The acknowledgement number in an incoming ack segment is decremented by 1 to retrieve the generated syn cookie. When the result of the function is equal to the acknowledgment value, the server detects the corresponding syn segment and thus concludes that the segment is valid. It will then open a full connection on the socket. Otherwise, the server does not allocate resources to open a connection, and no harm is done restaurant to the server. So unless the attacker knows the initial sequence number the server created, it cannot just simply send an ack to fool the server into thinking its valid as the server looks to detect the corresponding syn segment to the ack with an equal acknowledgment value. So even if the attacker knew the server used syn cookies to protect its information, the attacker will still not be able to create half-open or fully open connections by simply sending an ack to the target. It has no way of retrieving the cookie information or its sequence number as only the server knows it and creates.
We know that syn cookies are a way to defend systems against attackers. The advantage of syn cookies is that it hides a unique sequence number that only the server knows. It sends this packet with the cookie to the client where it does essay not remember the cookie or any state information. What if the attacker were to send an ack packet? Its impossible for the attacker to know the sequence number as only the server knows that information, and it gets passed on to the client where it erases that information from the server. A client will then return an ack segment. While the server reserves no memory of syn segments sent to the client earlier, it has ways of assessing the validity of a syn segment.
Direct Attack- a direct attack is when an attacker rapidly sends syn segments without a spoof of the ip source address. Its an easy attack to perform, but the attackers operating system must not respond to a synack segment because any acks responded to by the attacker allow the listener to move tcb out of syn-received, thus repelling the attack. Spoofing Attack- the attacker spoofs an ip address and forms and injects raw ip packets with valid tcp and ip headers. The attacker must not respond to synacks or the listener can back out of the attack. If an attacker attacks with multiple spoofed addresses, the attack becomes difficult for the listener to defend against as more complicated solutions are required to repel the attack. However, if the attacker only spoofs one ip address, its easy to defend against as the listener can just filter and detect the spoofed ip address. Distributed Attack- where the attacker uses botnets and drones to use direct attacks and spoofing attacks on a hosts operating systems. This attack is very difficult for a system to defend against, and drones can change their addresses which makes this attack increasingly difficult to defend against.
Homework, problems and Solutions Part
We will use the information provided in Section.5.6 to answer the questions. Its necessary for a server to use a special initial sequence number in the synack because this initial sequence number is a syn cookie. Syn cookies contain a secret number that only the server has knowledge of and it protects the information in the packet from a dos attack. Suppose an attacker knows that a target host uses syn cookies. Can the attacker create half-open or fully open connections by writing simply sending an ack packet to the target.
Why or why not? Suppose an attacker collects a large amount of initial sequence numbers sent by the server. Can the attacker cause the server to create many fully open connections by sending acks with those initial sequence numbers? To answer these questions, we will use the article defense Against tcp syn flooding Attacks. Syn flood attackers have a set of methods they can use to perform a syn flood attack. For instance, there are a number of attacks they can perform: direct attack, spoofing-based attack, distributed attack, and attack parameters.
Meanwhile, host b allocates a receive buffer denoted by RcvBuffer. The receive window rwnd is set to the amount of space in the buffer for Host B which is given by the following formula: rwnd RcvBuffer-LastByteRcvd- lastByteread, for Host A to avoid overflowing Host Bs allocated buffer, we have: LastByteRcvd-LastByteread RcvBuffer. Host B tells Host A how much space it has in its buffer by placing its current value of rwnd in the receive window of every segment it sends. So host B sets rwndRcvBuffer. Host a keeps track of the two variables LastByteSent and LastByteAcked whereas Host B must keep track of several connection-specific variables.
To avoid overflowing the buffer, the difference must be taken between LastByteSent-LastByteAcked- the unacknowledged data that Host A has sent into the connection. The unacknowledged data must be less than the value of rwnd to avoid overflowing the buffer. So: once host Bs rwnd0, host A must continue to send segments with one data byte. The segments will be acknowledged by host b, and the buffer will eventually begin to empty causing the acknowledgments rwnd value to be nonzero. (page 250-252 of text p29. Syn cookies were discussed in Section.5.6. Why is it necessary for the server to use a special initial sequence number in the synack?
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There is one tcp connection between the two hosts, and Host a is sending to host b an enormous file over this connection. Host A can send its application data into its tcp socket at a rate as high as 120 summary Mbps but Host B can read out of its tcp receive buffer at a maximum rate of 50 Mbps. Describe the effect of tcp flow control. Flow control is a service provided by tcp that matches the rate at which the sender is sending against the rate which the receiver is reading what was sent. Here, host a is sending its application data at a speed of up to 120 Mbps whereas Host b is reading the data at 50 Mbps max. Tcp allows the sender to maintain a variable called the receive window, which is used to allow the sender(Host mba A) a certain amount of buffer space that is available at the receiver(Host B). So host A gets an idea of how much buffer space is available at Host.
What about when the packet size is essay 1,500 bytes? 1,500 bytes 12,000 bits, let L denote the size of the packet and R denote the rate of the packet. The rate at which the packet is being transmitted is 109 bits/sec. We use the dtrans formula to calculate the value of L/R. Dtrans L/R (12000 bits/packet 109 bits/sec.0012 ms or 12 microseconds, we now plug this L/r value into the the formula for. The formula for t is: t rtt l/R.015.0012.0162. We then plug the value for t and (L/R) into the utilization formula: Usender 3(L/r rtt(L/R) 3(0.012.0162s.22222.98 W(0.0012.0162s, w (0.98/0.0012).0162. So.0162 is the window size where the size of the packet is 1,500 bytes where the utilization of the sender would be 98 percent. Host a and b are directly connected with a 100 Mbps link.
would the window size have to be for the channel utilization to be greater than 98 percent. Suppose that the size of a packet is 1,500 bytes, including header fields and data. Figure.17(page 217 The stop-and-wait protocol contains the utilization formula of: Usender (L/r rtt(L/R).008/30.008.00027, we modify this formula: Usender 3(L/r rtt(L/R).0024/30.008.00081, the number.0081 represents the efficiency. We want this value to equal.98, or 98 percent. Let W denote the size of the window. . So given. Roseines demonstration, we have the formula:.98 W (L/r rtt(L/R) where (L/R).008 seconds or 8 microseconds and rtt.008s.98 W(8.008. W (0.98/0.008 so the window size must.008 in order to have a 98 percent efficiency.
Each 16-bit word is added together as demonstrated below:, look at the last digit of the bottom two checksum values. Notice how the 0 changes to a 1 desk and the 1 changes to a 0 in the checksum values. There are changes in the checksum bits, which alters the checksum payload value- now to 6132. This signifies that there are bit errors. Since this happens twice, we have two bit errors. So the receiver cannot be absolutely certain that no bit errors have occurred. This is because of the way 16-bit checksum values are added together: the bits are binary- that is, they only use the values 0 and. Adding 0 and 0 together is 0, but adding 1 and 1 together. Also, adding 1 and 0 together.
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Suppose that the udp receiver computes the Internet checksum for the received udp segment and finds that it matches the value carried in the checksum field. Can the receiver be absolutely certain that no bit errors have occurred? To determine this, i will provide a demonstration. Roseine: In this demonstration, we have a checksum payload given of 6123. Whenever a digit is changed to a new digit, there is a bit error that occurs. Recall that a checksum as defined on page 202 (3.3.2 udp checksum) is used to detect errors. It determines where bits in the checksum have been changed. We have the checksum bits of the checksum payload. The checksum bits represent 16-bit words.