Networking fundamentals adp3rd (Semester)

UNIVERSITY OF CENTRAL PUNJAB
SUBMITTED TO:
SIR RAO RASHID
SUBMITTED BY:
ATTIQ UR REHMAN CHISHTI
ROLL NO:
43
NETWORKING FUNDAMENTALS

ADP3RD(SEMESTER)

Review Questions
Q1.What is the number of bits in an IPv4 address? What is the number of bits in
an
IPv6 address?
An IPv4 address is 32 bits long. An IPv6 address is 128 bits long.
Q2.What is dotted decimal notation in IPv4 addressing? What is the number of
bytes
in an IPv4 address represented in dotted decimal notation? What is hexadecimal


notation in IPv6 addressing? What is the number of digits in an IPv6 address
represented
in hexadecimal notation?

Q3.IPv4 addresses are usually written in decimal form with a decimal point (dot)
separating
the bytes. This is called dotted-decimal notation. Each address is 4 bytes.
IPv6 addresses are usually written in hexadecimal form with a colon separating the
bytes. This is called hexadecimal notation. Each address is 16 bytes or 32
hexadecimal

digits.
Q4.What are the differences between classful addressing and classless
addressing in IPv4?
Classful addressing assigns an organization a Class A, Class B, or Class C block
of addresses. Classless addressing assigns an organization a block of contiguous
addresses based on its needs.
Q5.List the classes in classful addressing and define the application of each
class (unicast,
multicast, broadcast, or reserve).

Classes A, B, and C are used for unicast communication. Class D is for multicast
communication and Class E addresses are reserved for special purposes.

Q6.Explain why most of the addresses in class A are wasted. Explain why a
medium-size
or large-size corporation does not want a block of class C addresses.
A block in class A address is too large for almost any organization. This means
most of the addresses in class A are wasted and not used. A block in class C is
probably too small for many organizations.

Q7.What is a mask in IPv4 addressing? What is a default mask in IPv4
addressing?
A mask in classful addressing is used to find the first address in the block when
one of the addresses is given. The default mask refers to the mask when there is no


subnetting or supernetting.

Q8.What is the network address in a block of addresses? How can we find the
network

address if one of the addresses in a block is given?
The network address in a block of addresses is the first address. The mask can be
ANDed with any address in the block to find the network address.

Q9.Briefly define subnetting and supemetting. How do the subnet mask and
supemet
mask differ from a default mask in classful addressing?
In subnetting, a large address block could be divide into several contiguous groups
and each group be assigned to smaller networks called subnets. In supernetting,
several small address blocks can be combined to create a larger range of addresses.
The new set of addresses can be assigned to a large network called a supernet. A
subnet mask has more consecutive 1s than the corresponding default mask. A
supernet mask has less consecutive 1s than the corresponding default mask.

Q10.How can we distinguish a multicast address in IPv4 addressing? How can we
do so
in IPv6 addressing?
Multicast addresses in IPv4 are those that start with the 1110 pattern. Multicast
addresses in IPv6 are those that start with the 11111111 pattern.
What is NAT? How can NAT help in address depletion?
Home users and small businesses may have created small networks with several
hosts and need an IP address for each host. With the shortage of addresses, this is a
serious problem. A quick solution to this problem is called network address translation
(NAT). NAT enables a user to have a large set of addresses internally and
one address, or a small set of addresses, externally. The traffic inside can use the
large set; the traffic outside, the small set

Exercises
11. What is the address space in each of the following systems?

