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Pdf 32 Data communication and networking forouzan 2nd edition. The main Unit of the book are as follows: PART 1: Overview Chapter 1. Data Communications and networking Fourth Edition Forouzan PPT Slides. to Data Communications and Networking, Tata McGraw Hill, 2nd Edition, Available in: Hardcover. Data Communications And Networking by Behrouz Forouzan provides a thorough introduction to the concepts that.

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He graduated from the University of California, Irvine. Design, implement and analyze simple computer networks. Differentiate between different types of computer networks iii. Such as frame relay and. SMDS were designed for todays almost- error-free digital lines. Hope you enjoy learning networking and clear your all concepts.

The five components of a data. This chapter provides an introduction to Computer networks and covers fundamental topics. Data Communication Networking Behrouz Forouzan. Services Provided at the Destination Computer Free trial Sign in. Using a bottom-up approach, Data Communications and Networking presents. Preface 1. A free tool that helps you create your own course website.

Jan 9, PART 3. Sep 15, 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: 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.

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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 We can calculate time based on the assumption of 10 Mbps data rate: 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 only relationship between the size of the codeword and dataword is the one based on the definition: 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.

The last example shows how a noise of small duration can affect so many bits if the data rate is high. The above shows three properties of the exclusive-or operation. 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.

Part c and d show that the distance between a codeword and itself is 0. 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: 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: 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: 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: 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: In this chapter, we discussed three protocols for noisy channels: 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. We can then say Event A: Sender Site: ACK 0 received. Event B: ACK 1 received. Receiver Site: Frame 0 received. Frame 1 received. Frame 0 arrives and ACK 1 is sent. ACK 1 arrives and Frame 1 is sent. Event C: Frame 1 arrives and ACK 0 is sent. Event D: ACK 0 arrives and Frame 0 is sent.

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: 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 Third Property: We can say: Station 1: Frame 1 for all four stations: 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. A layer-2 switch is an N-port bridge with additional sophistication that allows faster handling of packets. In a full-duplex Ethernet, each station can send data without the need to sense the line.

The rates are as follows: Standard Ethernet: We interpret each four-bit pattern as a hexadecimal digit. We then group the hexa- decimal digits with a colon between the pairs: The bytes are sent from left to right. However, the bits in each byte are sent from the least significant rightmost to the most significant leftmost. We have shown the bits with spaces between bytes for readability, but we should remember that that bits are sent without gaps.

The arrow shows the direction of movement. The first byte in binary is The least significant bit is 1. This means that the pattern defines a multicast address. A multicast address can be a destination address, but not a source address.

Therefore, the receiver knows that there is an error, and discards the packet. The minimum data size in the Standard Ethernet is 46 bytes. The maximum data size in the Standard Ethernet is bytes.

The data of bytes, therefore, must be split between two frames. The standard dictates that the first frame must carry the maximum possible number of bytes ; the second frame then needs to carry only 10 bytes of data it requires padding.

The follow- ing shows the breakdown: Data size for the first frame: The smallest Ethernet frame is 64 bytes and carries 46 bytes of data and possible padding. The largest Ethernet frame is bytes and carries bytes of data. We can then answer the question as follows: The smallest frame is 64 bytes or bits. A station with no-transition mobility is either stationary or moving only inside a BSS. All the subbands are used by one source at a given time. Sources contend with one another at the data link layer for access.

Network Allocation Vector NAV forces other stations to defer sending their data if one station acquires access. In other words, it provides the collision avoidance aspect. When a station sends an RTS frame, it includes the duration of time that it needs to occupy the channel. The stations that are affected by this transmission create a timer called a NAV.

A Bluetooth network is called a piconet. A scatternet is two or more piconets. The following shows the relationship: A Bluetooth primary and secondary can be connected by a synchronous connec- tion-oriented SCO link or an asynchronous connectionless ACL link. An ACL link is used when data integrity is more important than avoiding latency. The primary sends on the even-numbered slots; the secondary sends on the odd- numbered slots. If there is a collision, it will be detected, destroyed, and the frame will be resent.

