EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
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Chapter 4, Problem P18P
Program Plan Intro
Transmission Control Protocol (TCP):
- TCP is an internet protocol used for exchanging of data between the sender and the receiver. It is a secured protocol for transferring the data.
- When the sender sends data to the receiver, he/she waits for the acknowledgement from the receiver.
- If acknowledgement is received then the next data packet will be sent and the process continues till all the data packets are sent.
- If acknowledgement is not received then the receiver is assumed as hacker or intruder and the connection will be removed.
Network Address Translation (NAT):
- NAT is a method which is used to connect numerous numbers of computers to the internet using a single Internet Protocol (IP) address.
- It offers the home users who using multiple computers which are used by the family members at home to connect with the internet and small businesses a cheap and efficient internet connections.
- Usually, most of the companies use more than one router to connect with the internet.
- But after installing NAT, the mapping is kept only at a single router. Thus, it is essential to use a same router to connect with the internet.
- Multi-homing with more routers using NAT can be done by using the same router for a single connection and thus mapping is done properly.
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We will examine the effect of NATs on P2P applications in this problem. Suppose a peer with Arnold finds that a peer with Bernard username has a file to download. Suppose also that both Bernard and Arnold are behind a NAT. Try to formulate a technique that will permit Arnold to create a TCP link to Bernard without a special NAT application setup. If you have trouble developing such a strategy, please explain why.
In this problem we investigate whether either UDP or TCP provides a degree of
end-point authentication.
a. Consider a server that receives a request within a UDP packet and responds to
that request within a UDP packet (for example, as done by a DNS server). If a
client with IP address X spoofs its address with address Y, where will the server
send its response?
b. Suppose a server receives a SYN with IP source address Y, and after responding
with a SYNACK, receives an ACK with IP source address Y with the correct
acknowledgment number. Assuming the server chooses a random initial sequence
number and there is no "man-in-the-middle," can the server be certain that the
client is indeed at Y (and not at some other address X that is spoofing Y)?
We will examine the effect of NATs on P2P applications in this problem. Suppose a peer with Arnold finds that a peer with Bernard username has a file to download. Suppose also that both Bernard and Arnold are behind a NAT. Try to formulate a technique that will permit Arnold to create a TCP link to Bernard without a special NAT application setup. If you have trouble developing such a strategy, please explain why.qap
Chapter 4 Solutions
EBK COMPUTER NETWORKING
Ch. 4 - Lets review some of the terminology used in this...Ch. 4 - Prob. R2RQCh. 4 - Prob. R3RQCh. 4 - Prob. R4RQCh. 4 - Prob. R5RQCh. 4 - Prob. R6RQCh. 4 - Prob. R7RQCh. 4 - Prob. R8RQCh. 4 - Prob. R9RQCh. 4 - Prob. R10RQ
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Similar questions
- We explore whether either UDP or TCP offers any level of end-point authentication in this problem. a. Consider a server that accepts a request via UDP and responds via UDP (for example, as done by a DNS server). Where would the server give its answer if a client with IP address X spoofs it with address Y? b. Assume a server receives a SYN with IP source address Y and responds with SYNACK. The server then receives an ACK with IP source address Y and the right acknowledgmentamount. Assuming the server selects a random initial sequence number and there is no "man-in-the-middle," how can the server be confident that the recipient is really at Y (and not at any other address X that spoofs Y)?arrow_forwardThree-way handshake is used by a TCP client and a TCP server to establish a connection, as illustrated below: 1st: client:port1 -> server:port2, SYN 2nd: server:port2 -> client:port1, SYNACK 3rd: client:port1 -> server:port2, ACK When this client is performing scanning attacks, it will generated a large number of failed connections. In each failed connection, the three-way handshake fails to complete. People commonly use SYN together with the absence of its corresponding SYNACK in this same TCP session to identify whether this connection is failed. By investigating the failed connections, an engineer finds that in legitimate/benign cases, if the server does not return SYNACK to the client, the client will not send the ACK packet after SYNACK (e.g., the 3rd packet above). Therefore, this engineer suggests that we can count the failed connections based on the following rules without considering SYNACK: If a client:port1 sends…arrow_forwardThree-way handshake is used by a TCP client and a TCP server to establish a connection, as illustrated below: 1st: client:port1 -> server:port2, SYN 2nd: server:port2 -> client:port1, SYNACK 3rd: client:port1 -> server:port2, ACK When this client is performing scanning attacks, it will generated a large number of failed connections. In each failed connection, the three-way handshake fails to complete. People commonly use SYN together with the absence of its corresponding SYNACK in this same TCP session to identify whether this connection is failed. By investigating the failed connections, an engineer finds that in legitimate/benign cases, if the server does not return SYNACK to the client, the client will not send the ACK packet after SYNACK (e.g., the 3rd packet above). Therefore, this engineer suggests that we can count the failed connections based on the following rules without considering SYNACK:arrow_forward
