IP Addressing and Subnetting Workbook - Instructors Version 1.5.docx

1010100 10001111100 1011100101011100 101100011101001 1011110100011010 00001010010110010 1001010101100111 1111010101000101 1101001101010011 001010010101010 1010101000110010 010101001011000 110101100011010 11010100001011 001010100110 1001010010

IP Addressin an Subnettin Workbo Versi

Instructor’s Editi

11111110 10010101 00011011

1000

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 .

Class B

128 – 128 – 191 191 Leading bit pattern 10

Class D Class E

.

Host

.

Host

.

Host

.

Host

10000000.00000000.00000000.00000000 Network .

Class C

Host

192 – 192 – 223 223 Leading bit pattern 110

Network .

Host

11000000.00000000.00000000.00000000 Network .

Network

. Network

224 – 224 – 239

( Reserved for multicast )

240 – 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] Frederick County Career & Technology Center Cisco Networking Academy Frederick 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 Workbooks VLSM 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 you are giving everyone else else worldwide the answers. Yes, students look for answers answers this way. It also discourages others; myself included, from posting high quality materials. Inside Cover

Binary To Decimal Conversion 128 64 32 16 8 4 2 1 Answers

Scratch Area

1

1

1

1

0

1

1

0

0

0

0

1

0

0

1

1

1

0

0

0

0

0

0

1

246 19 129

1

0

0

1

0

0

1

0

146

0

1

1

1

0

1

1

1

119

128 16 2

64 32 16 146

1

1

1

1

1

1

1

1

255

4

2 1

1

1

0

0

0

1

0

1

197

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

00011011

27

10101010 01101111 11111000

248

119

00111110

170 111 62 3

00000011 00100000 32 01010101 85

11101101

237

11000000

192

Decimal To Binary Conversion Conversion 1

Use all 8 bits for each problem

128 64 32 16 8 4 2 1 = 255 Scratch Scratch Area ______________________ _____________ _____________ _____________ ______ 1 1 0 1 1 1 0 238 1 _______________

____________________ ______________ _______________ _____________ ______ 0 1 0 0 0 1 0 34 0 ______________ ______________________ _____________ _____________ _____________ ______ 11 0 _______________ 123 1 1

1

0

0 1 1 0 0 1 _____________________ _____________ ______________ _____________ ______ 50 0 _______________

1

1 1 1 1 1 1 _____________________ _____________ ______________ _____________ ______ 255 1 _______________

1 1 0 0 1 0 0 ____________________ ______________ _______________ _____________ ______ 200 0 ______________ 0 0 0 0 1 0 1 ____________________ ______________ _______________ _____________ ______ 10 0 ______________

2

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 ____________________ ______________ ______________ ______________ _______ 138 0 _____________ 0

1

1

0

1

0

1

1

____________________ ______________ ______________ ______________ _______ 13 1 _____________ 1

1

1

0

0

0

0

250

____________________ ______________ ______________ ______________ _______ 107 0 _____________ 0

1

1

1

0

0

1

224

____________________ ______________ ______________ ______________ _______ 114 0 _____________ 192 1 1 0 0 0 0 0 ____________________ ______________ ______________ ______________ _______ 172 0 _____________ 1 0 1 0 1 1 0 100 ____________________ ______________ ______________ ______________ _______ 0 _____________

119 0

1

1

0

0

1

0

57 ____________________ ______________ ______________ ______________ _______ 0 _____________ 98 0

1

1

1

0

1

1

179 ____________________ ______________ ______________ ______________ _______ 1 _____________ 2

3

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 ______________

Address Class Identification Address Class

4

10.250.1.1

_____ A

150.10.15.0

_____ B

192.14.2.0

_____ C C

148.17.9.1

_____ B

193.42.1.1

_____ C C

126.8.156.0

_____ A

220.200.23.1 _____ C C 230.230.45.58 230.230.45. 58 _____ D 177.100.18.4 _____ B 119.18.45.0

_____ A

249.240.80.78 249.240.80. 78 _____ E E 199.155.77.56 199.155.77. 56 _____ C C 117.89.56.45 _____ A 215.45.45.0 _____ C 199.200.15.0 _____ C C

95.0.21.90

_____ A

33.0.0.0 _____ A 158.98.80.0

_____ B

219.21.56.0

_____ C C

Network & Host Identification Circle the network

192.200.15.0

portion of these addresses:

177.100.18.4

95.0.21.90 33.0.0.0

119.18.45.0

158.98.80.0

209.240.80.78

217.21.56.0

199.155.77.56

10.250.1.1

117.89.56.45

150.10.15.0

215.45.45.0

192.14.2.0 5

148.17.9.1

155.25.169.227

193.42.1.1

192.15.155.2

126.8.156.0

123.102.45.254

220.200.23.1

148.17.9.155

Circle the host portion of these addresses:

100.25.1.1

10.15.123.50

195.0.21.98

171.2.199.31

25.250.135.46

198.125.87.177

171.102.77.77

223.250.200.222

55.250.5.5

17.45.222.45

218.155.230.14

126.201.54.231

10.250.1.1

191.41.35.112

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

6

__________________ ____________________________ ___________ _ 188 188 . 10 . 0 . 0

__________________ ____________________________ ___________ _ 10 10 188.10.18.2 255.255.0.0 10.10.48.80 255.255.255.0 192.149.24.191 255.255.255.0

. 10 . 48 . 0

__________________ ____________________________ ___________ _ 192 192 . 149 149 . 24 . 0

__________________ ____________________________ ___________ _ 150 150 . 203 203 . 0 . 0

150.203.23.19 255.255.0.0 10.10.10.10 255.0.0.0 186.13.23.110 255.255.255.0 223.69.230.250 255.255.0.0 200.120.135.15 255.255.255.0 27.125.200.151 255.0.0.0 199.20.150.35 255.255.255.0 191.55.165.135 255.255.255.0

__________________ ____________________________ ___________ _ 10 10 . 0. 0. 0

__________________ ____________________________ ___________ _ 186 186 . 13 . 23 . 0

__________________ ____________________________ ___________ _ 223 223 . 69 . 0 . 0

__________________ ____________________________ ___________ _ 200 200 . 120 120 . 135 . 0

__________________ ____________________________ ___________ _ 27 27 . 0. 0. 0

__________________ ____________________________ ___________ _ 199 199 . 20 . 150 . 0

28.212.250.254

7

____________________ _____________________________ _________ 191 191 . 55 . 165 165 . 0

___________________ _____________________________ __________ 28 28 . 212 . 0 . 0

255.255.0.0

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

8

___________________ _____________________________ __________ 0 . 0 . 18 . 2

___________________ _____________________________ __________ 0 . 0 . 0 . 80

188.10.18.2 255.255.0.0

___________________ _____________________________ __________ 0 . 0 . 0 . 11

10.10.48.80 255.255.255.0 222.49.49.11 255.255.255.0 128.23.230.19 255.255.0.0 10.10.10.10 255.0.0.0 200.113.123.11 255.255.255.0 223.169.23.20 255.255.0.0 203.20.35.215 255.255.255.0

