Master Of Engineering In Internetworking

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02 Nov 2017

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Lab Assignment #3

IPv4 Addressing

INWK 6111

SECTION – 1

Submitted By: Saair Ali Qureshi

B00618386

Question No.1

Binary

Decimal

MSC

128

+64

+32

+16

+8

+4

+2

LSB

+1

1

0

0

1

1

0

0

0

152

0

0

0

0

1

0

1

0

10

1

0

1

0

0

0

0

1

161

0

1

1

1

0

0

1

0

114

0

1

1

1

1

1

1

1

127

1

1

0

0

0

0

0

1

193

1

1

1

0

1

1

0

0

236

1

1

1

1

1

1

0

0

252

1

1

1

1

1

1

1

1

255

1

1

0

0

0

0

0

0

192

1

1

0

1

0

0

1

0

210

1

1

1

0

0

0

0

0

224

The following table shows example of binary numbers and their corresponding decimal values. Fill the rest of the table.

Question No.2

What is the maximum decimal value of an eight bit number?

WEIGHT

128

64

32

16

8

4

2

1

TOTAL = 255

BIT

1

1

1

1

1

1

1

1

Question No.3

Find the answer for the next two operations?

PART 1:

BINARY REPRESENTATION

1

1

0

0

1

1

0

0

1

1

0

1

0

1

1

1

1

1

0

0

0

1

0

0

AND

PART 2:

BINARY REPRESENTATION

0

1

1

1

1

1

0

0

0

0

0

1

0

1

1

1

0

0

0

1

0

1

0

0

AND

Question No.4

Find the network portion of the following IP address using their subnet masks?

Part 1:

DOTTED DECIMAL REPRESENTATION

126

26

67

11

255

255

248

0

126

26

64

0

AND

BINARY REPRESENTATION

| --------------------IP ADDRESS----------------------|

| 01111110 00011010 01000011 00001011 | IP ADDRESS

_AND_ | 11111111 11111111 11110000_00000000 |_SUBNET MASK______

| 01111110 00011010 01000000 00000000 | NETWORK PORTION

Part 2:

DOTTED DECIMAL REPRESENTATION

20

89

0

1

255

192

0

0

20

64

0

0

AND

BINARY REPRESENTATION

| --------------------IP ADDRESS----------------------|

| 00010100 01011011 00000000 00000001 | IP ADDRESS

_AND_ | 11111111 11000000 01000000_ 00000000 |_SUBNET MASK______

| 00010100 01000000 00000000 00000000 |NETWORK PORTION

Question No.5

Identify the class of each IP address and write the network and the host components.

Address

Class

Network

Host

172.16.1.1

B

172.16.0.0

0.0.1.1

245.233.1.1

E

Undefined

Undefined

15.0.16.9

A

15.0.0.0

0.0.16.9

180.25.0.1

B

180.25.0.0

0.0.0.1

125.125.125.125

A

125.0.0.0

0.125.125.125

192.168.6.1

C

192.168.6.0

0.0.0.1

10.10.10.0

A

10.0.0.0

0.10.10.0

Question No.6

How many network IDs and host IDs is available in classes A, B, and C?

0

Network (7 bits)

Host (24 bits)

Class A

Total Number of Networks = 2^7 = 128

Total Available Hosts = (2^24) – 2 = 16777214

1

0

Network (14 bits)

Host (16 bits)

Class B

Total Number of Networks = 2^14 = 16384

Total Available Hosts = (2^16) – 2 = 65534

1

1

0

Network (21 bits)

Host (8 bits)

Class C

Total Number of Networks = 2^21 = 2097152

Total Available Hosts = (2^8) – 2 = 62

Question No.7

Determine the subnet mask, the network ID, and the broadcast address in the following.

