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在Socket长连接通讯后为什么会出现错误:Unable to write data to the transport connection
anna_aiya
2005-06-20 04:04:50
在服务器与客户端通讯一段时间后,出现错误:Unable to write data to the transport connection
这是什么原因引起的?是由于服务器和客户端同时都在写数据吗?如何解决这个错误?
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在Socket长连接通讯后为什么会出现错误:Unable to write data to the transport connection
在服务器与客户端通讯一段时间后,出现错误:Unable to write data to the transport connection 这是什么原因引起的?是由于服务器和客户端同时都在写数据吗?如何解决这个错误?
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jingkingsoft
2005-06-21
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建议用UDP协议,他能穿过防火墙。
jingkingsoft
2005-06-21
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有可能一段时间内你没有往客户端发送数据
客户端路由把你的信息视为不安全的而过滤掉了。
dazhu2
2005-06-21
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你代码贴出来看看
anna_aiya
2005-06-21
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计算机网络第六版答案
Computer Networking: A Top-Down Approach, 6th Edition Solutions to Review Questions and Problems Version Date: May 2012 This document contains the solutions to review questions and problems for the 5th edition of Computer Networking: A Top-Down Approach by Jim Kurose and Keith Ross. These solutions are being made available to instructors ONLY. Please do NOT copy or distribute this document to others (even other instructors). Please do not post any solutions on a publicly-available Web site. We’ll be happy to provide a copy (up-to-date) of this solution manual ourselves to anyone who asks. Acknowledgments: Over the years, several students and colleagues have helped us prepare this solutions manual. Special thanks goes to HongGang Zhang, Rakesh Kumar, Prithula Dhungel, and Vijay Annapureddy. Also thanks to all the readers who have made suggestions and corrected errors. All material © copyright 1996-2012 by J.F. Kurose and K.W. Ross. All rights reserved Chapter 1 Review Questions There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, PDAs, Internet-connected game consoles, etc. From Wikipedia: Diplomatic protocol is commonly described as a set of international courtesy rules. These well-established and time-honored rules have made it easier for nations and people to live and work together. Part of protocol has always been the acknowledgment of the hierarchical standing of all present. Protocol rules are based on the principles of civility. Standards are im
port
ant for protocols so that people can create networking systems and products that interoperate. 1. Dial-up modem over telephone line: home; 2. DSL over telephone line: home or small office; 3. Cable to HFC: home; 4. 100 Mbps switched Ethernet: enterprise; 5. Wifi (802.11): home and enterprise: 6. 3G and 4G: wide-area wireless. HFC bandwidth is shared among the users. On the downstream channel, all packets emanate from a single source, namely, the head end. Thus, there are no collisions in the downstream channel. In most American cities, the current possibilities include: dial-up; DSL; cable modem; fiber-to-the-home. 7. Ethernet LANs have
trans
mission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps. 8. Today, Ethernet most commonly runs over twisted-pair copper wire. It also can run over fibers optic links. 9. Dial up modems: up to 56 Kbps, bandwidth is dedicated; ADSL: up to 24 Mbps downstream and 2.5 Mbps upstream, bandwidth is dedicated; HFC, rates up to 42.8 Mbps and upstream rates of up to 30.7 Mbps, bandwidth is shared. FTTH: 2-10Mbps upload; 10-20 Mbps download; bandwidth is not shared. 10. There are two popular wireless Internet access technologies today: Wifi (802.11) In a wireless LAN, wireless users
trans
mit/receive packets to/from an base station (i.e., wireless access point) within a radius of few tens of meters. The base station is typically connected to the wired Internet and thus serves to connect wireless users to the wired network. 3G and 4G wide-area wireless access networks. In these systems, packets are
trans
mitted over the same wireless infrastructure used for cellular telephony, with the base station thus being managed by a telecommunications provider. This provides wireless access to users within a radius of tens of kilometers of the base station. 11. At time t0 the sending host begins to
trans
mit. At time t1 = L/R1, the sending host completes
trans
mission and the entire packet is received at the router (no propagation delay). Because the router has the entire packet at time t1, it can begin to
trans
mit the packet to the receiving host at time t1. At time t2 = t1 + L/R2, the router completes
trans
mission and the entire packet is received at the receiving host (again, no propagation delay). Thus, the end-to-end delay is L/R1 + L/R2. 12. A circuit-switched network can guarantee a certain amount of end-to-end bandwidth for the duration of a call. Most packet-switched networks today (including the Internet) cannot make any end-to-end guarantees for bandwidth. FDM requires sophisticated analog hardware to shift signal into appropriate frequency bands. 13. a) 2 users can be sup
port
ed because each user requires half of the link bandwidth. b) Since each user requires 1Mbps when
trans
mitting, if two or fewer users
trans
mit simultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users
trans
mit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link. c) Probability that a given user is
trans
mitting = 0.2 d) Probability that all three users are
trans
mitting simultaneously = = (0.2)3 = 0.008. Since the queue grows when all the users are
trans
mitting, the fraction of time during which the queue grows (which is equal to the probability that all three users are
trans
mitting simultaneously) is 0.008. 14. If the two ISPs do not peer with each other, then when they send traffic to each other they have to send the traffic through a provider ISP (intermediary), to which they have to pay for carrying the traffic. By peering with each other directly, the two ISPs can reduce their payments to their provider ISPs. An Internet Exchange Points (IXP) (typically in a standalone building with its own switches) is a meeting point where multiple ISPs can connect and/or peer together. An ISP earns its money by charging each of the the ISPs that connect to the IXP a relatively small fee, which may depend on the amount of traffic sent to or received from the IXP. 15. Google's private network connects together all its
data
centers, big and small. Traffic between the Google
data
centers passes over its private network rather than over the public Internet. Many of these
data