___________________ _____________________________ __________ 0 . 0 . 230 . 19

___________________ _____________________________ __________ 0 . 10 . 10 . 10

___________________ _____________________________ __________ 0 . 0 . 0 . 11

___________________ _____________________________ __________ 0 . 0 . 23 . 20

___________________ _____________________________ __________ 0 . 0 . 0 . 215

117.15.2.51 255.0.0.0 199.120.15.135 255.255.255.0 191.55.165.135 255.255.255.0

___________________ _____________________________ __________ 0 . 15 . 2 . 51

___________________ _____________________________ __________ 0 . 0 . 0 . 135

48.21.25.54

9

____________________ _____________________________ _________ 0 . 0 . 0 . 135

____________________ _____________________________ _________ 0 . 0 . 25 . 54

255.255.0.0

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

10

__________________ ____________________________ ___________ _ 255 255 . 255 255 . 0 . 0

177.100.18.4 119.18.45.0 191.249.234.191 223.23.223.109 10.10.250.1 126.123.23.1 223.69.230.250 192.12.35.105

__________________ ____________________________ ___________ _ 255 255 . 0. 0. 0

__________________ ____________________________ ___________ _ 255 255 . 255 255 . 0 . 0

__________________ ____________________________ ___________ _ 255 255 . 255 . 255 . 0

__________________ ____________________________ ___________ _ 255 255 . 0. 0. 0

77.251.200.51 __________________ ____________________________ ___________ _ 255 255 189.210.50.1 . 0. 0. 0

88.45.65.35 __________________ ____________________________ ___________ _ 255 255 128.212.250.254 . 255 255 . 255 . 0

193.100.77.83 125.125.250.1

__________________ ____________________________ ___________ _ 255 255 . 255 255 . 255 . 0

1.1.10.50 __________________ ____________________________ ___________ _ 255 255 220.90.130.45 . 0. 0. 0

134.125.34.9 __________________ ____________________________ ___________ _ 255 255 95.250.91.99 11

. 255 . 0 . 0

___________________ _____________________________ __________ 255 255 . 0. 0. 0

____________________ _____________________________ _________ 255 255 . 255 . 0 . 0

___________________ _____________________________ __________ 255 255 . 255 . 255 . 0

___________________ _____________________________ __________ 255 255 . 0. 0. 0

___________________ _____________________________ __________ 255 255 . 0. 0. 0

___________________ _____________________________ __________ 255 255 . 255 . 255 . 0

___________________ _____________________________ __________ 255 255 . 255 . 0 . 0

___________________ _____________________________ __________ 255 255 . 0. 0. 0

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 12

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 255.0.0.0 Class B 255.255.0.0 Class C 255.255.255.0 ANDING Equations: 1 AND 1 = 1 1 AND 0 = 0 0 AND 1 = 0 0 AND 0 = 0 Sample:

What you see... IP Address: 192 . 100 . 10 . 33 What you can figure out in your head... Address Class: Class: C Network Portion: 192 . 100 . 10 . 33 Host 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 . 100 100 . 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 with the default default subnet subnet mask allows allows your your computer computer to figure figure out the network portion of the address.

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 13

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 . 0 Custom Subnet Mask: 255.255.255.240 Address Ranges: Ranges: 192.10.10.0

to 192.100.10.16 192.100.10.32 192.100.10.48 192.100.10.64 192.100.10.80 192.100.10.96 192.100.10.112 192.100.10.128 192.100.10.144 192.100.10.160 192.100.10.176 192.100.10.192 192.100.10.208 192.100.10.224 192.100.10.240

192.100.10.15 to 192.100.10.31 to 192.100.10.47 (Range in the sample below ) to 192.100.10.63 to 192.100.10.79 to 192.100.10.95 to 192.100.10.111 to 192.100.10.127 to 192.100.10.143 to 192.100.10.159 to 192.100.10.175 to 192.100.10.191 to 192.100.10.207 to 192.100.10.223 to 192.100.10.239 to 192.100.10.255

Sub Network 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)

Custom 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 . 1 1 1 1 0 0 0 0

(255 . 255 . 255

. 240)

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 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.

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

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

14

s

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

s

)

Number of hosts per subnet = 2

-2

h

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 2 or 2 x 2 x 2 = 8 subnets 3 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 would be 2 or 2 x 2 x 2 x 2 x 2 = 32 hosts. 5 When dealing with the number of hosts per subnet you 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 The number of hosts created by created by borrowing 2 leaving 6 bits is 2 - 2 or 6 bits is 2 or 2 x 2 = 4 2 2 x 2 x 2 x 2 x 2 x 2 = 64 - 2 = 62 subnets. usable hosts per subnet.

What about that second subnet formula: Number of subnets = 2 - 2

s

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 s ubnet . The subnet number for the zero subnet is is also the subnet number for the classful subnet address. The last range of addresses is the broadcas broadcastt s ubnet . The broadcast address for the last subnet in the broadcast subnet is is the same as the classful broadcast address.

Class C Address unsubnetted: 195. 223 . 50 . 0 195.223.50.0 to 195.223.50.255 Notice that the subnet and 15

Class C Address subnetted (2 bits borrowed):

broadcast addresses match.

195. 223 . 50 . 0 0 0 0 0 0 0 0 (Invalid range) (0) 195.223.50.0

to 195.223.50.63 (1) 195.223.50.64 to 195.223.50.127 (2) 195.223.50.128 to 195.223.50.191 (Invalid range) (3) 195.223.50.192 to 195.223.50.255

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 s

s

Use the 2 - 2 formula and don’t use the Use the 2 zero and broadcast ranges if... broadcast ranges if... Classful routing is used

formula and use the zero and

Classless routing or VLSM is used

RIP version 1 is used RIP version 2, EIGRP, or OSPF is used The no ip s ubnet zero command is The ip s ubnet zero command is configured on your router 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 = 2 formula.

16

s

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address

Problem 1 Number of needed subnets 14 Number of needed usable hosts 14 Network Address 192.10.10.0 __________ __________ Address class class C Default subnet mask _______________________________ 255 255 . 255 255 . 255 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 240 Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

16

__________________ ___________________ _

16

___________________ ___________________

Number of bits borrowed

14

___________________ ___________________

4

Show your work for Problem 1 in the space below.

Number of Subnets

-

2

4

Number of 256 128 64 32 16 8 8 16 32 64 128 256

4

2 - Hosts

128 64 32 16

8

4

2

1 - Binary values

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

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

12864

16-2 Observe the total number ofhosts.

Subtract 2 for the number of

17

Custom Subnet Masks 14

usable hosts.

Problem 2 1000 60 165.100.0.0 __________ __________ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 192 Total number of subnets

__________________ ___________________ _

1,024

Total number of host addresses Number of usable addresses

___________________ ___________________

64

___________________ ___________________


62

___________________ ___________________

10


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

165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 128

18

2

1

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address numbers to the left of the line tocreate the custom subnet mask.Add the binary value

32 64 168 +64 192

64-2 62 Observe the total

number ofhosts. Subtract 2 for the number of usable hosts.

4

Problem 3

/26bits used for the network and indicates the total number of

Network Address 148.75.0.0 /26

subnetwork address. All bits remaining remaining belongportion of the

Address class class __________ __________ B to the host portion of the address. Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 255 . 255 . 192 Total number of subnets ___________________ 1,024 Total number of host addresses Number of usable addresses

__________________ ___________________ _

64

___________________ ___________________

62


___________________ ___________________

10


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

19

Custom Subnet Masks 148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 128 64 +64 numbers to the left of the line tocreate the custom subnet mask.Add the binary value

32168 192

64-2 62 Observe the total number

ofhosts. Subtract 2 for the number of usable hosts.

