Computer Networks: A Comprehensive Study of TCP/IP, Ethernet, and Network Security
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This comprehensive study delves into the intricacies of computer networks, focusing on the TCP/IP protocol suite, Ethernet technology, and network security. It explores the fundamental concepts of data transmission, network addressing, and routing, while examining the role of protocols like TCP, UDP, and IP in enabling communication. The study also investigates the challenges of network security, including threats, vulnerabilities, and countermeasures, providing insights into securing networks against malicious attacks. Through detailed explanations, diagrams, and real-world examples, this document aims to provide a thorough understanding of computer networks and their essential components.
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SIT202
Computer Networks
Computer Networks
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Table of Contents
Question 1.............................................................................................................................................3
Question 2.............................................................................................................................................4
Question 3.............................................................................................................................................6
Question 4.............................................................................................................................................8
Question 5........................................................................................................................................... 10
Question 6........................................................................................................................................... 11
References........................................................................................................................................... 13
1
Question 1.............................................................................................................................................3
Question 2.............................................................................................................................................4
Question 3.............................................................................................................................................6
Question 4.............................................................................................................................................8
Question 5........................................................................................................................................... 10
Question 6........................................................................................................................................... 11
References........................................................................................................................................... 13
1
Question 1
Study the two processes that communicate over a TCP/IP on the network Ethernet with the
help of TCP protocol. Each layer of the TCP protocol will capture the payload and create a
PDU (Protocol Data Unit) when the data comes from the sender travels through the protocol
stack and then passes to the next below layer. Make a diagram that demonstrates the data
flow direction, all layers, peer layer communication, and identify the name of PDU and the
structure of the PDU components. Also, explain the steps of encapsulation of each layer.
When the data is sent to the physical layer, it is again constructed at the receiver end from the
digital signal. To reach the process which receives the protocol at the application layer,
progressing is done through the protocol stack and this complete method is called as de-
capsulation. Elaborate the stages of de-capsulation at the receiver side and also represents the
combined steps of both encapsulation diagram and decapsulation.
Above figure represents the all layers of TCP and OSI models and their structure, protocols,
peer layer communication, and explanation.
When the network gets a failure and we want that network still remains available, add the
redundant devices and links so that when working of one link stops or fails, the other
links give the path to provide communication. But this approach is used with the iteration
system because without using the iteration scheme, the broadcast will result in flood and
would create a loop and endlessly in the network and this will result in the spiking or
failure of the network. As shown in the below network diagram, host A of LAN1 receives
the frame from Host F of LAN2. Illuminate how will the MAC address table in the switch
and suffer thrashing and why would the frame be dispatched infinitely.
2
Study the two processes that communicate over a TCP/IP on the network Ethernet with the
help of TCP protocol. Each layer of the TCP protocol will capture the payload and create a
PDU (Protocol Data Unit) when the data comes from the sender travels through the protocol
stack and then passes to the next below layer. Make a diagram that demonstrates the data
flow direction, all layers, peer layer communication, and identify the name of PDU and the
structure of the PDU components. Also, explain the steps of encapsulation of each layer.
When the data is sent to the physical layer, it is again constructed at the receiver end from the
digital signal. To reach the process which receives the protocol at the application layer,
progressing is done through the protocol stack and this complete method is called as de-
capsulation. Elaborate the stages of de-capsulation at the receiver side and also represents the
combined steps of both encapsulation diagram and decapsulation.
Above figure represents the all layers of TCP and OSI models and their structure, protocols,
peer layer communication, and explanation.
When the network gets a failure and we want that network still remains available, add the
redundant devices and links so that when working of one link stops or fails, the other
links give the path to provide communication. But this approach is used with the iteration
system because without using the iteration scheme, the broadcast will result in flood and
would create a loop and endlessly in the network and this will result in the spiking or
failure of the network. As shown in the below network diagram, host A of LAN1 receives
the frame from Host F of LAN2. Illuminate how will the MAC address table in the switch
and suffer thrashing and why would the frame be dispatched infinitely.
2
Use the following diagrams and recreate the required diagrams to show and explain the
broadcast. Also represent the updates of the MAC table for switches, switch1 and switch2
and also the frame positions in both local area networks.