201.212.10.40/29

DOTTED DECIMAL REPRESENTATION

________________________________________

| 201.212.10.40 | IP ADDRESS

_AND_ | 255.255.255.248 |_SUBNET MASK______

| 210.212.10.40 | NETWORK PORTION

| 210.212.10.47 | BROADCAST ADDRESS

BINARY REPRESENTATION

______________________________________

| 11001001 11010100 00001010 00101|000 | IP ADDRESS

_AND_ | 11111111 11111111 11111111 11111|000 |_SUBNET MASK______

| 11001001 11010100 00001010 00101|000 | NETWORK PORTION

| 11001001 11010100 00001010 00101|111 | BROADCAST ADDRESS

115.16.193.6/21

DOTTED DECIMAL REPRESENTATION

___________________________________

| 115.16.193.6 | IP ADDRESS

_AND_ | 255.255.248.0 |_SUBNET MASK______

| 115.16.192.0 | NETWORK PORTION

| 115.16.199.255 | BROADCAST ADDRESS

BINARY REPRESENTATION

______________________________________

| 01110011 00010000 11000|001 00000110 | IP ADDRESS

_AND_ | 11111111 11111111 11111|000 00000000 |_SUBNET MASK______

| 01110011 00010000 11000|000 00000000 | NETWORK PORTION

| 01110011 00010000 11000|111 11111111 | BROADCAST ADDRESS

128.16.54.13/30

DOTTED DECIMAL REPRESENTATION

________________________________________

| 128.16.54.13 | IP ADDRESS

_AND_ | 255.255.255.252 | _SUBNET MASK______

| 128.16.54.12 | NETWORK PORTION

| 128.16.54.15 | BROADCAST ADDRESS

BINARY REPRESENTATION

______________________________________

| 10000000 00010000 00110110 000011|01 | IP ADDRESS

_AND_ | 11111111 11111111 11111111 111111|00 |_SUBNET MASK______

| 10000000 00010000 00110110 000011|00 | NETWORK PORTION

| 10000000 00010000 00110110 000011|11 | BROADCAST ADDRESS

153.50.6.27/25

DOTTED DECIMAL REPRESENTATION

________________________________________

| 153.50.6.27 | IP ADDRESS

_AND_ | 255.255.255.128 | _ SUBNET MASK______

| 153.50.6.0 | NETWORK PORTION

| 153.50.6.127 | BROADCAST ADDRESS

BINARY REPRESENTATION

______________________________________

| 10011001 00110010 00000110 0|0011011 | IP ADDRESS

_AND_ | 11111111 11111111 11111111 1|0000000 |_SUBNET MASK______

| 10011001 00110010 00000110 0|0000000 | NETWORK PORTION

| 10011001 00110010 00000110 0|1111111 | BROADCAST ADDRESS

Question No.8

Is it possible to further subnet the previous IP address? How many subnets can be created, if any? How many host in each?

Note: We can make subnet up to 30th bit in the 32-bit IP address.

201.212.10.40/29

In the case, we can borrow one more bit from the host part to make subnets. Following will be the total number subnets and usable host Ids.

CLASS C ADDRESS

-----------------NETWORK PORTION----------------- | SUBNET | HOST PORTION

11001001 11010100 00001010| 00101|000

11111111 11111111 11111111| 11111|000

Number of Subnets = 2^ (subnet bits) = 2^5 = 32 Subnets

Number of Usable Host ID = [2^ (host bits)] – 2 = (2^3) – 2 = 8-2 = 6 hosts / subnets

115.16.193.6/21

In the case, we can borrow 9 more bits from the host part to make subnets. Following will be the total number subnets and usable host Ids.

CLASS A ADDRESS

NETWORK PORTION |--------- SUBNET--------- -| HOST PORTION

01110011 | 00010000 11000|001 00000110

11111111 | 11111111 11111|000 00000000

Number of Subnets = 2^ (subnet bits) = 2^13 = 8192 Subnets

Number of Usable Host ID = [2^ (host bits)] – 2 = (2^12) – 2 = 4096-2 = 4094 hosts / subnets

128.16.54.13/30

In this scenario, we cannot borrow any bit from the host part to make the subnets as we at-least require 4 host address (i.e. 2 host for point to point communication, one for network address and one for broadcast address).

153.50.6.27/25

In the case, we can borrow 5 more bits from the host part to make subnet. Following will the total number subnets and usable host Ids.

CLASS B ADDRESS

----------------NETWORK PORTION--- |---- SUBNET------| HOST PORTION

10011001 00110010 | 00000110 0|0011011

11111111 11111111 | 11111111 1|0000000

Number of Subnets = 2^ (subnet bits) = 2^9 = 512 Subnets

Number of Usable Host ID = [2^ (host bits)] – 2 = (2^7) – 2 = 128-2 = 126 hosts / subnets

Question No.9

Which of the following address is invalid?

111.12.10.4/25

DOTTED DECIMAL REPRESENTATION

________________________________________

| 111.12.10.4 | IP ADDRESS

_AND_ | 255.255.255.128 |_ SUBNET MASK______

| 111.12.10.0 | NETWORK PORTION

| 111.12.10.1 | FIRST HOST ADDRESS

| 111.12.10.126 | LAST HOST ADDRESS

| 111.12.10.127 | BROADCAST ADDRESS

BINARY REPRESENTATION

_NETWORK PART___SUBNET PART_______HOST PART_________ __________

| 01101111 |00001100 00001010 0|0000100 | IP ADDRESS

_AND_ | 11111111 |11111111 11111111 1|0000000 |_SUBNET MASK______

| 01101111 |00001100 00001010 0|0000000 | NETWORK PORTION

| 01101111 |00001100 00001010 0|0000001 | FIRST HOST ADDRESS

| 01101111 |00001100 00001010 0|1111110 | LAST HOST ADDRESS

| 01101111 |00001100 00001010 0|1111111 | BROADCAST ADDRESS

Yes, The IP address 111.12.10.4 is a valid host address because it does not fall into either network address or broadcast address.

