Packet 1: 2 Packet 2: 3 Packet 3: 3 Packet 4: 2 Packet 1: 2, 70 Packet 2: 1, 45 Packet 3: 3, 11 Packet 4: 4, 41 In a datagram network, the destination addresses are unique. They cannot be duplicated in the routing table. In a virtual-circuit network, the VCIs are local. A VCI is unique only in rela- tionship to a port. In other words, the port, VCI combination is unique. This means that we can have two entries with the same input or output ports.
However, we cannot have two entries with the same port, VCI pair. When a packet arrives at a router in a datagram network, the only information in the packet that can help the router in its routing is the destination address of the packet. The table then is sorted to make the searching faster.
When a packet arrives at a switch in a virtual-circuit network, the pair input port, input VCI can uniquely determined how the packet is to be routed; the pair is the only two pieces of information in the packet that is used for routing.
The table in the virtual-circuit switch is sorted based on the this pair. However, since the number of port numbers is normally much smaller than the number of virtual circuits assigned to each port, sorting is done in two steps: first according to the input port number and second according to the input VCI.
However, we need to know that a regular multiplexer discussed in Chapter b. However, we need to know that a regular demultiplexer discussed in See Figure 8. Figure 8. Only four simultaneous connections are possible for each crossbar at the first stage. This means that the total number of simultaneous connections is Only six simultaneous connections are possible for each crossbar at the first stage. The number of cross- can be left unused.
Some of the input lines tion. We can see that there is no blocking involved because each 8 input line has 15 intermediate crossbars. With less than , cross- points we can design a three-stage switch. The total number of crosspoints is , We give two solutions. We first solve the problem using only crossbars and then we replace the cross- bars at the first and the last stage with TSIs.
We can replace the crossbar at the first and third stages with TSIs as shown in Figure 8. In other words, the input frame has 10 slots and the output frame has only 4 slots. The data in the first slot of all input TSIs are directed to the first switch, the output in the second slot are directed to the sec- ond switch, and so on. We can see the inefficiency in the first solution. Since the slots are separated in time, only one of the switches at the middle stage is active at each moment.
This means that, instead of 4 crossbars, we could have used only one with the same result. In this case we still need memory locations but only crosspoints. The telephone network is made of three major components: local loops, trunks, and switching offices.
The telephone network has several levels of switching offices such as end offices, tandem offices, and regional offices. A LATA is a small or large metropolitan area that according to the divestiture of was under the control of a single telephone-service provider.
These car- riers, sometimes called long-distance companies, provide communication services between two customers in different LATAs. Signaling System Seven SS7 is the protocol used to provide signaling services in the telephone network. It is very similar to the five-layer Internet model. Telephone companies provide two types of services: analog and digital. Dial-up modems use part of the bandwidth of the local loop to transfer data.
The latest dial-up modems use the V-series standards such as V. Telephone companies developed digital subscriber line DSL technology to pro- vide higher-speed access to the Internet. It uses a device called a digital sub- scriber line access multiplexer DSLAM at the telephone company site.
The traditional cable networks use only coaxial cables to distribute video infor- mation to the customers. The hybrid fiber-coaxial HFC networks use a combi- nation of fiber-optic and coaxial cable to do so.
To provide Internet access, the cable company has divided the available bandwidth of the coaxial cable into three bands: video, downstream data, and upstream data. The downstream-only video band occupies frequencies from 54 to MHz. The downstream data occupies the upper band, from to MHz. The upstream data occupies the lower band, from 5 to 42 MHz. The cable modem CM is installed on the subscriber premises.
The cable modem transmission system CMTS is installed inside the distribution hub by the cable company. It receives data from the Internet and passes them to the combiner, which sends them to the subscriber. Packet-switched networks are well suited for carrying data in packets. The end-to- end addressing or local addressing VCI occupies a field in each packet. Tele- phone networks were designed to carry voice, which was not packetized. A cir- cuit-switched network, which dedicates resources for the whole duration of the conversation, is more suitable for this type of communication.
The setup phase can be matched to the dialing process. After the callee responds, the data transfer phase here voice transfer phase starts. When any of the parties hangs up, the data transfer is terminated and the teardown phase starts. It takes a while before all resources are released. In a telephone network, the telephone numbers of the caller and callee are serving as source and destination addresses.
