COMPUTER SIXTH EDITION NETWORKING A Top-Down Approach James F. Kurose University of Massachusetts, Amherst Keith W. Ross Polytechnic Institute . Computer Networking: A Top Down Approach James, Keith – Book Review. Ijesrt Journal. [Kamath, 3(8): August, ] ISSN: top-down approach featuring the internet by jim kurose and keith ross. Computer Networking Kurose And Ross 7th Edition Pdf computer networking kurose and.

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Introduction to Computer Networks, Jim Kurose .. find links below to overheads (powerpoint files, compressed postscript and PDF format) for the textbook. Computer Networking: A Top-Down Approach 6th Edition. DOWNLOAD FULL PDF EBOOK here { }. . edition of Computer Networking: A Top-Down Approach by Jim Kurose and Keith Ross. By James Kurose, Keith Ross Click here for Download Ebook Computer Networking: A Top-Down Approach (7th Edition) By James Kurose.

This is additional information added in the Baggage layer if Figure 1. This information is used to ensure e. Without message segmentation, if bit errors are not tolerated, if there is a single bit error, the whole message has to be retransmitted rather than a single packet.

Without message segmentation, huge packets containing HD videos, for example are sent into the network. Routers have to accommodate these huge packets.

Smaller packets have to queue behind enormous packets and suffer unfair delays. Packets have to be put in sequence at the destination. Message segmentation results in many smaller packets. Since header size is usually the same for all packets regardless of their size, with message segmentation the total amount of header bytes is more.

Problem 32 Yes, the delays in the applet correspond to the delays in the Problem The propagation delays affect the overall end-to-end delays both for packet switching and message switching equally.

When a Skype user connected to the Internet calls an ordinary telephone, a circuit is established between a gateway and the telephone user over the circuit switched network. The skype user's voice is sent in packets over the Internet to the gateway. At the gateway, the voice signal is reconstructed and then sent over the circuit.

In the other direction, the voice signal is sent over the circuit switched network to the gateway. The gateway packetizes the voice signal and sends the voice packets to the Skype user. Chapter 2 Review Questions 1. The Web: HTTP; file transfer: FTP; remote login: Telnet; e-mail: BitTorrent protocol 2. Network architecture refers to the organization of the communication process into layers e.

Application architecture, on the other hand, is designed by an application developer and dictates the broad structure of the application e. The process which initiates the communication is the client; the process that waits to be contacted is the server.

In a P2P file-sharing application, the peer that is receiving a file is typically the client and the peer that is sending the file is typically the server. The IP address of the destination host and the port number of the socket in the destination process. You would use UDP. With TCP, a minimum of two RTTs are needed - one to set-up the TCP connection, and another for the client to send the request, and for the server to send back the reply.

One such example is remote word processing, for example, with Google docs. SSL operates at the application layer. A protocol uses handshaking if the two communicating entities first exchange control packets before sending data to each other. The applications associated with those protocols require that all application data be received in the correct order and without gaps. When the user first visits the site, the server creates a unique identification number, creates an entry in its back-end database, and returns this identification number as a cookie number.

During each subsequent visit and download , the browser sends the cookie number back to the site. Thus the site knows when this user more precisely, this browser is visiting the site. Web caching can reduce the delay for all objects, even objects that are not cached, since caching reduces the traffic on links.

Telnet is not available in Windows 7 by default. After issuing the command, you have established a TCP connection between your client telnet program and the web server. An example is given below: Since the index. FTP uses two parallel TCP connections, one connection for sending control information such as a request to transfer a file and another connection for actually transferring the file. Because the control information is not sent over the same connection that the file is sent over, FTP sends control information out of band.

Bob then transfers the message from his mail server to his host over POP3. Test mail Date: Sat, 19 May A sample mail message header Received: This header field indicates the sequence in which the SMTP servers send and receive the mail message including the respective timestamps. Message-id is a unique string assigned by the mail system when the message is first created.

This indicates the email address of the sender of the mail. This field indicates the email address of the receiver of the mail. This gives the subject of the mail if any specified by the sender. The date and time when the mail was sent by the sender. In the example, the sender sent the mail on 19th May , at time MIME version used for the mail. In the example, it is 1.

The type of content in the body of the mail message. This specifies the email address to which the mail will be sent if the receiver of this mail wants to reply to the sender. With download and delete, after a user retrieves its messages from a POP server, the messages are deleted. This poses a problem for the nomadic user, who may want to access the messages from many different machines office PC, home PC, etc.