4 1024 Subtract 2 for the total number of -2 subnets. subnets to get the usable number of 1,022

Problem 4 6 30 210.100.56.0 _______ Address class class C Default subnet mask _______________________________ 255 255 . 255 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 224 Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

8

__________________ ___________________ _

32

___________________ ___________________


30

___________________ ___________________

3


Number of

20

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address Number of Subnets - 2

256 128 64 32 16 8 4 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

8

4

2

1 - Binary values

210 . 100 . 56 . 0 0 0 0 0 0 0 0 128

21

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address

Problem 5 6 30 195.85.8.0 _______ Address class class C Default subnet mask _______________________________ 255 255 . 255 . 255 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 255 . 255 . 224 Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

8

__________________ ___________________ _

32

___________________ ___________________


30

___________________ ___________________

3


Number of Number of Subnets - 2

256 128 64 32 16 8 4 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

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

22

8

4

2

1 - Binary values

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address 128 32

8

-2 -2 30 6

Problem 6 126 131,070 118.0.0.0 _______ Address class class A Default subnet mask _______________________________ 255 255 . 0 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 254. 0 . 0 Total number of subnets Total number of host addresses

___________________ ___________________

128

___________________ ___________________

131,072

Number of usable addresses ___________________ 131,070 131,070 Number of bits borrowed

___________________ ___________________

7


Number of

23

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address 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 1 6

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

128 64 32 16 8 4 +2 -2

128 131,072

-2 254 126 131,070

Problem 7 2000 15 178.100.0.0 Address class class

__________ __________

B

Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 224 Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

2,048

__________________ ___________________ _

32

___________________ ___________________


30

___________________ ___________________

11

Show your work for Problem 7 in the space below. 24

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address Number of Hosts

Number of Subnets -

. 256 128

2 4

Binary values -

32 16

8

8

4

2

.

8 16 32 64 128 256

128 64

64 32 16

4

2

1

.

128

64 32 16

8

4

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

2

1

0

128 64 32 16 8 4

2,048 32 -2 -2

2,046 30

Problem 8 3 45 200.175.14.0 _______ Address class class C Default subnet mask _______________________________ 255 255 . 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 255 . 255 . 192 Total number of subnets Total number of host addresses

___________________ ___________________

4

__________________ ___________________ _

64

25

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address ___________________ ___________________ Number of usable addresses 62 Number of bits borrowed

___________________ ___________________

2



256 128 64 32 16 8 4 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

8

4

2

1 - Binary values

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

128 +64 240

4 -2 2

64 2 62

Problem 9 60 1,000 128.77.0.0 _______ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 252 252 . 0 Total number of subnets 26

___________________ ___________________

64

Custom Subnet Masks Number of needed subnets Number of needed usable hosts Network Address __________________ ___________________ _ Total number of host addresses 1,024 Number of usable addresses

___________________ ___________________

1,022


___________________ ___________________

6


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 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 128 64 32 16 8 +4 252

64 -2 62

1,024 -2 1,022

27

Custom Subnet Masks Number of needed subnets

Problem 10 Number of needed usable hosts 60 Network Address 198.100.10.0 _______ Address class class C Default subnet mask _______________________________ 255 255 . 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 192 Total number of subnets Total number of host addresses Number of usable addresses

__________________ ___________________ _

4

__________________ ___________________ _

64

___________________ ___________________


62

___________________ ___________________

2



256 128 64 32 16 8 4 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

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

28

8

4

2

1 - Binary values

Custom Subnet Masks 128 64 4 +64 -2 -2 192 62 2

Problem 11 250 Network Address 101.0.0.0 _______ Address class class A Default subnet mask _______________________________ 255 255 . 0 . 0. 0

Custom subnet mask _______________________________ 255 255 . 255 . 0 . 0

Total number of subnets

___________________ ___________________

256

Total number of host addresses ___________________ 65,536 65,536 Number of usable addresses ___________________ 65,534 65,534 Number of bits borrowed

___________________ ___________________

8


Number of

. 256 128 64 32 16 8

4

2

Hosts -

Number of

.

Binary values -128 64 32 16 8

4

2

1

Subnets

-

2

4

8

16 32 64 128 256

.

. 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

128 64 32 29

Custom Subnet Masks Number of needed subnets 16 8 4 2 +1 255

65,536 256 -2 -2 254 65,534

Problem 12 Number of needed subnets 5 Network Address 218.35.50.0 _______ Address class class C Default subnet mask _______________________________ 255 255 . 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 224 Total number of subnets Total number of host addresses Number of usable addresses

__________________ ___________________ _

8

__________________ ___________________ _

32

___________________ ___________________


30

___________________ ___________________

3



256 128 64 32 16 8 4 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

218 . 35 . 50 . 0 0 0 0 0 0 0 0 30

8

4

2

1 - Binary values

Custom Subnet Masks 128 64 +32 224

64 -2 62

4 -2 2

31

Custom Subnet Masks Problem 13 Number of needed usable hosts 25 Network Address 218.35.50.0 _______ Address class class C Default subnet mask _______________________________ 255 255 . 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 255 . 224 Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

8

__________________ ___________________ _

32

___________________ ___________________


30

___________________ ___________________

3



256 128 64 32 16 8 4 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

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

128

32

8

4

2

1 - Binary values

Custom Subnet Masks

Problem 14 Number of needed subnets 10 Network Address 172.59.0.0 _______ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 240 . 0 Total number of subnets Total number of host addresses

___________________ ___________________

16

__________________ ___________________ _

4,096


___________________ ___________________

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

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

33

Custom Subnet Masks 128

64

Problem 15 Number of needed usable hosts 50 Network Address 172.59.0.0 _______ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 192 Total number of subnets ___________________ 1,024 1,024 Total number of host addresses Number of usable addresses

__________________ ___________________ _

64

___________________ ___________________


62

___________________ ___________________

10


34

Custom Subnet Masks

Number of Hosts

Number of Subnets Binary values -

172

. 256 128 64 32 16

2 4

32 16

8

4

2

.

8 16 32 64 128 256 128 64

8

4

2

1

.

128

64 32 16

8

. 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0

4

2

1

0

128 64 32 16 8 64

-2

-2 62

1,024

1,022

Problem 16 Number of needed usable hosts 29 Network Address 23.0.0.0 _______ Address class class A Default subnet mask _______________________________ 255 255 . 0 . 0 . 0

35

Custom Subnet Masks Custom subnet mask _______________________________ 255 255 . 255 255 . 255 255 . 224 Total number of subnets ___________________ 524,288 524,288 Total number of host addresses Number of usable addresses Number of bits borrowed

__________________ ___________________ _

32

___________________ ___________________

30

___________________ ___________________

19


Number of . 256 128 64 32 16 8

Hosts -

Number of Subnets 4

8

16 32 64 128 256

2

.

.

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

128 64

36

32 -2 224

524,288 +32 30 524,286

-2

4

2

Subnetting Problem 1 Number of needed subnets 14 Number of needed usable hosts 14 Network Address 192.10.10.0 __________ __________ Address class class C Default subnet mask _______________________________ 255 255 . 255 . 255 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 255 . 255 . 240 Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

16

__________________ ___________________ _

16

___________________ ___________________


14

___________________ ___________________

4

What is the 4th subnet range? _______________________________________________ 192.10.10.48 192.10.10.48 to 192.10.10.63

What

is the subnet number for the 8th ___________________ ________________________ _____ 192 . 10 . 10 . 112 112

subnet?

