Learning Books. Contribute to arasty/books development by creating an account on GitHub. Computer networking: a top-down approach / James F. Kurose, Keith W. Ross. in this sixth edition, but we've also kept unchanged what we believe (and the. Computer Networking: A Top-Down Approach, 6th Edition Solutions to Review Standards are important for protocols so that people can create networking systems and The hashes for all of the blocks are saved in terney.infot file.
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We use the central limit theorem to approximate this probability. Similarly, we can find the delay caused by the second switch and the third link: Problem 11 Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Problem 12 The arriving packet must first wait for the link to transmit 4.
Since these bits are transmitted at 2 Mbps, the queuing delay is 27 msec. Thus, the average delay for the N packets is: Thus, the buffer is empty when a each batch of N packets arrive.
Thus, the average delay of a packet across all batches is the average delay within one batch, i. Problem 16 The total number of packets in the system includes those in the buffer and the packet that is being transmitted.
Let q procd denote the processing delay at the q th node. Let q propd be the propagation delay across the q th link. Problem 18 On linux you can use the command traceroute www. For those three measurements you can calculate the mean and standard deviation. Repeat the experiment at different times of the day and comment on any changes. Here is an example solution: The standard deviations are 0.
Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Traceroutes from www. In this example, there are 11 routers in the path at each of the three hours. Traceroute packets passed three ISP networks from source to destination. Problem 19 An example solution: Traceroutes from two different cities in France to New York City in United States a In these traceroutes from two different cities in France to the same destination host in United States, seven links are in common including the transatlantic link.
Traceroutes to two different cities in China from same host in United States c Five links are common in the two traceroutes. Problem 22 Probability of successfully receiving a packet is: The number of transmissions needed to be performed until the packet is successfully received by the client is a geometric random variable with success probability ps.
Thus, the average number of transmissions needed is given by: Then, the average number of re-transmissions needed is given by: The left hand side of the above inequality represents the time needed by the second packet to arrive at the input queue of the second link the second link has not started transmitting the second packet yet.
The right hand side represents the time needed by the first packet to finish its transmission onto the second link. If we send the second packet T seconds later, we will ensure that there is no queuing delay for the second packet at the second link if we have: Problem 25 a , bits b , bits c The bandwidth-delay product of a link is the maximum number of bits that can be in the link.
Problem 29 Recall geostationary satellite is 36, kilometers away from earth surface. 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. 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. 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.
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. 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 or n approximately packets.
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. A timeline trace for the sender and receiver recovering from a lost packet is shown below: Figure 2: Sender and receiver for Problem 3.
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. The receiver state indicates which sequence number the receiver is waiting for. Figure 3. Receiver side FSM for 3. Because the channel delays are variable and unknown, it is possible that A will send duplicate requests i.
To be able to detect duplicate request messages, the protocol will use sequence numbers. Here the requestor is waiting for a D0 data message from B. A timer is always running in this state. If the timer expires, A sends another R0 message, restarts the timer and remains in this state.
If A receives a D1 data message while in this state, it is ignored. Here the requestor is waiting for a D1 data message from B. If the timer expires, A sends another R1 message, restarts the timer and remains in this state. If A receives a D0 data message while in this state, it is ignored. The data supplier B has only two states: If B receives a R1 message, then it knows its D0 message has been received correctly. Suppose the receiver has received packet k-1, and has ACKed that and all other preceding packets.
Suppose next that none of the ACKs have been received at the sender. In this second case, the sender's window contains k-1 and the N packets up to and including k The sender's window is thus [k-N,k-1].
By these arguments, the senders window is of size 3 and begins somewhere in the range [k-N,k]. Because the sender has sent packets [k-N, k-1], it must be the case that the sender has already received an ACK for k-N Thus the range of in- flight ACK values can range from k-N-1 to k Problem 23 In order to avoid the scenario of Figure 3.
That is, the sequence number space must be large enough to fit the entire receiver window and the entire sender window without this overlap condition. So - we need to determine how large a range of sequence numbers can be covered at any given time by the receiver and sender windows. Suppose that the lowest-sequence number that the receiver is waiting for is packet m.
If none of those w ACKs have been yet received by the sender, then ACK messages with values of [m-w,m-1] may still be propagating back.
Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at t 0. At resends 1, 2, 3. At t 3 the receiver receives the duplicates and re-acknowledges 1, 2, 3. At t 4 the sender receives the ACKs that the receiver sent at t1 and advances its window to 4, 5, 6. At t 5 the sender receives the ACKs 1, 2, 3 the receiver sent at t 2. These ACKs are outside its window. By essentially the same scenario as in a.
