Q1. (a) What is the need of modulating a signal? Will it be a right approach to send the information as the signal itself?
Answer :  In the modulation process, two signals are used namely the modulating signal and the carrier. The modulating signal is nothing but the baseband signal or information signal while the carrier is a high frequency sinusoidal signal.
In the modulation process, some parameter of the carrier wave (such as amplitude, frequency or phase ) is varied in accordance with the modulating signal. This modulated signal is then transmitted by the transmitter.
The receiver demodultes the received modulated signal and gets the original information signal back. Thus, demodulation is exactly opposite to modulation.
In the process of modulation the carrier wave actually acts as carrier which carries the information signal from the transmitter to receiver.
Need of Modulation
The baseband signals can be transmitted directly, but the baseband transmission has many limitations which can be overcome using modulation.
In the process of modulation, the baseband signal is translated i.e. shifted from low frequency to high frequency. This frequency shift is proportional to the frequency of carrier.
Advantages of Modulation
Avoids mixing of signals  This is a point from the practical side of things. Suppose you are transmitting the baseband signal as it is to a receiver, say your friends phone. Just like you , there will be thousands of people of people in the city using their mobile phones.
There is no way to tell such signals apart and they will interfere with each other leading to a lot of noise in the system and a very bad output.
By using a carrier wave of high frequencies and allotting a band of frequencies to each message, there is no mixing up of signals and the received signals are absolutely perfect.
Reduction in the height of antenna  For the transmission of radio signals, the antenna height must be multiple of λ/4 ,where λ is the wavelength .
λ = c/f
c → is the velocity of light
f → is the frequency of the signal to be transmitted
The minimum antenna height required to transmit a baseband signal of f = 10 kHz is calculated as follows :
Minimum antenna height = λ/4 = c/4f = (3*10^{8})/(4*10*10^{3}) = 7500 meters i.e. 7.5 km
The antenna of this height is practically impossible to install.
Now, let us consider a modulated signal at f=1 MHz . The minimum antenna height is given by,
Minimum antenna height = λ/4 = c/4f = (3*10^{8})/(4*10*10^{6}) = 75 meters
This antenna can be easily installed practically . Thus, modulation reduces the height of the antenna.
Increase the range of communication  By using modulation to transmit the signals through space to long distances, we have removed the need for wires in the communication systems. The technique of modulation helped humans to become wireless.
Multiplexing is possible  Multiplexing is a process in which two or more signals can be transmitted over the same communication channel simultaneously. This is possible only with modulation.
Improves quality of reception  With frequency modulation (FM), and the digital communication techniques like PCM, the effect of noise is reduced to a great extent. This improves quality of reception.
Q1. (b) Explain techniques used in digital to analog modulation with the help of diagram
Answer :  Digitaltoanalog conversion is the process of changing one of the characteristics of an analog signal based on the information in digital data.
A sine wave is defined by three characteristics : amplitude, frequency, and phase. When we change anyone of these characteristics, we create a different version of that wave. So, by changing one characteristic of a simple electric signal, we can use it to represent digital data.
There are three mechanisms for modulating digital data into an analog signal : Amplitude Shift Keying (ASK), Frequency Shift Keying (FSK), and Phase Shift Keying (PSK). In addition, there is a fourth (and better) mechanism that combines changing both the amplitude and phase, called Quadrature Amplitude Modulation (QAM).
Amplitude Shift Keying (ASK)  In an ASK system, the binary symbol 1 is represented by transmitting a fixedamplitude carrier wave and fixed frequency for a bit duration of T seconds. If the signal value is 1 then the carrier signal will be transmitted; otherwise, a signal value of 0 will be transmitted.
The simplest and most common form of ASK operates as a switch, using the presence of a carrier wave to indicate a binary one and its absence to indicate a binary zero. This type of modulation is called onoff keying (OOK), and is used at radio frequencies to transmit Morse code (referred to as continuous wave operation).