18.145.192.1/17

DOTTED DECIMAL REPRESENTATION

________________________________________

| 18.145.192.1 | IP ADDRESS

_AND_ | 255.255.128.0 |_ SUBNET MASK______

| 18.145.128.0 | NETWORK PORTION

| 18.145.128.1 | FIRST HOST ADDRESS

| 18.145.255.254 | LAST HOST ADDRESS

| 18.145.255.255 | BROADCAST ADDRESS

BINARY REPRESENTATION

_NETWORK PART___SUBNET PART_______HOST PART_________ __________

| 00010100 |10010001 1|1000000 00000001 | IP ADDRESS

_AND_ | 11111111 |11111111 1|0000000 00000000 |_SUBNET MASK______

| 00010100 |10010001 1|0000000 00000000 | NETWORK PORTION

| 00010100 |10010001 1|0000000 00000001 | FIRST HOST ADDRESS

| 00010100 |10010001 1|1111111 11111110 | LAST HOST ADDRESS

| 00010100 |10010001 1|1111111 11111111 | BROADCAST ADDRESS

Yes, The IP addresses 18.145.192.1is a valid host address because it does not fall into either network address or broadcast address.

125.23.13.11/31

We can not further subnet the subnetted IP address 125.23.13.11/31 as we atleast require 2 bit to represent 4 hosts (i.e to assign IP address to 2 hosts and one for network and one for broadcast).

22.0.0.2/9

DOTTED DECIMAL REPRESENTATION

_______________________________________

| 22.0.0.2 | IP ADDRESS

_AND_ | 255.128.0.0 |_ SUBNET MASK______

| 22.0.0.0 | NETWORK PORTION

| 22.0.0.1 | FIRST HOST ADDRESS

| 22.127.255.254 | LAST HOST ADDRESS

| 22.127.255.255 | BROADCAST ADDRESS

BINARY REPRESENTATION

_NETWORK PART___SUBNET PART_______HOST PART_________ __________

| 00010110 |0|0000000 00000000 00000010 | IP ADDRESS

_AND_ | 11111111 |1|0000000 00000000 00000000 |_SUBNET MASK______

| 00010110 |0|0000000 00000000 00000000 | NETWORK PORTION

| 00010110 |0|0000000 00000000 00000001 | FIRST HOST ADDRESS

| 00010110 |0|1111111 11111111 11111110 | LAST HOST ADDRESS

| 00010110 |0|1111111 11111111 11111111 | BROADCAST ADDRESS

Yes, The IP addresses 22.0.02 is a valid host address because it does not fall into either network address or broadcast address.

Question No.10

What is the minimum and maximum number of subnet bits in a Class A address? Give the subnets of both cases.

Minimum Subnets from Class A Address 10.128.0.1/9

CLASS A ADDRESS

NETWORK PORTION |-- SUBNET-- HOST PORTION

00001010 | 1|0000000 00000000 00000000

11111111 | 1|0000000 00000000 00000000

Number of Subnets = 2^ (subnet bits) = 2^1 = 2 Subnets

Number of Usable Host ID = [2^ (host bits)] – 2 = (2^24) – 2 = 8388608-2 = 8388606 hosts / subnets

Maximum Subnets from Class A Address 10.128.192.4/30

CLASS A ADDRESS

NETWORK PORTION |--------------------- SUBNET------------------- | HOST PORTION

00001010 | 10000000 11000000 000001|00

11111111 | 11111111 11111111 111111|00

Number of Subnets = 2^ (subnet bits) = 2^22 = 4194304 Subnets

Number of Usable Host ID = [2^ (host bits)] – 2 = (2^2) – 2 = 4-2 = 2 hosts / subnets

Question No.10

Using the address 172.16.0.0/16 assign subnet addresses to the network in the following figure. The serial links should be use masks of 30-bit length. Choose any mask length for the other networks. Network clouds are connected to each other by routers.

172.16.64.0/18

19111212221

172.16.128.8/30

172.16.128.16/30

172.16.128.4/30

172.16.196.0/22

172.16.200.0/22

172.16.208.0/22

172.16.216.12/30

192.16.128.20/30

172.16.216.8/30

172.16.216.4/30

172.16.216.16/30

192.16.128.24/30

172.16.199.0/25

172.16.198.128/25

172.16.198.0/25

172.16.197.128/25

172.16.197.0/25

172.16.196.128/25



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