centers are located in, or close to, lower tier ISPs. Therefore, when Google delivers content to a user, it often can bypass higher tier ISPs. What motivates content providers to create these networks? First, the content provider has more control over the user experience, since it has to use few intermediary ISPs. Second, it can save money by sending less traffic into provider networks. Third, if ISPs decide to charge more money to highly profitable content providers (in countries where net neutrality doesn't apply), the content providers can avoid these extra payments. 16. The delay components are processing delays,
trans
mission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable. 17. a) 1000 km, 1 Mbps, 100 bytes b) 100 km, 1 Mbps, 100 bytes 18. 10msec; d/s; no; no 19. a) 500 kbps b) 64 seconds c) 100kbps; 320 seconds 20. End system A breaks the large file into chunks. It adds header to each chunk, thereby generating multiple packets from the file. The header in each packet includes the IP address of the destination (end system B). The packet switch uses the destination IP address in the packet to determine the outgoing link. Asking which road to take is analogous to a packet asking which outgoing link it should be forwarded on, given the packet’s destination address. 21. The maximum emission rate is 500 packets/sec and the maximum
trans
mission rate is 350 packets/sec. The corresponding traffic intensity is 500/350 =1.43 > 1. Loss will eventually occur for each experiment; but the time when loss first occurs will be different from one experiment to the next due to the randomness in the emission process. 22. Five generic tasks are error control, flow control, segmentation and reassembly, multiplexing, and
connection
setup. Yes, these tasks can be duplicated at different layers. For example, error control is often provided at more than one layer. 23. The five layers in the Internet protocol stack are – from top to bottom – the application layer, the
trans
port
layer, the network layer, the link layer, and the physical layer. The principal responsibilities are outlined in Section 1.5.1. 24. Application-layer message:
data
which an application wants to send and passed onto the
trans
port
layer;
trans
port
-layer segment: generated by the
trans
port
layer and encapsulates application-layer message with
trans
port
layer header; network-layer
data
gram: encapsulates
trans
port
-layer segment with a network-layer header; link-layer frame: encapsulates network-layer
data
gram with a link-layer header. 25. Routers process network, link and physical layers (layers 1 through 3). (This is a little bit of a white lie, as modern routers sometimes act as firewalls or caching components, and process
Trans
port
layer as well.) Link layer switches process link and physical layers (layers 1 through2). Hosts process all five layers. 26. a) Virus Requires some form of human interaction to spread. Classic example: E-mail viruses. b) Worms No user replication needed. Worm in infected host scans IP addresses and
port
numbers, looking for vulnerable processes to infect. 27. Creation of a botnet requires an attacker to find vulnerability in some application or system (e.g. exploiting the buffer overflow vulnerability that might exist in an application). After finding the vulnerability, the attacker needs to scan for hosts that are vulnerable. The target is basically to compromise a series of systems by exploiting that particular vulnerability. Any system that is part of the botnet can automatically scan its environment and propagate by exploiting the vulnerability. An im
port
ant property of such botnets is that the originator of the botnet can remotely control and issue commands to all the nodes in the botnet. Hence, it becomes possible for the attacker to issue a command to all the nodes, that target a single node (for example, all nodes in the botnet might be commanded by the attacker to send a TCP SYN message to the target, which might result in a TCP SYN flood attack at the target). 28. Trudy can pretend to be Bob to Alice (and vice-versa) and partially or completely modify the message(s) being sent from Bob to Alice. For example, she can easily change the phrase “Alice, I owe you $1000” to “Alice, I owe you $10,000”. Furthermore, Trudy can even drop the packets that are being sent by Bob to Alice (and vise-versa), even if the packets from Bob to Alice are encrypted. Chapter 1 Problems Problem 1 There is no single right answer to this question. Many protocols would do the trick. Here's a simple answer below: Messages from ATM machine to Server Msg name purpose -------- ------- HELO Let server know that there is a card in the ATM machine ATM card
trans
mits user ID to Server PASSWD User enters PIN, which is sent to server BALANCE User requests balance WITHDRAWL User asks to withdraw money BYE user all done Messages from Server to ATM machine (display) Msg name purpose -------- ------- PASSWD Ask user for PIN (password) OK last requested operation (PASSWD, WITHDRAWL) OK ERR last requested operation (PASSWD, WITHDRAWL) in ERROR AMOUNT sent in response to BALANCE request BYE user done, display welcome screen at ATM Correct operation: client server HELO (userid) --------------> (check if valid userid) <------------- PASSWD PASSWD --------------> (check password) <------------- AMOUNT WITHDRAWL --------------> check if enough $ to cover withdrawl (check if valid userid) <------------- PASSWD PASSWD --------------> (check password) <------------- AMOUNT WITHDRAWL --------------> check if enough $ to cover withdrawl <------------- BYE Problem 2 At time N*(L/R) the first packet has reached the destination, the second packet is stored in the last router, the third packet is stored in the next-to-last router, etc. At time N*(L/R) + L/R, the second packet has reached the destination, the third packet is stored in the last router, etc. Continuing with this logic, we see that at time N*(L/R) + (P-1)*(L/R) = (N+P-1)*(L/R) all packets have reached the destination. Problem 3 a) A circuit-switched network would be well suited to the application, because the application involves long sessions with predictable smooth bandwidth requirements. Since the
trans
mission rate is known and not bursty, bandwidth can be reserved for each application session without significant waste. In addition, the overhead costs of setting up and tearing down
connection
s are amortized over the lengthy duration of a typical application session. b) In the worst case, all the applications simultaneously
trans
mit over one or more network links. However, since each link has sufficient bandwidth to handle the sum of all of the applications'
data
rates, no congestion (very little queuing) will occur. Given such generous link capacities, the network does not need congestion control mechanisms. Problem 4 Between the switch in the upper left and the switch in the upper right we can have 4