The ASK technique is commonly used to transmit digital data over optical fiber.
Frequency Shift Keying (FSK)  In FSK, the frequency of the carrier signal is varied according to the digital signal changes. The frequency of the modulated signal is constant for the duration of one signal element, but changes for the next signal element if the data element changes. Both amplitude and phase remain constant for all signal elements.
Phase Shift Keying (PSK)  In PSK, the phase of the carrier is varied to represent two or more different signal elements. Both amplitude and frequency remain constant when the phase changes. The simplest PSK is binary PSK, in which we have only two signal elements, one with a phase of 0°, and the other with a phase of 180°. The modulation is accomplished by varying the sine and cosine inputs at a precise time.. It is widely used for wireless LANs, RFID and Bluetooth communication. The following figure gives a conceptual view of PSK.
Quadrature Amplitude Modulation (QAM)  QAM utilises both amplitude and phase components to provide a form of modulation that is able to provide high levels of spectrum usage efficiency.
Q2. (a) Discuss the different approaches to circuit switching? Why it is suitable for voice Transmission? What are its limitations ?
Answer :  A circuitswitched communication system involves three phases : circuit establishment (setting up dedicated links between the source and destination); data transfer (transmitting the data between the source and destination); and circuit disconnect (removing the dadicated links).
In circuit switching network dedicated channel has to be established before the call is made between users. The channel is reserved between the users till the connection is active. For half duplex communication, one channel is allocated and for full duplex communication, two channels are allocated. It is mainly used for voice communication.Reason behind suitable for Voice Transmission
Limitations of Circuit Switch
Q2. (b) How does multistage switching overcome these limitations ?
Answer :  Multistage interconnection networks (MINs) consist of more than one stages of small interconnection elements called switching elements and links interconnecting them. A MIN normally connects N inputs to N outputs and is referred as an N × N MIN. The parameter N is called the size of the network.
Multistage Interconnect Network can be classified into three types :
Nonblocking  A nonblocking network can connect any idle input to any idle output, regardless of the connections already established across the network.
Rearrangeable nonblocking  This type of network can establish all possible connections between inputs and outputs by rearranging its existing connections.
Blocking  This type of network cannot realize all possible connections between inputs and outputs. This is because a connection between one free input to another free output is blocked by an existing connection in network.
Omega Network
An Omega network consists of multiple stages of 2*2 switching elements. Each input has a dedicated connection to an output. An N*N omega network has log(N) number of stages and N/2 number of switching elements in each stage for perfect shuffle between stages.
Example  8*8 Omega Network
Clos Network
A Clos network uses 3 stages to switch from N inputs to N outputs. In the first stage, there are r= N/n crossbar switches and each switch is of size n*m. In the second stage there are m switches of size r*r and finally the last stage is mirror of first stage with r switches of size m*n. A clos network will be completely nonblocking if m >= 2n1.
Example  9*9 Clos Network
We assume that the value of n=3 and m=2n. So, r=N/n =9/3 =3 and m=2*3 =6.
From the above diagram we can say that multistage switching overcome the limitations of circuit switching because of multiple path present in the network.
Q4. Define checksum and write the algorithm for computing the checksum ?
Answer :  COMING SOON
Q5. Define flow error control and piggybacking concepts. Show the operation of Stop & Wait ARQ with the help of an illustration. Also illustrate the outcome of Stop & Wait ARQ in the following scenarios.
Answer : 
Flow Control  When a data frame (Layer2 data) is sent from one host to another over a single medium, it is required that the sender and receiver should work at the same speed. That is, sender sends at a speed on which the receiver can process and accept the data. If sender is sending too fast the receiver may be overloaded, and data may be lost.
