IP addressing and subnetting PDF

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Conversion from Binary to Decimal: 10010010 128 64 32 1 0 0 128 + 16 + 2 = 146

16 1

8 0

4 0

2 1

1 0

8 0

4 1

2 1

1 1

01110111 128 64 32 16 0 1 1 1 64 + 32 + 16 + 4 + 2 + 1 = 119

Conversion from Decimal to Binary 238 = 128 + 64 + 32 + 8 + 4 + 2 128 1

64 1

32 1

16 0

8 1

4 1

2 1

1 0

32 1

16 0

8 0

4 0

2 1

1 0

34 = 32 + 2 128 0

64 0

Class A  N.H.H.H 15.32.12.234  Network Address : 15.0.0.0 Or 15.32.12.234 AND 255.0.0.0 (default subnet mask for class A)

15.32.12.234  00001111.00100000.0000110.11101010 255.0.0.0

11111111.00000000. 00000000.00000000

15.0.0.0

00001111.00000000.00000000.00000000

The default subnet mask for class A is 255.0.0.0 or /8 That is: 15.32.12.234 /8 Is equal to 15.32.12.234, 255.0.0.0

Class B  N.N.H.H 150.132.102.23  Network Address : 150.132.0.0 Or 150.132.102.23 AND 255.255.0.0 (default subnet mask for class B)

150.132.102.234 10010110.10000100.01100110.11101010 255.255.0.0

11111111.11111111.00000000.00000000

150.132.0.0

10010110. 10000100.0000110.00000000

The default subnet mask for class B is 255.255.0.0 or /16 That is: 150.132.102.234 /16 Is equal to 150.132.102.234, 255.255.0.0

Class C  N.N.N.H 225.132.129.216  Network Address : 225.132.129.0 Or 225.132.129.216 AND 255.255.255.0 (default subnet mask for class C) 225.132.129.21611100001.10000100.10000001.11101010 255.255.255.0  11111111.11111111.11111111.00000000 225.132.129.0  11100001.10000100.10000001.00000000

The default subnet mask for class C is 255.255.255.0 or /24 That is: 225.132.129.216 /24 Is equal to 225.132.129.216, 255.255.255.0

IP Address Subnetting

IP Addresses – Network Address:

No sub-netting Class A : N.H.H.H Class B: N.N.H.H Class C: N.N.N.H

With Sub-netting Class A: N.snH’.H.H, Class B: N.N.snH’.H, Class C: N.N.N.snH’

Class A subnetting Ex 1: 93.101.177.80 / 8  not sub-netted address default subnet mask 255.0.0.0 Ex 2: 93.101.177.80 / 11  sub-netted address subnet mask : 255.11100000.00000000.00000000  255.224.0.0

Ex 3: 93.101.177.80 / 21  sub-netted address subnet mask : 255.11111111.111 00000.00000000  255.255.240.0

Class B subnetting Ex 1: 123.347.198.220 /16  not sub-netted address default subnet mask 255.255.0.0 Ex 2: 123.347.198.220 /22  sub-netted address subnet mask : 255.255.11111100.00000000  255.255.252.0

Class C subnetting Ex 1: 193.91.167.9 /8  not sub-netted address default subnet mask 255.255.255.0 Ex 2: 193.91.167.9 / 28  sub-netted address subnet mask : 255.255.255.11110000  255.255.255.240

Example: A company with 8 departments. Each department needs its own network. Each department has less than 30 users.

1st Solution: One network for each department.

192.168.10.0  network of department 1 192.168.11.0  network of department 2 192.168.12.0  network of department 3 192.168.13.0  network of department 4 192.168.14.0  network of department 5 192.168.15.0  network of department 6 192.168.16.0  network of department 7 192.168.17.0  network of department 8

Each network provides 256 IP. Each department needs less than 30 IPs  Waste of IPs.

2nd Solution: one network divided into 8 sub nets. Each sub net has 32 IP. Class C: N.N.N.H 192.168.10.0  192.168.10.00000000 192.168.10.0 /27 or 192.168.10.0 , 255.255.255.224

198.168.10.000 00000  sub net 0 198.168.10.001 00000  sub net 1 198.168.10.010 00000  sub net 2 198.168.10.011 00000  sub net 3 198.168.10.100 00000  sub net 4 198.168.10.101 00000  sub net 5 198.168.10.110 00000  sub net 6 198.168.10.111 00000  sub net 7 The details of some of these subnets will be explained below:

Sub net 0: Find:  Network address  Broadcast address  (Usable/assignable/Valid) IP addresses