a. A system with 8-bit addresses
b. A system with 16-bit addresses
c. A system with 64-bit addresses
A:
a. 28 = 256
b. 216 = 65536
c. 264 = 1.846744737 × 1019
12. An address space has a total of 1024 addresses. How many bits are needed to
represent
an address?
A:
2x = 1024 → x = log21024 = 10
13. An address space uses the three symbols 0, 1, and 2 to represent addresses.
If each address is made of 10 symbols, how many addresses are available in this
system?
A:
310 = 59,049
14. Change the following IP addresses from dotted-decimal notation to binary
notation.
a. 114.34.2.8
b. 129.14.6.8
c. 208.34.54.12
d. 238.34.2.1
A:
a. 01110010 . 00100010. 00000010. 00001000
b. 10000001 00001110 00000110 00001000
c. 11010000 00100010 00110110 00001100
d. 11101110 00100010 00000010 00000001
15. Change the following IP addresses from binary notation to dotted-decimal

notation.
a. 01111111 11110000 01100111 01111101
b. 10101111 11000000 11111000 00011101
c. 11011111 10110000 00011111 01011101
d. 11101111 11110111 11000111 00011101
A:
a. 127.240.103.125
b. 175.192.240.29
c. 223.176.31.93
d. 239.247.199.29
16. Find the class of the following IP addresses.


a. 208.34.54.12
b. 238.34.2.1
c. 114.34.2.8
d. 129.14.6.8
A:
a. Class C (first byte is between 192 and 223)
b. Class D (first byte is between 224 and 239)
c. Class A (first byte is between 0 and 127)
d. Class B (first byte is between 128 and 191)
17. Find the class of the following IP addresses.
a. 11110111 11110011 10000111 11011101
b. 10101111 11000000 11110000 00011101
c. 11011111 10110000 00011111 01011101
d. 11101111 11110111 11000111 00011101
A:
a. Class E (first four bits are 1s)
b. Class B (first bit is 1 and second bit is 0)

c. Class C (first two bits are 1s and the third bit is 0)
d. Class D (first three bits are 1s and the fourth bit is 0)
18. Find the netid and the hostid of the following IP addresses.
a. 114.34.2.8
b. 132.56.8.6
c. 208.34.54.12
A:
a. netid: 114 hostid: 34.2.8
b. netid: 132.56 hostid: 8.6
c. netid: 208.34.54 hostid: 12
19. In a block of addresses, we know the IP address of one host is 25.34.12.56/16.
What are the first address (network address) and the last address (limited
broadcast
address) in this block?
A:
With the information given, the first address is found by ANDing the host address
with the mask 255.255.0.0 (/16).
Host Address: 25 . 34 . 12 . 56
Mask (ANDed): 255 . 255 . 0 . 0
Network Address (First): 25 . 34 . 0 . 0

The last address can be found by ORing the host address with the mask complement
0.0.255.255.


Host Address: 25 . 34 . 12 . 56
Mask Complement (ORed): 0 . 0 . 255 . 255
Last Address: 25 . 34 . 255 . 255
However, we need to mention that this is the largest possible block with 216
addresses. We can have many small blocks as long as the number of addresses

divides this number.
20. In a block of addresses, we know the IP address of one host is
182.44.82.16/26.
What are the first address (network address) and the last address in this block?
A:
With the information given, the first address is found by ANDing the host address
with the mask 255.255.255.192 (/26).
Host Address: 182 . 44 . 82 . 16
Mask (ANDed): 255 . 255 . 255 . 192
Network Address (First): 182 . 44 . 82 . 0
The last address can be found by ORing the host address with the mask complement
0.0.0.63.
Host Address: 182 . 44 . 82 . 16
Mask Complement (ORed): 0 . 0 . 0 . 63
Last Address: 182 . 44 . 82 . 63
However, we need to mention that this is the largest possible block with 26
addresses. We can have several small blocks as long as the number of addresses
divides this number.
21. An organization is granted the block 16.0.0.0/8. The administrator wants to
create
500 fixed-length subnets.
a. Find the subnet mask.
b. Find the number of addresses in each subnet.
c. Find the first and last addresses in subnet 1.
d. Find the first and last addresses in subnet 500.
A:
a. log2500 = 8.95 Extra 1s = 9 Possible subnets: 512 Mask: /17 (8+9)
b. 232−17 = 215 = 32,768 Addresses per subnet
c. Subnet 1: The first address in the this address is the beginning address of the
block or 16.0.0.0. To find the last address, we need to write 32,767 (one less

than the number of addresses in each subnet) in base 256 (0.0.127.255) and add
it to the first address (in base 256).