Error Control  When dataframe is transmitted, there is a probability that dataframe may be lost in the transit or it is received corrupted. In both cases, the receiver does not receive the correct dataframe and sender does not know anything about any loss. In such case, both sender and receiver are equipped with some protocols which helps them to detect transit errors such as loss of dataframe. Hence, either the sender retransmits the dataframe or the receiver may request to resend the previous dataframe.
Requirements for error control mechanism :
There are three types of techniques available which Datalink layer may deploy to control the errors by Automatic Repeat Requests (ARQ) :
Piggybacking
Piggybacking is the process which appends acknowledgement of frame with the data frame. Piggybacking process can be used if Sender and Receiver both have some data to transmit. This will increase the overall efficiency of transmission.
The three principles governing piggybacking when the station X wants to communicate with station Y are :
If station X has both data and acknowledgment to send, it sends a data frame with the ACK field containing the sequence number of the frame to be acknowledged.
If station X has only an acknowledgment to send, it waits for a finite period of time to see whether a data frame is available to be sent. If a data frame becomes available, then it piggybacks the acknowledgment with it. Otherwise, it sends an ACK frame.
If station X has only a data frame to send, it adds the last acknowledgment with it. The station Y discards all duplicate acknowledgments. Alternatively, station X may send the data frame with the ACK field containing a bit combination denoting no acknowledgment.
Stop and Wait ARQ
In this method of flow control, the sender sends a single frame to receiver & waits for an acknowledgment.
The next frame is sent by sender only when acknowledgment of previous frame is received.
This process of sending a frame & waiting for an acknowledgment continues as long as the sender has data to send.
To end up the transmission sender transmits end of transmission (EOT) frame.
The main advantage of stop & wait protocols is its accuracy. Next frame is transmitted only when the first frame is acknowledged. So there is no chance of frame being lost.
The main disadvantage of this method is that it is inefficient. It makes the transmission process slow. In this method single frame travels from source to destination and single acknowledgment travels from destination to source. As a result each frame sent and received uses the entire time needed to traverse the link. Moreover, if two devices are distance apart, a lot of time is wasted waiting for ACKs that leads to increase in total transmission time.
When ACK is lost  Sequence Number on data packets help to solve the problem of delayed acknowledgement. Consider the acknowledgement sent by the receiver gets lost. The sender retransmits the same data packet after its timer goes off. This prevents the occurrence of deadlock. The Sequence Number on the data packet helps the receiver to identify the duplicate data packet. Receiver discards the duplicate packet and resends the same acknowledgement.
When Frame is lost  After sending a data packet to the receiver, sender starts the time out timer. If the data packet gets acknowledged before the timer expires, sender stops the time out timer. If the timer goes off before receiving the acknowledgement, sender retransmits the same data packet. After retransmission, sender resets the timer. This prevents the occurrence of deadlock.
When ACK timeout occurs
Sender sends a data packet with Sequence Number 0 to the receiver.
Receiver receives the data packet correctly and expects the next data packet with Sequence Number 1 by sending acknowledgement ACK1 to the Sender.
Acknowledgement ACK1 sent by the receiver gets lost on the way.
Sender retransmits the same data packet with Sequence Number 0 when time out occurs.
Receiver receives the data packet and discovers it is the duplicate packet. It expects the data packet with Sequence Number 1 but receiving the data packet with Sequence Number 0. It discards the duplicate data packet and resends acknowledgement ACK1.
Q6. (a) How does MACAW work? Show diagrammatically. What are the added features in MACAW compared to MACA ?
Answer :  Multiple Access with Collision Avoidance (MACA) is a slotted media access control protocol used in wireless LAN data transmission to avoid collisions caused by the hidden station problem and to simplify exposed station problem.
The basic idea of MACA is a wireless network node makes an announcement before it sends the data frame to inform other nodes to keep silent. When a node wants to transmit, it sends a signal called RequestToSend (RTS) with the length of the data frame to send. If the receiver allows the transmission, it replies the sender a signal called ClearToSend (CTS) with the length of the frame that is about to receive.
