16.4.0 to 1172.16.4.2551 0 1 172.16.5.0 to 1172.16.5.2551 1 0 172.16.6.0 to 1172.16.6.2551 1 1 172.16.7.0 to 1172.16.7.2550 0 0 172.16.8.0 to 1172.16.8.2550 0 1 172.16.9.0 to 1172.16.9.2550 1 0 172.16.10.0 to 1172.16.10.2550 1 1 172.16.11.0 to 1172.16.11.2551 0 0 172.16.12.0 to 1172.16.12.2551 0 1 172.16.13.0 to 1172.16.13.2551 1 0 172.16.14.0 to 1172.16.14.2551 1 1 172.16.15.0 to 1172.16.15.255

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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

172. 16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

.

4x..25

1

225 .x.2 5 .

56.25 .(Round up to 57) .

.

.

.

.

64

Practical Subnetting 4Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnet s , and allow enough extra subnets and hosts for 70% growth in all areas. Circle each subnet on the graphic and answer the questions below.


F0/0 S0/0/0Router A

S0/0/1

S0/0/1

F0/0 Router B

S0/0/0

DallasF0/1

Router C F0/0

150 Hosts New York

Washington D.C.220 Hosts

325 Hosts

Address class

Custom subnet mask








______________B_______________

_______2__5_5__.2__5_5_._2_4__0_.0________

____5_____

_+___4_____

_=___9_____

___3_2__5___

_+__2_2__8___

_=__5_5__3___


IP address range for New York 1

_3__5_._1_2_6_._0_.0__t_o__1_3_5__.1_2__6_.1_5__.2__5_5

IP address range for Washington D. C. 1_3_5__.1_2__6_.1_6__.0__t_o__1_3_5_._1_2_6_._3_1_._2_5_5

IP address range for Dall

as1_3_5__.1_2_6__.3_2__.0__t_o__1_3_5__.1_2_6__.4_7_.

_2_5_ 5

IP address range for Router Ato Router B serial connection

1_3_5__.1_2_6__.4_8_._0__t_o__1_3_5_._1_2_6_._6_3_._2_5

_ 5

IP address range for Router Ato Router C serial connection 1_3_5__.1_2_6__.6_4__.0__t_o__1_3_5__.1_2_6__.7_9_._2_5_5

1 0 0 135.126.192.0 to 135.126.207.2551 0 1 135.126.208.0 to 135.126.223.2551 1 0 135.126.224.0 to 135.126.239.2551 1 1 135.126.240.0 to 135.126.1255.255

(12)

1(13)

1(14)

1(15)

1

.

.

.

0 0 0 135.126.128.0 to 135.126.143.2550 0 1 135.126.144.0 to 135.126.159.2550 1 0 135.126.160.0 to 135.126.175.255

(11) 1 . 0 1 1 135.126.176.0 to 135.126.191.255

(8) 1(9) 1(10)

1

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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

135. 126 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

(0) . 0 135.126.0.0 to 135.126.15.255(1) 1 135.126.16.0 to 135.126.31.255(2) 1 0 135.126.32.0 to 135.126.47.255(3) 1 1 135.126.48.0 to 135.126.63.255(4) 1 0 0 135.126.64.0 to 135.126.79.255(5) 1 0 1 135.126.80.0 to 135.126.95.255(6) 1 1 0 135.126.96.0 to 135.126.111.255(7) 1 1 1 135.126.112.0 to 135.126.127.255

.

...

66

Practical Subnetting 5Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subne t , and allow enough extra subnets and hosts for 100% growth in all areas. Circle each subnet on the graphic and answer the questions below.