These are used only during the setup dialing and teardown hanging up phases. The delay can be attributed to the fact that some telephone companies use satellite networks for overseas communication. In these case, the signals need to travel sev- eral thousands miles earth station to satellite and satellite to earth station.
See Figure 9. Figure 9. SDSL e. VDSL The DSL technology is based on star topology with the hub at the telephone office. The local loop connects each customer to the end office. This means that there is no sharing; the allocated bandwidth for each customer is not shared with neigh- bors. The data rate does not depend on how many people in the area are transfer- ring data at the same time. The cable modem technology is based on the bus or rather tree topology.
The cable is distributed in the area and customers have to share the available band- width. This means if all neighbors try to transfer data, the effective data rate will be decreased. In a single bit error only one bit of a data unit is corrupted; in a burst error more than one bit is corrupted not necessarily contiguous. Redundancy is a technique of adding extra bits to each data unit to determine the accuracy of transmission. In forward error correction, the receiver tries to correct the corrupted codeword; in error detection by retransmission, the corrupted message is discarded the sender needs to retransmit the message.
A linear block code is a block code in which the exclusive-or of any two code- words results in another codeword.
A cyclic code is a linear block code in which the rotation of any codeword results in another codeword. The Hamming distance between two words of the same size is the number of differences between the corresponding bits. The Hamming distance can easily be found if we apply the XOR operation on the two words and count the number of 1s in the result. The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words.
The single parity check uses one redundant bit for the whole data unit. In a two- dimensional parity check, original data bits are organized in a table of rows and columns. The parity bit is then calculated for each column and each row. The remainder is always one bit smaller than the divisor. The degree of the generator polynomial is one less than the size of the divisor.
For example, the CRC generator with the polynomial of degree 32 uses a bit divisor. The degree of the generator polynomial is the same as the size of the remainder length of checkbits. For example, CRC with the polynomial of degree 32 creates a remainder of 32 bits.
In this arithmetic, when a number needs more than n bits, the extra bits are wrapped and added to the number. In this arithmetic, the complement of a number is made by inverting all bits. At least three types of error cannot be detected by the current checksum calcula- tion. First, if two data items are swapped during transmission, the sum and the checksum values will not change.
Third, if one or more data items is changed in such a way that the The value of a checksum can be all 0s in binary. This happens when the value of the sum after wrapping becomes all 1s in binary. It is almost impossible for the value of a checksum to be all 1s. For this to happen, the value of the sum after wrapping must be all 0s which means all data units must be 0s.
First, the result of XORing two equal patterns is an all-zero pattern part b. Second, the result of XORing of any pattern with an all-zero pattern is the original non-zero pattern part c. Third, the result of XORing of any pattern with an all-one pattern is the complement of the original non-one pattern. The codeword for dataword 10 is This codeword will be changed to if a 3-bit burst error occurs. This pattern is not one of the valid codewords, so the receiver detects the error and discards the received pattern.
This pattern is not one of the valid codewords, so the receiver discards the received pattern. The code is not linear. We check five random cases. All are in the code. We show the dataword, the codeword, the corrupted codeword, and the interpreta- tion of the receiver for each case: a.
Comment: The above result does not mean that the code can never detect three errors. The last two cases show that it may happen that three errors remain unde- tected.
We show the dataword, codeword, the corrupted codeword, the syndrome, and the interpretation of each case: a. C 7,4 cannot correct two errors. C 7,4 cannot correct three errors. If we rotate one bit, the result is , which is in the code. If we rotate two bits, the result is , which is in the code. And so on. We use trial and error to find the right answer: a. To detect single bit errors, a CRC generator must have at least two terms and the coefficient of x0 must be nonzero.
It has more than one term and the coefficient of x0 is 1. It can detect a single-bit error. It will detect all burst errors of size 8 or less. This means 8 out of burst errors of size 9 c.
Burst errors of size 9 are detected most of the time, but they slip by with proba- are left undetected. This means 4 out of burst errors of size 15 d. Burst errors of size 15 are detected most of the time, but they slip by with prob- are left undetected.