In the download and keep configuration, messages are not deleted after the user retrieves the messages. This can also be inconvenient, as each time the user retrieves the stored messages from a new machine, all of non-deleted messages will be transferred to the new machine including very old messages.

You should be able to see the sender's IP address for a user with an. But you will not be able to see the sender's IP address if the user uses a gmail account. It is not necessary that Bob will also provide chunks to Alice. Alice has to be in the top 4 neighbors of Bob for Bob to send out chunks to her; this might not occur even if Alice provides chunks to Bob throughout a second interval. Recall that in BitTorrent, a peer picks a random peer and optimistically unchokes the peer for a short period of time.

Therefore, Alice will eventually be optimistically unchoked by one of her neighbors, during which time she will receive chunks from that neighbor. The overlay network in a P2P file sharing system consists of the nodes participating in the file sharing system and the logical links between the nodes. An overlay network does not include routers.

Mesh DHT: The advantage is in order to a route a message to the peer with ID that is closest to the key, only one hop is required; the disadvantage is that each peer must track all other peers in the DHT. Circular DHT: With the UDP server, there is no welcoming socket, and all data from different clients enters the server through this one socket.

With the TCP server, there is a welcoming socket, and each time a client initiates a connection to the server, a new socket is created. If the TCP server is not running, then the client will fail to make a connection.

For the UDP application, the client does not initiate connections or attempt to communicate with the UDP server immediately upon execution Transfer parameter commands: Service commands: Problem 3 Application layer protocols: The Host: This information is not contained in an HTTP message anywhere. So there is no way to tell this from looking at the exchange of HTTP messages alone.

The browser type information is needed by the server to send different versions of the same object to different types of browsers. Problem 5 a The status code of and the phrase OK indicate that the server was able to locate the document successfully. The reply was provided on Tuesday, 07 Mar The server agreed to a persistent connection, as indicated by the Connection: Sections 8.

From the server's point of view, the connection is being closed while it was idle, but from the client's point of view, a request is in progress.

The average time is the average size of the object divided by R: Thus, the average access delay is. The total average response time is therefore. Thus the average access delay is. The response time is approximately zero if the request is satisfied by the cache which happens with probability. So the average response time is. Thus the average response time is reduced from 3.

Problem 10 Note that each downloaded object can be completely put into one data packet. Let Tp denote the one-way propagation delay between the client and the server.

First consider parallel downloads using non-persistent connections. Thus, the total time needed to receive all objects is given by: The total time needed is given by: Tp is therefore negligible compared with transmission delay. Thus, we see that persistent HTTP is not significantly faster less than 1 percent than the non-persistent case with parallel download.

Problem 11 a Yes, because Bob has more connections, he can get a larger share of the link bandwidth.

Problem 12 Server. The From: Problem 14 SMTP uses a line containing only a period to mark the end of a message body. A host sends the message to an MTA. We see that this spam message follows a chain of MTAs.

An honest MTA should report where it receives the message. By maintaining a file that lists the messages retrieved during earlier sessions, the client can use the UIDL command to determine which messages on the server have already been seen. Problem 17 a C: COM from www. NET from ww. Problem 19 a The following delegation chain is used for gaia. NET ns1. Among all returned edu DNS servers, we send a query to the first one.

NET any gaia. Among all three returned authoritative DNS servers, we send a query to the first one.

The Web server that appears most frequently in the DNS caches is the most popular server. This is because if more users are interested in a Web server, then DNS requests for that server are more frequently sent by users. Thus, that Web server will appear in the DNS caches more frequently. For a complete measurement study, see: Craig E. If cnn. Otherwise, the query time is large. Problem 22 For calculating the minimum distribution time for client-server distribution, we use the following formula: Combining these two gives: Equation 2 We can similarly show that: Combining Equation 2 and Equation 3 gives the desired result.

Also have each peer i forward the bits it receives to each of the N-1 peers at rate ri. The aggregate forwarding rate by peer i is N-1 ri.

Thus the aggregate forwarding rate of peer i is less than its link rate ui. By assumption Each peer i forwards the bits arriving at rate ri to each of the other N-1 peers. We now provide that here. We know from section 2. Problem 25 There are N nodes in the overlay network. Problem 26 Yes. His first claim is possible, as long as there are enough peers staying in the swarm for a long enough time. Bob can always receive data through optimistic unchoking by other peers. His second claim is also true.