What is the subnet broadcast address for the 13th subnet? ________________________ 192 . 10 10 . 10 . 207 207 What are the assignable addresses for the 9th subnet?______________________________________ 192.10.10.129 to 192.10.10.142 37

Subnetting Number of needed subnets Show your work for Problem 1 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

192. 10 . 10 . 0 0 0 0 0 0 0 0 (0) (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15)

0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

192.10.10.0 to 192.10.10.15 192.10.10.16 to 192.10.10.31 192.10.10.32 to 192.10.10.47 192.10.10.48 to 192.10.10.63 192.10.10.64 to 192.10.10.79 192.10.10.80 to 192.10.10.95 192.10.10.96 to 192.10.10.111 192.10.10.111 192.10.10.112 to 192.10.10.127 192.10.10.127 192.10.10.128 to 192.10.10.143 192.10.10.143 192.10.10.144 to 192.10.10.159 192.10.10.159 192.10.10.160 to 192.10.10.175 192.10.10.175 192.10.10.176 to 192.10.10.191 192.10.10.191 192.10.10.192 to 192.10.10.207 192.10.10.207 192.10.10.208 to 192.10.10.223 192.10.10.224 to 192.10.10.239 192.10.10.240 to 192.10.10.255

128 64 Custom subnet

Usable subnets

mask

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

38

Usable hosts

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

Problem 2 1000 Number of needed usable hosts 60 Network Address 165.100.0.0 __________ __________ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 255 . 255 . 192 Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

1,024

__________________ ___________________ _

64

___________________ ___________________


62

___________________ ___________________

10


What

is the subnet number for the 6th ___________________ ________________________ _____ 165 . 100 . 1 . 64

subnet?

What is the subnet broadcast address for the 6th subnet? ___________________ ________________________ _____ 165 . 100 . 1 . 127 127 What are the assignable addresses for the subnet?______________________________________ 165.100.2.1

9th to

165.100.0.62 39

Show your work for

Problem 3 2 Network Address 195.223.50.0 __________ __________ Address class class C Default subnet mask _______________________________ 255 255 . 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 192

Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

4

__________________ ___________________ _

64

___________________ ___________________


62

___________________ ___________________

2

What is the 3rd subnet range? ____________________ ______________________________ _____________________ _________________ ______ 195.223.50.128 195.223.50.128 195.223.50.191

What

is the subnet number for the 2nd subnet? ____________________ ________________________ ____ 195.223.50.64 195.223.50.64

What is the subnet broadcast address for the 1st subnet? ________________________ 195.223.50.63 195.223.50.63 What are the assignable addresses for the 3rd

41

Subnetting Number of needed subnets subnet?______________________________________ 195.223.50.129 195.223.50.129 195.223.50.190

Problem 3 in the space below. Number of Number of 256 128 64 32 16 8 4 Subnets - 2 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

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

0 1 1 1

8

4

2

1 - Binary values

195. 223 . 50 . 0 0 0 0 0 0 0 0 195.223.50.0 195.223.50. 0 to 195.223.50.63 195.223.50. 63 195.223.50.64 to 195.223.50.127 0 195.223.50.128 to 195.223.50.191 1 195.223.50.192 to 195.223.50.255

-2 62

Problem 4

64 128 +64 192

750 Network Address 190.35.0.0 __________ __________ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 192


42

___________________ ___________________

1,024

Show your work for Total number of host addresses Number of usable addresses

__________________ ___________________ _

64

___________________ ___________________


62

___________________ ___________________

10

What is the 15th subnet range? ___________________ ____________________________ ___________________ __________________ _________ _ 190.35.3.128 190.35.3.128 to 190.35.3.191

What is the subnet number for the 13th subnet? ____________________ ________________________ ____ 190.35.3.0 190.35.3.0 What is the subnet broadcast address for the 10th subnet? ___________________ ________________________ _____ 190.35.2.127 190.35.2.127 What are the assignable addresses for the 6th subnet? ____________________ _____________________________ __________________ _________ 190.35.1.65 190.35.1.65 to 190.35.1.126

43

Problem 4 in the space below. Subnetting

Number of needed subnets

44

Show your work for

Problem 5 Number of needed usable hosts 6 Network Address 126.0.0.0 __________ __________ Address class class A Default subnet mask _______________________________ 255 255 . 0 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 248 Total number of subnets ___________________ 2,097,152 2,097,152 Total number of host addresses Number of usable addresses

________________ ________________

8

___________________ ___________________


6

___________________ ___________________

21

What is the 2nd subnet range? _______________________________________________ 126.0.0.8 126.0.0.8 to 126.0.0.15

What

is the subnet number for the ___________________ ________________________ _____ 126.0.0.32 126.0.0.32

5th

subnet?

What is the subnet broadcast address for the 7th subnet? ____________________ ________________________ ____ 126.0.0.55 126.0.0.55 What are the assignable addresses for the 10th subnet?______________________________________ 126.0.0.73 to 126.0.0.78

45

Problem 5 in the space below. Subnetting

46

Show your work for

Problem 6 Number of needed subnets10 subnets 10 Network Address192.70.10.0 Address192.70.10.0 __________ Address class class C Default subnet mask_______________________________ 255 255 . 255 255 . 255 . 0 Custom subnet mask_______________________________ 255 255 . 255 . 255 . 240 ___________________ ___________________ Total number of subnets 16 ___________________ ___________________ Total number of host addresses 16 ___________________ ___________________ Number of usable addresses 14 ___________________ ___________________ Number of bits borrowed 4

What is the 9th subnet range? 192.70.10.128 to 192.70.10.143 ___________________ _____________________________ ____________________ __________________ ________ What


4th

subnet?

What is the subnet broadcast address for the 12th subnet? __________________ ________________________ ______ 192.70.10.191 192.70.10.191 What are the assignable addresses for the 10th subnet? 192.70.10.145 192.70.10. 145 to ____________________________ _________________ __________ __ 192.70.10.158 ___________________ Problem 6 in the space below. Number of Number of

256 128 64 32 16

8

4

2 - Hosts

47

Subnetting Subnets

- 2

4

8 16 32 64 128 256 128 64 32 16 8 4 2

192 . 70 . 10 . 0 0 0 0 0 0 0 0 (0) (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15)

0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 - Binary values

192.70.10.0 192.70.10. 0 to 192.70.10.15 192.70.10. 15 192.70.10.16 192.70.10. 16 to 192.70.10.31 192.70.10. 31 0 192.70.10.32 to 192.70.10.47 1 192.70.10.48 to 192.70.10.63 0 0 192.70.10.64 to 192.70.10.79 0 1 192.70.10.80 to 192.70.10.95 1 0 192.70.10.96 to 192.70.10.111 192.70.10.111 1 1 192.70.10.112 to 192.70.10.127 0 0 0 192.70.10.128 192.70.10. 128 to 192.70.10.143 192.70.10.1 43 0 0 1 192.70.10.144 192.70.10. 144 to 192.70.10.159 192.70.10.1 59 0 1 0 192.70.10.160 192.70.10. 160 to 192.70.10.175 192.70.10. 175 0 1 1 192.70.10.176 192.70.10. 176 to 192.70.10.191 192.70.10.1 91 1 0 0 192.70.10.192 192.70.10. 192 to 192.70.10.0207 192.70.10.0 207 1 0 1 192.70.10.208 192.70.10. 208 to 192.70.10.223 192.70.10.2 23 1 1 0 192.70.10.224 192.70.10. 224 to 192.70.10.239 192.70.10.2 39 1 1 1 192.70.10.240 192.70.10. 240 to 192.70.10.255 192.70.10.2 55

128 +64 240

16 2 14

Problem 7 48

Show your work for Network Address 10.0.0.0 /16 Address class class __________ __________ A Default subnet mask _______________________________ 255 255 . 0 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 0 . 0 Total number of subnets ___________________ 256 256 Total number of host addresses ___________________ 65,536 65,536 Number of usable addresses ___________________ 65,534 65,534 Number of bits borrowed ___________________ 8


What


6th

subnet?