F0/0F0/1

Science Room 10 Hosts

Tech Ed Lab 18 Hosts

English classroom 15 Hosts Art Classroom

12 Hosts

Address class

Custom subnet mask








______________C_______________

______2__5_5_._2_5__5_._2_5_5__.1_9__2______

____2_____

_+___2_____

_=___4_____

___3__0____

_+__3__0____

_=__6__0____


IP address range for Router F0/0 Port

IP address range for Router F0/1 Port

__2_1_0_._1_5_._1_0_._0__t_o_2__1_0_.1_5__.1_0__.6_3__

_2_1_0_._1_5_._1_0_._6_4__t_o_2

__1_0_._1_5_.1_0__.1_2__7

68


Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

210. 15 . 10 . 0 0 0 0 0 0 0 0

(0) 0 210.15.10.0 to 210.15.10.63(1) 1 210.15.10.64 to 210.15.10.127(2) 1

0 210.15.10.128 to 210.15.10.191(3) 1 1 210.15.10.192 to 210.15.10.255

67

68


Router A

F0/0

S0/0/0

S0/0/1

IP Address 10.0.0.0

S0/0/1

S0/0/0 Router B

F0/1

Technology Building 320 Hosts

S0/0/0Art & Drama

S0/0/1

Router CAdministration

75 Hosts F0/0 F0/135 Hosts

Science Building 225 Hosts

Address class

Custom subnet mask




______________A_______________

_________2_5__5_._2_4_0__.0_._0_________

_____7____

_+____2____

_=____9____


IP address range for Technology

IP address range for Science

IP address range for Arts & Drama

IP Address range Administration



to Router C serial connection

IP address range for Router B to Router C serial connection

__1_0_._0_._0_.0__t_o__1_0_._1_5

_._2_5__5_._2_5_5___

__1_0_._1_6_.0__.0__t_o__1_0_.3__1_.2__5_5_._2_5__5__

_1_0__.3_2__.0__.0__t_o__1_0_._4_7_.2__5_5__.2__5_5__

__1_0_.4__8_.0_._0__t_o__1_0_._6

_3_.2__5_5__.2_5__5__

__1_0_.6__4_.0__.0__t_o__1_0_._7

_9_.2__5_5__.2_5__5__

_1_0__.8_0__.0_._0__t_o__1_0_.9__5_.2__5_5__.2__5_5__

_1_0__.9_6_._0_._0__t_o_1_0__.1_1_1_._2_5_5__.2__5_5__

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Number of

Hosts -. 256 128 64 32 16 8 4 2

Number of

Subnets - 2 4 8 16 32 64 128 256 . .

Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

10. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0


1 0 1 0 10.160.0.0 to 10.175.255.255(11) 1 0 1 1 10.176.0.0 to 10.191.255.255(12)

1 1 0 0 10.192.0.0 to 10.207.255.255(13)

1 1 0 1 10.208.0.0 to 10.223.255.255(14)

1 1 1 0 10.224.0.0 to 10.239.255.255(15)

1 1 1 1 10.240.0.0 to 10.255.255.255

70

Practical Subnetting 7Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subn e t , and allow enough extra subnets and hosts for 125% growth in all areas. Circle each subnet on the graphic and answer the questions below.

IP Address 177.135.0.0S0/0/0

Router A

F0/0S0/0/0 F0/0

Router B

F0/1

Marketing 75 Hosts

Administration33 Hosts Sales

255 Hosts

Research 135 Hosts

Deployment 63 Hosts

Address class

Custom subnet mask








______________B_______________

_______2__5_5__.2_5__5_._2_5__2_.0________

____4_____

_+___5_____

_=___9_____

___3_6__3___

_+__4_5__4___

_=__8_1_7____


IP address range for Router A Port F0/0

IP address range for Research

IP address range for Deployment


_1_7_7_._1_3_5__.0_._0__t_o_1_7__7_.1_3_5__.3_._2_5__5_

_1_7_7__.1_3_5__.4_._0__t_o_1__7_

7_.1_3_5__.7_._2_5__5_

_

1_7_7_._1_3_5_._8_.0__t_o__1_7_7

_.1__3_5_._1_1_.2_5__5_to Router B serial connection 1_7_7_._1_3_5__.1_2_._0__t_o_1_7_7__.1_3_5__.1_5__.2_5__5

(0)(1)(2)(3)(4)(5)(6)(7)(8) 1(9) 1

(10) 1(11) 1(12) 1(13) 1(14) 1(15) 1

0 177.135.0.0 to 177.135.3.2551 177.135.4.0 to 177.135.7.255

1 0 177.135.8.0 to 177.135.11.2551 1 177.135.12.0 to 177.135.15.255


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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

177.135 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

.

.

.

.

.

.

.

.

.

72

Practical Subnetting 8Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number subnet s , and allow enough extra subnets and hosts for 85% growth in all areas. Circle each subnet on the graphic and answer the questions below.