connection
s. Similarly we can have four
connection
s between each of the 3 other pairs of adjacent switches. Thus, this network can sup
port
up to 16
connection
s. We can 4
connection
s passing through the switch in the upper-right-hand corner and another 4
connection
s passing through the switch in the lower-left-hand corner, giving a total of 8
connection
s. Yes. For the
connection
s between A and C, we route two
connection
s through B and two
connection
s through D. For the
connection
s between B and D, we route two
connection
s through A and two
connection
s through C. In this manner, there are at most 4
connection
s passing through any link. Problem 5 Tollbooths are 75 km apart, and the cars propagate at 100km/hr. A tollbooth services a car at a rate of one car every 12 seconds. a) There are ten cars. It takes 120 seconds, or 2 minutes, for the first tollbooth to service the 10 cars. Each of these cars has a propagation delay of 45 minutes (travel 75 km) before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 47 minutes. The whole process repeats itself for traveling between the second and third tollbooths. It also takes 2 minutes for the third tollbooth to service the 10 cars. Thus the total delay is 96 minutes. b) Delay between tollbooths is 8*12 seconds plus 45 minutes, i.e., 46 minutes and 36 seconds. The total delay is twice this amount plus 8*12 seconds, i.e., 94 minutes and 48 seconds. Problem 6 a) seconds. b) seconds. c) seconds. d) The bit is just leaving Host A. e) The first bit is in the link and has not reached Host B. f) The first bit has reached Host B. g) Want km. Problem 7 Consider the first bit in a packet. Before this bit can be
trans
mitted, all of the bits in the packet must be generated. This requires sec=7msec. The time required to
trans
mit the packet is sec= sec. Propagation delay = 10 msec. The delay until decoding is 7msec + sec + 10msec = 17.224msec A similar analysis shows that all bits experience a delay of 17.224 msec. Problem 8 a) 20 users can be sup
port
ed. b) . c) . d) . We use the central limit theorem to approximate this probability. Let be independent random variables such that . “21 or more users” when is a standard normal r.v. Thus “21 or more users” . Problem 9 10,000 Problem 10 The first end system requires L/R1 to
trans
mit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay of dproc; after receiving the entire packet, the packet switch connecting the first and the second link requires L/R2 to
trans
mit the packet onto the second link; the packet propagates over the second link in d2/s2. Similarly, we can find the delay caused by the second switch and the third link: L/R3, dproc, and d3/s3. Adding these five delays gives dend-end = L/R1 + L/R2 + L/R3 + d1/s1 + d2/s2 + d3/s3+ dproc+ dproc To answer the second question, we simply plug the values into the equation to get 6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec. Problem 11 Because bits are immediately
trans
mitted, the packet switch does not introduce any delay; in particular, it does not introduce a
trans
mission delay. Thus, dend-end = L/R + d1/s1 + d2/s2+ d3/s3 For the values in Problem 10, we get 6 + 20 + 16 + 4 = 46 msec. Problem 12 The arriving packet must first wait for the link to
trans
mit 4.5 *1,500 bytes = 6,750 bytes or 54,000 bits. Since these bits are
trans
mitted at 2 Mbps, the queuing delay is 27 msec. Generally, the queuing delay is (nL + (L - x))/R. Problem 13 The queuing delay is 0 for the first
trans
mitted packet, L/R for the second
trans
mitted packet, and generally, (n-1)L/R for the nth
trans
mitted packet. Thus, the average delay for the N packets is: (L/R + 2L/R + ....... + (N-1)L/R)/N = L/(RN) * (1 + 2 + ..... + (N-1)) = L/(RN) * N(N-1)/2 = LN(N-1)/(2RN) = (N-1)L/(2R) Note that here we used the well-known fact: 1 + 2 + ....... + N = N(N+1)/2 It takes seconds to
trans
mit the packets. Thus, the buffer is empty when a each batch of packets arrive. Thus, the average delay of a packet across all batches is the average delay within one batch, i.e., (N-1)L/2R. Problem 14 The
trans
mission delay is . The total delay is Let . Total delay = For x=0, the total delay =0; as we increase x, total delay increases, approaching infinity as x approaches 1/a. Problem 15 Total delay . Problem 16 The total number of packets in the system includes those in the buffer and the packet that is being
trans
mitted. So, N=10+1. Because , so (10+1)=a*(queuing delay +
trans
mission delay). That is, 11=a*(0.01+1/100)=a*(0.01+0.01). Thus, a=550 packets/sec. Problem 17 There are nodes (the source host and the routers). Let denote the processing delay at the th node. Let be the
trans
mission rate of the th link and let . Let be the propagation delay across the th link. Then . Let denote the average queuing delay at node . Then . Problem 18 On linux you can use the command traceroute www.targethost.com and in the Windows command prompt you can use tracert www.targethost.com In either case, you will get three delay measurements. 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: Traceroutes between San Diego Super Computer Center and www.poly.edu The average (mean) of the round-trip delays at each of the three hours is 71.18 ms, 71.38 ms and 71.55 ms, respectively. The standard deviations are 0.075 ms, 0.21 ms, 0.05 ms, respectively. In this example, the traceroutes have 12 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours. Traceroute packets passed through four ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Traceroutes from www.stella-net.net (France) to www.poly.edu (USA). The average round-trip delays at each of the three hours are 87.09 ms, 86.35 ms and 86.48 ms, respectively. The standard deviations are 0.53 ms, 0.18 ms, 0.23 ms, respectively. In this example, there are 11 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours. Traceroute packets passed three ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Problem 19 An example solution: Traceroutes from two different cities in France to New York City in United States In these traceroutes from two different cities in France to the same destination host in United States, seven links are in common including the
trans
atlantic link. In this example of traceroutes from one city in France and from another city in Germany to the same host in United States, three links are in common including the
trans
atlantic link. Traceroutes to two different cities in China from same host in United States Five links are common in the two traceroutes. The two traceroutes diverge before reaching China Problem 20 Throughput = min{Rs, Rc, R/M} Problem 21 If only use one path, the max throughput is given by: . If use all paths, the max throughput is given by . Problem 22 Probability of successfully receiving a packet is: ps= (1-p)N. The number of