Let us consider that a transmitting station A has data frame to send to a receiving station B. The operation works as follows :
Any node that receives CTS frame knows that it is close to the receiver, therefore, cannot transmit a frame.
Any node that receives RTS frame but not the CTS frame knows that is not close to the receiver to interfere with it, So it is free to transmit data.
WLAN data transmission collisions may still occur, and the MACA for Wireless (MACAW) is introduced to extend the function of MACA. It requires nodes sending acknowledgements after each successful frame transmission, as well as the additional function of Carrier Sense.
Q7. How does a bridge operate in different LAN environments? What are the problems encountered in building a bridge between the various 802 LANs ? Discuss.
Answer :  COMING SOON
Q8. Write Dijkstra and Bellman Fords shortest path routing algorithms and make a comparison between the two algorithms.
Answer : 
Shortest Path Algorithm Formula 
if (d[u] + c(u, v) < d[v]) then d[v] = d[u] + c(u, v)
d[u] represent the distance of vertex u
c(u, v) represent the cost of edge which connect the vertices u and v
d[v] represent the distance of vertex v
Dijkstra’s Algorithm
Dijkstra’s algorithm is an algorithm for finding the shortest paths between nodes in a graph.
Process of Dijkstra’s Algorithm 
Create a set "SPT Set" (Shortest Path Tree Set) that keeps track of vertices included in shortest path tree, i.e., whose minimum distance from source is calculated and finalized. Initially, this set is empty.
Assign a distance value to all vertices in the input graph. Initialize all distance values as INFINITE. Assign distance value as 0 for the source vertex so that it is picked first.
While sptSet doesn’t include all vertices 
Dijkstra’s Algorithm Example 
Implement Dijkstra’s algorithm to the following graph and find the shortest path from the node A to the remaining nodes.
Start from vertex A and update distance values of its adjacent vertices which is B, C and E.
Initial STP Set { A }.
Update the distance values of adjacent vertices of A. The distance value of vertex B, C and E becomes 2, 2 and 7 respectively.
Pick the vertex with minimum distance value and not already included in SPT Set. The vertex B is picked and added to SPT Set.
So SPT Set now becomes { A , B }.
Update the distance values of adjacent vertices of B. The distance value of vertex D is 6.
Again pick the vertex with minimum distance value and not already included in SPT Set. The vertex C is picked and added to SPT Set.
So SPT Set now becomes { A , B , C }.
There is no undiscover adjacent vertices of C, but the distance value of vertex E is require to change from 7 to 5.
Next vertex E is picked and added to SPT Set.
So SPT Set now becomes { A , B , C , E }.
Update the distance values of adjacent vertices of E. The distance value of vertex F is 9.
We repeat the above steps until SPT Set doesn’t include all vertices of given graph.
Finally, we get the following Shortest Path Tree (SPT) { A , B , C , E , D , F }.
Bellman Ford’s Algorithm
Bellman Ford algorithm helps us find the shortest path from a vertex to all other vertices of a weighted graph. It is similar to Dijkstra’s algorithm but it can work with graphs in which edges can have negative weights.
Negative weight edges can create negative weight cycles i.e. a cycle which will reduce the total path distance by coming back to the same point. Dijkstra’s Algorithm can not able to detect such a cycle and give an incorrect result.
Bellman Ford algorithm works by overestimating the length of the path from the starting vertex to all other vertices. Then it iteratively relaxes those estimates by finding new paths that are shorter than the previously overestimated paths.
In Bellman Ford’s algorithm shortest path contains at most (N  1) edges, because the shortest path could not have a cycle. N represent number of vertices present in the graph.
Bellman Ford’s Algorithm Example 
Implement Bellman Ford’s algorithm to the following graph and find the shortest path from the node A to the remaining nodes.
Initialize all distances as infinite, except the distance to source itself. Total number of vertices in the graph is 6, so all edges must be processed 5 times.
Let all edges are processed in following order : (A, B), (A, C), (A, D), (B, E), (C, B), (C, E), (D, C), (D, F), (E, G), (F, G).
First Iteration
Step 1   Distance value of vertex A = 0 Cost of edge (A, B) = 6 Distance value of vertex B = ∞. Since (0+6) < ∞, therefore, the distance value of vertex B is change from ∞ to 6. 