IP Address 192.168.1.0F0/0 S0/0/0

Router A S0/0/1 F0/1

F0/0 Router B

New York 8 Hosts

Boston 5 Hosts

Research & Development 8 Hosts

Address class

Custom subnet mask








______________C_______________

______2_5__5_._2_5_5__.2__5_5__.2_2__4______

____3_____

_+___3_____

_=___6_____

____1_3____

_+___1_2____

_=___2_5____


IP address range for Router A F0/0 IP address range for New York


__1_9_2__.1_6__8_.1_._0__to__1_9_2__.1_6_8_._1_.3__1__

__1_9_2_.1__6_8_.1_._3_2__t_o__1_9_2_._1_6_8_.1_._6_3__

_1_9__2_.1_6__8_.1_._6_4__t_o__1_9_2_._1_6_8_.1_._9_5__

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Number of

Number of 256 128 64 32 16 8 4 2 - Hosts

Subnets - 2 4 8 16 32 64 128 256

128 64 32 16 8 4 2 1 - Binary values

192. 168 . 1 . 0 0 0 0 0 0 0 0


73

Practical Subnetting 9Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subn e t , and allow enough extra subnets andhosts for 15% growth in all areas. Circle each subnet on the graphic and answer the questions below.

Router A

IP Address 148.55.0.0S0/0/0

S0/0/1 F0/1

S0/0/1 F0/0 Router B

S0/0/0

Router C

Dallas 1500 Hosts

F0/0

Router D S0/0/0S0/0/1

Ft. Worth 2300 Hosts Address class

Custom subnet mask

______________B_______________

_______2__5_5__.2__5_5_._2_4__0_.0

________








____5_____

_+___1_____

_=___6_____

__2__3_0_0___

_+__3_4__5___

_=_2__6_4_5___


IP address range for Ft. Worth

IP address range for Dallas


IP address range for Router A to Router C serial connection

IP address range for Router C74 to Router D serial connection

_1_4__8_.5__5_.0__.0_.__t_o_1_4__8_.5__5_._1_5_.2__5_5_

_1_4_8_._5_5__.1_6_._0_._t_o__1

_4_8_._5_5__.3_1_._2_5__5

_1_4_8_.5__5_._3_2_._0_._t_o__1

_4_8_._5_5__.4__7_.2_5__5

_1_4_8_._5_5_._4_8_._0_._t_o__1

_4_8_._5_5__.6_3__.2_5__5

_1_4_8_._5_5_._6_4_._0_._t_o__1

_4_8__.5_5__.7_9__.2_5__5

(11) 1 . 0 1 1 148.55.176.0 to 148.55.191.255(12)

1 .1 0 0 148.55.192.0 to 148.55.207.2551 0 1 148.55.208.0 to 148.55.223.2551 1 0 148.55.224.0 to 148.55.239.2551 1 1 148.55.240.0 to 148.55.255.255

(13)

1 .(14)

1 .(15)

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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

148. 55 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

(0) . 0 148.55.0.0 to 148.55.15.255(1) 1 148.55.16.0 to 148.55.31.255(2) 1 0 148.55.32.0 to 148.55.47.255(3) 1 1 148.55.48.0 to 148.55.63.255(4) 1 0 0 148.55.64.0 to 148.55.79.255(5) 1 0 1 148.55.80.0 to 148.55.95.255(6) 1 1 0 148.55.96.0 to 148.55.111.255(7) 1 1 1 148.55.112.0 to 148.55.127.255(8) 1 . 0 0 0 148.55.128.0 to 148.55.143.255(9) 1 . 0 0 1 148.55.144.0 to 148.55.159.255(10)

1 . 0 1 0 148.55.160.0 to 148.55.175.255

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Sales115 Hosts


S0/0/0

Marketing 56 Hosts

F0/0F0/0 Router A S0/0/1

Router B

F0/1

Management 25 Hosts

Research 35 Hosts

Address class

Custom subnet mask








______________B_______________

______2_5__5_._2_5__5_.2__5_5__.2__4_0______

_____4____

_+____5____

_=____9____

___1_4_0____

_+__1_5_4____

_=__2_9_4____


IP address range for Sales/Managemnt


IP address range for

Research

IP address range for Router A to Router B serial connection _

_1_7_2__.1_6_._0_.0__t_o__1_7_2__.1_6_._1_5_._2_5_5__

__1_7_2_._1_6_.1_6__.0__t_o__1_7_2_._1_6_.3_1__.2_5__5_

_1_7__2_.1_6__.3_2__.0__t_o__1_7_2_._1_6_.4__7_.2__5_5_

__1_7_2_.1_6__.4_8__.0__t_o__1_7_2_._1_6_.6__3_.2__5_5_

(0)(1)(2)(3)(4)(5)(6)(7)(8) 1(9) 1(10)

1(11) 1(12)

1(13)

1(14)

1(15)

1

0 172.16.0.0 to 172.16.15.2551 172.16.16.0 to 172.16.31.255

1 0 172.16.32.0 to 172.16.47.2551 1 172.16.48.0 to 172.16.63.255


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Number of Hosts - . 256 128 64 32 16 8 4 2


Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

172.16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

.