trans
missions 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
trans
missions needed is given by: 1/ps . Then, the average number of re-
trans
missions needed is given by: 1/ps -1. Problem 23 Let’s call the first packet A and call the second packet B. If the bottleneck link is the first link, then packet B is queued at the first link waiting for the
trans
mission of packet A. So the packet inter-arrival time at the destination is simply L/Rs. If the second link is the bottleneck link and both packets are sent back to back, it must be true that the second packet arrives at the input queue of the second link before the second link finishes the
trans
mission of the first packet. That is, L/Rs + L/Rs + dprop = L/Rs + dprop + L/Rc Thus, the minimum value of T is L/Rc L/Rs . Problem 24 40 terabytes = 40 * 1012 * 8 bits. So, if using the dedicated link, it will take 40 * 1012 * 8 / (100 *106 ) =3200000 seconds = 37 days. But with FedEx overnight delivery, you can guarantee the
data
arrives in one day, and it should cost less than $100. Problem 25 160,000 bits 160,000 bits The bandwidth-delay product of a link is the maximum number of bits that can be in the link. the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football field s/R Problem 26 s/R=20000km, then R=s/20000km= 2.5*108/(2*107)= 12.5 bps Problem 27 80,000,000 bits 800,000 bits, this is because that the maximum number of bits that will be in the link at any given time = min(bandwidth delay product, packet size) = 800,000 bits. .25 meters Problem 28 t
trans
+ tprop = 400 msec + 80 msec = 480 msec. 20 * (t
trans
+ 2 tprop) = 20*(20 msec + 80 msec) = 2 sec. Breaking up a file takes longer to
trans
mit because each
data
packet and its corresponding acknowledgement packet add their own propagation delays. Problem 29 Recall geostationary satellite is 36,000 kilometers away from earth surface. 150 msec 1,500,000 bits 600,000,000 bits Problem 30 Let’s suppose the passenger and his/her bags correspond to the
data
unit arriving to the top of the protocol stack. When the passenger checks in, his/her bags are checked, and a tag is attached to the bags and ticket. This is additional information added in the Baggage layer if Figure 1.20 that allows the Baggage layer to implement the service or separating the passengers and baggage on the sending side, and then reuniting them (hopefully!) on the destination side. When a passenger then passes through security and additional stamp is often added to his/her ticket, indicating that the passenger has passed through a security check. This information is used to ensure (e.g., by later checks for the security information) secure
trans
fer of people. Problem 31 Time to send message from source host to first packet switch = With store-and-forward switching, the total time to move message from source host to destination host = Time to send 1st packet from source host to first packet switch = . . Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch = Time at which 1st packet is received at the destination host = . After this, every 5msec one packet will be received; thus time at which last (800th) packet is received = . It can be seen that delay in using message segmentation is significantly less (almost 1/3rd). Without message segmentation, if bit errors are not tolerated, if there is a single bit error, the whole message has to be re
trans
mitted (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 31.The propagation delays affect the overall end-to-end delays both for packet switching and message switching equally. Problem 33 There are F/S packets. Each packet is S=80 bits. Time at which the last packet is received at the first router is sec. At this time, the first F/S-2 packets are at the destination, and the F/S-1 packet is at the second router. The last packet must then be
trans
mitted by the first router and the second router, with each
trans
mission taking sec. Thus delay in sending the whole file is To calculate the value of S which leads to the minimum delay, Problem 34 The circuit-switched telephone networks and the Internet are connected together at "gateways". 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 The Web: HTTP; file
trans
fer: FTP; remote login: Telnet; e-mail: SMTP; BitTorrent file sharing: BitTorrent protocol Network architecture refers to the organization of the communication process into layers (e.g., the five-layer Internet architecture). Application architecture, on the other hand, is designed by an application developer and dictates the broad structure of the application (e.g., client-server or P2P). The process which initiates the communication is the client; the process that waits to be contacted is the server. No. 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 UDP, the
trans
action can be completed in one roundtrip time (RTT) - the client sends the
trans
action request into a UDP
socket
, and the server sends the reply back to the client's UDP
socket
. 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. However, because Google docs runs over the Internet (using TCP), timing guarantees are not provided. a) Reliable
data
trans
fer TCP provides a reliable byte-stream between client and server but UDP does not. b) A guarantee that a certain value for throughput will be maintained Neither c) A guarantee that
data
will be delivered within a specified amount of time Neither d) Confidentiality (via encryption) Neither SSL operates at the application layer. The SSL
socket
takes unencrypted
data
from the application layer, encrypts it and then passes it to the TCP
socket
. If the application developer wants TCP to be enhanced with SSL, she has to include the SSL code in the application. A protocol uses handshaking if the two communicating entities first exchange control packets before sending
data
to each other. SMTP uses handshaking at the application layer whereas HTTP does not. The applications associated with those protocols require that all application
data
be received in the correct order and without gaps. TCP provides this service whereas UDP does not. When the user first visits the site, the server creates a unique identification number, creates an entry in its back-end
data