Step 2   Distance value of vertex A = 0 Cost of edge (A, C) = 5 Distance value of vertex C = ∞. Since(0+5) < ∞, therefore, the distance value of vertex C is change from ∞ to 5. 

Step 3   Distance value of vertex A = 0 Cost of edge (A, D) = 5 Distance value of vertex D = ∞. Since(0+5) < ∞, therefore, the distance value of vertex D is change from ∞ to 5. 

Step 4   Distance value of vertex B = 6 Cost of edge (B, E) = 1 Distance value of vertex E = ∞. Since{6+(1)} < ∞, therefore, the distance value of vertex E is change from ∞ to 5. 

Step 5   Distance value of vertex C = 5 Cost of edge (C, B) = 2 Distance value of vertex B = 6. Since{5+(2)} < 6, therefore, the distance value of vertex B is change from 6 to 3. 

Step 6   Distance value of vertex C = 5 Cost of edge (C, E) = 1 Distance value of vertex E = 5. Since(5+1) > 5, therefore, the distance value of vertex E remain same. 

Step 7   Distance value of vertex D = 5 Cost of edge (D, C) = 2 Distance value of vertex C = 5. Since {5+(2)} < 5, therefore, the distance value of vertex C is change from 5 to 3. 

Step 8   Distance value of vertex D = 5 Cost of edge (D, F) = 1 Distance value of vertex F = ∞. Since {5+(1)} < ∞, therefore, the distance value of vertex F is change from ∞ to 4. 

Step 9   Distance value of vertex E = 5 Cost of edge (E, G) = 3 Distance value of vertex G = ∞. Since (5+3) < ∞, therefore, the distance value of vertex G is change from ∞ to 8. 

Step 10   Distance value of vertex F = 4 Cost of edge (F, G) = 3 Distance value of vertex G = 8. Since (4+3) < 8, therefore, the distance value of vertex G is change from 8 to 7. 

Second Iteration
Step 1   Distance value of vertex A = 0 Cost of edge (A, B) = 6 Distance value of vertex B = 3. Since (0+6) > 3, therefore, the distance value of vertex B remain same. 

Step 2   Distance value of vertex A = 0 Cost of edge (A, C) = 5 Distance value of vertex C = 3. Since(0+5) > 3, therefore, the distance value of vertex C is remain same. 

Step 3   Distance value of vertex A = 0 Cost of edge (A, D) = 5 Distance value of vertex D = 5. Since(0+5) = 5, therefore, the distance value of vertex D is remain same. 

Step 4   Distance value of vertex B = 3 Cost of edge (B, E) = 1 Distance value of vertex E = 5. Since{3+(1)} < 5, therefore, the distance value of vertex E is change from 5 to 2. 

Step 5   Distance value of vertex C = 3 Cost of edge (C, B) = 2 Distance value of vertex B = 3. Since{3+(2)} < 3, therefore, the distance value of vertex B is change from 3 to 1. 

Step 6   Distance value of vertex C = 3 Cost of edge (C, E) = 1 Distance value of vertex E = 2. Since(3+1) > 2, therefore, the distance value of vertex E remain same. 

Step 7   Distance value of vertex D = 5 Cost of edge (D, C) = 2 Distance value of vertex C = 3. Since {5+(2)} = 3, therefore, the distance value of vertex C is remain same. 

Step 8   Distance value of vertex D = 5 Cost of edge (D, F) = 1 Distance value of vertex F = 4. Since {5+(1)} = 4, therefore, the distance value of vertex F is remain same. 

Step 9   Distance value of vertex E = 2 Cost of edge (E, G) = 3 Distance value of vertex G = 7. Since (2+3) < 7, therefore, the distance value of vertex G is change from 7 to 5. 

Step 10   Distance value of vertex F = 4 Cost of edge (F, G) = 3 Distance value of vertex G = 5. Since (4+3) > 5, therefore, the distance value of vertex G is remain same. 

Third Iteration
Step 1   Distance value of vertex A = 0 Cost of edge (A, B) = 6 Distance value of vertex B = 3. Since (0+6) > 3, therefore, the distance value of vertex B remain same. 

Step 2   Distance value of vertex A = 0 Cost of edge (A, C) = 5 Distance value of vertex C = 3. Since(0+5) > 3, therefore, the distance value of vertex C is remain same. 

Step 3   Distance value of vertex A = 0 Cost of edge (A, D) = 5 Distance value of vertex D = 5. Since(0+5) = 5, therefore, the distance value of vertex D is remain same. 

Step 4   Distance value of vertex B = 1 Cost of edge (B, E) = 1 Distance value of vertex E = 2. Since{1+(1)} < 2, therefore, the distance value of vertex E is change from 2 to 0. 