- Consider a client connecting to a web server via a router as shown in Fig.Q2. Client A sends a request to the server to retrieve a 7.5 Mbytes file. Given that the segment size is 50 Kbytes, the round trip time (RTT) between the server and client is 10 ms, the initial slow-start threshold is 16 and the client's buffer always has a storage space of 1 Mbytes. Assume that TCP Reno is used, there is no loss during transmission and the headers of protocols are ignored. 400 Mbps 200 Mbps 400 Mbps Link a Link b Link c Client Web Server Fig.Q2 (a) Describe how the value of sending window changes as a function of time (in units of RTT) during the whole connection time. 2.arrow_forward3. Consider a simple application-level protocol built on top of UDP that allows a client to retrieve a file from a remote server residing at a well-known address. The client first sends a request with a file name, and the server responds with a sequence of data packets containing different parts of the requested file. To ensure reliability and sequenced delivery, client and server use a stop-and-wait protocol. Ignoring the obvious performance issue, do you see a problem with this protocol? Think carefully about the possibility of processing crashing. State conclusion: Provide Proof: Solution:arrow_forwardConsider a simple application-level protocol built on top of UDP that allows a client to retrieve a file from a remote server residing at a well-known address. The client first sends a request with file name, and the server responds with a sequence of data packets containing different parts of the requested file. To ensure reliability and sequenced delivery, client and server use a stop-and-wait protocol. Ignoring the obvious performance issue, do you see a problem with this protocol? Think carefully about the possibility of processes crashing.arrow_forward
- Kerberos is a protocol that is based around Needham-Schroeder protocol for many to many authentications. Now answer the following questions. (Use necessary diagrams to justify your answers) i) Explain why the password of the user is not sent over the network and instead session keys are generated and shared in the Kerberos protocol. ii) The information in a TGT (Ticket Granting Ticket) is encrypted so the client cannot access the information in the TGT. However, all information in the ticket is already known to the client. Why is it still necessary to encrypt it? iii) Describe the working mechanism of how a ticket is generated between the client and server by the TGS (Ticket Granting Server) and how it is used for client-server communication.arrow_forwardConsider a TCP connection between Host A and Host B and the transmission of a large file from A to B. If the buffer of host B is significantly smaller than the size of the buffer of host A and the file itself, what would be outcome?arrow_forwardUsing a TCP SYN spoofing attack, the attacker aims to flood the table of TCP connection requests on a system so that it is unable to respond to legitimate connection requests. Consider a server system with a table for 256 connection requests. This system will retry sending the SYN-ACK packet five times when it fails to receive an ACK packet in response, at 30 second intervals, before purging the request from its table. Assume that no additional countermeasures are used against this attack and that the attacker has filled this table with an initial flood of connection requests. a. At what rate must the attacker continue to send TCP connection requests to this system in order to ensure that the table remains full? b. Assuming that the TCP SYN packet is 40 bytes in size (ignoring framing overhead), how much bandwidth does the attacker consume to continue this attack?arrow_forward
- Assume that a client can use UDP to obtain a file from a distant server at a known address. The client initiates the request for a file name, and the server responds with a sequence of data packets containing various file components. To ensure delivery reliability, the client and server use a stop-and-wait protocol. Is there anything else wrong with this protocol aside apparent performance issues? Consider the possibility of process failure.arrow_forwardR6Fragmentation of an IP datagram takes place if its size is larger than the MTU of the subnet over which the datagram will be routed. Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose a datagram is fragmented into four fragments. The first three fragments arrive, but the last one is delayed. Eventually the timer goes off and the three fragments in the receiver's memory are discarded. A little later, the last fragment stumbles in. What should be done with it?arrow_forwardTCP is a connection-oriented protocol. This means that... а. there is a direct physical connection between the two endpoints of a TCP session. O b. both of the endpoints communicating over TCP keep information about the state of the connection. с. both endpoints of a TCP session have to connect to a third party before data can be sent or received. O d. a TCP sender can start sending data to the receiver before the three-way handshake takes place.arrow_forward
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