base, and returns this identification number as a cookie number. This cookie number is stored on the user’s host and is managed by the browser. During each subsequent visit (and purchase), 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 bring the desired content “closer” to the user, possibly to the same LAN to which the user’s host is connected. 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. to make it available, go to Control Panel, Programs and Features, Turn Windows Features On or Off, Check Telnet client. To start Telnet, in Windows command prompt, issue the following command > telnet webserverver 80 where "webserver" is some webserver. After issuing the command, you have established a TCP
connection
between your client telnet program and the web server. Then type in an HTTP GET message. An example is given below: Since the index.html page in this web server was not modified since Fri, 18 May 2007 09:23:34 GMT, and the above commands were issued on Sat, 19 May 2007, the server returned "304 Not Modified". Note that the first 4 lines are the GET message and header lines inputed by the user, and the next 4 lines (starting from HTTP/1.1 304 Not Modified) is the response from the web server. FTP uses two parallel TCP
connection
s, one
connection
for sending control information (such as a request to
trans
fer a file) and another
connection
for actually
trans
ferring 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. The message is first sent from Alice’s host to her mail server over HTTP. Alice’s mail server then sends the message to Bob’s mail server over SMTP. Bob then
trans
fers the message from his mail server to his host over POP3. 17. Received: from 65.54.246.203 (EHLO bay0-omc3-s3.bay0.hotmail.com) (65.54.246.203) by mta419.mail.mud.yahoo.com with SMTP; Sat, 19 May 2007 16:53:51 -0700 Received: from hotmail.com ([65.55.135.106]) by bay0-omc3-s3.bay0.hotmail.com with Microsoft SMTPSVC(6.0.3790.2668); Sat, 19 May 2007 16:52:42 -0700 Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC; Sat, 19 May 2007 16:52:41 -0700 Message-ID: Received: from 65.55.135.123 by by130fd.bay130.hotmail.msn.com with HTTP; Sat, 19 May 2007 23:52:36 GMT From: "prithula dhungel" To: prithula@yahoo.com Bcc: Subject: Test mail Date: Sat, 19 May 2007 23:52:36 +0000 Mime-Version: 1.0 Content-Type: Text/html; format=flowed Return-Path: prithuladhungel@hotmail.com Figure: 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. In this example there are 4 “Received:” header lines. This means the mail message passed through 5 different SMTP servers before being delivered to the receiver’s mail box. The last (forth) “Received:” header indicates the mail message flow from the SMTP server of the sender to the second SMTP server in the chain of servers. The sender’s SMTP server is at address 65.55.135.123 and the second SMTP server in the chain is by130fd.bay130.hotmail.msn.com. The third “Received:” header indicates the mail message flow from the second SMTP server in the chain to the third server, and so on. Finally, the first “Received:” header indicates the flow of the mail messages from the forth SMTP server to the last SMTP server (i.e. the receiver’s mail server) in the chain. Message-id: The message has been given this number BAY130-F26D9E35BF59E0D18A819AFB9310@phx.gbl (by bay0-omc3-s3.bay0.hotmail.com. Message-id is a unique string assigned by the mail system when the message is first created. From: This indicates the email address of the sender of the mail. In the given example, the sender is “prithuladhungel@hotmail.com” To: This field indicates the email address of the receiver of the mail. In the example, the receiver is “prithula@yahoo.com” Subject: This gives the subject of the mail (if any specified by the sender). In the example, the subject specified by the sender is “Test mail” Date: The date and time when the mail was sent by the sender. In the example, the sender sent the mail on 19th May 2007, at time 23:52:36 GMT. Mime-version: MIME version used for the mail. In the example, it is 1.0. Content-type: The type of content in the body of the mail message. In the example, it is “text/html”. Return-Path: This specifies the email address to which the mail will be sent if the receiver of this mail wants to reply to the sender. This is also used by the sender’s mail server for bouncing back undeliverable mail messages of mailer-daemon error messages. In the example, the return path is “prithuladhungel@hotmail.com”. 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
trans
ferred to the new machine (including very old messages). Yes an organization’s mail server and Web server can have the same alias for a host name. The MX record is used to map the mail server’s host name to its IP address. You should be able to see the sender's IP address for a user with an .edu email address. 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 30-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. There is a logical link (an “edge” in graph theory terms) from node A to node B if there is a semi-permanent TCP
connection
between A and B. 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: the advantage is that each peer needs to track only a few other peers; the disadvantage is that O(N) hops are needed to route a message to the peer that is closest to the key. 25. File Distribution Instant Messaging Video Streaming Distributed Computing 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. Thus, to sup
port
n simultaneous
connection
s, the server would need n+1
socket
s. For the TCP application, as soon as the client is executed, it attempts to initiate a TCP
connection
with the server. 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
connection
s (or attempt to communicate with the UDP server) immediately upon execution Chapter 2 Problems Problem 1 a) F b) T c) F d) F e) F Problem 2 Access control commands: USER, PASS, ACT, CWD, CDUP, SMNT, REIN, QUIT.
Trans
fer parameter commands:
PORT
, PASV, TYPE STRU, MODE. Service commands: RETR, STOR, STOU, APPE, ALLO, REST, RNFR, RNTO, ABOR, DELE, RMD, MRD, PWD, LIST, NLST, SITE, SYST, STAT, HELP, NOOP. Problem 3 Application layer protocols: DNS and HTTP
Trans
port
layer protocols: UDP for DNS; TCP for HTTP Problem 4 The document request was http://gaia.cs.umass.edu/cs453/index.html. The Host : field indicates the server's name and /cs453/index.html indicates the file name. The browser is running HTTP version 1.1, as indicated just before the first pair. The browser is requesting a persistent
connection
, as indicated by the
Connection
: keep-alive. This is a trick question. 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. One would need information from the IP
data
