Introduction To Computer Networks

Published Date: 02 Nov 2017

INWK6111

Tariq Ayub

B00603054

Group 4

Submitted in partial fulfilment of the requirements for the degree of

MASTER OF ENGINEERING

Major Subject: Internetworking

DALHOUSIE UNIVERSITY

Halifax, Nova Scotia

February, 2013

ï¿½ Copyright by Tariq Ayub, 2013

this page is

left blank

intentionally

Dalhousie University

Faculty of Engineering

Internetworking

The undersigned hereby certify that they have read and award a pass in INWK 6111 for the course entitled " Introduction to Computer Networks " by Tariq Ayub - B00603054 in partial fulfilment of the requirements for the degree of Master of Engineering.

___________________________

DALHOUSIE UNIVERSITY

INTERNETWORKING PROGRAM

AUTHORITY TO DISTRIBUTE REPORT

Title:

Serial Connections,Ipv4 Addressing, VLANs / Ethernet Networks And RIP

The Internetworking Program may make available or authorise others to make available individual photo / microfilm or soft copies of this report without restrictions after

February 15, 2013.

The author attests that permission has been obtained for the use of any copyrighted material appearing in this report (other than brief excerpts requiring only proper acknowledgement in scholarly writing) and that all such use is clearly acknowledged.

Full Name of Author: Tariq Ayub

Signature of Author: _________________________

Date: _________________________

iii

LIST OF CONTENTS

LIST OF TABLES................................................................................................................................vi

LIST OF FIGURES............................................................................................................................vii

LIST OF SYMBOLS AND ABBREVIATIONS..............................................................................viii

ACKNOWLEDGEMENTS............................................................................................................... ix

EXECUTIVE SUMMARY...................................................................................................................x

1 INTRODUCTION..............................................................................................................................1

1.1 SERIAL CONNECTIONS

1.2 PHYSICAL LAYER PARAMETERS

1.3 DATA LINK LAYER PARAMETERS

1.4 NETWORK LAYER PARAMETERS

1.5 FIRST SET OF RESULTS FROM THE LABORATORY WORK

1.6 SECOND SET OF RESULTS FROM THE LABORATORY WORK

1.7 ASYNCHRONOUS LINES

1.7.1 CONFIGURATION OF ASYNCHRONOUS INTERFACE

1.8 THIRD SET OF RESULTS FROM THE LABORATORY WORK

1.9 CONCLUSION

2 IP ADDRESS Ipv4

2.1 INTRODUCTION

2.2 BINARY OPERATIONS

2.3 NETWORK ADDRESSES

2.4 NETWORK Ids AND HOST Ids

2.5 SUBNET MASK

2.6 VARIABLE-LENGTH SUBNET MASK (VLSM)

2.6.1 GIVEN SCENARIO AND ITS SOLUTION USING VLSM

2.7 CONCLUSION

LIST OF TABLES

LIST OF FIGURES

1 INTRODUCTION

Before going into details about laboratory assignments, tasks given, solutions and what have we learnt from them? It is better to get familiar with the background information about the Lab topics such as serial connections, IPv4 addressing, VLANs / Ethernet networks and RIP networks. It will help in understanding the issues which arose during the Lab work in designing, implementing and configuring different devices. Here are descriptions about some technologies related to lab work, the commands used to implemented those technologies and the results which came out after performing different configurations. These are necessary to know in order to get the idea about our learning during Laboratory assignments. There is also a brief information about the different commands we used.

1.1 Serial Connections:-

Most Wide Area Network (WAN) technologies like X.25, Frame Relay, Asynchronous Transfer Mode (ATM), Fibre Channels, leased lines and Integrated Services Digital Network (ISDN) are made to use different types of connections including serial connections. Every above mentioned technologies have their own physical interface standards. A synchronous transmission is the mode of the most serial connections. A DB-60 V.35 connector is the common interface for these connections. Normally this interface is known as the serial interface. The cables with DB-60 connector at both the ends or in some cases Winchester-type connector is used at the one end is required for this interface.