198.168.10.000 00000  198.168.10.0  Network address 198.168.10.000 00001  198.168.10.1  usable IP 198.168.10.000 00010  198.168.10.2  usable IP 







198.168.10.000 11110  198.168.10.30  usable IP 198.168.10.000 11111  198.168.10.31  Broadcast address

Usable Range: 198.168.10.1  198.168.10.30

Sub net 1: Find:  Network address  Broadcast address  (Usable/assignable/Valid) IP addresses

198.168.10.001 00000  198.168.10.32  Network address 198.168.10.001 00001  198.168.10.33  usable IP 198.168.10.001 00010  198.168.10.34  usable IP 







198.168.10.001 11110  198.168.10.62  usable IP 198.168.10.001 11111  198.168.10.63  Broadcast address Usable Range: 198.168.10.33  198.168.10.62

Sub net 6: Find  Network address  Broadcast address  (Usable/assignable/Valid) IP addresses

198.168.10.110 00000  198.168.10.192  Network address 198.168.10.110 00001  198.168.10.193  usable IP 198.168.10.110 00010  198.168.10.194  usable IP 







198.168.10.110 11110  198.168.10.222  usable IP 198.168.10.110 11111  198.168.10.223  Broadcast address Usable Range: 198.168.10.193  198.168.10.222

Sub net 7: Find  Network address  Broadcast address  (Usable/assignable/Valid) IP addresses

198.168.10.111 00000  198.168.10.224  Network address 198.168.10.111 00001  198.168.10.225  usable IP 198.168.10.111 00010  198.168.10.226  usable IP 

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

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198.168.10.111 11110  198.168.10.254  usable IP 198.168.10.111 11111  198.168.10.255  Broadcast address Usable Range: 198.168.10.225  198.168.10.254

Which of the following IP addresses is a (valid/assignable/usable) address. That is, the IP address is neither a network address nor a broadcast address. Q1: 192.168.10.111/27 ( Class C) 192.168.10.11  198.168.10.000 001011 It is a valid IP address. Q2: 200.171.103.208/28 (Class C) 200.171.103.208  200.171.103.1101 0000 It is an invalid IP address … it is a network address

Q3: 176.54.123.11/ 20 (Class B) 176.54.123.11 176.54. 0111 1011.00001011 It is a valid IP address.

Q4: 79.213.127.255/17 (Class A) 79.213.127.255  79.11010101.0 1111111.11111111 It is an invalid IP address … it is a broadcast address

Exercise 1: Subnet the Class B network 147.10.0.0 / 29 147.10.00000000.00000 000 Q: How any subnets ? Q: How many hosts per subnet ? Q: Is this a usable IP address: 147.10.231.144 / 29

Exercise 2: Subnet the Class A network 28.0.0.0/20 28.00000000.0000 0000.00000000 Q: How any subnets ? Q: How many hosts per subnet ? Q: Is this a usable IP address: 28.101.153.144 / 20

Ex: A company has 6 departments as follows: Department A has 30 users, Department B has 16 users, Department C has 60 users, Department D has 120 users, Department E has 13 users, Department F has 6 users.

Q1: Can we use a class C Network :200.130.15.0 ? 6 networks needs 3 bits for subnetting: 200.130.15.0

200.130.15.000 00000

Q 2: Can we use a class B network : 147.123.0.0 ? What subnet mask should be used ? 147.123.000 000000.00000000

Please refer to : IP Addressing and Subnetting Workbook

Problem 8 (page 15) Number of needed usable subnets Number of needed usable hosts Network Address Address class Default subnet mask

1 45 200.175.14.0 C 255.255.255.0

200.175.14.00 000000 Number of bits borrowed Custom subnet mask Total number of subnets Number of usable subnets Total number of host addresses Number of usable addresses

2 255.255.255.192 4 2 64 60

Problem 9 (page 16)

Number of needed usable subnets Number of needed usable hosts Network Address Address class Default subnet mask

60 1000 128.77.0.0 B 255.255.0.0

128.77.000000 00.00000000 Number of bits borrowed Custom subnet mask Total number of subnets Number of usable subnets Total number of host addresses Number of usable addresses

6 255.255.252.0 64 62 1024 1022

Problem 10 (page 17)

Number of needed usable subnets Number of needed usable hosts Network Address Address class Default subnet mask

Not given 60 198.100.10.0 C 255.255.255.0

198.100.10.00 000000 Number of bits borrowed Custom subnet mask Total number of subnets Number of usable subnets Total number of host addresses Number of usable addresses

2 255.255.255.192 4 2 64 62

Problem 11 (page 18)