Note that with a window size of 1, SR, GBN, and the alternating bit protocol are functionally equivalent. The window size of 1 precludes the possibility of out-of- order packets within the window. A cumulative ACK is just an ordinary ACK in this situation, since it can only refer to the single packet within the window.
Problem 25 a Consider sending an application message over a transport protocol. UDP, on the other hand, encapsulates in a segment whatever the application gives it; so that, if the application gives UDP an application message, this message will be the payload of the UDP segment.
Thus, with UDP, an application has more control of what data is sent in a segment. UDP does not have delays due to flow control and congestion control. Rather, it increments by the number of bytes of data sent. Problem 27 a In the second segment from Host A to B, the sequence number is , source port number is and destination port number is Still, host A sends data into the receive buffer faster than Host B can remove data from the buffer.
The receive buffer fills up at a rate of roughly 40Mbps. On average, the long-term rate at which Host A sends data to Host B as part of this connection is no more than 60Mbps. Half-open connections are not possible since a server using SYN cookies does not maintain connection variables and buffers for any connection before full connections are established. For establishing fully open connections, an attacker should know the special initial sequence number corresponding to the spoofed source IP address from the attacker.
This sequence number requires the "secret" number that each server uses. Since the attacker does not know this secret number, she cannot guess the initial sequence number. Problem 30 a If timeout values are fixed, then the senders may timeout prematurely. Thus, some packets are re-transmitted even they are not lost. But there might be one potential problem. Queuing delay might be very large, similar to what is shown in Scenario 1.
Suppose the source sends packet P1, the timer for P1 expires, and the source then sends P2, a new copy of the same packet. Finally suppose that shortly after transmitting P2 an acknowledgment for P1 arrives. The source will mistakenly take this acknowledgment as an acknowledgment for P2 and calculate an incorrect value of SampleRTT.
Problem 34 At any given time t, SendBase — 1 is the sequence number of the last byte that the sender knows has been received correctly, and in order, at the receiver. The actually last byte received correctly and in order at the receiver at time t may be greater if there are acknowledgements in the pipe. The actual last byte SendBase or if there are other acknowledgements in the pipe. The designers of the triple duplicate ACK scheme probably felt that waiting for two packets rather than 1 was the right tradeoff between triggering a quick retransmission when needed, but not retransmitting prematurely in the face of packet reordering.
Problem 37 a GoBackN: A sends 9 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re- sent segments 2, 3, 4, and 5. B sends 8 ACKs. Selective Repeat: A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re- sent segments 2.
B sends 5 ACKs. They are 4 ACKS with sequence number 1, 3, 4, 5. And there is one ACK with sequence number 2. They are 4 ACKS with sequence number 2. There is one ACK with sequence numbers 6. This is because TCP uses fast retransmit without waiting until time out. With increased loss, even a larger fraction of the packets leaving the queue will be retransmissions.
If there was a timeout, the congestion window size would have dropped to 1.
When loss is detected during transmission round 16, the congestion windows size is Hence the threshold is 21 during the 18th transmission round. When loss is detected during transmission round 22, the congestion windows size is Hence the threshold is 14 taking lower floor of Thus packet 70 is sent in the 7th transmission round. Thus the new values of the threshold and window will be 4 and 7 respectively. So, the total number is Problem 41 Refer to Figure 5.
In Figure 5 a , the ratio of the linear decrease on loss between connection 1 and connection 2 is the same - as ratio of the linear increases: In this case, the throughputs never move off of the AB line segment. In Figure 5 b , the ratio of the linear decrease on loss between connection 1 and connection 2 is 2: That is, whenever there is a loss, connection 1 decreases its window by twice the amount of connection 2.
We see that eventually, after enough losses, and subsequent increases, that connection 1's throughput will go to 0, and the full link bandwidth will be allocated to connection 2. Figure 5: Lack of TCP convergence with linear increase, linear decrease Problem 42 If TCP were a stop-and-wait protocol, then the doubling of the time out interval would suffice as a congestion control mechanism.
However, TCP uses pipelining and is therefore not a stop-and-wait protocol , which allows the sender to have multiple outstanding unacknowledged segments. The doubling of the timeout interval does not prevent a TCP sender from sending a large number of first-time-transmitted packets into the network, even when the end-to-end path is highly congested. Also, because there is no loss and acknowledgements are returned before timers expire, TCP congestion control does not throttle the sender.
However, the process in host A will not continuously pass data to the socket because the send buffer will quickly fill up. Problem 45 a The loss rate, L , is the ratio of the number of packets lost over the number of packets sent. In a cycle, 1 packet is lost. Recall the window size increases by one in each RTT. Problem 47 Let W denote max window size. Let S denote the buffer size.