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.

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Valid and Non-Valid IP Addresses

Using the material in this workbook identify which of the addresses below are correct and usable. If they are not usable addresses explain why.

IP Address: 0.230.190.192 _T__h_e_n_e_t_w__o_r_k__I_D___c_a_n_n_o_t__b_e__0__. __ Subnet Mask: 255.0.0.0 ________________________________ Reference Page Inside Front Cover

IP Address: 192.10.10.1 _O__K_____________________________ Subnet Mask: 255.255.255.0 ________________________________ Reference Pages 28-29

IP Address: 245.150.190.10 _2__4_5__i_s__r_e_s_e_r_v_e_d___f_o_r___________ Subnet Mask: 255.255.255.0 _e_x_p_e_r_i_m__e_n_t_a_l_u_s_e_._______________ Reference Page Inside Front Cover

IP Address: 135.70.191.255 _T__h_is__i_s__t_h_e__b_r_o_a_d__c_a_s_t__a_d_d_r__e_s_s_ Subnet Mask: 255.255.254.0 _f_o_r__t_h_i_s__r_a_n_g_e_._________________ Reference Pages 48-49

IP Address: 127.100.100.10 _1_2__7__i_s_r__e_s_e_r_v_e_d__f_o_r__l_o_o_p_b_a__c_k__ Subnet Mask: 255.0.0.0 _t_e_s_t_i_n_g_.________________________ Reference Pages Inside Front Cover


IP Address: 200.10.10.128 _T_h_i_s_i_s_t_h_e__s_u_b_n_et__a_d_d_r_e_s_s__fo_r__t_h_e__ Subnet Mask: 255.255.255.224 _3_r_d__u_s_a_b_le__r_a_n_ge__o_f__2_0_0_._1_0_._1_0_.0____Reference Pages 54-55


IP Address: 190.35.0.10 _T_h_is__a_d_d_r_e_s_s_i_s_t_a_k_e_n_f_r_o_m__th_e__f_ir_s_t__ Subnet Mask: 255.255.255.192

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_r_a_ng_e__f_or__t_h_is__s_ub_n_e_t__w_h_ic_h__is__in_v_a_l_id_._ Reference Pages 34-35

IP Address: 218.35.50.195 _T_h_i_s__h_a_s__a__c_l_a_s_s__B__s_u_b_n__e_t______ Subnet Mask: 255.255.0.0 _m_a__s_k_.__________________________ Reference Page Inside Front Cover

IP Address: 200.10.10.175 /22 _A__c_l_a_s_s__C__a_d__d_r_e_s_s__m__u_s_t__u_s_e__a__Reference Pages 54-55 and/or Inside Front Cover

_m_i_n_i_m__u_m__o_f__2__4__b_i_t_s_.___________

IP Address: 135.70.255.255 _T_h_i_s__i_s__a__b_r_o_a_d_c_a__s_t_a__d_d_r_e_s_s__. __ Subnet Mask: 255.255.224.0 ________________________________ Reference Pages 48-49

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IP Address Breakdown/24 /25 /26 /27 /28 /29 /30

8+8+8255.255.255.0

256 Hosts

8+8+8+1255.255.255.128

128 Hosts

8+8+8+2255.255.255.192

64 Hosts

8+8+8+3255.255.255.224

32 Hosts

8+8+8+4255.255.255.240

16 Hosts

8+8+8+5255.255.255.248

8 Hosts

8+8+8+6255.255.255.252

4 Hosts

0-255

0-127

0-63

0-150-7

0-34-7

8-158-11

12-15

16-3116-23

16-1920-23

24-3124-2728-31

32-4732-39

32-3536-39

40-4740-4344-47

48-6348-55

48-5152-55

56-6356-5960-63

64-127

64-7964-71

64-6768-71

72-7972-7576-79

80-9580-87

80-8384-87

88-9588-9192-95

96-11196-103

96-99100-103

104-111104-107108-111

112-127112-119

112-115116-119

120-127120-123124-127

128-255

128-191

128-143128-135

128-131132-135

136-143136-139140-143

144-159144-151

144-147148-151

152-159152-155156-159

160-17516-167

160-163164-167

168-175168-171172-175

176-191176-183

176-179180-183

184-191184-187188-191

192-255

192-207192-199

192-195196-199

200-207200-203204-207

208-223208-215

208-211212-215

216-223216-219220-223

224-239224-231

224-227228-231

232-239232-235236-239

240-255240-247

240-243244-247

248-255248-251252-255

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Visualizing Subnets Using The Box Method

The box method is the simplest way to visualize the breakdown of subnets and addresses into smaller sizes.