Question 2
When there is one or more than one layer2 path among the endpoints of the systems, there
will be arising of an L2 loop inside that system. A broadcast storm is created by the loop and
it multicasts, and switches forwarded this broadcast storm at each port and will constantly
produce the message of the broadcast that overflows the network. When the frames are sent
to the iterative type of topology, it can continue to act in a looping manner as the header of
layer2 does not have transistor-transistor logic (TTL). To resolve this problem of
continuously looping, STP (Spanning Tree Protocol) is used.
3
broadcast. Also represent the updates of the MAC table for switches, switch1 and switch2
and also the frame positions in both local area networks.
Question 2
When there is one or more than one layer2 path among the endpoints of the systems, there
will be arising of an L2 loop inside that system. A broadcast storm is created by the loop and
it multicasts, and switches forwarded this broadcast storm at each port and will constantly
produce the message of the broadcast that overflows the network. When the frames are sent
to the iterative type of topology, it can continue to act in a looping manner as the header of
layer2 does not have transistor-transistor logic (TTL). To resolve this problem of
continuously looping, STP (Spanning Tree Protocol) is used.
3
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The numbers that are drawn over the arrows represent the frame path or the traffic that travels
through the whole networks.
Host F sends the frame having no TTL value. So when the frame is received by the Hub 2, it
will overflow it to the first and second switch which are represented by number 1 and 2 as
shown in the above diagram. Now both the switch 1 and switch 2 tries to determine the MAC
(Media Access Control) address of host F on the respective port and make the entire active
ports flooded. After learning the MAC of F, both switches will send this to Hub 1 and this
Hub 1 added broadcast the address to both switch 1 and switch 2 respectively and also to
those ports that are active. Now both the switches will cram the MAC of host F on both end
ports. The hubs and other switches repeat the same procedures and create the loop F2 and this
will cause instability to the switch.
4
through the whole networks.
Host F sends the frame having no TTL value. So when the frame is received by the Hub 2, it
will overflow it to the first and second switch which are represented by number 1 and 2 as
shown in the above diagram. Now both the switch 1 and switch 2 tries to determine the MAC
(Media Access Control) address of host F on the respective port and make the entire active
ports flooded. After learning the MAC of F, both switches will send this to Hub 1 and this
Hub 1 added broadcast the address to both switch 1 and switch 2 respectively and also to
those ports that are active. Now both the switches will cram the MAC of host F on both end
ports. The hubs and other switches repeat the same procedures and create the loop F2 and this
will cause instability to the switch.
4
Question 3
As shown in the above figure, as Alice first time sends data to the Bob, he will make the
frame as it does not have the media access control address of the Bob.
S-MAC: L6, D-MAC: S-IP: N6, D-IP: N5
When the data is delivered to Bob by Alice, he will accomplish the AND process in between
the subnet and D-IP and determines if the data goes to the same network or not. If the data
does not belong to the same network, he will try to resolve the ARP (Address Resolution
Protocol) for the defaulting gateway.
S-MAC: L6, D-MAC: FFFF: FFFF: FFFF, Transmitter IP: N6, Objective IP: N2,
Transmitter MAC: L6, Objective MAC: 0000:0000:0000
When the ARP (Address Resolution Protocol) request is received by the R1, the L2 header
will be opened and checks the objective address. Because it has its own address, it will reply
back to its own MAC (Media Access Control) address. After responding to its address, the
completion process of the frame gets finished and S-MAC: L6, D-MAC: L2, S-IP: N6, D-IP:
N5 will be forwarded to the R1. As R1 receives the frame, D-MAC will be checked by its
own data by opening it and then re-checks the IP (Internet Protocols) that whether there is the
availability of IP in the routing table or not. If it is available, it is transmitted to the R2 and if
5
As shown in the above figure, as Alice first time sends data to the Bob, he will make the
frame as it does not have the media access control address of the Bob.
S-MAC: L6, D-MAC: S-IP: N6, D-IP: N5
When the data is delivered to Bob by Alice, he will accomplish the AND process in between
the subnet and D-IP and determines if the data goes to the same network or not. If the data
does not belong to the same network, he will try to resolve the ARP (Address Resolution
Protocol) for the defaulting gateway.