For simplicity, suppose TCP sender sends data packets in a round by round fashion, with each round corresponding to a RTT. If the window size reaches W, then a loss occurs. Let Tp denote the one-way propagation delay between the sender and the receiver. Problem 48 a Let W denote the max window size. In order to speed up the window increase process, we can increase the window size by a much larger value, instead of increasing window size only by one in each RTT.
Problem 49 1. Thus C1 adjusts its window size after 50 msec, but C2 adjusts its window size after msec. Assume that whenever a loss event happens, C1 receives it after 50msec and C2 receives it after msec. We further have the following simplified model of TCP. After each RTT, a connection determines if it should increase window size or not. For C1, we compute the average total sending rate in the link in the previous 50 msec. If that rate exceeds the link capacity, then we assume that C1 detects loss and reduces its window size.
But for C2, we compute the average total sending rate in the link in the previous msec. If that rate exceeds the link capacity, then we assume that C2 detects loss and reduces its window size.
Note that it is possible that the average sending rate in last 50msec is higher than the link capacity, but the average sending rate in last msec is smaller than or equal to the link capacity, then in this case, we assume that C1 will experience loss event but C2 will not. The following table describes the evolution of window sizes and sending rates based on the above assumptions. If we look at the above table, we can see a cycle every msec, e.
Problem 51 a Similarly as in last problem, we can compute their window sizes over time in the following table. Both C1 and C2 have the same window size 2 after msec. Their max window size is 2. Thus, the link is not fully utilized recall we assume this link has no buffer. One possible way to break the synchronization is to add a finite buffer to the link and randomly drop packets in the buffer before buffer overflow.
This will cause different connections cut their window sizes at different times. Problem 52 Note that W represents the maximum window size. This is given by: Loss rate L is given by: Note that this is different from TCP which has average throughput: From the TCP throughput 1. A disadvantage of using these values is that they may be no longer accurate. Problem 55 a The server will send its response to Y. Even if the attacker were to send an appropriately timed TCP ACK segment, it would not know the correct server sequence number since the server uses random initial sequence numbers.
A network-layer packet is a datagram. Datagram-based network layer: Additional function of VC-based network layer: Routing is about determining the end-to-routes between sources and destinations. Yes, both use forwarding tables. For descriptions of the tables, see Section 4. Single packet: Flow of packets: ABR does not provide any of these services. With the shadow copy, the forwarding lookup is made locally, at each input port, without invoking the centralized routing processor.
Such a decentralized approach avoids creating a lookup processing bottleneck at a single point within the router. Switching via memory; switching via a bus; switching via an interconnection network. An interconnection network can forward packets in parallel as long as all the packets are being forwarded to different output ports.
If the rate at which packets arrive to the fabric exceeds switching fabric rate, then packets will need to queue at the input ports. If this rate mismatch persists, the queues will get larger and larger and eventually overflow the input port buffers, causing packet loss. Packet loss can be eliminated if the switching fabric speed is at least n times as fast as the input line speed, where n is the number of input ports. Assuming input and output line speeds are the same, packet loss can still occur if the rate at which packets arrive to a single output port exceeds the line speed.
If this rate mismatch persists, the queues will get larger and larger and eventually overflow the output port buffers, causing packet loss. Note that increasing switch fabric speed cannot prevent this problem from occurring. HOL blocking: Sometimes the a packet that is first in line at an input port queue must wait because there is no available buffer space at the output port to which it wants to be forwarded.
When this occurs, all the packets behind the first packet are blocked, even if their output queues have room to accommodate them. HOL blocking occurs at the input port. They have one address for each interface. Students will get different correct answers for this question. The 8-bit protocol field in the IP datagram contains information about which transport layer protocol the destination host should pass the segment to. Typically the wireless router includes a DHCP server. IPv6 has a fixed length header, which does not include most of the options an IPv4 header can include.
Even though the IPv6 header contains two bit addresses source and destination IP address the whole header has a fixed length of 40 bytes only. Several of the fields are similar in spirit. Traffic class, payload length, next header and hop limit in IPv6 are respectively similar to type of service, datagram length, upper-layer protocol and time to live in IPv4.
Yes, because the entire IPv6 datagram including header fields is encapsulated in an IPv4 datagram. Link state algorithms: Computes the least-cost path between source and destination using complete, global knowledge about the network.