grams (that carried the TCP segment that carried the HTTP GET request) to answer this question. Mozilla/5.0. The browser type information is needed by the server to send different versions of the same object to different types of browsers. Problem 5 The status code of 200 and the phrase OK indicate that the server was able to locate the document successfully. The reply was provided on Tuesday, 07 Mar 2008 12:39:45 Greenwich Mean Time. The document index.html was last modified on Saturday 10 Dec 2005 18:27:46 GMT. There are 3874 bytes in the document being returned. The first five bytes of the returned document are : connection, as indicated by the
Connection
: Keep-Alive field Problem 6 Persistent
connection
s are discussed in section 8 of RFC 2616 (the real goal of this question was to get you to retrieve and read an RFC). Sections 8.1.2 and 8.1.2.1 of the RFC indicate that either the client or the server can indicate to the other that it is going to close the persistent
connection
. It does so by including the
connection
-token "close" in the
Connection
-header field of the http request/reply. HTTP does not provide any encryption services. (From RFC 2616) “Clients that use persistent
connection
s should limit the number of simultaneous
connection
s that they maintain to a given server. A single-user client SHOULD NOT maintain more than 2
connection
s with any server or proxy.” Yes. (From RFC 2616) “A client might have started to send a new request at the same time that the server has decided to close the "idle"
connection
. 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.” Problem 7 The total amount of time to get the IP address is . Once the IP address is known, elapses to set up the TCP
connection
and another elapses to request and receive the small object. The total response time is Problem 8 . . Problem 9 The time to
trans
mit an object of size L over a link or rate R is L/R. The average time is the average size of the object divided by R: = (850,000 bits)/(15,000,000 bits/sec) = .0567 sec The traffic intensity on the link is given by =(16 requests/sec)(.0567 sec/request) = 0.907. Thus, the average access delay is (.0567 sec)/(1 - .907) .6 seconds. The total average response time is therefore .6 sec + 3 sec = 3.6 sec. The traffic intensity on the access link is reduced by 60% since the 60% of the requests are satisfied within the institutional network. Thus the average access delay is (.0567 sec)/[1 – (.4)(.907)] = .089 seconds. The response time is approximately zero if the request is satisfied by the cache (which happens with probability .6); the average response time is .089 sec + 3 sec = 3.089 sec for cache misses (which happens 40% of the time). So the average response time is (.6)(0 sec) + (.4)(3.089 sec) = 1.24 seconds. Thus the average response time is reduced from 3.6 sec to 1.24 sec. 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
connection
s. Parallel downloads would allow 10
connection
s to share the 150 bits/sec bandwidth, giving each just 15 bits/sec. Thus, the total time needed to receive all objects is given by: (200/150+Tp + 200/150 +Tp + 200/150+Tp + 100,000/150+ Tp ) + (200/(150/10)+Tp + 200/(150/10) +Tp + 200/(150/10)+Tp + 100,000/(150/10)+ Tp ) = 7377 + 8*Tp (seconds) Now consider a persistent HTTP
connection
. The total time needed is given by: (200/150+Tp + 200/150 +Tp + 200/150+Tp + 100,000/150+ Tp ) + 10*(200/150+Tp + 100,000/150+ Tp ) =7351 + 24*Tp (seconds) Assuming the speed of light is 300*106 m/sec, then Tp=10/(300*106)=0.03 microsec. Tp is therefore negligible compared with
trans
mission 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 Yes, because Bob has more
connection
s, he can get a larger share of the link bandwidth. Yes, Bob still needs to perform parallel downloads; otherwise he will get less bandwidth than the other four users. Problem 12 Server.py from
socket
im
port
* server
Port
=12000 server
Socket
=
socket
(AF_INET,SOCK_STREAM) server
Socket
.bind(('',server
Port
)) server
Socket
.listen(1)
connection
Socket
, addr = server
Socket
.accept() while 1: sentence =
connection
Socket
.recv(1024) print 'From Server:', sentence, '\n' server
Socket
.close() Problem 13 The MAIL FROM: in SMTP is a message from the SMTP client that identifies the sender of the mail message to the SMTP server. The From: on the mail message itself is NOT an SMTP message, but rather is just a line in the body of the mail message. Problem 14 SMTP uses a line containing only a period to mark the end of a message body. HTTP uses “Content-Length header field” to indicate the length of a message body. No, HTTP cannot use the method used by SMTP, because HTTP message could be binary
data
, whereas in SMTP, the message body must be in 7-bit ASCII format. Problem 15 MTA stands for Mail
Trans
fer Agent. A host sends the message to an MTA. The message then follows a sequence of MTAs to reach the receiver’s mail reader. We see that this spam message follows a chain of MTAs. An honest MTA should re
port
where it receives the message. Notice that in this message, “asusus-4b96 ([58.88.21.177])” does not re
port
from where it received the email. Since we assume only the originator is dishonest, so “asusus-4b96 ([58.88.21.177])” must be the originator. Problem 16 UIDL abbreviates “unique-ID listing”. When a POP3 client issues the UIDL command, the server responds with the unique message ID for all of the messages present in the user's mailbox. This command is useful for “download and keep”. 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: dele 1 C: retr 2 S: (blah blah … S: ………..blah) S: . C: dele 2 C: quit S: +OK POP3 server signing off b) C: retr 2 S: blah blah … S: ………..blah S: . C: quit S: +OK POP3 server signing off C: list S: 1 498 S: 2 912 S: . C: retr 1 S: blah ….. S: ….blah S: . C: retr 2 S: blah blah … S: ………..blah S: . C: quit S: +OK POP3 server signing off Problem 18 For a given input of domain name (such as ccn.com), IP address or network administrator name, the whois
data
base can be used to locate the corresponding registrar, whois server, DNS server, and so on. NS4.YAHOO.COM from www.register.com; NS1.MSFT.NET from ww.register.com Local Domain: www.mindspring.com Web servers : www.mindspring.com 207.69.189.21, 207.69.189.22, 207.69.189.23, 207.69.189.24, 207.69.189.25, 207.69.189.26, 207.69.189.27, 207.69.189.28 Mail Servers : mx1.mindspring.com (207.69.189.217) mx2.mindspring.com (207.69.189.218) mx3.mindspring.com (207.69.189.219) mx4.mindspring.com (207.69.189.220) Name Servers: itchy.earthlink.net (207.69.188.196) scratchy.earthlink.net (207.69.188.197) www.yahoo.com Web Servers: www.yahoo.com (216.109.112.135, 66.94.234.13) Mail Servers: a.mx.mail.yahoo.com (209.191.118.103) b.mx.mail.yahoo.com (66.196.97.250) c.mx.mail.yahoo.com (68.142.237.182, 216.39.53.3) d.mx.mail.yahoo.com (216.39.53.2) e.mx.mail.yahoo.com (216.39.53.1) f.mx.mail.yahoo.com (209.191.88.247, 68.142.202.247) g.mx.mail.yahoo.com (209.191.88.239, 206.190.53.191) Name Servers: ns1.yahoo.com (66.218.71.63) ns2.yahoo.com (68.142.255.16) ns3.yahoo.com (217.12.4.104) ns4.yahoo.com (68.142.196.63) ns5.yahoo.com (216.109.116.17) ns8.yahoo.com (202.165.104.22) ns9.yahoo.com (202.160.176.146) www.hotmail.com Web Servers: www.hotmail.com (64.4.33.7, 64.4.32.7) Mail Servers: mx1.hotmail.com (65.54.245.8, 65.54.244.8, 65.54.244.136) mx2.hotmail.com (65.54.244.40, 65.54.244.168, 65.54.245.40) mx3.hotmail.com (65.54.244.72, 65.54.244.200, 65.54.245.72) mx4.hotmail.com (65.54.244.232, 65.54.245.104, 65.54.244.104) Name Servers: ns1.msft.net (207.68.160.190) ns2.msft.net (65.54.240.126) ns3.msft.net (213.199.161.77) ns4.msft.net (207.46.66.126) ns5.msft.net (65.55.238.126) d) The yahoo web server has multiple IP addresses www.yahoo.com (216.109.112.135, 66.94.234.13) e) The address range for Polytechnic University: 128.238.0.0 – 128.238.255.255 f) An attacker can use the whois