1.2 Physical Layer Parameters:

In WAN technologies the service provider side is called Data Circuit-terminating Equipment (DCE) network and the customer side is called a Data Terminal Equipment (DTE). Most of the connections are made to work between these two. Only DCE node provides the signal timing.

Interface can be verified by using the following command:

Router# show controller serial <interface number>

If one wants to determine the bit rate of the line, the clock rate command is . Used. The standard serial interface supports bit rates up to 2Mbps, which is also the default rate. Other types of serial interfaces support bit rates up to 52Mbps.

During the Lab work we used the following command to change the bit rate.

Router(config-if)# clock rate <rate in bps>

We can change the bandwidth using bandwidth command.

Router(config-if)# bandwidth <kbps>

The actual bandwidth of the connection remains unchanged by this command. But the higher layers get the information of changed bandwidth.

When the activation of interface is required no shutdown command is used.

The status of the interfaces is very useful information during the

configuration, the following command is used to get this information but in

the EXEC mode.

show ip interface brief

Changes to the configuration of serial lines only implemented when the interface is shutdown then reactivated by no shutdown command. Non Return to Zero (NRZ) is the default data-encoding scheme implemented in standard serial lines. NRZ Inverted encoding scheme can also be used. A successful communication between the two ends of the serial link is only possible if the same encoding scheme is implemented at the ends. Desired encoding scheme can be implemented using following command. Obviously at both the ends.

Router(config-if)# nrzi-encoding

To activate default scheme , this command is used:

Router(config-if)# no nrzi-encoding

1.3 Data Link Layer Parameters:

Many data link protocols like High Level Data Link Control (HDLC), Synchronous data Link Control (SDLC), Point-to-Point Protocol (PPP), LAPB, etc. are used to encapsulated the data moving across the serial line. Out of above mentioned data link protocols, HDLC is the most used protocol. Cisco has its own standards for implementation of its HDLC, which is also proprietary and has compatibility issues with other's implementations. List of supported protocol can be viewed using the encapsulation command.

Encapsulation command is also used to change the required protocol.

Router(config-if)# encapsulation ppp

1.4 Network Layer Parameters:

By definition when two network nodes have direct connection with each other, it referred as a point-to-point connection. In such connections there is only one path for the data to flow that is form source to destination so a network layer address is not seemed necessary in this case. However, in a complex network, especially for the routing purpose it is useful for each interface to have unique address while configuring serial links. For example, We can assign an IP address to serial interface using IP address command as follows.

Router(config-if)# ip address <address> <mask>

As we know that serial interfaces don't need an IP address as it is a point to point connection. In such situation serial interfaces can ï¿½borrowï¿½ an IP address from another interface to use briefly. This borrowing process can be done using the command ip unnumbered.

The following commands are useful in monitoring interface.

Router# show run

Router# show run interface serial <interface number>

Router# show controller serial <interface number>

Router# show interface serial <interface number>

1.5 First set of Results from the Laboratory Work:

a) List of all serial interfaces in the assigned router.

Assigned Router 4R1

Serial Interfaces

ï¿½ S 0/0/0

ï¿½ S 0/0/1

b) Visual inspection of a V.35 cable with a Winchester connector.

The Winchester does not connect to a DTE device or even a DCE directly,

the DB-60 to Winchester male connector connects to a DB-60 winchester

female connector which connects one end to the DTE device and the other

end of winchester to Db-60 male converter connects to a DCE device.

There is no connector on the DTE or DCE that accepts the Winchester

connector, the Winchester connector is sometimes connected to a patch

panel.

c) High Speed Serial Interface port.

It is developed by Cisco Systems and T3plus Networking primarily for use

in WAN router connections. But it is widely used to connect routers and

switches on on local area networks with the high-speed lines of a wide area

network. It is a short-distance communications interface used where the

distance between the devices is fifty feet or less and can attain data rate up

to 52 Mbps.

d) Default bandwidth for a T1 line in Cisco router.