He can even write a small scheduling program to make the different hosts ask for different chunks of the file. This is actually a kind of Sybil attack in P2P networks. Problem 27 Peer 3 learns that peer 5 has just left the system, so Peer 3 asks its first successor Peer 4 for the identifier of its immediate successor peer 8. Peer 3 will then make peer 8 its second successor. Next, peer 5 sends this predecessor and successor information back to 6. Peer 6 can now join the DHT by making peer 8 its successor and by notifying peer 5 that it should change its immediate successor to 6.

Problem 29 For each key, we first calculate the distances using d k,p between itself and all peers, and then store the key in the peer that is closest to the key that is, with smallest distance value. Problem 30 Yes, randomly assigning keys to peers does not consider the underlying network at all, so it very likely causes mismatches. Such mismatches may degrade the search performance.

For example, consider a logical path p1 consisting of only two logical links: Suppose that there is another logical path p2 from A to C consisting of 3 logical links: It might be the case that A and B are very far away physically and separated by many routers , and B and C are very far away physically and separated by many routers.

But In other words, a shorter logical path may correspond to a much longer physical path. A TCP connection will not be made. Errors will occur.

Problem 32 In the original program, UDPClient does not specify a port number when it creates the socket. In this case, the code lets the underlying operating system choose a port number. UDPServer needs to know the client port number so that it can send packets back to the correct client socket. Thus UDP server will work with any client port number, including UDPServer therefore does not need to be modified.

The advantage is that you will you potentially download the file faster. The Problem 34 For an application such as remote login telnet and ssh , a byte-stream oriented protocol is very natural since there is no notion of message boundaries in the application. When a user types a character, we simply drop the character into the TCP connection. In other applications, we may be sending a series of messages that have inherent boundaries between them.

Since TCP does not have a mechanism to indicate the boundaries, the application must add the indications itself, so that receiving side of the application can distinguish one message from the next. If each message were instead put into a distinct UDP segment, the receiving end would be able to distinguish the various messages without any indications added by the sending side of the application.

Problem 35 To create a web server, we need to run web server software on a host. Many vendors sell web server software. However, the most popular web server software today is Apache, which is open source and free. Over the years it has been highly optimized by the open- source community.

Computer Networking: A Top-Down Approach, 6th Edition

Problem 36 The key is the infohash, the value is an IP address that currently has the file designated by the infohash. Chapter 3 Review Questions 1. At the sender side, STP accepts from the sending process a chunk of data not exceeding bytes, a destination host address, and a destination port number. STP adds a four-byte header to each chunk and puts the port number of the destination process in this header. STP then gives the destination host address and the resulting segment to the network layer.

The network layer delivers the segment to STP at the destination host. STP then examines the port number in the segment, extracts the data from the segment, and passes the data to the process identified by the port number.

At the sender side, STP accepts a chunk of data not exceeding bytes, a destination host address, a source port number, and a destination port number. STP creates a segment which contains the application data, source port number, and destination port number. It then gives the segment and the destination host address to the network layer.

After receiving the segment, STP at the receiving host gives the application process the application data and the source port number.

If You're an Educator

For sending a letter, the family member is required to give the delegate the letter itself, the address of the destination house, and the name of the recipient. The delegate then puts the letter in an envelope and writes the address of the destination house on the envelope. At the receiving side, the delegate receives the letter from the mail service, takes the letter out of the envelope, and takes note of the recipient name written at the top of the letter. The delegate then gives the letter to the family member with this name.

No, the mail service does not have to open the envelope; it only examines the address on the envelope. Source port number y and destination port number x.

Also, some applications do not need the reliable data transfer provided by TCP. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however. Yes, both segments will be directed to the same socket. For each received segment, at the socket interface, the operating system will provide the process with the IP addresses to determine the origins of the individual segments.

Each connection socket is identified with a four-tuple: Thus, the requests from A and B pass through different sockets. The identifier for both of these sockets has 80 for the destination port; however, the identifiers for these sockets have different values for source IP addresses.

Sequence numbers are required for a receiver to find out whether an arriving packet contains new data or is a retransmission. To handle losses in the channel. Hence, the packet is retransmitted. A timer would still be necessary in the protocol rdt 3. If the round trip time is known then the only advantage will be that, the sender knows for sure that either the packet or the ACK or NACK for the packet has been lost, as compared to the real scenario, where the ACK or NACK might still be on the way to the sender, after the timer expires.

However, to detect the loss, for each packet, a timer of constant duration will still be necessary at the sender. After the timeout, sender retransmitted the lost packet and receiver delivered the buffered packets to application in correct order. First segment: False, it is set to half of the current value of the congestion window.