What is the subnet broadcast address for the 2nd subnet? ____________________ ________________________ ____ 10.1.255.255 10.1.255.255 What are the assignable addresses for the 9th subnet?______________________________________ 10.8.0.1 to 10.8.255.254 Problem 7 in the space below.

49

Subnetting

Problem 8 Number of needed subnets 5 Network Address 172.50.0.0

50

Show your work for __________ __________ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 255 . 224 224 . 0 Total number of subnets Total number of host addresses Number of usable addresses

___________________ ___________________

8

__________________ ___________________ _

8,192

___________________ ___________________


8,190

___________________ ___________________

3


What


5th

subnet?

What is the subnet broadcast address for the 6th subnet? ____________________ ________________________ ____ 172.50.191.255 172.50.191.255 What are the assignable addresses for the 3rd subnet?______________________________________ 172.50.64.1 to 172.50.95.254 Problem 8 in the space below.

51

Subnetting

Problem 9 52

Show your work for Number of needed usable hosts 28 Network Address 172.50.0.0 __________ __________ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 224 Total number of subnets ___________________ 2,048 2,048 Total number of host addresses Number of usable addresses

__________________ ___________________ _

32

___________________ ___________________


30

___________________ ___________________

11

What is the 2nd subnet range? 172.50.0.32 to 172.50.0.63 ___________________ _____________________________ ____________________ __________________ ________ What


10th

subnet?

What is the subnet broadcast address for the 4th subnet? ________________________ 172.50.0.127 172.50.0.127 What are the assignable addresses for the 6th subnet? 172.50.0.161 to 172.50.0.190 __________________ ______________________________ ____________________ ________ Problem 9 in the space below.

53

Subnetting

54

Show your work for

Problem 10 Number of needed subnets 45 Network Address 220.100.100.0 __________ __________ Address class class C Default subnet mask _______________________________ 255 255 . 255 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 252 Total number of subnets Total number of host addresses Number of usable addresses

__________________ ___________________ _

64

__________________ ___________________ _

4

___________________ ___________________


2

___________________ ___________________

6

What is the 5th subnet range? ___________________ _____________________________ ____________________ __________________ ________ 220.100.100.16 220.100.100.16 to 220.100.100.19

What

is the subnet number for the 4th ___________________ ________________________ _____ 220.100.100.12 220.100.100.12

subnet?

What is the subnet broadcast address for the 13th subnet? __________________ ________________________ ______ 220.100.100.51 220.100.100.51 What are the assignable addresses for the 12th subnet?______________________________________ 220.100.100.45 220.100.100. 45 to 220.100.100.46

55

Subnetting Problem 10 in the space below.

Problem 11

56

Show your work for Number of needed usable hosts 8,000 Network Address 135.70.0.0 __________ __________ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 224 . 0 Total number of subnets Total number of host addresses Number of usable addresses

__________________ ___________________ _

8

__________________ ___________________ _

8,192

___________________ ___________________


8,190

___________________ ___________________

3

What is the 6th subnet range? _______________________________________________ 135.70.160.0 to 135.70.191.255

What

is the subnet number for the 7th __________________ ________________________ ______ 135.70.192.0 135.70.192.0

subnet?

What is the subnet broadcast address for the 3rd subnet? ________________________ 135.70.95.255 135.70.95.255 What are the assignable addresses for the 5th subnet?______________________________________ 135.70.128.1 to 135.70.159.254

57

Subnetting Problem 11 in the space below.

Problem 12 Number of needed usable hosts 45

58

Show your work for Network Address 198.125.50.0 __________ __________ Address class class C Default subnet mask _______________________________ 255 255 . 255 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 192


4

__________________ ___________________ _

Total number of host addresses Number of usable addresses

___________________ ___________________

64

___________________ ___________________


62

___________________ ___________________

2

What is the 2nd subnet range? 198.125.50.64 to 98.125.50.127 ____________________ ______________________________ _____________________ _________________ ______ What

is the subnet number for the 2nd subnet? ___________________ ________________________ _____ 198.125.50.64 198.125.50.64

What is the subnet broadcast address for the 4th subnet? ________________________ 198.125.50.255 198.125.50.255 What are the assignable addresses for the 3rd subnet? 198.125.50.129 198.125.50.1 29 to _______________________________ ___________________ _______ 198.125.50.190 ___________________

59

Subnetting Problem 12 in the space below. below. Number of Number of Subnets - 2

256 128 64 32 16 8 4 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

8

4

2

1 - Binary values

198 . 125 . 50 . 0 0 0 0 0 0 0 0 (0) (1) (2) (3)

1 1

0 1 0 1

198.125.50.0

to to 198.125.50.63

198.125.50.64

to to 198.125.50.127

198.125.50.128

198.125.50.191

198.125.50.192

198.125.50.255

Problem 13 Network Address 165.200.0.0 165.200.0.0 /26 __________ __________ Address class class B Default subnet mask _______________________________ 255 255 . 255 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 192

Total number of subnets ___________________ 1,024 1,024 Total number of host addresses Number of usable addresses

64

___________________ ___________________


60

__________________ ___________________ _

62

___________________ ___________________

10

Show your work for What is the 10th subnet range? _______________________________________________ 165.200.2.64 to 165.200.2.127

What


subnet?

What is the subnet broadcast address for the 1023rd subnet? ________________________ 165.200.255.191 165.200.255.191 What are the assignable addresses for the 1022nd subnet?______________________________________ 165.200.255.65 to 165.200.255.126

61

Show your work for

Show your work for

Problem 14 Number of needed usable hosts 16 Network Address 200.10.10.0 __________ __________ Address class class C Default subnet mask _______________________________ 255 255 . 255 255 . 255 . 0 Custom subnet mask _______________________________ 255 255 . 255 . 255 . 224 Total number of subnets Total number of host addresses Number of usable addresses

__________________ ___________________ _

8

__________________ ___________________ _

32

___________________ ___________________


30

___________________ ___________________

3

What is the 7th subnet range? ___________________ _____________________________ _____________________ __________________ _______ 200.10.10.192 200.10.10.192 to 200.10.10.223

What


subnet?

What is the subnet broadcast address for the 4th subnet? ________________________ 200.10.10.127 200.10.10.127 What are the assignable addresses for the 6th subnet?______________________________________ 200.10.10.161 200.10.10. 161 to 200.10.10.190

63

Subnetting Problem 14 in the space below. 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

200 . 10 . 10 . 0 0 0 0 0 0 0 0 (0) (1) (2) (3) (4) (5) (6) (7)

128 64 +32 224

0 1 1 1 1 1 1 1

200.10.10.0 to 200.10.10.32 200.10.10.3 2 to 0 200.10.10.64 1 200.10.10.96 0 0 200.10.10.128 200.10.10. 128 0 1 200.10.10.160 200.10.10. 160 1 0 200.10.10.192 200.10.10. 192 1 1 200.10.10.224 200.10.10. 224

200.10.10.31 200.10.10.63 200.10.10. 63 to 200.10.10.95 to 200.10.10.127 200.10.10.127 to 200.10.10.159 200.10.10. 159 to 200.10.10.191 200.10.10. 191 to 200.10.10.223 200.10.10. 223 to 200.10.10.255 200.10.10. 255

32 -2 30

Problem 15 Network Address 93.0.0.0 \19 __________ __________ Address class class A Default subnet mask _______________________________ 255 255 . 0 . 0 . 0 Custom subnet mask _______________________________ 255 255 . 255 255 . 224 . 0 Total number of subnets ___________________ 2,048 2,048

64

Show your work for Total number of host addresses

__________________ ___________________ _

8,192


___________________ ___________________

11

What is the 15th subnet range? __________________ ____________________________ ____________________ __________________ _________ _ 93.1.192.0 93.1.192.0 to 93.1.223.255

What


9th

subnet?