It detects all single-bit error. It will detect all burst errors of size 32 or less. This means out of burst c. Burst errors of size 33 are detected most of the time, but they are slip by with errors of size 33 are left undetected. This means out of burst d. Burst errors of size 55 are detected most of the time, but they are slipped with errors of size 55 are left undetected.
We need to add all bits modulo-2 XORing. However, it is simpler to count the number of 1s and make them even by adding a 0 or a 1.
We have shown the parity bit in the codeword in color and separate for emphasis. Figure Checksum at the sender site b. Checksum at the receiver site one caught error d. Checksum at the receiver site two errors. In part a, we calculate the checksum to be sent 0x2E32 b. In part b, there is no error in transition. The receiver recalculates the checksum to be all 0x The receiver correctly assumes that there is no error. In part c, there is one single error in transition.
The receiver calculates the checksum to be 0FFFD. The receiver correctly assumes that there is some error and discards the packet. In part d, there are two errors that cancel the effect of each other. The receiver calculates the checksum to be 0x The receiver erroneously assumes that there is no error and accepts the packet.
This is an example that shows that the checksum may slip in finding some types of errors. This example shows that the checksum can be all 0s. It can be all 1s only if all data items are all 0, which means no data at all. The two main functions of the data link layer are data link control and media access control.
Data link control deals with the design and procedures for commu- nication between two adjacent nodes: node-to-node communication. Media access control deals with procedures for sharing the link.
The data link layer needs to pack bits into frames. Framing divides a message into smaller entities to make flow and error control more manageable. In a byte-oriented protocol, data to be carried are 8-bit characters from a coding system.
Character-oriented protocols were popular when only text was exchanged by the data link layers. In a bit-oriented protocol, the data section of a frame is a sequence of bits. Bit-oriented protocols are more popular today because we need to send text, graphic, audio, and video which can be better represented by a bit pat- tern than a sequence of characters.
Character-oriented protocols use byte-stuffing to be able to carry an 8-bit pattern that is the same as the flag. Byte-stuffing adds an extra character to the data section of the frame to escape the flag-like pattern.
Bit-oriented protocols use bit-stuffing to be able to carry patterns similar to the flag. Bit-stuffing adds an extra bit to the data section of the frame whenever a sequence of bits is similar to the flag. Flow control refers to a set of procedures used to restrict the amount of data that the sender can send before waiting for acknowledgment. Error control refers to a set of procedures used to detect and correct errors. In this chapter, we discussed two protocols for noiseless channels: the Simplest and the Stop-and-Wait.
The second uses pipelining, the first does not. In the first, we need to wait for an acknowledgment for each frame before sending the next one. In the second we can send several frames before receiving an acknowledgment. If a frame is lost or damaged, all outstanding frames sent before that frame are resent. In the Selective- Repeat ARQ protocol we avoid unnecessary transmission by sending only the frames that are corrupted or missing.
HDLC is a bit-oriented protocol for communication over point-to-point and multi- point links. PPP is a byte-oriented protocol used for point-to-point links. Piggybacking is used to improve the efficiency of bidirectional transmission. When a frame is carrying data from A to B, it can also carry control information about frames from B; when a frame is carrying data from B to A, it can also carry control information about frames from A.
We give a very simple solution. We write two very simple algorithms. We assume that a frame is made of a one- byte beginning flag, variable-length data possibly byte-stuffed , and a one-byte ending flag; we ignore the header and trailer.
We also assume that there is no error during the transmission. Algorithm We assume that a frame is made of an 8-bit flag , variable-length data possibly bit-stuffed , and an 8-bit ending flag ; we ignore header and trailer.
Note that when the algorithm exits from the loop, there are six bits of the ending flag in the buffer, which need to be removed after the loop. A five-bit sequence number can create sequence numbers from 0 to The sequence number in the Nth packet is N mod This means that the th packet has the sequence number mod 32 or 5.
See Algorithm Note that we have assumed that both events request and arrival have the same priority. Note that in this algorithm, we assume that the arrival of a frame by a site also means the acknowledgment of the previous frame sent by the same site. This is a very simple implementation in which we assume that both sites always have data to send.
Event B: Receiver Site: Frame 1 received. Since there are no lost or damaged frames and the round trip time is less than the time-out, each frame is sent only once.