First address in subnet 1: 16 . 0 . 0 . 0
Number of addresses: 0 . 0 . 127 . 255
Last address in subnet 1: 16 . 0 . 127 . 255


d. Subnet 500:
Note that the subnet 500 is not the last possible subnet; it is the last subnet used
by the organization. To find the first address in subnet 500, we need to add
16,351,232 (499 × 32678) in base 256 (0. 249.128.0) to the first address in subnet
1. We have 16.0.0.0 + 0.249.128.0 = 16.249.128.0. Now we can calculate
the last address in subnet 500.
First address in subnet 500: 16 . 249 . 128 . 0
Number of addresses: 0 . 0 . 127 . 255
Last address in subnet 500: 16 . 249 . 255 . 255
22. An organization is granted the block 130.56.0.0/16. The administrator wants to
create 1024 subnets.
a. Find the subnet mask.
b. Find the number of addresses in each subnet.
c. Find the first and last addresses in subnet 1.
d. Find the first and last addresses in subnet 1024.
A:
a. log21024 = 10 Extra 1s = 10 Possible subnets: 1024 Mask: /26
b. 232− 26 = 64 Addresses per subnet
c. Subnet 1:
The first address is the beginning address of the block or 130.56.0.0. To find the
last address, we need to write 63 (one less than the number of addresses in each
subnet) in base 256 (0.0.0.63) and add it to the first address (in base 256).

First address in subnet 1: 130 . 56 . 0 . 0
Number of addresses: 0 . 0 . 0 . 63
Last address in subnet 1: 130 . 56 . 0 . 63
d. Subnet 1024:
To find the first address in subnet 1024, we need to add 65,472 (1023 × 64) in
base 256 (0.0.255.92) to the first address in subnet 1. We have 130.56.0.0. +
0.0.255.192 = 130.56.255.192. Now we can calculate the last address in subnet
500 as we did for the first address.
First address in subnet 1024: 130 . 56 . 255 . 192
Number of addresses: 0 . 0 . 0 . 63
Last address in subnet 1024: 130 . 56 . 255 . 255
23. An organization is granted the block 211.17.180.0/24. The administrator wants
to
create 32 subnets.
a. Find the subnet mask.
b. Find the number of addresses in each subnet.
c. Find the first and last addresses in subnet 1.
d. Find the first and last addresses in subnet 32.
A:
a. log232 = 5 Extra 1s = 5 Possible subnets: 32 Mask: /29 (24 + 5)


b. 232− 29 = 8 Addresses per subnet
c. Subnet 1:
The first address is the beginning address of the block or 211.17.180.0. To find
the last address, we need to write 7 (one less than the number of addresses in
each subnet) in base 256 (0.0.0.7) and add it to the first address (in base 256).
First address in subnet 1: 211 . 17 . 180 . 0
Number of addresses: 0 . 0 . 0 . 7
Last address in subnet 1: 211 . 17 . 180 . 7