Step 5   Distance value of vertex C = 3 Cost of edge (C, B) = 2 Distance value of vertex B = 1. Since{3+(2)} = 1, therefore, the distance value of vertex B is remain same. 

Step 6   Distance value of vertex C = 3 Cost of edge (C, E) = 1 Distance value of vertex E = 0. Since(3+1) > 0, therefore, the distance value of vertex E remain same. 

Step 7   Distance value of vertex D = 5 Cost of edge (D, C) = 2 Distance value of vertex C = 3. Since {5+(2)} = 3, therefore, the distance value of vertex C is remain same. 

Step 8   Distance value of vertex D = 5 Cost of edge (D, F) = 1 Distance value of vertex F = 4. Since {5+(1)} = 4, therefore, the distance value of vertex F is remain same. 

Step 9   Distance value of vertex E = 0 Cost of edge (E, G) = 3 Distance value of vertex G = 5. Since (0+3) < 5, therefore, the distance value of vertex G is change from 5 to 3. 

Step 10   Distance value of vertex F = 4 Cost of edge (F, G) = 3 Distance value of vertex G = 3. Since (4+3) > 3, therefore, the distance value of vertex G is remain same. 

After third iteration no changes are happen in this graph.
Q9. (a) Discuss general principle of congestion control and the mechanisms used in congestion control in packet switched network.
Answer :  Congestion is an important issue that can arise in packet switched network. Congestion is a situation in Communication Networks in which too many packets are present in a part of the subnet, performance degrades. Congestion in a network may occur when the load on the network (i.e. the number of packets sent to the network) is greater than the capacity of the network (i.e. the number of packets a network can handle.). Network congestion occurs in case of traffic overloading.
Causing of Congestion
The input traffic rate exceeds the capacity of the output lines. If suddenly, a stream of packet start arriving on three or four input lines and all need the same output line. In this case, a queue will be built up. If there is insufficient memory to hold all the packets, the packet will be lost. Increasing the memory to unlimited size does not solve the problem. This is because, by the time packets reach front of the queue, they have already timed out (as they waited the queue). When timer goes off source transmits duplicate packet that are also added to the queue. Thus same packets are added again and again, increasing the load all the way to the destination.
The router's buffer is too limited.
Congestion in a subnet can occur if the processors are slow. Slow speed CPU at routers will perform the routine tasks such as queuing buffers, updating table etc slowly. As a result of this, queues are built up even though there is excess line capacity.
Congestion is also caused by slow links. This problem will be solved when high speed links are used. But it is not always the case. Sometimes increase in link bandwidth can further deteriorate the congestion problem as higher speed links may make the network more unbalanced. If a route does not have free buffers, it start ignoring/discarding the newly arriving packets. When these packets are discarded, the sender may retransmit them after the timer goes off. Such packets are transmitted by the sender again and again until the source gets the acknowledgement of these packets. Therefore multiple transmissions of packets will force the congestion to take place at the sending end.
How to correct the Congestion Problem
Congestion control mechanisms are divided into two categories, one category prevents the congestion from happening and the other category removes congestion after it has taken place. These two categories are :
Open loop congestion control policies are applied to prevent congestion before it happens. The congestion control is handled either by the source or the destination.
Window Policy  To implement window policy, selective repeat method is used for congestion control.
The gobackn protocol works well if errors are less, but if the line is poor it wastes a lot of bandwidth on retransmitted frames. Selective Repeat attempts to retransmit only those packets that are actually lost.
Acknowledgement Policy  Since acknowledgement are also the part of the load in network, the acknowledgment policy imposed by the receiver may also affect congestion. Several approaches can be used to prevent congestion related to acknowledgment :
The receiver should send acknowledgement for N packets rather than sending acknowledgement for a single packet.
The receiver should send a acknowledgment only if it has to sent a packet or a timer expires.
Discarding Policy  A router may discard less sensitive packets when congestion is likely to happen. Such a discarding policy may prevent congestion and at the same time may not harm the integrity of the transmission.
Retransmission Policy  The sender retransmits a packet, if it feels that the packet it has sent is lost or corrupted. This retransmission may increase the congestion in the network. To prevent congestion, retransmission timers must be designed to prevent congestion and also able to optimize efficiency.
Admission Policy  In admission policy a mechanism should be used to prevent congestion. Switches in a flow should first check the resource requirement of a network flow before transmitting it further. If there is a chance of a congestion or there is a congestion in the network, router should deny establishing a virtual network connection to prevent further congestion.