Here, we have a special situation. Although no frame is damaged or lost, the sender sends each frame twice. The reason is that the the acknowledgement for each frame reaches the sender after its timer expires. The sender thinks that the frame is lost. In this case, only the first frame is resent; the acknowledgment for other frames arrived on time.
We need to send frames. We ignore the overhead due to the header and trailer. In the worst case, we send the a full window of size 7 and then wait for the We ignore the overhead due to the header and trailer. In the worst case, we send the a full window of size 4 and then wait for the We ignore the overhead due to the header and trailer. The three categories of multiple access protocols discussed in this chapter are ran- dom access, controlled access, and channelization.
In random access methods, no station is superior to another station and none is assigned the control over another. Each station can transmit when it desires on the condition that it follows the predefined procedure. In controlled access methods, the stations consult one another to find which sta- tion has the right to send.
A station cannot send unless it has been authorized by other stations. We discuss three popular controlled-access methods: reservation, polling, and token passing. Channelization is a multiple-access method in which the available bandwidth of a link is shared in time, frequency, or through code, between different stations.
In random access methods, there is no access control as there is in controlled access methods and there is no predefined channels as in channelization. Each station can transmit when it desires. This liberty may create collision. In a random access method, there is no control; access is based on contention.
In a controlled access method, either a central authority in polling or other stations in reservation and token passing control the access. Random access methods have less administration overhead. On the other hand, controlled access method are collision free.
In a random access method, the whole available bandwidth belongs to the station that wins the contention; the other stations needs to wait. In a channelization method, the available bandwidth is divided between the stations.
If a station does not have data to send, the allocated channel remains idle. In a controlled access method, the whole available bandwidth belongs to the sta- tion that is granted permission either by a central authority or by other stations. If a station does not have data to send the allocated channel remains idle. We do not need a multiple access method in this case. The local loop provides a dedicated point-to-point connection to the telephone office. We do need a multiple access, because a channel in the CATV band is normally shared between several neighboring customers.
The cable company uses the ran- dom access method to share the bandwidth between neighbors. If we let ns to be the number of stations and nfs to be the number of frames a station can send per sec- ond. This means that either the data rate must be very high or the frames must be very small.
If we let ns to be the number of stations and nfs to be the number of frames a station can send per second. We can first calculate Tfr and G, and then the throughput. Let us find the relationship between the minimum frame size and the data rate. In Example Let us find the relationship between the collision domain maximum length of the network and the data rate. When the data rate is increased, the distance or maximum length of network or col- lision domain is decreased proportionally.
We calculate the maximum distance based on the above proportionality relationship. The reason is that with the Mbps, the minimum number of bits requirement is feasible only when the maxi- mum distance between stations is less than or equal to meters as we will see in Chapter See Figure The reason is that we need to send 16 extra polls. The preamble is a bit field that provides an alert and timing pulse.
It is added to the frame at the physical layer and is not formally part of the frame. SFD is a one- byte field that serves as a flag. An NIC provides an Ethernet station with a 6-byte physical address. Most of the physical and data-link layer duties are done by the NIC. A multicast address identifies a group of stations; a broadcast address identifies all stations on the network.
A unicast address identifies one of the addresses in a group. A bridge can raise the bandwidth and separate collision domains. An instant messaging client communicates with an IM server application. Once a user is online, the server application can monitor connections so that multiple pre-identified clients can be notified and decide to participate in real-time messaging.
IM may include video or audio. Video exchange, of course, requires cameras. Underlying this application requires a full-duplex connection between destination and host. Compare and contrast the application architecture for videoconferencing with the architecture for e-mail. Videoconferencing must deliver real-time services demanding high capacity data transfer for both image and voice transmission.
Specialized hardware and even rooms may be required. E-mail messages typically without large attachments are relatively small by comparison, can be received by any Internet-capable computer, and do not have to be consumed in real time. The best architecture for email can depend on how one wants to use e-mail. If a person wants to be able to access their e-mail from anywhere, then Web-based is best. If the person wants professional backup and storage within an organization, then two-tier client-server is best.
If the person wants storage of e-mail strictly under their control and they also want to be able to access their e-mail files off-line when there is a network service interruption, then host-based is best.