Start with a square. The whole square is a single subnet comprised of 256addresses.

/24255.255.255.0

256 Hosts 1 Subnet

Split the box in half and you get two subnets with 128 addresses,

/25255.255.255.128

128 Hosts 2 Subnets

Divide the box into quarters and you get four subnets with 64 addresses,

/26255.255.255.192

64 Hosts4 Subnets

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Split each individual square and you get eight subnets with 32 addresses,

/27255.255.255.224

32 Hosts8 Subnets

Split the boxes in half again and you get sixteen subnets with sixteenaddresses,

/28255.255.255.240

16 Hosts16 Subnets

The next split gives you thirty two subnets with eight addresses,

/29255.255.255.248

8 Hosts32 Subnets

The last split gives sixty four subnets with four addresses each,

/30255.255.255.252

4 Hosts64 Subnets

CIDR# of Bits

BorrowedSubnet Mask

Total # of Subnets

Total # of Hosts

Usable # of Hosts

/24 0 255.255.255.0 1 256 254/25 1 255.255.255.128 2 128 126/26 2 255.255.255.192 4 64 62/27 3 255.255.255.224 8 32 30/28 4 255.255.255.240 16 16 14/29 5 255.255.255.248 32 8 6/30 6 255.255.255.252 64 4 2

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Class A Addressing Guide

CIDR# of Bits

BorrowedSubnetMask

Total # ofSubnets

Total # ofHosts

Usable # ofHosts

/8 0 255.0.0.0 1 16,777,216 16,777,214/9 1 255.128.0.0 2 8,388,608 8,388,606

/10 2 255.192.0.0 4 4,194,304 4,194,302/11 3 255.224.0.0 8 2,097,152 2,097,150/12 4 255.240.0.0 16 1,048,576 1,048,574/13 5 255.248.0.0 32 524,288 524,286/14 6 255.252.0.0 64 262,144 262,142/15 7 255.254.0.0 128 131,072 131,070/16 8 255.255.0.0 256 65,536 65,534/17 9 255.255.128.0 512 32,768 32,766/18 10 255.255.192.0 1,024 16,384 16,382/19 11 255.255.224.0 2,048 8,192 8,190/20 12 255.255.240.0 4,096 4,096 4,094/21 13 255.255.248.0 8,192 2,048 2,046/22 14 255.255.252.0 16,384 1,024 1,022/23 15 255.255.254.0 32,768 512 510/24 16 255.255.255.0 65,536 256 254/25 17 255.255.255.128 131,072 128 126/26 18 255.255.255.192 262,144 64 62/27 19 255.255.255.224 524,288 32 30/28 20 255.255.255.240 1,048,576 16 14/29 21 255.255.255.248 2,097,152 8 6/30 22 255.255.255.252 4,194,304 4 2

Class B Addressing Guide

CIDR# of Bits

BorrowedSubnetMask

Total # ofSubnets

Total # ofHosts

Usable # ofHosts

/16 0 255.255.0.0 1 65,536 65,534/17 1 255.255.128.0 2 32,768 32,766/18 2 255.255.192.0 4 16,384 16,382/19 3 255.255.224.0 8 8,192 8,190/20 4 255.255.240.0 16 4,096 4,094/21 5 255.255.248.0 32 2,048 2,046/22 6 255.255.252.0 64 1,024 1,022/23 7 255.255.254.0 128 512 510/24 8 255.255.255.0 256 256 254/25 9 255.255.255.128 512 128 126/26 10 255.255.255.192 1,024 64 62/27 11 255.255.255.224 2,048 32 30/28 12 255.255.255.240 4,096 16 14/29 13 255.255.255.248 8,192 8 6/30 14 255.255.255.252 16,384 4 2

Class C Addressing Guide

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Inside Cover

Ip Addressing and Subnetting Workbook -Desarrollado Todo El Documento PDF (1)

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