data
base and nslookup tool to determine the IP address ranges, DNS server addresses, etc., for the target institution. By analyzing the source address of attack packets, the victim can use whois to obtain information about domain from which the attack is coming and possibly inform the administrators of the origin domain. Problem 19 The following delegation chain is used for gaia.cs.umass.edu a.root-servers.net E.GTLD-SERVERS.NET ns1.umass.edu(authoritative) First command: dig +norecurse @a.root-servers.net any gaia.cs.umass.edu ;; AUTHORITY SECTION: edu. 172800 IN NS E.GTLD-SERVERS.NET. edu. 172800 IN NS A.GTLD-SERVERS.NET. edu. 172800 IN NS G3.NSTLD.COM. edu. 172800 IN NS D.GTLD-SERVERS.NET. edu. 172800 IN NS H3.NSTLD.COM. edu. 172800 IN NS L3.NSTLD.COM. edu. 172800 IN NS M3.NSTLD.COM. edu. 172800 IN NS C.GTLD-SERVERS.NET. Among all returned edu DNS servers, we send a query to the first one. dig +norecurse @E.GTLD-SERVERS.NET any gaia.cs.umass.edu umass.edu. 172800 IN NS ns1.umass.edu. umass.edu. 172800 IN NS ns2.umass.edu. umass.edu. 172800 IN NS ns3.umass.edu. Among all three returned authoritative DNS servers, we send a query to the first one. dig +norecurse @ns1.umass.edu any gaia.cs.umass.edu gaia.cs.umass.edu. 21600 IN A 128.119.245.12 The answer for google.com could be: a.root-servers.net E.GTLD-SERVERS.NET ns1.google.com(authoritative) Problem 20 We can periodically take a snapshot of the DNS caches in the local DNS servers. 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. Wills, Mikhail Mikhailov, Hao Shang “Inferring Relative Popularity of Internet Applications by Actively Querying DNS Caches”, in IMC'03, October 2729, 2003, Miami Beach, Florida, USA Problem 21 Yes, we can use dig to query that Web site in the local DNS server. For example, “dig cnn.com” will return the query time for finding cnn.com. If cnn.com was just accessed a couple of seconds ago, an entry for cnn.com is cached in the local DNS cache, so the query time is 0 msec. Otherwise, the query time is large. Problem 22 For calculating the minimum distribution time for client-server distribution, we use the following formula: Dcs = max {NF/us, F/dmin} Similarly, for calculating the minimum distribution time for P2P distribution, we use the following formula: Where, F = 15 Gbits = 15 * 1024 Mbits us = 30 Mbps dmin = di = 2 Mbps Note, 300Kbps = 300/1024 Mbps. Client Server N 10 100 1000 u 300 Kbps 7680 51200 512000 700 Kbps 7680 51200 512000 2 Mbps 7680 51200 512000 Peer to Peer N 10 100 1000 u 300 Kbps 7680 25904 47559 700 Kbps 7680 15616 21525 2 Mbps 7680 7680 7680 Problem 23 Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of a rate of us/N. Note that this rate is less than each of the client’s download rate, since by assumption us/N ≤ dmin. Thus each client can also receive at rate us/N. Since each client receives at rate us/N, the time for each client to receive the entire file is F/( us/N) = NF/ us. Since all the clients receive the file in NF/ us, the overall distribution time is also NF/ us. Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of dmin. Note that the aggregate rate, N dmin, is less than the server’s link rate us, since by assumption us/N ≥ dmin. Since each client receives at rate dmin, the time for each client to receive the entire file is F/ dmin. Since all the clients receive the file in this time, the overall distribution time is also F/ dmin. From Section 2.6 we know that DCS ≥ max {NF/us, F/dmin} (Equation 1) Suppose that us/N ≤ dmin. Then from Equation 1 we have DCS ≥ NF/us . But from (a) we have DCS ≤ NF/us . Combining these two gives: DCS = NF/us when us/N ≤ dmin. (Equation 2) We can similarly show that: DCS =F/dmin when us/N ≥ dmin (Equation 3). Combining Equation 2 and Equation 3 gives the desired result. Problem 24 Define u = u1 + u2 + ….. + uN. By assumption us <= (us + u)/N Equation 1 Divide the file into N parts, with the ith part having size (ui/u)F. The server
trans
mits the ith part to peer i at rate ri = (ui/u)us. Note that r1 + r2 + ….. + rN = us, so that the aggregate server rate does not exceed the link rate of the server. 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. We have (N-1)ri = (N-1)(usui)/u = (us + u)/N Equation 2 Let ri = ui/(N-1) and rN+1 = (us – u/(N-1))/N In this distribution scheme, the file is broken into N+1 parts. The server sends bits from the ith part to the ith peer (i = 1, …., N) at rate ri. Each peer i forwards the bits arriving at rate ri to each of the other N-1 peers. Additionally, the server sends bits from the (N+1) st part at rate rN+1 to each of the N peers. The peers do not forward the bits from the (N+1)st part. The aggregate send rate of the server is r1+ …. + rN + N rN+1 = u/(N-1) + us – u/(N-1) = us Thus, the server’s send rate does not exceed its link rate. The aggregate send rate of peer i is (N-1)ri = ui Thus, each peer’s send rate does not exceed its link rate. In this distribution scheme, peer i receives bits at an aggregate rate of Thus each peer receives the file in NF/(us+u). (For simplicity, we neglected to specify the size of the file part for i = 1, …., N+1. We now provide that here. Let Δ = (us+u)/N be the distribution time. For i = 1, …, N, the ith file part is Fi = ri Δ bits. The (N+1)st file part is FN+1 = rN+1 Δ bits. It is straightforward to show that F1+ ….. + FN+1 = F.) The solution to this part is similar to that of 17 (c). We know from section 2.6 that Combining this with a) and b) gives the desired result. Problem 25 There are N nodes in the overlay network. There are N(N-1)/2 edges. 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 run a client on each host, let each client “free-ride,” and combine the collected chunks from the different hosts into a single file. 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. Problem 28 Peer 6 would first send peer 15 a message, saying “what will be peer 6’s predecessor and successor?” This message gets forwarded through the DHT until it reaches peer 5, who realizes that it will be 6’s predecessor and that its current successor, peer 8, will become 6’s 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): ABC, where A and B are neighboring peers, and B and C are neighboring peers. Suppose that there is another logical path p2 from A to C (consisting of 3 logical links): ADEC. 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 it may be the case that A, D, E, and C are all very close physically (and all separated by few routers). In other words, a shorter logical path may correspond to a much longer physical path. Problem 31 If you run TCPClient first, then the client will attempt to make a TCP