Employers may choose to use client-server architecture for email access within the organization and Web-based architecture for access to the same system for those times when employees are outside the company at home, at another business, or on travel. Some experts argue that thin-client client-server architectures are really host-based architectures in disguise and suffer from the same old problems. Do you agree? While thin client have substantially less application logic than thick client, they have sufficient application logic as, for example, a Web browser possibly with Java applets to participate in a client-server relationship.
The older host-based terminals did not even have this much application logic. While thin-client use today reflects some level of return to a more centralized approach, the client is likely served by multiple servers and even multiple tiers , rather than a single large host server as in the past.
Thus, the two approaches are similar, but not exact, from a technological design perspective. Mini-Cases I. The tool will communicate with the DRUB server to select data to analyze. The alternatives are shown in the text as Figures client-based and two-tier clientserver.
Client-based is simple; however all data must travel to the client for processing, thus giving the potential for speed delays over the network. Client-server provides processing on the server, which could be an advantage if the data to be processed also resides on the server; yet, because it involves heterogeneous software, this can be a significant disadvantage in terms of interoperability.
One vendor is offering an SMTP-based two-tier client server architecture. The second vendor is offering a Web-based email architecture. The network administrator at DRUB can then configure and manage the email user accounts, etc himself.
The second solution is to use one of the cloud-based providers and completely outsource the company email. Two examples are Gmail and GoDaddy. With each of these services, the DRUB will pay a monthly fee for one of the providers to configure and manage the mail servers for them. A few of these include: In-house advantages: control, potentially lower cost In-house disadvantages: more work to do, potentially less expertise available Outsourcing advantage: potentially lower cost, better service, less work to do in-house Outsourcing disadvantages: loss of control, potentially higher costs III.
Will the new videoconferencing software and hardware work as simply as e-mail, or will it be IM all over again? Prepare a response to his questions. E-mail standards enable it to be used easily between companies. Lack of IM standards means that several competing protocols exist. This problem will be overcome as commercial interests push for standardization or development of middleware that enables disparate systems to talk to one another.
The same pattern of innovation will likely happen that is typical of all new technologies. There are a variety of ways the Internet could help Mr. He could develop a website where his paintings would be featured and some additional information about the artist would be displayed for potential buyers to learn and explore more about the works.
Further, the website could include an e-commerce function whereby visitors would be able to order prints of paintings they like. The site could help Mr. Ling track customer tastes so that he can better understand the types of paintings that sell well, thus allowing the business to develop while matching future production to the strongest market for the paintings.
Prepare a brief management summary on the technical essential aspects of the Internet and the World Wide Web and how they work. Remember, the audience is not technical. He is confused about the relationship between the World Wide Web and the Internet and often states that they are the same. Please be sure to explain this in your summary. One thing to remember and to emphasize with your students is the point that this question brings out, namely that there is great confusion between what constitutes the Internet and what exactly is the World Wide WEB.
The majority of my students are surprised to learn that there is a major difference. The other WWW is but one, albeit now the largest, applications that are available to run over the Internet.
The Internet actually began without the WWW. It was a bit level, text based network that evolved in a largely DOS based world. President Coone is particularly intrigued with the potential of the Internet, but he and the other members of management are not exactly sure what or how NDAS can use them to improve its competitive edge. Present some alternatives. Plans call for first offerings to be services to Britain, France, and Germany, with later expansion to South America. President Coone wants you to involve Bob Jones in your work on the Internet.
The Internet is global these days. Carrier access providers, known as ISPs, can provide a company like NDAS with a number of network transport options to connect each of their sites, both domestic and international to the Internet. Once connected there are many applications that NDAS could consider. NDAS could put in place an Intranet at their headquarters. Each of the other sites could access any of t he applications running on the Intranet via the Internet.
NDAS could for example have a standard Email application system and make this available to everyone in the company. It is common these days for companies to do this. The type of WEB Site applications typically can run from simple apps like providing essential information about the company to having elaborate back end database applications supporting things like order entry and making electronic payments.
The Marketing departments of companies these days are involved in the development of corporate WEB Sites because of the tremendous reach that a WEB Site can have in the marketplace. The image projected by the quality of the WEB Site will be a critical factor to its success.
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