S-MAC: L6, D-MAC: FFFF: FFFF: FFFF, Transmitter IP: N6, Objective IP: N2,
Transmitter MAC: L6, Objective MAC: 0000:0000:0000
When the ARP (Address Resolution Protocol) request is received by the R1, the L2 header
will be opened and checks the objective address. Because it has its own address, it will reply
back to its own MAC (Media Access Control) address. After responding to its address, the
completion process of the frame gets finished and S-MAC: L6, D-MAC: L2, S-IP: N6, D-IP:
N5 will be forwarded to the R1. As R1 receives the frame, D-MAC will be checked by its
own data by opening it and then re-checks the IP (Internet Protocols) that whether there is the
availability of IP in the routing table or not. If it is available, it is transmitted to the R2 and if
5
not, then it gets released. Now for R2, performs the AND operation and resolved ARP. The
packet now created is:
S-MAC: L3, D-MAC: L4, S-IP: N6, D-IP: N5.
Checks the routing table and sends the packets to R2 and then this R2 opens the second
header L2 which then examines the third header L3 and identified the destination address of
internet protocol which is directly connected to the R2. If the connection established is the
first time, that means the ARP is fixed for the MAC address of Bob.
S-MAC: L3, D-MAC: L5, S-IP: N1, D-IP: N5
The frame is now received by the Bob and he will check if the destination address of MAC is
its self or not. If the address is of its own then open the second header L2 and examines the
destination address of IP addresses. Now it will encapsulate and accept the data and transmit
the response back to it.
6
packet now created is:
S-MAC: L3, D-MAC: L4, S-IP: N6, D-IP: N5.
Checks the routing table and sends the packets to R2 and then this R2 opens the second
header L2 which then examines the third header L3 and identified the destination address of
internet protocol which is directly connected to the R2. If the connection established is the
first time, that means the ARP is fixed for the MAC address of Bob.
S-MAC: L3, D-MAC: L5, S-IP: N1, D-IP: N5
The frame is now received by the Bob and he will check if the destination address of MAC is
its self or not. If the address is of its own then open the second header L2 and examines the
destination address of IP addresses. Now it will encapsulate and accept the data and transmit
the response back to it.
6
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Question 4
1. Explain the importance of PPP with respect to wide area networks (WANs)?
A PPP (Point to Point) protocol connection is a is common layer 2 protocol among all types
of wide area network connections and it is used to connect the local area network to the
service provider of wide area networks. It is also termed as a leased-line connection or serial
connection as the lines are rented usually from the telephone lines and are used by the
company. Generally, the objective of PPP connection is to send the data packets of the third
layer across the point-to-point link of data link layer. It enables the real-time utilization of
many protocols of network layers such as IPC (Internet Protocol control), CSC (Cisco
Systems Control) protocol, ACP (AppleTalk Control Protocol), and NIP (Novell IPX Control
Protocol). Also, there are many benefits of using PPP with respect to the wide area network.
It transfers a large amount of information in minimum time and hence provides great speed
with a secure connection. The information shared over the PPP in wide area networks are
secure as limited users access this protocol and it is distinctively accessible also Jorgensen
(2018).
2. Explain how PPP enables users to exchange data in WAN by using sub-protocols such as
LCP, PAP, CHAP, NCP, IPCP.
PPP (Point to Point) protocol allows the users to exchanges the information in a wide area
network through sub-protocols. It delivers transport over ISDN (Integrated Service Digital
Network), ATM, optical links and frame relay. It contains four main components which are
V.24, EIA/TIA-232-C, V.35, and integrated digital service network (ISDN). Data
encapsulation is the method for encapsulating the datagrams of multiple protocols. Diverse
protocols of network-layers are transported simultaneously and encapsulated on the same
link. The sub-protocols that are used in PPP for exchanging the information in WANs are
Carlson (2011):
LCP: LCP means Link Control Protocol that is used to negotiate the connection. It
establishes, organizes and check the connections at the data link layer. It can handle the
variable size of data packets and is flexible in detecting the iterative-back link,
configuration bugs, and termination of the link.