d. Subnet 32:
To find the first address in subnet 32, we need to add 248 (31 × 8) in base 256
(0.0.0.248) to the first address in subnet 1. We have 211.17.180.0 + 0.0.0.248 or
211.17.180.248. Now we can calculate the last address in subnet 32 as we did
for the first address.
First address in subnet 32: 211 . 17 . 180 . 248
Number of addresses: 0 . 0 . 0 . 7
Last address in subnet 32: 211 . 17 . 180 . 255
24. Write the following masks in slash notation (In).
a. 255.255.255.0
b. 255.0.0.0
c. 255.255.224.0
d. 255.255.240.0
A:
a. The mask 255.255.255.0 has 24 consecutive 1s → slash notation: /24
b. The mask 255.0.0.0 has 8 consecutive 1s → slash notation:/8
c. The mask 255.255.224.0 has 19 consecutive 1s → slash notation:/19
d. The mask 255.255.240.0 has 20 consecutive 1s → slash notation:/20
25. Find the range of addresses in the following blocks.
a. 123.56.77.32/29
b. 200.17.21.128/27
c. 17.34.16.0/23
d. 180.34.64.64/30
A:
a. The number of address in this block is 232−29 = 8. We need to add 7 (one less)
addresses (0.0.0.7 in base 256) to the first address to find the last address.
From: 123 . 56 . 77 . 32
0.0.0.7
To: 123 . 56 . 77 . 39
b. The number of address in this block is 232−27 = 32. We need to add 31 (one less)

addresses (0.0.0.31 in base 256) to the first address to find the last address.
From: 200 . 17 . 21 . 128
0 . 0 . 0 . 31
To: 200 . 17 . 21 . 159


The number of address in this block is 232−23 = 512. We need to add 511 (one
less) addresses (0.0.1.255 in base 256) to the first address to find the last
address.
From: 17 . 34 . 16 . 0
0 . 0 . 1 . 255
To: 17 . 34 . 17 . 255
d. The number of address in this block is 232−30 = 4. We need to add 3 (one less)
addresses (0.0.0.3 in base 256) to the first address to find the last address.
From: 180 . 34 . 64 . 64
0.0.0.3
To: 180 . 34 . 64 . 67
26. An ISP is granted a block of addresses starting with 150.80.0.0/16. The ISP
wants
to distribute these blocks to 2600 customers as follows.
a. The first group has 200 medium-size businesses; each needs 128 addresses.
b. The second group has 400 small businesses; each needs 16 addresses.
c. The third group has 2000 households; each needs 4 addresses.
Design the subblocks and give the slash notation for each subblock. Find out
how
many addresses are still available after these allocations.
A:
The total number of addresses in this block is 232-16 = 65536. The ISP can divide
this large block in several ways depending on the predicted needs of its customers
in the future. We assume that the future needs follow the present pattern. In other

words, we assume that the ISP will have customers that belong to one of the
present groups. We design four ranges: group 1, group 2, group 3, and one reserved
range of addresses as shown in figure.

Group 1
In the first group, we have 200 businesses. We augment this number to 256 (the


next number after 200 that is a power of 2) to let 56 more customers of this kind
in the future. The total number of addresses is 256 × 128 = 32768. For this group,
each customer needs 128 addresses. This means the suffix length is log2128 = 7.
The prefix length is then 32 − 7 = 25. The addresses are:
1st customer: 150.80.0.0/25 to 150.80.0.127/25
2nd customer: 150.80.0.128/25 to 150.80.0.255/25
... ... ...
200th customer: 150.80.99.128/25 to 150.80.99.255/25
Unused addresses 150.80.100.0 to 150.80.127.255
Total Addresses in group 1 = 256 × 128 = 32768 Used = 200 × 128 = 25600.
Reserved: 7168, which can be assigned to 56 businesses of this size.
Group 2
In the second group, we have 400 business. We augment this number to 512 (the
next number after 400 that is a power of 2) to let 112 more customer of this kind in
the future. The total number of addresses is = 512 × 16 = 8192. For this group,
each customer needs 16 addresses. This means the suffix length is 4 log216 = 4.
The prefix length is then 32 − 4 = 28. The addresses are:
1st customer: 150.80.128.0/28 to 150.80.128.15/28
2nd customer: 150.80.128.16/28 to 150.80.128.31/28
... ... ...
400th customer: 150.80.152.240/28 to 150.80.152.255/28
Unused addresses 150.80.153.0 to 150.80.159.255