Distance-vector routing: The calculation of the least-cost path is carried out in an iterative, distributed manner. A node only knows the neighbor to which it should forward a packet in order to reach given destination along the least-cost path, and the cost of that path from itself to the destination. Routers are organized into autonomous systems ASs. Within an AS, all routers run the same intra-AS routing protocol.
The problem of scale is solved since an router in an AS need only know about routers within its AS and the subnets that attach to the AS. To route across ASes, the inter-AS protocol is based on the AS graph and does not take individual routers into account. Each AS has administrative autonomy for routing within an AS. The advertisement tells D that it can get to z in 11 hops by way of A.
However, D can already get to z by way of B in 7 hops. Therefore, there is no need to modify the entry for z in the table. If, on the other hand, the advertisement said that A were only 4 hops away from z by way of C, then D would indeed modify its forwarding table. With OSPF, a router periodically broadcasts routing information to all other routers in the AS, not just to its neighboring routers.
A RIP advertisement sent by a router contains information about all the networks in the AS, although this information is only sent to its neighboring routers. Among ASs, policy issues dominate. It may well be important that traffic originating in a given AS not be able to pass through another specific AS. Similarly, a given AS may want to control what transit traffic it carries between other ASs. Within an AS, everything is nominally under the same administrative control and thus policy issues a much less important role in choosing routes with in AS.
Within an AS, scalability is less of a concern. For one thing, if a single administrative domain becomes too large, it is always possible to divide it into two ASs and perform inter-AS routing between the two new ASs. Because inter-AS routing is so policy oriented, the quality for example, performance of the routes used is often of secondary concern that is, a longer or more costly route that satisfies certain policy criteria may well be taken over a route that is shorter but does not meet that criteria.
Indeed, we saw that among ASs, there is not even the notion of cost other than AS hop count associated with routes. Within a single AS, however, such policy concerns are of less importance, allowing routing to focus more on the level of performance realized on a route. A subnet is a portion of a larger network; a subnet does not contain a router; its boundaries are defined by the router and host interfaces.
A prefix is the network portion of a CDIRized address; it is written in the form a. Routers use the AS-PATH attribute to detect and prevent looping advertisements; they also use it in choosing among multiple paths to the same prefix. N-way unicast has a number of drawbacks, including: T; controlled flooding: T; spanning-tree: F b uncontrolled flooding: F; spanning-tree: False IGMP is a protocol run only between the host and its first-hop multicast router.
IGMP allows a host to specify to the first-hop multicast router the multicast group it wants to join. It is then up to the multicast router to work with other multicast routers i. In a group-shared tree, all senders send their multicast traffic using the same routing tree.
With source-based tree, the multicast datagrams from a given source are routed over s specific routing tree constructed for that source; thus each source may have a different source-based tree and a router may have to keep track of several source- based trees for a given multicast group.
Chapter 4 Problems Problem 1 a With a connection-oriented network, every router failure will involve the routing of that connection. Moreover, all of the routers on the initial path that are downstream from the failed node must take down the failed connection, with all of the requisite signaling involved to do this.
With a connectionless datagram network, no signaling is required to either set up a new downstream path or take down the old downstream path. We have seen, however, that routing tables will need to be updated e. We have seen that with distance vector algorithms, this routing table change can sometimes be localized to the area near the failed router. Thus, a datagram network would be preferable. Interestingly, the design criteria that the initial ARPAnet be able to function under stressful conditions was one of the reasons that datagram architecture was chosen for this Internet ancestor.
That is, the router must have per-session state in the router. This is possible in a connection-oriented network, but not with a connectionless network. Thus, a connection-oriented VC network would be preferable. This is due to the various packet headers needed to route the datagrams through the network.
But in VC architecture, once all circuits are set up, they will never change. Thus, the signaling overhead is negligible over the long run. In this manner, it is not possible that there are fewer VCs in progress than without there being any common free VC number. Thus, a VC will likely have a different VC number for each link along its path. Problem 3 For a VC forwarding table, the columns are: For a datagram forwarding table, the columns are: Destination Address, Outgoing Interface.
Problem 4 a Data destined to host H3 is forwarded through interface 3 Destination Address Link Interface H3 3 b No, because forwarding rule is only based on destination address.
Router B. There are four links. One example combination is 10,00,00, Problem 6 In a virtual circuit network, there is an end-to-end connection in the sense that each router along the path must maintain state for the connection; hence the terminology connection service. In a connection-oriented transport service over a connectionless network layer, such as TCP over IP, the end systems maintain connection state; however the routers have no notion of any connections; hence the terminology connection-oriented service.