What is the subnet broadcast address for the 7th subnet? ________________________ 93.0.223.255 93.0.223.255 What are the assignable addresses for the 12th subnet?______________________________________ 93.1.96.1 to 93.1.127.254 Problem 15 in the space below. below.

65

Subnetting

66

Practical Subnetting 1 Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnets , and allow enough extra subnets and hosts for 100 % growth growth in both areas. Circle each subnet on the graphic and answer the questions below. IP Address 172.16.0.0 F0/0 Router A

S0/0/1

S0/0/0

F0/1 Router B

F0/0

Marketing 24 Hosts

Management 15 Hosts

Reasearch 60 Hosts

Address class class __________________ _____________________________ ___________ B Minimum number of subnets needed _________ 4

+

Extra subnets required for 100% growth _________ 4 ( Round up to the next whole number )

=

Total number of subnets needed _________ 100% growth in the largest subnet _________

8

+

( Round up to the next whole number )

Total number of address 120 needed for the largest subnet _________

=

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

IP address range for Research _____________________________ 172.16.0.0 to 172.31.255

IP address range for Marketing _____________________________ 172.16.32.0 to 172.63.255

67

IP address range for Management __________________ ____________________________ ___________ _ 172.16.64.0 172.16.64.0 to 172.95.255 IP address range for Router A to Router B serial connection _____________________________ 172.16.96.0 172.16.96.0 to 172.127.255

68

Show your work for Practical Subnetting 1 in the space below.

Practical Subnetting 2

69

Based 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

Custom subnet mask _____________________________ 255.255.255.224 255.255.255.224 Minimum number of subnets needed _________ 5 Extra subnets required for 30% growth _________ +

2


Total number of subnets needed _________ =

7

Number of host addresses in the largest subnet group _________ 20 20 Number of addresses needed for 30% growth in the largest subnet _________ +

6


Total number of address needed for the largest subnet _________ = 26 Start with the first subnet and arrange your sub-networks from the largest group to the smallest.

IP address range for Tech ____________________ _____________________________ _________ 135.126.0.0 135.126.0.0 to 135.126.0.31

Ed

IP address range for English _____________________________ 135.126.0.32 to 135.126.0.63 70

IP address range for Science ___________________ _____________________________ __________ 135.126.0.64 135.126.0.64 to 135.126.0.95 IP address range for Router A to Router B serial connection 135.126.0.96 to ____________________ _____________________________ _________ 135.126.0.127 IP address range for Router A to Router B serial connection 135.126.0.128 _____________________________ __________ 135.126.0.159 ___________________

to

71

Problem 2 in the space below.

72

Show your work for

Practical Subnetting 3 Based on the information in the graphic shown, design a classfull network addressing that will supply theminimum number of hosts per subnet, and allow enough extra subnets and hosts for 25% growth in all areas. Circle each subnet on the graphic and answer the questions below. IP Address 172.16.0.0 F0/0

Administrative Administrative 30 Hosts

S0/0/1 F0/0

Router A

F0/1

S0/0/0

Router B

Sales 185 Hosts

Marketing 50 Hosts

B Address class class ___________ ___________________ __________________ __________ 255.255.255.0 Custom subnet mask ___________________ ___________ __________________ __________ scheme

Minimum number of subnets needed _________ 4 Extra subnets required for 25% growth _________ + 1 ( Round up to the next whole number )

Total number of subnets needed _________ = 5 Number of host addresses in the largest subnet group _________ 185 185 Number of addresses needed for 25% growth in the largest subnet _________ + 47 ( Round up to the next whole number )


IP address range for Sales _____________________________ 172.16.0.0 to 172.16.0.255 73

IP address range for Marketing _____________________________ 172.16.1.0 to 172.16.1.255

IP address range for Administrative ___________________ _____________________________ __________ 172.16.2.0 172.16.2.0 to 172.16.2.255 IP address range for Router A to Router B serial connection _____________________________ 172.16.3.0 172.16.3.0 to 172.16.3.255

74


75

Based on the information in the graphic shown, design a network addressing scheme that will supply the

Practical Subnetting 4 minimum number of subnets , and allow enough extra subnets and hosts for 70% growth in all areas. Circle each subnet on the graphic and answer the questions below.

Custom subnet mask _____________________________ 255.255.240.0 255.255.240.0 Minimum number of subnets needed _________ 5 Extra subnets required for 70% growth _________ + 4 ( Round up to the next whole number )


9

Number of host addresses in the largest subnet group _________ 325 325 Number of addresses needed for 70% growth in the largest subnet _________ + 228 ( Round up to the next whole number )


IP address range for New York 135.126.0.0 to _____________________________ __________ 135.126.15.255 ___________________ 76

Show your work for IP address range for Washington D. _____________________________ __________ 135.126.31.255 ___________________

C.

135.126.16.0

to

IP address range for Dallas 135.126.32.0 to _____________________________ __________ 135.126.47.255 ___________________ IP address range for Router A to Router B serial connection 135.126.48.0 to _____________________________ __________ 135.126.63.255 ___________________ IP address range for Router A to Router C serial connection 135.126.64.0 to _____________________________ __________ 135.126.79.255 ___________________

77




minimum number of hosts per subnet , and allow enough extra subnets and hosts for 100% growth in all areas. Circle each subnet on the graphic and answer the questions below.

78

Show your work for


Total number of subnets needed _________ = 4 in the largest subnet groupNumber of host addresses _________ 30

100Number of addresses needed for% growth in the largest subnet _________ + 30 ( Round up to the next whole number )

needed for the largest subnetTotal number of address _________ = 60 Start with the first subnet and arrange your sub-networks from the largest group to the smallest.

IP address range for Router F0/0 ___________________ _____________________________ __________ 210.15.10.0 210.15.10.0 to 210.15.10.63

Port

79


IP address range for Router F0/1 Port ___________________ ____________________________ __________ _ 210.15.10.64 210.15.10.64 to 210.15.10.127 Problem 5 in the space below. Number of Number of 256 128 64 32 16 8 4 Subnets - 2 4 8 16 32 64 128 256

2 - Hosts 128 64 32 16

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

80

0 1 1 1

210. 15 . 10 . 0 0 0 0 0 0 0 0 210.15.10.0 to 210.15.10.63 210.15.10.64 to 210.15.10.127 0 210.15.10.128 to 210.15.10.191 1 210.15.10.192 to 210.15.10.255

8

4

2

1 - Binary values

Show your work for

Practical Subnetting 6 minimum number of subnets , and allow enough extra subnets and hosts for 20% growth in all areas. Circle each subnet on the graphic and answer the questions below.

Address class class _________________ ___________________________ ____________ __ A Custom subnet mask _____________________________ 255.240.0.0 255.240.0.0 Minimum number of subnets needed _________ 7 7

+

Extra subnets required for 20% growth _________

2


=

Total number of subnets needed _________

9

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

IP address range for Technology _____________________________ 10.0.0.0 to 10.15.255.255

IP address range for ____________________ _____________________________ _________ 10.16.0.0 10.16.0.0 to 10.31.255.255 10.31.255.255

Science

IP address range for Arts & Drama _____________________________ 10.32.0.0 to 10.47.255.255 10.47.255.255

IP Address range Administration _____________________________ 10.48.0.0 10.48.0.0 to 10.63.255.255

IP address range for Router A 81


to Router B serial connection ___________________ _____________________________ __________ 10.64.0.0 10.64.0.0 to 10.79.255.255 10.79.255.255 IP address range for Router A to Router C serial connection ___________________ ____________________________ __________ _ 10.80.0.0 10.80.0.0 to 10.95.255.255 10.95.255.255 IP address range for Router B to Router C serial connection ___________________ ____________________________ __________ _ 10.96.0.0 10.96.0.0 to 10.111.255.255 10.111.255.255 Problem 6 in the space below.