Total Addresses in group 2 = 512 × 16 = 8192 Used = 400 × 16 = 6400
Reserved: 1792, which can be assigned to 112 businesses of this size.
Group 3
In the third group, we have 2000 households. We augment this number to 2048
(the next number after 2000 that is a power of 2) to let 48 more customer of this
kind in the future. The total number of addresses is = 2048 × 4 = 8192. For this
group, each customer needs 4 addresses. This means the suffix length is 2 log24 =
2. The prefix length is then 32 − 2 = 30. The addresses are:
1st customer: 150.80.160.0/30 to 150.80.160.3/30
2nd customer: 150.80.160.4/30 to 150.80.160.7/30
... ... ...
2000th customer: 150.80.191.60/30 to 150.80.191.63/30
Unused addresses 150.80.191.64 to 150.80.191.255
Total Addresses in group 3 = 2048 × 4 = 8192 Used = 2000 × 4 = 8000
Reserved: 192, which can be assigned to 48 households.
Reserved Range
In the reserved range, we have 16384 address that are totally unused.
27. An ISP is granted a block of addresses starting with 120.60.4.0/22. The ISP
wants to distribute these blocks to 100 organizations with each organization


receiving just eight addresses. Design the subblocks and give the slash notation
for each subblock. Find out how many addresses are still available after these
allocations.
A:
The site has 232−22 = 210 = 1024 from 120.60.4.0/22 to 120.60.7.255/22 addresses.
One solution would be to divide this block into 128 8-address sub-blocks as shown
in Figure 19.2. The ISP can assign the first 100 sub-blocks to the current customers
and keep the remaining 28 sub-blocks. Of course, this does not mean the future
customer have to use 8-address subblocks. The remaining addresses can later be

divided into different-size sub-blocks (as long as the three restrictions mentioned
in this chapter are followed). Each sub-block has 8 addresses. The mask for each
sub-block is /29 (32 − log28). Note that the mask has changed from /22 (for the
whole block) to /29 for each subblock because we have 128 sub-blocks (27 = 128).
Sub-blocks:
1st subnet: 120.60.4.0/29 to 120.60.4.7/29
2nd subnet: 120.60.4.8/29 to 120.60.4.15/29
... ... ...
32nd subnet: 120.60.4.248/29 to 120.60.4.255/29
33rd subnet: 120.60.5.0/29 to 120.60.5.7/29
... ... ...
64th subnet: 120.60.5.248/29 to 120.60.5.255/29
... ... ...
99th subnet: 120.60.7.16/29 to 120.60.7.23/29
100th subnet: 120.60.7.24/29 to 120.60.7.31/29
1024 − 800 = 224 addresses left (from 120.60.7.31 to 120.60.7.155)
28. An ISP has a block of 1024 addresses. It needs to divide the addresses among
1024 customers. Does it need subnetting?
A:
Each customer has only 1 address and, therefore, only one device. Since we
defined a network as 2 or more connected devices, this is not a network

29. Show the shortest form of the following addresses.
a. 2340: lABC:119A:AOOO:0000:0000:0000:0000
b. OOOO:OOAA:OOOO:OOOO:OOOO:OOOO: 119A:A231
c. 2340:0000:0000:0000:0000: 119A:AOO1:0000
d. 0000:0000:0000:2340:0000:0000:0000:0000
A:

a. 2340:1ABC:119A:A000::0
b. 0:AA::119A:A231
c. 2340::119A:A001:0
d. 0:0:0:2340::0
30. Show the original (unabbreviated) form of the following addresses.
a. 0::0
b.O:AA::O
c. 0: 1234::3
d. 123::1:2
A:
a. 0000:0000:0000:0000:0000:0000:0000:0000
b. 0000:00AA:0000:0000:0000:0000:0000:0000
c. 0000:1234:0000:0000:0000:0000:0000:0003
d. 0123:0000:0000:0000:0000:0000:0001:0002
31. What is the type of each of the following addresses?
a. FE80::12
b. FECO: :24A2
c. FF02::0
d. 0::01
A:
a. Link local address
b. Site local address
c. Multicast address (permanent, link local)
d. Loopback address
32. What is the type of each of the following addresses?
a. 0::0
b. 0: :FFFF:O:O
c. 582F:1234::2222
d. 4821::14:22
e. 54EF::A234:2