See Table Table An amplifier amplifies the signal, as well as noise that may come with the signal, whereas a repeater regenerates the signal, bit for bit, at the original strength. Bridges have access to station physical addresses and can forward a packet to the appropriate segment of the network.

In this way, they filter traffic and help control congestion. If a bridge is added or deleted from the system, reconfigura- tion of the stations is unnecessary.

A signal can only travel so far before it becomes corrupted. A repeater regenerates the original signal; the signal can continue to travel and the LAN length is thus extended. A hub is a multiport repeater. A forwarding port forwards a frame that it receives; a blocking port does not. In a bus backbone, the topology of the backbone is a bus; in a star backbone, the topology is a star. A VLAN saves time and money because reconfiguration is done through software. Physical reconfiguration is not necessary.

Members of a VLAN can send broadcast messages with the assurance that users in other groups will not receive these messages. A VLAN creates virtual workgroups. Each workgroup member can send broadcast messages to others in the workgroup.

This eliminates the need for multicasting and all the overhead messages associated with it. Stations can be grouped by port number, MAC address, IP address, or by a com- bination of these characteristics.

We have sorted the table based on the physical address to make the searching faster. We made bridge B1 the root. Although any router is also a bridge, replacing bridges with routers has the follow- ing consequences: Routers are more expensive than bridges. Routers operate at the first three-layers; bridges operates at the first two layers. Routers are not designed to provide direct filtering the way the bridges do. A router needs to search a routing table which is normally longer and more time consuming than a filtering table.

A router needs to decapsulate and encapsulate the frame and change physical addresses in the frame because the physical addresses in the arriving frame define the previous node and the current router; they must be changed to the physical addresses of the current router and the next hop. A bridge does not change the physical addresses. Changing addresses, and other fields, in the frame means much unnecessary overhead.

A filtering table is based on physical addresses; a routing table is based on the logical addresses. We have shown the network, the graph, the spanning tree, and the blocking ports. Network b. Spanning tree d.

Blocking ports A router has more overhead than a bridge. A router process the packet at three layers; a bridge processes a frame at only two layers. A router needs to search a routing table for finding the output port based on the best route to the final destina- tion; A bridge needs only to consult a filtering table based on the location of sta- tions in a local network.

A routing table is normally longer than a filtering table; searching a routing table needs more time than searching a filtering table. A router changes the physical addresses; a bridge does not. A bridge has more overhead than a repeater. A bridge processes the packet at two layers; a repeater processes a frame at only one layer. A bridge needs to search a table and find the forwarding port as well as to regenerate the signal; a repeater only regenerates the signal.

In other words, a bridge is also a repeater and more ; a repeater is not a bridge. A gateway has more overhead than a router.

A gateway processes the packet at five layers; a router processes a packet at only three layers. A gateway needs to worry about the format of the packet at the transport and application layers; a router does not. In other words, a gateway is also a router but more ; a router is not a gateway. A gateway may need to change the port addresses and application addresses if the gateway connects two different systems together; a router does not change these addresses.

A mobile switching center coordinates communications between a base station and a telephone central office. A mobile switching center connects cells, records call information, and is respon- sible for billing.

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A high reuse factor is better because the cells that use the same set of frequencies are farther apart separated by more cells. In a hard handoff, a mobile station communicates with only one base station.

In a soft handoff, a mobile station communicates with two base stations at the same time. GSM is a European standard that provides a common second-generation technol- ogy for all of Europe. CDMA encodes each traffic channel using one of the rows in the Walsh table. The three orbit types are equatorial, inclined, and polar.

A GEO satellite has an equatorial orbit since the satellite needs to remain fixed at a certain spot above the earth. A footprint is the area on earth at which the satellite aims its signal. A satellite orbiting in a Van Allen belt would be destroyed by the charged parti- cles. Therefore, satellites need to orbit either above or below these belts.

Transmission from the earth to the satellite is called the uplink. Transmission from the satellite to the earth is called the downlink. GPS is a satellite system that provides land and sea navigation data for vehicles and ships. The system is also used for clock synchronization. The main difference between Iridium and Globalstar is the relaying mechanism.

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