Q9. (b) Explain the implementation of token bucket traffic shaper with the help of a diagram ?
Answer :  The token bucket algorithm is based on an analogy of a fixed capacity bucket into which tokens (normally representing a unit of bytes or a single packet of predetermined size) are added at a fixed rate. When a packet is ready to be send, it is first checked whether the bucket contains sufficient tokens or not. If sufficient tokens are present in the bucket, the appropriate numbers of tokens [equivalent to the length of the packet in bytes] are removed and the packet is passed for transmission. Else if there is a deficiency of tokens, then the packet has to wait in a queue.
Q10. Explain with the help of an example and a diagram how the congestion controls algorithm (slow start algorithm) work at transport layer.
Answer :  TCP slow start is an algorithm which balances the speed of a network connection. Slow start gradually increases the amount of data transmitted until it finds the network’s maximum carrying capacity.
One of the most common ways to optimize the speed of a connection is to increase the speed of the link (i.e. increase the amount of bandwidth). However, any link can become overloaded if a device tries to send out too much data. Overloading a link is known as congestion, and it can result in slow communications or even data loss.
Slow start prevents a network from becoming congested by regulating the amount of data that’s sent over it. It negotiates the connection between a sender and receiver by defining the amount of data that can be transmitted with each packet, and slowly increases the amount of data until the network’s capacity is reached. This ensures that as much data is transmitted as possible without clogging the network.
Slow Start Process Step by Step
A sender attempts to communicate to a receiver. The sender’s initial packet contains a small congestion window, which is determined based on the sender’s maximum window.
The receiver acknowledges the packet and responds with its own window size. If the receiver fails to respond, the sender knows not to continue sending data.
After receiving the acknowledgement, the sender increases the next packet’s window size. The window size gradually increases until the receiver can no longer acknowledge each packet, or until either the sender or the receiver’s window limit is reached.
Once a limit has been determined, slow start’s job is done. Other congestion control algorithms take over to maintain the speed of the connection.
Q11. (a) Define digital signature and explain its benefits.
Answer :  A digital signature is an electronic signature that can be used to authenticate the identity of the sender of a message or the signer of a document, and to ensure that the original content of the message or document that has been sent is unchanged. Digital signatures are easily transportable, cannot be imitated by someone else, and can be automatically timestamped. A digital signature can be used with any kind of message, whether it is encrypted or plaintext.
The Digital Signatures require a key pair called the Public and Private Keys. Just as physical keys are used for locking and unlocking, in cryptography, the equivalent functions are encryption and decryption. The private key is kept confidential with the owner usually on a secure media like crypto smart card or crypto token. The public key is shared with everyone. Information encrypted by a private key can only be decrypted using the corresponding public key.
Digital Signature Versus Handwritten Signatures
An ink signature can be easily replicated from one document to another by copying the image manually or electronically. Digital Signatures cryptographically bind an electronic identity to an electronic document and the digital signature cannot be copied to another document. Further, paper contracts often have the ink signature block on the last page, allowing previous pages to be replaced after the contract has been signed. Digital signatures on the other hand compute the hash or digest of the complete document and a change of even one bit in the previous pages of the document will make the digital signature verification fail.
Benefits of Digital Signature
Reduce Cost  There are a lot of direct savings from switching to a paperless process, including the cost of paper, ink, printer maintenance and shipping costs. You will also notice a lot of indirect savings, including the time saved that would’ve been spent filing documents, rekeying data, searching for lost documents or tracking down a contract that’s been lost in the mail.
Get Paid Faster  Because it’s so fast and easy to sign documents online, you’re sure to see faster contract turnaround. It’s also easy to quickly execute contracts that have multiple signers. After the first person signs, the electronic signature software automatically routes documents to the next signer in the workflow. And when you get documents signed in minutes, you can get paid faster than ever before.
Enhance Customer Relationships  Your customers are used to doing business online, and they’ve come to expect businesses they work with to provide online services. With digital signature software, your customers can sign contracts online with nothing to download or install. As long as they have an Internet connection, they can sign documents no matter where life takes them. This service adds value for your customers by making it fast and easy to do business with your company.