Problem 7 a No, you can only transmit one packet at a time over a shared bus. Problem 8 a n-1 D b n-1 D c 0 Problem 9 The minimal number of time slots needed is 3. The scheduling is as follows. Slot 1: Slot 2: Largest number of slots is still 3. Actually, based on the assumption that a non-empty input queue is never idle, we see that the first time slot always consists of sending X in the top input queue and Y in either middle or bottom input queue, and in the second time slot, we can always send two more datagram, and the last datagram can be sent in third time slot.
Actually, if the first datagram in the bottom input queue is X, then the worst case would require 4 time slots. Problem 10 a Prefix Match Link Interface 00 0 1 2 1 3 otherwise 3 b Prefix match for first address is 5th entry: Also, label D, E, F for the upper-right, bottom, and upper- left interior subnets, respectively.
Each fragment except the last one will be of size bytes including IP header. The last datagram will be of size bytes including IP header. The offsets of the 4 fragments will be 0, 85, , Note that here there is no fragmentation — the source host does not create datagrams larger than bytes, and these datagrams are smaller than the MTUs of the links.
Problem 21 a Home addresses: As each host generates a sequence of IP packets with sequential numbers and a distinct very likely, as they are randomly chosen from a large space initial identification number ID , we can group IP packets with consecutive IDs into a cluster. The number of clusters is the number of hosts behind the NAT.
For more practical algorithms, see the following papers. Problem 23 It is not possible to devise such a technique. In order to establish a direct TCP connection between Arnold and Bernard, either Arnold or Bob must initiate a connection to the other.
Problem 24 y-x-u, y-x-v-u, y-x-w-u, y-x-w-v-u, y-w-u, y-w-v-u, y-w-x-u, y-w-x-v-u, y-w-v-x-u, y-z-w-u, y-z-w-v-u, y-z-w-x-u, y-z-w-x-v-u, y-z-w-v-x-u, Problem 25 x to z: At each iteration, a node exchanges distance tables with its neighbors. Thus, if you are node A, and your neighbor is B, all of B's neighbors which will all be one or two hops from you will know the shortest cost path of one or two hops to you after one iteration i.
Now consider if c x,w changes. Connecting two nodes with a link is equivalent to decreasing the link weight from infinite to the finite weight. There is no increasing in values. If no updating, then no message will be sent out. Thus, D x is non-increasing. Since those costs are finite, then eventually distance vectors will be stabilized in finite steps. The following table shows the routing converging process. Assume that at time t0, link cost change happens. At time t1, y updates its distance vector and informs neighbors w and z.
If we continue the iterations shown in the above table, then we will see that, at t27, z detects that its least cost to x is 50, via its direct link with x. At t29, w learns its least cost to x is 51 via z.
At t30, y updates its least cost to x to be 52 via w. Finally, at time t31, no updating, and the routing is stabilized. Problem 36 The chosen path is not necessarily the shortest AS-path.
Recall that there are many issues to be considered in the route selection process. It is very likely that a longer loop-free path is preferred over a shorter loop-free path due to economic reason. For example, an AS might prefer to send traffic to one neighbor instead of another neighbor with shorter AS distance. Consider a BitTorrent file sharing network in which peer 1, 2, and 3 are in stub networks W, X, and Y respectively. This is equivalent to B forwarding data that is finally destined to stub network Y.
A should advise to C only one route, A-V. C receives AS paths: In this manner, when Y has a datagram that is destined to an IP that can be reached through Z, Y will have the option of sending the datagram through Z. However, if Z advertizes routes to Y, Y can re-advertize those routes to X.
Therefore, in this case, there is nothing Z can do from preventing traffic from X to transit through Z. Problem 45 The 32 receives are shown connected to the sender in the binary tree configuration shown above. With network-layer broadcast, a copy of the message is forwarded over each link exactly once. With unicast emulation, the sender unicasts a copy to each receiver over a path with5 hops. A topology in which all receivers are in a line, with the sender at one end of the line, will have the largest disparity between the cost of network-layer broadcast and unicast emulation.
Other solutions are possible, but in these solutions, B can not route to either C or D from A. This center-based tree is different from the minimal spanning tree shown in the figure. Problem 50 The center-based tree for the topology shown in the original figure connects t to v; u to v; w to v; x to v; and y to v all directly. And z connected to v via x. This center-based tree is different from the minimal spanning tree.
After 3 steps, 12 copies are transmitted, and so on. For example, an application may periodically multicast its identity to all other group members in an application-layer message.
Problem 54 A simple application-layer protocol that will allow all members to know the identity of all other members in the group is for each instance of the application to send a multicast message containing its identity to all other members.