NCP: NCP stands for Network Control Protocol which is mainly used for the
encapsulation of traffic. It creates and configures the different protocol of the network
layer and established their uses simultaneously. Some more NCPs are:
7
1. Explain the importance of PPP with respect to wide area networks (WANs)?
A PPP (Point to Point) protocol connection is a is common layer 2 protocol among all types
of wide area network connections and it is used to connect the local area network to the
service provider of wide area networks. It is also termed as a leased-line connection or serial
connection as the lines are rented usually from the telephone lines and are used by the
company. Generally, the objective of PPP connection is to send the data packets of the third
layer across the point-to-point link of data link layer. It enables the real-time utilization of
many protocols of network layers such as IPC (Internet Protocol control), CSC (Cisco
Systems Control) protocol, ACP (AppleTalk Control Protocol), and NIP (Novell IPX Control
Protocol). Also, there are many benefits of using PPP with respect to the wide area network.
It transfers a large amount of information in minimum time and hence provides great speed
with a secure connection. The information shared over the PPP in wide area networks are
secure as limited users access this protocol and it is distinctively accessible also Jorgensen
(2018).
2. Explain how PPP enables users to exchange data in WAN by using sub-protocols such as
LCP, PAP, CHAP, NCP, IPCP.
PPP (Point to Point) protocol allows the users to exchanges the information in a wide area
network through sub-protocols. It delivers transport over ISDN (Integrated Service Digital
Network), ATM, optical links and frame relay. It contains four main components which are
V.24, EIA/TIA-232-C, V.35, and integrated digital service network (ISDN). Data
encapsulation is the method for encapsulating the datagrams of multiple protocols. Diverse
protocols of network-layers are transported simultaneously and encapsulated on the same
link. The sub-protocols that are used in PPP for exchanging the information in WANs are
Carlson (2011):
LCP: LCP means Link Control Protocol that is used to negotiate the connection. It
establishes, organizes and check the connections at the data link layer. It can handle the
variable size of data packets and is flexible in detecting the iterative-back link,
configuration bugs, and termination of the link.
NCP: NCP stands for Network Control Protocol which is mainly used for the
encapsulation of traffic. It creates and configures the different protocol of the network
layer and established their uses simultaneously. Some more NCPs are:
7
ACP (AppleTalk Control Protocol)
CSC (Cisco Systems Control) Protocol
IPC (Internet Protocol Control) Protocol
CCP (Compression Control Protocol)
NCP (Novell IPX Control Protocol)
CHAP: CHAP stands for Challenge Handshake Authentication Protocol and is another
sub-protocol of point-to-point (PPP) protocol which is used to encrypt the data by using
an MD5 hash. Basically, three steps are performed in CHAP sub-protocol:
After all phases of LCP gets finished and CHAT is assigned between both devices,
the senders send the authenticate challenge message to the peer.
The peer replies to the sender with the value calculated through a hash function called
MD5 (Message Digest 5).
The authenticator evaluates and checks the response and matches with its own
calculations of the normal hash value. If the value gets matched, the connection is
successful otherwise terminate the connection.
PAP: PAP stands for Password Authentication Protocol that directs all the information in
the plain or ciphertexts. It provides a simple approach for the remote node to generate its
identity with the help of two-way handshake.. after the establishment phase of PPP gets
completed, the username and password are repeatedly deliver by the remote node in the
clear text across the link till the process of acknowledgment not completed.
8
CSC (Cisco Systems Control) Protocol
IPC (Internet Protocol Control) Protocol
CCP (Compression Control Protocol)
NCP (Novell IPX Control Protocol)
CHAP: CHAP stands for Challenge Handshake Authentication Protocol and is another
sub-protocol of point-to-point (PPP) protocol which is used to encrypt the data by using
an MD5 hash. Basically, three steps are performed in CHAP sub-protocol:
After all phases of LCP gets finished and CHAT is assigned between both devices,
the senders send the authenticate challenge message to the peer.
The peer replies to the sender with the value calculated through a hash function called
MD5 (Message Digest 5).
The authenticator evaluates and checks the response and matches with its own
calculations of the normal hash value. If the value gets matched, the connection is
successful otherwise terminate the connection.
PAP: PAP stands for Password Authentication Protocol that directs all the information in
the plain or ciphertexts. It provides a simple approach for the remote node to generate its
identity with the help of two-way handshake.. after the establishment phase of PPP gets
completed, the username and password are repeatedly deliver by the remote node in the
clear text across the link till the process of acknowledgment not completed.
8
Question 5
1. Explain why modulation of a digital signal is required for transmission in digital
cellular telephony?