Upgrade Document Security  In the paper world, you secure your documents by putting them in a locked file cabinet. You may even put that file cabinet in a locked room. But even with those precautions, someone could break in and tamper with documents. The only evidence you’d have would be a broken lock (if that), and then you would have to guess which file was altered.
This type of document security is a liability for your company, and it’s unnecessary. With advanced electronic signature software (a type of electronic signature called a “Digital Signature”), you can safeguard your documents with a high level of document security and evidence. Each signature is protected with a tamperevident seal, which alerts you if any part of the document is changed after signing. Signed documents also come with a highly detailed log of events of the document’s lifecycle. Using this evidence, you can see when each person signed the document, which signers downloaded a copy of the finished document, and much more.
Q11. (b) What kind of a model is being used in India to provide public key infrastructure related services. (i.e. management of public keys). Elaborate
Answer :  A public key infrastructure (PKI) is a set of roles, policies, hardware, software and procedures needed to create, manage, distribute, use, store and revoke digital certificates and manage publickey encryption. The purpose of a PKI is to facilitate the secure electronic transfer of information for a range of network activities such as ecommerce, internet banking and confidential email.
Public key cryptography is a cryptographic technique that enables entities to securely communicate on an insecure public network, and reliably verify the identity of an entity via digital signatures.
A public key infrastructure (PKI) is a system for the creation, storage, and distribution of digital certificates which are used to verify that a particular public key belongs to a certain entity. The PKI creates digital certificates which map public keys to entities, securely stores these certificates in a central repository and revokes them if needed.
A PKI consists of
Q12. Discuss the implementation of Kerberos mechanism.
Answer :  Kerberos is a ticketingbased authentication system, based on the use of symmetric keys. Kerberos uses tickets to provide authentication to resources instead of passwords. This eliminates the threat of password stealing via network sniffing. One of the biggest benefits of Kerberos is its ability to provide single signon (SSO). Once you log into your Kerberos environment, you will be automatically logged into other applications in the environment.
To help provide a secure environment, Kerberos makes use of Mutual Authentication. In Mutual Authentication, both the server and the client must be authenticated. The client knows that the server can be trusted, and the server knows that the client can be trusted. This authentication helps prevent maninthemiddle attacks and spoofing. Kerberos is also time sensitive. The tickets in a Kerberos environment must be renewed periodically or they will expire.
When a user needs to access a service protected by Kerberos, the Kerberos protocol process can be divided into two phases  User Identity Authentication and Service Access.
User Identity Authentication  User authentication is a process of checking validity of identity information provided by users in the Kerberos authentication service. Identity information can be user names and passwords or information that can provide real identities in other forms. If user information passes the validity check, the Kerberos authentication service returns a valid TicketGranting Ticket (TGT) token, proving that the user has passed identity authentication. The user uses the TGT in the subsequent Service Access process.
Service Access  When the user needs to access a service, the user requests the TicketGranting Service (TGS) from the Kerberos server based on the TGT obtained in the first phase, providing the name of the service to be accessed. TGS checks the TGT and information about the service to be accessed. After the information passes the check, TGS returns a ServiceGranting Ticket (SGT) token to the user. The user requests the component service based on the SGT and related user authentication information. The component service decrypts the SGT information in a symmetrical way and finishes user authentication. If the user passes the authentication process, the user can successfully access related resources of the service.
Kerberos in Windows Systems
Kerberos is very prevalent in the Windows environment. In fact, Windows 2000 and later use Kerberos as the default method of authentication. When you install your Active Directory domain, the domain controller is also the Key Distribution Center. In order to use Kerberos in a Windows environment, your client system must be a part of the Windows domain. Kerberos is used when accessing file servers, Web servers, and other network resources. When you attempt to access a Web server, Windows will try to sign you in using Kerberos. If Kerberos authentication does not work, then the system will fall back to NTLM authentication.
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