82

Show your work for


83

Based 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



9



IP address range for Router A ___________________ _____________________________ __________ 177.135.0.0 177.135.0.0 to 177.135.3.255 177.135.3.255

Port

F0/0

IP address range for Research _____________________________ 177.135.4.0 to 177.135.7.255 177.135.7.255

IP address range for Deployment ___________________ _____________________________ __________ 177.135.8.0 177.135.8.0 to 177.135.11.255 177.135.11.255 84

Show your work for IP address range for Router A to Router B serial connection 177.135.12.0 to _____________________________ __________ 177.135.15.255 ___________________ Problem 7 in the space below.

Practical Subnetting 8 growth in all areas. Circle each subnet on the graphic and answer the questions below.

F0/0 Router A

IP Address 192.168.1.0 S0/0/0 S0/0/1 F0/0

F0/1 Router B

New York 8 Hosts Boston 5 Hosts

Research & Development 8 Hosts

C

Address class class ____________ ____________________ _________________ _________

minimum number subnets , and allow enough extra subnets and hosts for 85%

85



3



6



IP address range for Router A F0/0 ____________________ _____________________________ _________ 192.168.1.0 192.168.1.0 to 192.168.1.31 IP address range for New York _____________________________ 192.168.1.32 192.168.1.32 to 192.168.1.63

IP address range for Router A to Router B serial connection _____________________________ 192.168.1.64 192.168.1.64 to 192.168.1.95

Problem 8 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

192. 168 . 1 . 0 0 0 0 0 0 0 0 (0) (1) (2) (3) (4) (5) 86

0 1 1 0 1 1 100 1 01

192.168.1.0 192.168.1.32 192.168.1.64 192.168.1.96 192.168.1.128 192.168.1.160 192.168.1.160

to to to to to to

192.168.1.31 192.168.1.63 192.168.1.95 192.168.1.127 192.168.1.127 192.168.1.159 192.168.1.159 192.168.1.1191 192.168.1.1191

Show your work for (6) (7)

1 1 0 192.168.1.192 192.168.1.1 92 to 192.168.1.223 192.168.1. 223 1 1 1 192.168.1.224 192.168.1. 224 to 192.168.1.255 192.168.1. 255

87


Practical Subnetting 9 minimum number of hosts per subne t , and allow enough extra subnets and hosts for 15% growth in all areas. Circle each subnet on the graphic and answer the questions below.

Router A

IP Address 148.55.0.0 S0/0/0 S0/0/1

S0/0/1

F0/0

F0/1 Router B

S0/0/0

Dallas 1500 Hosts

Router C

F0/0 Router D

Ft. Worth 2300 Hosts

S0/0/1

S0/0/0

B

Address class class _____________ ___________________________ ________________ __ 255.255.240.0

Custom subnet mask _____________ ___________________________ ________________ __ 5

Minimum number of subnets needed _________ Extra subnets required for 15% growth _________ +

1



6


Total number of address needed for the largest subnet _________ =2645 Start with the first subnet and arrange your sub-networks from the largest group to the smallest.

IP address range for Ft. Worth _____________________________ 148.55.0.0. 148.55.0.0. to 148.55.15.255

88

Show your work for IP address range for Dallas _____________________________ 148.55.16.0. 148.55.16.0. to 148.55.31.255

IP address range for ___________________ _____________________________ __________ 148.55.32.0. 148.55.32.0. Router B serial connection IP address range for ___________________ ____________________________ __________ _ 148.55.48.0. 148.55.48.0. Router C serial connection IP address range for ___________________ ____________________________ __________ _ 148.55.64.0. 148.55.64.0. Router D serial connection

Router to

148.55.47.255

Router to

148.55.63.255

Router to

148.55.79.255

A to A to C to

89




minimum number of subnets , and allow enough extra subnets and hosts for

90

Show your work for


5



9



IP address range for Sales/Managemnt ___________________ _____________________________ __________ 172.16.0.0 172.16.0.0 to 172.16.15.255 172.16.15.255 IP address range for Marketing _____________________________ 172.16.16.0 to 172.16.31.255

IP address range for Research _____________________________ 172.16.32.0 to 172.16.47.255 91


IP address range for Router A to Router B serial connection _____________________________ 172.16.48.0 172.16.48.0 to 172.16.63.255

92


Valid and Non-Valid IP Addresses

93

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: ___________________ _____________________________ _____________ ___ The The network 0.230.190.192 ID cannot be 0. Subnet Mask: 255.0.0.0 ___________________ _____________________________ _____________ ___ Reference Page Inside Front Cover

IP Address: 192.10.10.1 ___________________ _____________________________ _____________ ___ OK OK Subnet Mask: ___________________ _____________________________ _____________ ___ 255.255.255.0 ___________________ _____________________________ _____________ ___ 245 245 is Reference Pages 28-29 IP Address: 245.150.190.10 Subnet Mask: 255.255.255.0 Reference Page Inside Front Cover

IP Address: 135.70.191.255 Subnet Mask: 255.255.254.0 Reference Pages 48-49

reserved for

___________________ _____________________________ _____________ ___ experimental experimental use.

___________________ _____________________________ _____________ ___ This This is the broadcast address

___________________ _____________________________ _____________ ___ for this range.

IP Address: ___________________ _____________________________ _____________ ___ 127 127 127.100.100.10 reserved for loopback Subnet Mask: 255.0.0.0 ___________________ _____________________________ _____________ ___ testing. testing. Reference Pages Inside Front Cover

is

IP Address: 93.0.128.1 ___________________ _____________________________ _____________ ___ OK OK Subnet Mask: ___________________ _____________________________ _____________ ___ 255.255.224.0 Reference Pages 56-57



IP Address: 190.35.0.10

___________________ _____________________________ _____________ ___ This This

is

the

subnet address for the

___________________ _____________________________ _____________ ___ 3

rd

usable

range of 200.10.10.0 200.10.10. 0

___________________ _____________________________ _____________ ___ OK OK ___________________ _____________________________ _____________ ___ ___________________ _____________________________ _____________ ___ This This taken

from

the

address is first

Subnet Mask: 255.255.255.192

____________________ ______________________________ ____________ __ range range

for this

subnet which is invalid.

Reference Pages 34-35

IP Address: 218.35.50.195 Subnet Mask: 255.255.0.0 Reference Page Inside Front Cover

IP Address: 200.10.10.175 200.10.10.175 /22

____________________ ______________________________ ____________ __ This This has a class B subnet

____________________ ______________________________ ____________ __ mask. mask. ____________________ ______________________________ ____________ __ A address

must

use

class

C a

____________________ ______________________________ ____________ __ minimum minimum of

Reference Pages 54-55 and/or Inside Front Cover

_____________________________ ___________ ___ This 24 bits. _____________________ This IP Address: 135.70.255.255 Subnet Mask: 255.255.224.0

Reference Pages 48-49

78

is a broadcast address.

____________________ ______________________________ ____________ __

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 256 addresses.