Digital modulation technique is used to improve the communication quality. The digital
signal consists of 0s and 1s. It delivers more information capability, high security of data, and
availability of a system with the great eminence communication. The main use of modulation
of the digital signal is to transfer a bit stream of digital mode at the high frequency over an
analog signal. This enables to send the signals produced in the digital circuit around the
physical medium. There should be a requirement of the channel to communicate such type of
signal with unlimited bandwidth that actually does not occur. There is finite bandwidth in all
types of bandwidth and they do not handle these types of rectangle pulse. So if we want to
transmit signals, we should modulate the signals in the digital form to send it at the correct
bit-rate. The deviation in the forte, quality, or pitch of the voice happens with the modulation
of excellent voice. There are some more benefits of modulating the digital signal for the
transmission of data in the digital cellular telephony such as:
The decrease in the antenna height
Increases in the communication range
Progresses the quality of transmission
Elude mixing of signals
There is the possibility of multiplexing
2. Briefly explain the digital transmission on a bandpass channel
Digital to analog and analog to digital conversion:
In analog to digital conversion, there is the conversion of low passband signal to the bandpass
analog signal. While in digital to analog conversion, there is conversion of digital data into
the analog data signal. To transmit the data from one system to another system, data must be
converted into the analog form. But today, radio broadcasters and television are switching
from the analog to digital signal. The reason behind this transformation is that digital signals
have a finite set of values, and digital signal values are in discrete and square form. To
transmit the digital signals, there are two different methodologies. One is baseband
transmission and other is broadband transmission Comer (2015).
Baseband transmission means transmitting the digital signal over the channel without
converting the digital signal to analog signal. For the transmission of data, it involves a low
pass channel. In low pass channel, bandwidth initiates from the zero level. In BPT (Baseband
9
1. Explain why modulation of a digital signal is required for transmission in digital
cellular telephony?
Digital modulation technique is used to improve the communication quality. The digital
signal consists of 0s and 1s. It delivers more information capability, high security of data, and
availability of a system with the great eminence communication. The main use of modulation
of the digital signal is to transfer a bit stream of digital mode at the high frequency over an
analog signal. This enables to send the signals produced in the digital circuit around the
physical medium. There should be a requirement of the channel to communicate such type of
signal with unlimited bandwidth that actually does not occur. There is finite bandwidth in all
types of bandwidth and they do not handle these types of rectangle pulse. So if we want to
transmit signals, we should modulate the signals in the digital form to send it at the correct
bit-rate. The deviation in the forte, quality, or pitch of the voice happens with the modulation
of excellent voice. There are some more benefits of modulating the digital signal for the
transmission of data in the digital cellular telephony such as:
The decrease in the antenna height
Increases in the communication range
Progresses the quality of transmission
Elude mixing of signals
There is the possibility of multiplexing
2. Briefly explain the digital transmission on a bandpass channel
Digital to analog and analog to digital conversion:
In analog to digital conversion, there is the conversion of low passband signal to the bandpass
analog signal. While in digital to analog conversion, there is conversion of digital data into
the analog data signal. To transmit the data from one system to another system, data must be
converted into the analog form. But today, radio broadcasters and television are switching
from the analog to digital signal. The reason behind this transformation is that digital signals
have a finite set of values, and digital signal values are in discrete and square form. To
transmit the digital signals, there are two different methodologies. One is baseband
transmission and other is broadband transmission Comer (2015).
Baseband transmission means transmitting the digital signal over the channel without
converting the digital signal to analog signal. For the transmission of data, it involves a low
pass channel. In low pass channel, bandwidth initiates from the zero level. In BPT (Baseband
9
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Pulse Transmission), a data stream is symbolized in the method of distinct pulse-amplitude-
modulated (PAM) signal which is transmitted above the low-pass channel. It contains the
band of frequencies that is in adjacent to the zero frequency. In DPT (Digital Passband
Transmission) signal, there is modulation of the receiving stream on the carrier wave with the
fixed level frequency and transferred it through the band-pass channel. The baseband
transmission required the modulation of the signal when it transfers the digital signal on the
channel without changing from digital signal to the analog signal Akaiwa (2015).
Question 6
As shown in the above figure, there are total nine collision domain that are identified marked
in the form of a circle with the numbers marking on it.