A:
a. Unspecified address
b. Mapped address
c. Provider based address with the address registered through INTERNIC (North
American registry).
d. Provider based address with the address registered through RIPNIC (European
registry).
e. Provider based address with the address registered through APNIC (Asian/
Pacific registry).
33. Show the provider prefix (in hexadecimal colon notation) of an address
assigned
to a subscriber if it is registered in the United States with ABC1 as the provider
identification.


A:
58ABC1
34. Show in hexadecimal colon notation the IPv6 address
a. Compatible to the IPv4 address 129.6.12.34
b. Mapped to the IPv4 address 129.6.12.34
A:
a. 0000:0000:0000:0000:0000:0000:8106:0C22 or 0::8106:C22
b. 0000:0000:0000:0000:0000:FFFF:8106:0C22 or 0::FFFF:8106:C22
35. Show in hexadecimal colon notation
a. The link local address in which the node identifier is 0:: 123/48
b. The site local address in which the node identifier is 0:: 123/48
A:
a. FE80:0000:0000:0000:0000:0000:0000:0123 or FE80::123
b. FEC0:0000:0000:0000:0000:0000:0000:0123 or FEC0::123
36. Show in hexadecimal colon notation the permanent multicast address used in

a link
local scope.
A:
FF02: < Group ID >
37. A host has the address 581E: 1456:2314:ABCD:: 1211. If the node
identification is
48 bits, find the address of the subnet to which the host is attached.
A:
The node identifier is 0000:0000:1211. Assuming a 32-bit subnet identifier, the
subnet address is 581E:1456:2314:ABCD:0000 where ABCD:0000 is the subnet
identifier.

Research Activities:
Find the block of addresses assigned to your organization or institution.
A:
The Internet Assigned Numbers Authority (IANA) has reserved the following three
blocks of the IP address space for private internets:
10.0.0.0

- 10.255.255.255 (10/8 prefix)

172.16.0.0

- 172.31.255.255 (172.16/12 prefix)

192.168.0.0

- 192.168.255.255 (192.168/16 prefix)

We will refer to the first block as "24-bit block", the second as "20-bit block", and to
the third as "16-bit" block. Note that (in pre-CIDR notation) the first block is nothing but a
single class A network number, while the second block is a set of 16 contiguous class B
network numbers, and third block is a set of 256 contiguous class C network numbers.
An enterprise that decides to use IP addresses out of the address space defined in
this document can do so without any coordination with IANA or an Internet registry. The
address space can thus be used by many enterprises. Addresses within this private
address space will only be unique within the enterprise, or the set of enterprises which
choose to cooperate over this space so they may communicate with each other in their
own private internet.
The address space can thus be used by many enterprises. Addresses within this private
address space will only be unique within the enterprise, or the set of enterprises which
choose to cooperate over this space so they may communication with each in their own
private internet.

Some people argue that we can consider the whole address space as one single
block in which each range of addresses is a sub block to this single block.
Elaborate
on this idea. What happens to subnetting if we accept this concept?
A:
We can consider the whole address space as single block, but there exits some
problems as:
Network becomes too large and complex
Involves limited network
Non centralized network
Complex routing process

So , in order to revolve these problems, subnetting is introduces as :
The routing tables become much shorter by using the subnet addresses and this means
that the data flow also becomes faster.

By using subnet addresses the entire IP address system becomes centralized, he
network administrators’ can assign IP numbers to specific departments in an
organization so that when they can easily identify the departments on the network using
the same numbers. Subnet also divides the bigger network into smaller individual
network.
Also it removes physical barriers and since the networks now are individually identified
all the networks can be connected through routers.