connection
with a non-existent server process. A TCP
connection
will not be made. UDPClient doesn't establish a TCP
connection
with the server. Thus, everything should work fine if you first run UDPClient, then run UDPServer, and then type some input into the keyboard. If you use different
port
numbers, then the client will attempt to establish a TCP
connection
with the wrong process or a non-existent process. 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. With the additional line, when UDPClient is executed, a UDP
socket
is created with
port
number 5432 . UDPServer needs to know the client
port
number so that it can send packets back to the correct client
socket
. Glancing at UDPServer, we see that the client
port
number is not “hard-wired” into the server code; instead, UDPServer determines the client
port
number by unraveling the
data
gram it receives from the client. Thus UDP server will work with any client
port
number, including 5432. UDPServer therefore does not need to be modified. Before: Client
socket
= x (chosen by OS) Server
socket
= 9876 After: Client
socket
= 5432 Problem 33 Yes, you can configure many browsers to open multiple simultaneous
connection
s to a Web site. The advantage is that you will you potentially download the file faster. The disadvantage is that you may be hogging the bandwidth, thereby significantly slowing down the downloads of other users who are sharing the same physical links. 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. For example, when one SMTP mail server sends another SMTP mail server several email messages back to back. 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 Call this protocol Simple
Trans
port
Protocol (STP). At the sender side, STP accepts from the sending process a chunk of
data
not exceeding 1196 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. The segment now has two header fields: a source
port
field and destination
port
field. At the sender side, STP accepts a chunk of
data
not exceeding 1192 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. No, the
trans
port
layer does not have to do anything in the core; the
trans
port
layer “lives” in the end systems. 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 clearly
write
s the recipient’s name on the top of the letter. The delegate then puts the letter in an envelope and
write
s the address of the destination house on the envelope. The delegate then gives the letter to the planet’s mail service. 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. An application developer may not want its application to use TCP’s congestion control, which can throttle the application’s sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCP’s congestion control. Also, some applications do not need the reliable
data
trans
fer provided by TCP. Since most firewalls are configured to block UDP traffic, using TCP for video and voice traffic lets the traffic though the firewalls. Yes. The application developer can put reliable
data
trans
fer 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. For each persistent
connection
, the Web server creates a separate “
connection
socket
”. Each
connection
socket
is identified with a four-tuple: (source IP address, source
port
number, destination IP address, destination
port
number). When host C receives and IP
data
gram, it examines these four fields in the
data
gram/segment to determine to which
socket
it should pass the payload of the TCP segment. Thus, the requests from A and B pass through different
socket
s. The identifier for both of these
socket
s has 80 for the destination
port
; however, the identifiers for these
socket
s have different values for source IP addresses. Unlike UDP, when the
trans
port
layer passes a TCP segment’s payload to the application process, it does not specify the source IP address, as this is implicitly specified by the
socket
identifier. Sequence numbers are required for a receiver to find out whether an arriving packet contains new
data
or is a re
trans
mission. To handle losses in the channel. If the ACK for a
trans
mitted packet is not received within the duration of the timer for the packet, the packet (or its ACK or NACK) is assumed to have been lost. Hence, the packet is re
trans
mitted. A timer would still be necessary in the protocol rdt 3.0. 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. The packet loss caused a time out after which all the five packets were re
trans
mitted. Loss of an ACK didn’t trigger any re
trans
mission as Go-Back-N uses cumulative acknowledgements. The sender was
unable
to send sixth packet as the send window size is fixed to 5. When the packet was lost, the received four packets were buffered the receiver. After the timeout, sender re
trans
mitted the lost packet and receiver delivered the buffered packets to application in correct order. Duplicate ACK was sent by the receiver for the lost ACK. The sender was
unable
to send sixth packet as the send win
SmtpClient发送邮件报
Unable
to read
data
from the
trans
port
connection
: The
connection
was closed
Unable
to read
data
from the
trans
port
connection
: The
connection
was closed
Socket
编程系列之7:RCF远程调用框架编程实战
Socket
编程系列之7:RCF远程调用框架编程实战
Socket
编程系列之7:RCF远程调用框架编程实战,本课程包括Windows搭建RCF环境及Helloworld案例、Linux开发必备之GCC与CMake、Linux编译RCF及跨平台案例实战、基于RCF框架编程原理流程步骤解析、基于RCF框架的异步调用案例实战、基于RCF框架的发布订阅案例实战等。-------------------------------------------------------注意,本系列
Socket
编程系列初步计划包括如下课程:
Socket
编程系列之1:Linux-API网络编程入门实战
Socket
编程系列之2:Windows-API网络编程入门实战
Socket
编程系列之3:Libevent网络编程入门实战
Socket
编程系列之4:Libcurl网络编程入门实战
Socket
编程系列之5:Boost-ASIO网络编程入门实战
Socket
编程系列之6:Libuv网络编程入门实战
Socket
编程系列之7:RCF远程调用框架编程实战
hbase遇到的坑及解决办法
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socket
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Hbase thrift 写数据Broken pipe问题
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