Default Bandwidth is 1.544 Mbps.

e) Encoding schemes used in Cisco routers. List of two such schemes and the

serial interfaces for which they are used.

Nonreturn-to-zero Line ï¿½coding

Nonreturn-to-zero Inverted Line-coding

Serial Interfaces in Cisco Routers:

ï¿½ High-Speed Serial Interface

ï¿½ Low-Speed Serial Interface

ï¿½ Synchronous Serial Interface

f) List of data link protocols that are available in the standard serial interface.

By using the encapsulation command.

- Point-to-Point Protocol (PPP)

- High Level Data Link Control (HDLC)

- Synchronous Data Link Control (SDLC)

- SDLC primary and secondary

- Link Access Procedures, D channel (LAPB)

- Block Serial Tunneling (BSTUN)

- Serial Line IP (SLIP)

- Frame Relay

- Simple Traversal of UDP through NAT (STUN)

- Switched Multi-megabit Data Service (SMDS)

- X25

g) Is it possible to connect Cisco router with Nortel router using HDLC? Can

you do that with PPP?

The Cisco version of HDLC is not compatible with another router vendor,

so we cannot use the HDLC on a cisco router with another vendor. We can

use the Point to Point Protocol (PPP) to connect the Cisco router with the

Nortel router because the PPP is not vendor specific.

1.6 Second Set of Results from the Laboratory Work:

a) Cisco HDLC versus Standard HDLC:

ï¿½ cHDLC uses alternate framing structure

ï¿½ It has a frame to identify network protocol

Itï¿½s hard for Analyzer to identify the traffic in the network.

b) Basically there are not two different control fields. First one is the real control field which contains the destination address where frame has to be sent. The other control field is actually Frame Check Sequence Field. It is an error detecting code which uses CRC(cyclic redundancy check) scheme.

c) The packet format for PPP protocol.

Name Number of Bytes Description.

Flag........................... 1........................................... indicates frame's begin or end

Address...................... 1........................................... broadcast address

Control....................... 1........................................... control byte

Protocol..................... 1 or 2.................................... l in information field

Information................ variable (0 or more)...............datagram

Padding...................... variable (0 or more)...............optional filler

FCS ï¿½....................... 2 (or 4).................................. error check

1.7 Asynchronous Lines

In modern networks, asynchronous transmission mode is not much used. This mode is normally used in following situations.

- to connect a PC to a printer (an EIA/TIA-232 cable),

- to connect a PC to Cisco router (a console port or a Modem),

- to connect a PC to the Internet (dial-up service).

- a backup connection between two nodes

Command used to see all asynchronous lines

Router# show line

* 0 CTY - - - - - 0 1 0/0 -

65 AUX 9600/9600 - - - - - 0 0 0/0 -

66 VTY - - - - - 0 0 0/0 -

67 VTY - - - - - 0 0 0/0 -

68 VTY - - - - - 0 0 0/0 -

69 VTY - - - - - 0 0 0/0 -

70 VTY - - - - - 0 0 0/0 -

Configuration of the line aux 0 is done as below:

Router(config)#line aux 0

Router(config-line)# modem InOut //Drop Outbound connections to the modem are allowed in this command.

Router(config-line)# speed 38400// Set the speed for transmit and receive

Router(config-line)# flowcontrol hardware // RTS/CTS flow control

1.7.1 Configuring an asynchronous interface involves four steps:

1) Establish an asynchronous interface by following command.. The value of line number depends on the class of router.

Router(config)# interface async 65 <or async 1, async 129>

2) Pick the required data link protocol.

Router(config-if)# encapsulation ppp

3) Enable dedicated mode

Router(config-if)# async mode dedicated

4) Assign an IP address

Router(config-if)# ip address 10.yx.0.x 255.255.0.0

no shutdown command is used to activate the line.

To see configuration use following command

Router# show ip interface brief

Router# show interface async 65

Router# show line 65

Router# ping 10.yx.0.10

1.8 Third set of Results from the Laboratory Work:

a)Why some transmission lines are called ï¿½asynchronousï¿½?