This protocol sends message in- band, since the multicast channel is used to distribute the identification messages as well as multicast data from the application itself.
The use of the in-band signaling makes use of the existing multicast distribution mechanism, leading to a very simple design. The transportation mode, e. Although each link guarantees that an IP datagram sent over the link will be received at the other end of the link without errors, it is not guaranteed that IP datagrams will arrive at the ultimate destination in the proper order. With IP, datagrams in the same TCP connection can take different routes in the network, and therefore arrive out of order.
TCP is still needed to provide the receiving end of the application the byte stream in the correct order. Also, IP can lose packets due to routing loops or equipment failures.
TCP is also full duplex. There will be a collision in the sense that while a node is transmitting it will start to receive a packet from the other node. Slotted Aloha: Token ring: It waits In polling, a discussion leader allows only one participant to talk at a time, with each participant getting a chance to talk in a round-robin fashion.
A participant is only allowed to talk if the participant is holding the wine glass. When a node transmits a frame, the node has to wait for the frame to propagate around the entire ring before the node can release the token. An ARP query is sent in a broadcast frame because the querying host does not which adapter address corresponds to the IP address in question.
No it is not possible. The three Ethernet technologies have identical frame structures. In We can string the N switches together. The first and last switch would use one port for trunking; the middle N-2 switches would use two ports. Chapter 5 Problems Problem 1 Problem 2 Suppose we begin with the initial two-dimensional parity matrix: The parity of row 2 is now correct! The parity of columns 2 and 3 is wrong, but we can't detect in which rows the error occurred!
It is clear that if we divide K by G, then the reminder is not zero. Thus, a sequence not necessarily contiguous of odd-number bit errors cannot be divided by 11, thus it cannot be divided by G.
The length of a polling round is The number of bits transmitted in a polling round is NQ. Forwarding table in E determines that the datagram should be routed to interface The adapter in E creates and Ethernet packet with Ethernet destination address Router 2 receives the packet and extracts the datagram.
The forwarding table in this router indicates that the datagram is to be routed to Router 2 then sends the Ethernet packet with the destination address of and source address of via its interface with IP address of The process continues until the packet has reached Host B. This ARP response packet is carried by an Ethernet frame with Ethernet destination address Problem 15 a No. Thus, E will not send the packet to the default router R1. Ethernet frame from E to F: Ethernet frame from E to R1: And it learns that A resides on Subnet 1 which is connected to S1 at the interface connecting to Subnet 1.
And, S1 will update its forwarding table to include an entry for Host A. Problem 16 Lets call the switch between subnets 2 and 3 S2. That is, router R1 between subnets 2 and 3 is now replaced with switch S2.
Thus, E will not send the packet to S2. This query packet will be re-broadcast by switch 1, and eventually received by Host B. Ethernet frame from E to S2: Yes, router S2 also receives this ARP request message, and S2 will broadcast this query packet to all its interfaces.
Problem 17 Wait for 51, bit times. For 10 Mbps, this wait is So A incorrectly thinks that its frame was successfully transmitted without a collision. Thus A and B do not collide. Thus the factor appearing in the exponential backoff algorithm is sufficiently large. Source MAC address: Problem 24 Each departmental hub is a single collision domain that can have a maximum throughput of Mbps.
Hence, if the three collision domains and the web server and mail server send out data at their maximum possible rates of Mbps each, a maximum total aggregate throughput of Mbps can be achieved among the 11 end systems. Problem 25 All of the 11 end systems will lie in the same collision domain. In this case, the maximum total aggregate throughput of Mbps is possible among the 11 end sytems.
The subnet mask is The IP addresses for those three computers from left to right in CS department are: The first one is for the subnet of EE department, and the second one is for the subnet of CS department. Suppose This means that each frame that comes from subnet Host A first encapsulates the IP datagram destined to Once the router receives the frame, then it passes it up to IP layer, which decides that the IP datagram should be forwarded to subnet Then the router encapsulates the IP datagram into a frame and sends it to port 1.
Note that this frame has an Once Host B receives this frame, it will remove the Problem 29 in out out label label dest interf. A 2 D 0 R2 R1 in out out label label dest interf.
You computer first creates a special IP datagram destined to A DHCP server on the Ethernet also gives your computer a list of IP addresses of first- hop routers, the subnet mask of the subnet where your computer resides, and the addresses of local DNS servers if they exist.
Your computer first will get the IP address of the Web page you would like to download. Once your computer has the IP address of the Web page, then it will send out the HTTP request via the first-hop router if the Web page does not reside in a local Web server. Your computer sends the Ethernet frames destined to the first-hop router. Once the router receives the frames, it passes them up into IP layer, checks its routing table, and then sends the packets to the right interface out of all of its interfaces.