Consider the following timing diagram. Note that a, b, c are distinct. To host A: To detect errors, the receiver adds the four words the three original words and the checksum.

If the sum contains a zero, the receiver knows there has been an error. All one-bit errors will be detected, but two-bit errors can be undetected e. Problem 4 a Adding the two bytes gives Problem 5 No, the receiver cannot be absolutely certain that no bit errors have occurred.

Computer Networking: A Top-Down Approach (4th Edition)

This is because of the manner in which the checksum for the packet is calculated. If the corresponding bits that would be added together of two bit words in the packet were 0 and 1 then even if these get flipped to 1 and 0 respectively, the sum still remains the same.

Hence, the 1s complement the receiver calculates will also be the same. This means the checksum will verify even if there was transmission error. However, the ACK is corrupted. When the rdt2. However, the receiver is waiting for a packet with sequence number 0 and as shown in the home work problem always sends a NAK when it doesn't get a packet with sequence number 0.

Hence the sender will always be sending a packet with sequence number 1, and the receiver will always be NAKing that packet. Neither will progress forward from that state. Problem 7 To best answer this question, consider why we needed sequence numbers in the first place. We saw that the sender needs sequence numbers so that the receiver can tell if a data packet is a duplicate of an already received data packet.

In the case of ACKs, the sender does not need this info i. A duplicate ACK is obvious to the rdt3. Problem 8 The sender side of protocol rdt3. We have seen that the introduction of timeouts adds the possibility of duplicate packets into the sender-to-receiver data stream. However, the receiver in protocol rdt. Receiver-side duplicates in rdt 2.

Hence the receiver in protocol rdt2. Problem 9 Suppose the protocol has been in operation for some time. The scenarios for corrupted data and corrupted ACK are shown in Figure 1. If the timeout event occurs, the most recently transmitted packet is retransmitted. Let us see why this protocol will still work with the rdt2. In this case, the receiver never received the previous transmission and, from the receiver's viewpoint, if the timeout retransmission is received, it looks exactly the same as if the original transmission is being received.

The receiver will eventually retransmit the packet on a timeout. But a retransmission is exactly the same action that if an ACK is garbled.

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Thus the sender's reaction is the same with a loss, as with a garbled ACK. The rdt 2. Problem 11 If the sending of this message were removed, the sending and receiving sides would deadlock, waiting for an event that would never occur. Now, the ender is awaiting an ACK of some sort from the receiver, and the receiver is waiting for a data packet form the sender — a deadlock!

Problem 12 The protocol would still work, since a retransmission would be what would happen if the packet received with errors has actually been lost and from the receiver standpoint, it never knows which of these events, if either, will occur. To get at the more subtle issue behind this question, one has to allow for premature timeouts to occur. In this case, if each extra copy of the packet is ACKed and each received extra ACK causes another extra copy of the current packet to be sent, the number of times packet n is sent will increase without bound as n approaches infinity.

On the other hand, if data is being sent often, then recovery under a NAK-only scheme could happen quickly. Problem 15 It takes 12 microseconds or 0. In order for the sender to be busy 98 percent of the time, we must have Problem 16 Yes.

This actually causes the sender to send a number of pipelined data into the channel. Here is one potential problem. If data segments are lost in the channel, then the sender of rdt 3. Data packets have a data field and carry a two-bit sequence number. That is, the valid sequence numbers are 0, 1, 2, and 3.

ACK messages carry the sequence number of the data packet they are acknowledging. The FSM for the sender and receiver are shown in Figure 2. In this figure, we assume that the seqnum is initially 0, and that the sender has sent the first Wait: A timeline trace for the sender and receiver recovering from a lost packet is shown below: Figure 2: Sender and receiver for Problem 3. Packet 0 drops send packet 1 receive packet 1 buffer packet 1 send ACK 1 receive ACK 1 timeout resend packet 0 receive packet 0 deliver pair 0,1 send ACK 0 receive ACK 0 Problem 19 This problem is a variation on the simple stop and wait protocol rdt3.

Because the channel may lose messages and because the sender may resend a message that one of the receivers has already received either because of a premature timeout or because the other receiver has yet to receive the data correctly , sequence numbers are needed.

As in rdt3. The sender and receiver FSM are shown in Figure 3. Computer Networking Manual Pdf computer networking a top down approach 5th edition solutions manual is for all of you who want to know about the computer networking. This book written.

Computer networking — James f. Kurose, Keith w.

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Thus, the average delay for the N packets is: A the requestor has 4 states: Bob then transfers the message from his mail server to his host over POP3.