Switches: A switch is the computer device works on the network that is used to connect the
multiple devices in together manner with the help of switch packet switched to send and
10
modulated (PAM) signal which is transmitted above the low-pass channel. It contains the
band of frequencies that is in adjacent to the zero frequency. In DPT (Digital Passband
Transmission) signal, there is modulation of the receiving stream on the carrier wave with the
fixed level frequency and transferred it through the band-pass channel. The baseband
transmission required the modulation of the signal when it transfers the digital signal on the
channel without changing from digital signal to the analog signal Akaiwa (2015).
Question 6
As shown in the above figure, there are total nine collision domain that are identified marked
in the form of a circle with the numbers marking on it.
Switches: A switch is the computer device works on the network that is used to connect the
multiple devices in together manner with the help of switch packet switched to send and
10
process and receive the data. There is a single collision area in the switch for each interface.
In the above figure, there are three connections of links, so there are three collision areas.
Hub: Hub is also part of network communication which is basically a node and used to
transmit the data to all computer systems. It is a central connection node for all the systems.
There is one L1 device which has one collision area.
Router: A router is used in the network connection to forward the data packets among the
computer networks. It performs and controls the traffic pointing purpose on the network.
There is no specific collision domain of the router. It only generates many broadcast domains.
Bridges: A bridge is also a network connection device that delivers interconnection with the
other bridge networks of the similar protocol. Two different networks can be connected
through a bridge device and provide connection among them. It is almost similar to the
switches and it also has one collision area as per the interface.
o Switch 1 = two collision area
o Switch 2 = three Collision area
o Switch 3 = two Collision area
o Hub 1 = one Collision area
o Hub 2= one Collision area
o Hub 3= one Collision area
o Hub4= one Collision area
o Hub5= one Collision area
o Bridges= three Collision area
11
In the above figure, there are three connections of links, so there are three collision areas.
Hub: Hub is also part of network communication which is basically a node and used to
transmit the data to all computer systems. It is a central connection node for all the systems.
There is one L1 device which has one collision area.
Router: A router is used in the network connection to forward the data packets among the
computer networks. It performs and controls the traffic pointing purpose on the network.
There is no specific collision domain of the router. It only generates many broadcast domains.
Bridges: A bridge is also a network connection device that delivers interconnection with the
other bridge networks of the similar protocol. Two different networks can be connected
through a bridge device and provide connection among them. It is almost similar to the
switches and it also has one collision area as per the interface.
o Switch 1 = two collision area
o Switch 2 = three Collision area
o Switch 3 = two Collision area
o Hub 1 = one Collision area
o Hub 2= one Collision area
o Hub 3= one Collision area
o Hub4= one Collision area
o Hub5= one Collision area
o Bridges= three Collision area
11
References
Jorgensen J. W. (2018) INTELLECTUAL VENTURES. CONTROL
PROTOCOL/INTERNET PROTOCOL (TCP/IP) PACKET-CENTRIC WIRELESS POINT
TO MULTI-POINT (PtMP) TRANSMISSION SYSTEM ARCHITECTURE.
Sarkar, S. K., Basavaraju, T. G., & Puttamadappa, C. (2016). Ad hoc mobile wireless
networks: principles, protocols, and applications. CRC Press.
Comer, D. (2015). Computer networks and internets. Pearson.
Akaiwa, Y. (2015). Introduction to digital mobile communication. John Wiley & Sons.
Carlson, J., & Eastlake 3rd, D. (2011). PPP Transparent Interconnection of Lots of Links
(TRILL) Protocol Control Protocol (No. RFC 6361).
12
Jorgensen J. W. (2018) INTELLECTUAL VENTURES. CONTROL
PROTOCOL/INTERNET PROTOCOL (TCP/IP) PACKET-CENTRIC WIRELESS POINT
TO MULTI-POINT (PtMP) TRANSMISSION SYSTEM ARCHITECTURE.
Sarkar, S. K., Basavaraju, T. G., & Puttamadappa, C. (2016). Ad hoc mobile wireless
networks: principles, protocols, and applications. CRC Press.
Comer, D. (2015). Computer networks and internets. Pearson.
Akaiwa, Y. (2015). Introduction to digital mobile communication. John Wiley & Sons.
Carlson, J., & Eastlake 3rd, D. (2011). PPP Transparent Interconnection of Lots of Links
(TRILL) Protocol Control Protocol (No. RFC 6361).
12
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