/24

255.255.255.0 256 Hosts 1 Subnet Split the box in half and you get two subnets with 128 addresses,

/25 255.255.255.128 128 Hosts 2 Subnets

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

/26 255.255.255.192

64 Hosts 4 Subnets 80

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

/27 255.255.255.224 32 Hosts 8 Subnets Split the boxes in half again and you get sixteen subnets with sixteen addresses,

/28 255.255.255.240 16 Hosts 16 Subnets The next split gives you thirty two subnets with eight addresses,

/29

81

255.255.255.248 8 Hosts 32 Subnets The last split gives sixty four subnets with four addresses each,

/30 255.255.255.252 4 Hosts 64 Subnets Class A Addressing Guide ________ __ CIDR CIDR ________ __/8 ________ __/9 ________ __/10 ________ __/11 ________ __/12 ________ __/13 ________ __/14 ________ __/15 ________ __/16 ________ __/17 ________ ___/18 ________ __/19 ________ __/20 ________ __/21 ________ __/22 ________ ___/23 ________ ___/24 ________ ___/25 ________ __/26

# of Bits

Subnet

Total # of

Total # of

Usable # of

_____________ _ Borrowed Borrowed _____________ _0 _____________ _1 _____________ __2 _____________ _3 _____________ _4 _____________ _5 _____________ _6 _____________ __7 _____________ __8 _____________ _9 _____________ __10 _____________ _11 _____________ __12 _____________ __13 _____________ _14 _____________ _15 _____________ __16 _____________ _17 _____________ __18

_____________________ ___ Mask Mask _____________________ ___255.0.0.0 _____________________ ___255.128.0.0 _____________________ ____255.192.0.0 _____________________ ___255.224.0.0 _____________________ ___255.240.0.0 _____________________ ____255.248.0.0 _____________________ ___255.252.0.0 _____________________ ____255.254.0.0 _____________________ _____255.255.0.0 _____________________ ____255.255.128.0 _____________________ _____255.255.192.0 _____________________ ___255.255.224.0 _____________________ ____255.255.240.0 _____________________ ____255.255.248.0 _____________________ ____255.255.252.0 _____________________ _____255.255.254.0 _____________________ _____255.255.255.0 _____________________ _____255.255.255.128 _____________________ _____255.255.255.192

_____________ ___ Subnets Subnets _____________ __1 _____________ __2 _____________ ___4 _____________ ___8 _____________ __16 _____________ ___32 _____________ ___64 _____________ ___128 _____________ ___256 _____________ __512 _____________ ___1,024 _____________ ___2,048 _____________ ___4,096 _____________ ___8,192 _____________ ___16,384 _____________ ____32,768 _____________ ___65,536 _____________ ____131,072 _____________ ___262,144

____________ __ Hosts Hosts ____________ __16,777,216 ____________ __8,388,608 ____________ ___4,194,304 ____________ __2,097,152 ____________ __1,048,576 ____________ __524,288 ____________ __262,144 ____________ ___131,072 ____________ ___65,536 ____________ __32,768 ____________ ___16,384 ____________ __8,192 ____________ ___4,096 ____________ __2,048 ____________ __1,024 ____________ __512 ____________ ___256 ____________ __128 ____________ ___64

_____________ ___ Hosts Hosts _____________ ___16,777,214 _____________ ___8,388,606 _____________ ____4,194,302 _____________ ___2,097,150 _____________ ___1,048,574 _____________ ____524,286 _____________ ___262,142 _____________ ____131,070 _____________ ____65,534 _____________ ____32,766 _____________ _____16,382 ____________ ____8,190 _____________ ____4,094 _____________ ____2,046 _____________ ____1,022 _____________ _____510 _____________ _____254 _____________ _____126 ____________ _____62

________ ___/27 ________ ___/28 ________ __/29 /30

_____________ _19 _____________ _20 _____________ _21 22

_____________________ _____255.255.255.224 _____________________ _____255.255.255.240 _____________________ ____255.255.255.248 255.255.255.252

_____________ ____524,288 _____________ ____1,048,576 _____________ ___2,097,152 4,194,304

____________ __32 ____________ __16 ____________ __8 4

_____________ _____30 _____________ _____14 _____________ ____6 2

Class B Addressing Guide ________ __ CIDR CIDR ________ __/16 ________ __/17 ________ __/18 ________ __/19 ________ __/20 ________ __/21 ________ __/22 ________ __/23 ________ __/24 ________ __/25 ________ ___/26 ________ __/27 ________ __/28

# of Bits

Subnet

_____________ _ Borrowed Borrowed _____________ _0 _____________ _1 _____________ __2 _____________ _3 _____________ _4 _____________ _5 ____________ __6 _____________ __7 _____________ __8 _____________ _9 _____________ __10 _____________ _11 _____________ _12

_____________________ ___ Mask Mask _____________________ ___255.255.0.0 _____________________ ___255.255.128.0 _____________________ ____255.255.192.0 _____________________ ___255.255.224.0 _____________________ ___255.255.240.0 _____________________ ____255.255.248.0 _____________________ ___255.255.252.0 _____________________ ____255.255.254.0 _____________________ _____255.255.255.0 _____________________ ____255.255.255.128 _____________________ _____255.255.255.192 _____________________ ___255.255.255.224 _____________________ ___255.255.255.240

Total # of _____________ ___ Subnets Subnets _____________ __1 _____________ __2 _____________ ___4 _____________ ___8 _____________ __16 _____________ ___32 _____________ ___64 _____________ ___128 _____________ ___256 _____________ __512 _____________ ___1,024 _____________ ___2,048 _____________ ___4,096

Total # of

Usable # of

____________ _ Hosts Hosts ____________ _65,536 ____________ _32,768 ____________ ___16,384 ____________ _8,192 ____________ _4,096 ____________ __2,048 ____________ _1,024 ____________ ___512 ____________ __256 ____________ __128 ____________ ___64 ____________ _32 ____________ _16

_____________ ____ Hosts Hosts _____________ ____65,534 _____________ ____32,766 _____________ ____16,382 _____________ ____8,190 _____________ ____4,094 _____________ ____2,046 _____________ ____1,022 _____________ ____510 _____________ _____254 _____________ ____126 _____________ _____62 _____________ ____30 _____________ ____14

________ _____________ _____________________ ____255.255.255.248 __/29 _13 14 255.255.255.252 /30

_____________ ___8,192 16,384

____________ __8 4

_____________ ____6 2

Class C Addr essing Gui de ________ __ CIDR CIDR ________ __/24 ________ __/25 ________ __/26 ________ __/27 ________ __/28 ________ __/29 /30

# of Bits

Subnet

_____________ _ Borrowed Borrowed _____________ _0 _____________ _1 _____________ __2 _____________ _3 _____________ _4 _____________ _5 6

_____________________ ___ Mask Mask _____________________ ___255.255.255.0 _____________________ ___255.255.255.128 _____________________ ____255.255.255.192 _____________________ ___255.255.255.224 _____________________ ___255.255.255.240 _____________________ ____255.255.255.248 255.255.255.252

82 Inside Cover

Total # of _____________ ___ Subnets Subnets _____________ __1 _____________ __2 _____________ ___4 _____________ ___8 _____________ __16 _____________ ___32 64

Total # of

Usable # of

____________ __ Hosts Hosts ____________ __256 ____________ __128 ____________ ___64 ____________ __32 ____________ __16 ____________ __8 4

_____________ ___ Hosts Hosts _____________ ___254 _____________ ___126 _____________ ____62 _____________ ___30 _____________ ___14 _____________ ____6 2

IP Addressing and Subnetting Workbook - Instructors Version 1.5.docx

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