Then your IP packets will be routed through the Internet until they reach the Web server. Those IP packets follow IP routes and finally reach your first-hop router, and then the router will forward those IP packets to your computer by encapsulating them into Ethernet frames. Similarly, there are four links between second and fourth tier-2 switches, together providing 40 Gbps for the traffic from racks to Thus the total aggregate bandwidth is 80 Gbps, and the value per flow rate is 1 Gbps.
So the host-to-host bit rate will be 0. Problem 33 a Both email and video application uses the fourth rack for 0. Let's assume that the fourth rack has all the data and software needed for both the email and video applications.
With the topology of Figure 5. From part b, both are using the fourth rack for no more than. Chapter 6 Review Questions 1. In infrastructure mode of operation, each wireless host is connected to the larger network via a base station access point. If not operating in infrastructure mode, a network operates in ad-hoc mode.
In ad-hoc mode, wireless hosts have no infrastructure with which to connect. In the absence of such infrastructure, the hosts themselves must provide for services such as routing, address assignment, DNS-like name translation, and more. Path loss is due to the attenuation of the electromagnetic signal when it travels through matter. Multipath propagation results in blurring of the received signal at the receiver and occurs when portions of the electromagnetic wave reflect off objects and ground, taking paths of different lengths between a sender and receiver.
Interference from other sources occurs when the other source is also transmitting in the same frequency range as the wireless network. APs transmit beacon frames. The beacon frames permit nearby wireless stations to discover and identify the AP.
False 7. False 9. Initially the switch has an entry in its forwarding table which associates the wireless station with the earlier AP. The frame is received by the switch. This forces the switch to update its forwarding table, so that frames destined to the wireless station are sent via the new AP. Any ordinary Bluetooth node can be a master node whereas access points in This allows the base station to make best use of the wireless medium.
A node can remain connected to the same access point throughout its connection to the Internet hence, not be mobile. A mobile node is the one that changes its point of attachment into the network over time. Since the user is always accessing the Internet through the same access point, she is not mobile. A permanent address for a mobile node is its IP address when it is at its home network. A care-of-address is the one its gets when it is visiting a foreign network. The COA is assigned by the foreign agent which can be the edge router in the foreign network or the mobile node itself.
The home network in GSM maintains a database called the home location register HLR , which contains the permanent cell phone number and subscriber profile information about each of its subscribers. The HLR also contains information about the current locations of these subscribers. The visited network maintains a database known as the visitor location register VLR that contains an entry for each mobile user that is currently in the portion of the network served by the VLR.
VLR entries thus come and go as mobile users enter and leave the network. After association, there is a virtual link between the new station and the AP. Suppose the new station associates with AP1. When the new station sends a frame, it will be addressed to AP1. Although AP2 will also receive the frame, it will not process the frame because the frame is not addressed to it. Thus, the two ISPs can work in parallel over the same channel.
However, the two ISPs will be sharing the same wireless bandwidth. For Thus, the maximum aggregate transmission rate for the two ISPs is 22 Mbps for Problem 6 Suppose that wireless station H1 has long frames to transmit. H1 may be an AP that is forwarding an MP3 to some other wireless station. Suppose initially H1 is the only station that wants to transmit, but that while half-way through transmitting its first frame, H2 wants to transmit a frame.
Before transmitting, H2 will sense that the channel is busy, and therefore choose a random backoff value. Now suppose that after sending its first frame, H1 returns to step 1; that is, it waits a short period of times DIFS and then starts to transmit the second frame.
Thus, H1 should get to transmit all of its frames before H2 has a chance to access the channel. On the other hand, if H1 goes to step 2 after transmitting a frame, then it too chooses a random backoff value, thereby giving a fair chance to H2. Thus, fairness was the rationale behind this design choice. Problem 7 A frame without data is 32 bytes long. This solution is not fair since only A is getting served. Now, to fulfill the given fairness requirement, we have the following condition: Problem 11 a No.
All the routers might not be able to route the datagram immediately. This is because the Distance Vector algorithm as well as the inter-AS routing protocols like BGP is decentralized and takes some time to terminate. So, during the time when the algorithm is still running as a result of advertisements from the new foreign network, some of the routers may not be able to route datagrams destined to the mobile node.
This might happen when one of the nodes has just left a foreign network and joined a new foreign network. In this situation, the routing entries from the old foreign network might not have been completely withdrawn when the entries from the new network are being propagated. Problem 13 Because datagrams must be first forward to the home agent, and from there to the mobile, the delays will generally be longer than via direct routing.