Number of needed usable subnets Number of needed usable hosts Network Address Address class Default subnet mask

250 Not given 101.0.0.0 A 255.0.0.0

101.00000000.00000000.00000000 Number of bits borrowed Custom subnet mask Total number of subnets Number of usable subnets Total number of host addresses Number of usable addresses

8 255.255.0.0 256 254 2^16 2^16 - 2

Problem 15 Number of needed usable subnets Number of needed usable hosts Network Address Address class Default subnet mask

Not given 50 172.59.0.0 B 255.255.0.0

172.59.00000000.00 000000 Number of bits borrowed Custom subnet mask Total number of subnets Number of usable subnets Total number of host addresses Number of usable addresses

10 255.255.255.192 1024 1022 64 62

Problem 1 (page 24) Number of needed usable subnets Number of needed usable hosts Network Address Address class Default subnet mask

14 14 192.10.10.0 C 255.255.255.0

192.10.10.0000 0000 Number of bits borrowed Custom subnet mask Total number of subnets Number of usable subnets Total number of host addresses Number of usable addresses

192.10.10.0000 0000  sub net 0 192.10.10.0001 0000  sub net 1 ( first usable subnet) 192.10.10.0011 0000  sub net 2 ( second usable subnet) 

192.10.10.1111 0000  sub net 15

What is the 3rd usable subnet range ? 192.10.10.0011 0000  192.10.10.48  

192.10.10.0011 1111

 192.10.10.63

Answer:  192.10.10.48 - 192.10.10.63

4 255.255.255.240 16 14 16 14

What is the subnet number for the 7th usable subnet? 192.10.10.0111 0000  192.10.10.112  

192.10.10.0111 1111  192.10.10.127 Answer:  192.10.10.112

What is the subnet broadcast address for the 12th usable subnet ? 192.10.10.1100 0000  192.10.10.112  

192.10.10.1100 1111  192.10.10.207

Answer:  192.10.10.207

What are the assignable addresses for the 8th usable subnet? 192.10.10.1000 0000  192.10.10.128 192.10.10.1000 0001  192.10.10.129  

192.10.10.1000 1110  192.10.10.142 192.10.10.1000 1111  192.10.10.143 Answer:  192.10.10.129 - 192.10.10.142

Problem 9 (page 40) Number of needed usable subnets

Not given

Number of needed usable hosts Network Address Address class Default subnet mask

28 172.50. 0.0 B 255.255.0.0

172.50.00000000.000 00000 Number of bits borrowed Custom subnet mask Total number of subnets Number of usable subnets Total number of host addresses Number of usable addresses

11 255.255.255.224 2048 2046 32 30

172.50.00000000.000 00000  sub net 0 172.50.00000000.001 00000  sub net 1 (first usable subnet) 172.50.00000000.010 00000  sub net 2 (second usable subnet) 

172.50.11111111.111 00000  sub net 2047

What is the 1st usable subnet range ? 172.50.00000000.001 00000

 175.50.0.32

 

172.50.00000000.001 11111

 172.50.0.63

Answer:  175.50.0.32 - 175.50.0.63

What is the subnet number for the 9th usable subnet? 172.50.00000001.001 00000

 172.50.1.32

 

172.50.00000001.001 11111  172.50.1.63 Answer:  175.50.1.32 What is the subnet broadcast address for the 3rd usable subnet ? 172.50.00000000.011 00000

 172.50.0.96

 

172.50.00000000.011 11111

 172.50.0.127

Answer:  172.50.0.127

What are the assignable addresses for the 5th usable subnet? 172.50.00000000.101 00000

 172.50.0.160

172.50.00000000.101 00001

 172.50.0.161



172.50.00000000.101 11110  172.50.0.190 172.50.00000000.101 11111

 172.50.0.191

Answer:  198.10.10.160 - 198.10.10.190

Classless Inter-Domain Routing (CIDR) For IPv4, CIDR notation is an alternative to the older system of representing networks by their starting address and the subnet mask, both written in dotdecimal notation.  subnet portion of address of arbitrary length  address format: a.b.c.d/x, where x is # bits in subnet portion of address

200.23.16.0/23

subnet part

host part

11001000 00010111 0001000 000000000 For example: 

192.168.100.14/24 represents the IPv4 address 192.168.100.14 and its associated routing prefix 192.168.100.0, or equivalently, mask 255.255.255.0



the IPv4 block 192.168.100.0/22 represents the 1024 IPv4 addresses from 192.168.100.0 to 192.168.103.255.

192.168.100.0/22 192.168.100.0  192.168.011001 00.000000000   192.168.103.255  192.168.011001 11.11111111...