The reason is because transmission from transmitter to receiver is done intermittently, this means transmission starts and stops. Data is transmitted one character at a time, this transmission is characterized by start and stop bits to tell the receiver the status of transmission.

b)EIA/TIA-232 Cable Bit Rate:

The EIA/TIA 232 operates up to 115.2kbps

c)Various data link protocols for asynchronous transmission:

- Block Serial Tunneling (BSTUN)

- Serial Line IP (SLIP)

- Point to Point Protocol (PPP)

The default is the Serial Line IP (SLIP)

d)Which show command gives information about the speed of the line?

The command is as per below :

Show line

Use Router (Config) # ï¿½show line 129ï¿½ or ï¿½show Interface async 129ï¿½ Router 3640

e)Which show command shows the type of parity used?

ï¿½show line(n)ï¿½, where n specifies the line number

Use Router (Config) # ï¿½show line 129ï¿½ or ï¿½show Interface async 129ï¿½ Router 3640

1.9 Conclusion:

The above information is basically related to LAB Assignment# 2 which is about serial connections between different devices in different environments. Though I had studied serial connections before but the knowledge I acquired here is more practical and advance. I learnt advance information about asynchronous, synchronous mode. I studied different parameters of Physical layer, Datalink layer, Network layer and how they are implement through different commands.

2 IP Addresses (Ipv4)

2.1 Introduction

An internet protocol (IP) is an address which uniquely identify the device on the network for communication. An IP address is a 32 bit binary number which is represented by the 4 decimal values, each value is of 8 bits and the ranges from 0 to 255.

2.2 Binary Operations

IP addresses are written in a dotted decimal notation. Each decimal number represents one octet (eight bits).

Binary 10111101000010010000000100110001

Dotted decimal notation 189.9.1.49

Examples of binary numbers and their corresponding decimal

Binary Decimal

MSB

128

+64

+32

+16

+2 LSB

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

Table 1

The maximum decimal value of an eight-bit number is 255.

i.e. 128+64+32+16+8+4+2+1 = 255

The AND operation is defined as follows:

0 AND 0 = 0

0 AND 1 = 0

1 AND 0 = 0

1 AND 1 = 1

Examples 1 1 0 0 1 1 0 0

AND 1 1 0 1 0 1 1 1

= 1 1 0 0 0 1 0 0

0 1 1 1 1 1 0 0

AND 0 0 0 1 0 1 1 1

= 0 0 0 1 0 1 0 0

Network portion of the following IP address using their subnet masks:

126 . 26 . 67 . 11

AND 255 . 255 . 248 . 0

= 126 . 26 . 64 . 0

Binary calculation of above solution

0 1 1 1 1 1 1 0 . 0 0 0 1 1 0 1 0 . 0 1 0 0 0 0 1 1 . 0 0 0 0 1 0 1 1

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

= 0 1 1 1 1 1 1 0 . 0 0 0 1 1 0 1 0 . 0 1 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

Decimal values 126 26 64 0

20 . 89 . 0 . 1

AND 255 . 192 . 64 . 0

= 20 . 64 . 0 . 0

Binary calculation of above solution

0 0 0 1 0 1 0 0 . 0 1 0 1 1 0 0 1 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 1

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

= 0 1 1 1 1 1 1 0 . 0 0 0 1 1 0 1 0 . 0 1 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

Decimal values 20 64 0 0

2.3 Network addresses in each classes:

1 to 126 ï¿½ Class A address

128 to 191 ï¿½ Class B address

192 to 223 ï¿½ Class C address

224 to 239 ï¿½ Class D address

Address Class Network Host

172.16.1.1 B 172.16.0.0 0.0.1.1

245.233.1.1 E - -

15.0.16.9 A 15.0.0.0 0.1.1.1

180.25.0.1 B 180.25.0.0 0.0.1.1

125.125.125.125 A 125.0.0.0 0.1.1.1

192.168.16.1 C 192.168.16.0 0.0.0.1

10.10.10.0 A 10.0.0.0 0.1.1.1

Table 2

2.4 Network IDs and host IDs available in classes A,

B, and C

Classes A B C

Network IDs 8

2 16

2 24

Host IDs 24

2 16

2 8

Table 3

2.5 Subnet Mask

Sub netting enables us to get the most out of 32 bits. Either we can get

more hosts or more networks by using the method.