Note that it is possible, however, that the direct delay from the correspondent to the mobile i. It would depend on the delays on these various path segments. Note that indirect routing also adds a home agent processing e. Problem 14 First, we note that chaining was discussed at the end of section 6. The foreign agent at C must notify the foreign agent at B that the mobile is no longer resident in B but in fact is resident in C and has the specified COA in C. Note that when the mobile goes offline i.
This teardown must also be done through signaling messages. The foreign agent in A can remove its state about the mobile, since it is no longer in A. The foreign agent in B can remove its state about the mobile, since it is no longer in B.
When the mobile goes offline or returns to its home network, the datagram-forwarding state maintained by the foreign agent in C must be removed.
Problem 15 Two mobiles could certainly have the same care-of-address in the same visited network. Indeed, if the care-of-address is the address of the foreign agent, then this address would be the same.
Once the foreign agent decapsulates the tunneled datagram and determines the address of the mobile, then separate addresses would need to be used to send the datagrams separately to their different destinations mobiles within the visited network.
Chapter 7 Review Questions 1. Spatial Redundancy: It is the redundancy within a given image. Intuitively, an image consists of mostly white space has a high degree of redundancy and can be efficiently compressed without significantly sacrificing image quality. Temporal Redundancy reflects repetition from image to subsequent image.
If, for example, an image and the subsequent image are exactly the same, there is no reason to re-encode the subsequent image; it is instead more efficient simply to indicate during encoding that the subsequent image is exactly the same.
If the two images are very similar, it may be not efficient to indicate how the second image differs from the first, rather than re-encode the second image. Quantizing a sample into levels means 10 bits per sample. The resulting rate of the PCM digital audio signal is Kbps.
In this class of applications, the underlying medium is prerecorded video, such as a movie, a television show, or a prerecorded sporting event. These prerecorded videos are played on servers, and users send requests to the servers to view the videos on demand. Many internet companies today provide streaming video, including YouTube, Netflix, and Hulu. Conversational Voice- and Video-over-IP: Conversational video is similar except that it includes the video of the participants as well as their voices.
Conversational voice and video are widely used in the Internet today, with the Internet companies like Skype and Google Talk boasting hundreds of millions of daily users. Streaming Live Audio and Video: These applications allow users to receive a live radio or television transmission over the Internet. Today, thousands of radio and television stations around the world are broadcasting content over the internet. UDP Streaming: HTTP Streaming: The server then sends the video file, within an HTTP response message, as quickly as possible, that is, as quickly as TCP congestion control and flow control will allow.
In Dynamic Adaptive Streaming over HTTP, the video is encoded several different versions, with each version having a different bit rate and, correspondingly, a different quality level. The client dynamically requests the chunks of video segments of a few seconds in length from the different versions.
When the amount of available bandwidth is high, the client naturally selects chunks from a high-rate version; and when the available bandwidth is low, it naturally selects from a low-rate version.
The three significant drawbacks of UDP Streaming are: Due to unpredictable and varying amount of available bandwidth between server and client, constant-rate UDP streaming can fail to provide continuous play out. It requires a media control server, such as an RTSP server, to process client-to- server interactivity requests and to track client state for each ongoing client session.
On the client side, the client application reads bytes from the TCP receive buffer and places the bytes in the client application buffer.
Enter Deep: This philosophy is to enter deep into the access networks of Internet Service Providers, by deploying server clusters in access ISPs all over the world. Bring Home: Geographically closest cluster selection and real time measurements selection can find a good cluster with respect to LDNS.
IP anycast chooses good cluster with respect to client itself. Load on the cluster — clients should not be directed to overload clusters. ISP delivery cost — the clusters may be chosen so that specific ISPs are used to carry CDN-to-client traffic, taking into account the different cost structures in the contractual relationships between ISPs and cluster operators.
End-to-end delay is the time it takes a packet to travel across the network from source to destination. Delay jitter is the fluctuation of end-to-end delay from packet to the next packet. A packet that arrives after its scheduled play out time cannot be played out. Therefore, from the perspective of the application, the packet has been lost. First scheme: Second scheme: Interleaving does not increase the bandwidth requirements of a stream.
RTP streams in different sessions: The role of a SIP registrar is to keep track of the users and their corresponding IP addresses which they are currently using. Each SIP registrar keeps track of the users that belong to its domain. In this regard, its role is similar to that of an authoritative name server in DNS. In non-preemptive priority queuing, the transmission of a packet is not interrupted once it has begun.