Determination of the subnet mask, the network ID, and the broadcast

address in the following examples.

a)- 201.212.10.40 / 29

Subnet mask = 255.255.255.248

Network ID = 201.212.10.40

Broadcast address = 201.212.10.47

Binary calculation

1 1 0 0 1 0 0 1 . 1 1 0 1 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 1 0 0 0

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

= 1 1 0 0 1 0 0 1 . 1 1 0 1 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 1 0 0 0

Decimal values 201 212 10 40

b)- 115.16.193.6 / 21

Subnet mask = 255.255.248.0

Network ID = 115.16.192.0

Broadcast address = 115.16.199.255

Binary calculation

0 1 1 1 0 0 1 1 . 0 0 0 1 0 0 0 0 . 1 1 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

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

= 0 1 1 1 0 0 1 1 . 0 0 0 1 0 0 0 0 . 1 1 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

Decimal values 115 16 192 0

c)- 128.16.54.13 / 30

Subnet mask = 255.255.255.252

Network ID = 128.16.54.12

Broadcast address = 128.16.54.15

Binary calculation

1 0 0 0 0 0 0 0 . 0 0 0 1 0 0 0 0 . 0 0 1 1 0 1 1 0 . 0 0 0 0 1 1 0 1

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

= 1 0 0 0 0 0 0 0 . 0 0 0 1 0 0 0 0 . 1 1 1 1 0 1 1 0 . 0 0 0 0 1 1 0 0

Decimal values 128 16 54 12

d)- 153.50.6.27 / 25

Subnet mask = 255.255.255.128

Network ID = 153.50.6.0

Broadcast address = 153.50.6.127

Binary calculation

1 0 0 1 1 0 0 1 . 0 0 1 1 0 0 1 0 . 0 0 0 0 0 1 1 0 . 0 0 0 1 1 0 1 1

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

= 1 0 0 1 1 0 0 1 . 0 0 1 1 0 0 1 0 . 0 0 0 0 0 1 1 0 . 0 0 0 0 0 0 0 0

Decimal values 153 50 6 0

2.6 Variable-length Subnet Masks (VLSM)

By using VLSM technique a network administrator can distribute given IP

address space into subnets of variable sizes

2.6.1 Scenario and its solution using VLSM:

Using the address 172.16.0.0/16, assign subnet addresses to the network

in the following figure. The serial links should use masks of 30-bit length.

Choose any mask length for the other networks. Network clouds are

connected to each other by routers.

Figure 1

Solution-

According to given diagram there are 15 different networks and 14 serial links,

which are also networks. Starting assigning subnet addresses from the top.

a) For 500 Hosts

Network address = 172.16.0.0 / 23

Range of the addresses = 172.16.0.0 ï¿½ 172.16.1.255

First host = 172.16.0.1

Last host = 172.16.1.254

Broadcast address = 172.16.1.255

Number of hosts = 2^n-2=2^9-2=510

b) For 300 Hosts

Network address = 172.16.2.0 / 23

Range of the addresses = 172.16.2.0 ï¿½ 172.16.3.255

First host = 172.16.2.1

Last host = 172.16.3.254

Broadcast address = 172.16.3.255

Number of hosts = 2^n-2=2^9-2=510

c) For 300 Hosts

Network address = 172.16.4.0 / 23

Range of the addresses = 172.16.4.0 ï¿½ 172.16.5.255

First host = 172.16.4.1

Last host = 172.16.5.254

Broadcast address = 172.16.5.255

Number of hosts = 2^n-2=2^9-2=510

d) For 250 Hosts

Network address = 172.16.6.0 / 24

Range of the addresses = 172.16.6.0 ï¿½ 172.16.6.255

First host = 172.16.6.1

Last host = 172.16.6.254

Broadcast address = 172.16.6.255

Number of hosts = 2^n-2=2^8-2=254

e) For 200 Hosts

Network address = 172.16.7.0 / 24

Range of the addresses = 172.16.7.0 ï¿½ 172.16.7.255

First host = 172.16.7.1

Last host = 172.16.7.254

Broadcast address = 172.16.7.255

Number of hosts = 2^n-2=2^8-2=254

f) For 120 Hosts

Network address = 172.16.8.0 / 25

Range of the addresses = 172.16.8.0 ï¿½ 172.16.8.127

First host = 172.16.8.1

Last host = 172.16.8.126

Broadcast address = 172.16.8.127

Number of hosts = 2^n-2=2^7-2=126

g) For 120 Hosts

Network address = 172.16.8.128 / 25

Range of the addresses = 172.16.8.128 ï¿½ 172.16.8.255

First host = 172.16.8.129

Last host = 172.16.8.254

Broadcast address = 172.16.8.255

Number of hosts = 2^n-2=2^7-2=126

h) For 100 Hosts

Network address = 172.16.9.0 / 25

Range of the addresses = 172.16.9.0 ï¿½ 172.16.9.127

First host = 172.16.9.1

Last host = 172.16.9.126

Broadcast address = 172.16.9.127

Number of hosts = 2^n-2=2^7-2=126

i) For 100 Hosts

Network address = 172.16.9.128/25

Range of the addresses = 172.16.9.128 ï¿½ 172.16.9.255

First host = 172.16.9.129

Last host = 172.16.9.254

Broadcast address = 172.16.9.255

Number of hosts = 2^n-2=2^7-2=126

For bottom networks we have 4 branch networks. Having 2 sub branches in two

networks and 3 branches in the two networks.

30 hosts each for 3 branch networks.

j) For 30 Hosts

Network address = 172.16.10.0/27

Range of the addresses = 172.16.10.0 ï¿½ 172.16.10.32

First host = 172.16.10.1

Last host = 172.16.10.31

Broadcast address = 172.16.10.32

Number of hosts = 2^n-2=2^5-2=30

k) For 30 Hosts

Network address = 172.16.11.0/27

Range of the addresses = 172.16.11.0 ï¿½ 172.16.11.31

First host = 172.16.11.1

Last host = 172.16.11.30

Broadcast address = 172.16.11.31

Number of hosts = 2^n-2=2^5-2=30

l) For 30 Hosts

Network address = 172.16.12.0/27

Range of the addresses = 172.16.12.0 ï¿½ 172.16.12.31

First host = 172.16.12.1

Last host = 172.16.12.30

Broadcast address = 172.16.12.31

Number of hosts = 2^n-2=2^5-2=30

20 hosts each for 3 branch networks.

m) For 20 Hosts

Network address = 172.16.13.0/27

Range of the addresses = 172.16.13.0 ï¿½ 172.16.13.31

First host = 172.16.13.1

Last host = 172.16.13.30

Broadcast address = 172.16.13.31

Number of hosts = 2^n-2=2^5-2=30

m) For 20 Hosts

Network address = 172.16.14.0/27

Range of the addresses = 172.16.14.0 ï¿½ 172.16.14.31

First host = 172.16.14.1

Last host = 172.16.14.30

Broadcast address = 172.16.14.31

Number of hosts = 2^n-2=2^5-2=30

n) For 20 Hosts

Network address = 172.16.15.0/27

Range of the addresses = 172.16.15.0 ï¿½ 172.16.15.31

First host = 172.16.15.1

Last host = 172.16.15.30

Broadcast address = 172.16.15.31

Number of hosts = 2^n-2=2^5-2=30

For Network segments

1) Network address = 172.16.16.0 / 30

Range of the addresses = 172.16.16.0 ï¿½ 172.16.16.3

First host = 172.16.16.1

Last host = 172.16.16.2

Broadcast address = 172.16.16.3