# Problem 1 Suppose that the client A initiates a TCP connection to a Web server whose name is S. More or less simultaneously,

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1 Note: Here you have the original answer proposed by working groups. Spelling mistakes from original answers have not been corrected. Sometimes teacher note is included in red. Problem 1 Suppose that the client A initiates a TCP connection to a Web server whose name is S. More or less simultaneously, the client B also initiates a TCP connection to S. a) Indicate possible source and destination port number for: 1. Segments sent from A to S 2. Segments sent from B to S 3. Segments sent from S to A 4. Segments sent from S to B b) If A and B are in different hosts, could source port number from A to S be the same that from B to S? c) What if client process A and B are in the same host? Solution proposed by Group 1: a.1) We choose as Source port value of A 9157 arbitrary within the free ports range, and the destination port of S as 80.That value is the default value for HTTP protocol. a.2) Value of Source Port of B: 9158, and destination port of S:80. a.3) Value of Source Port of S: 80, and destination port of A: a.4) Value of Source Port of S: 80, and destination port of B: b) Yes, host A and host B have the possibility to have the same port number because the source port number. There is more information, like the IP to see that both entities are different. c) If process A and B are in the same host, each one must have its own port number. Problem 2 a) What are the source and destination port number values in the segments flowing from the server back to the clients processes? b) What are the IP addresses (source and destination) of N_PDUs carrying these transport layer segments? Solution proposed by Group 2: a) P1 (ip: A, port: 9557) accesses to server (ip: C, port: 80). b) SP=80 client >SP=9157 DP=9157 D IP=C D IP=A S IP=A S IP=C DP=80 DP=80 SP=9157 D IP=B S IP=C Problem 3 UDP and TCP use 1s complement for their checksums. Suppose you have the following three 16 bit words: a) What is the 1s complement of the sum of these 16 bit words? b) Receiver adds the three words to the received checksum. If the result of the addition, in binary, contains some zero, the receiver realizes there is some error in some bit, is it right? 1/8

2 c) Is it possible that a 1 bit error will go undetected? d) Propose a particular example for a non detectable error. Solution proposed by Group 3: Checksum is a 16 bits word to check if there is any error in the receiver data. In our case, the receiver has the following three words: We have to sum these words and do the 1s complement to the result: = = a) 1s complement: b) If we sum this last result and the checksum from the transmitter, if the result is , that means that the data received is correct, and if there is any 0 means that there is any error in the data. c) If there is a 1 bit error, the checksum will always detect it. d) But it`s possible that 2 or more bits error couldn t be detected by the checksum, because if the 2 bits wrong are in the same position in the checksum, the result of the sum will be the same and the error wont be detected. Example: Problem 4 Pipeline protocols improve the performance of stop and wait protocols. Suppose a Tx and a Rx with a 1Gbps link, 30ms RTT, 1500 bytes PDUs and header size equal to zero (negligible size in comparison to the data size). How many data PDUs has Tx to have in flight for channel utilization is 95%? Solution proposed by Group 4: The channel utilisation for stop a wait protocols is: For pipeline protocols we have to take into account that we will send several packets before receiving any ACK from the receiver. This means that the channel utilisation is: In the particular case proposed in the exercise we have to calculate how many PDUs has the transmitter to have inflight for a 95% channel utilisation: /8

3 0, ,95~2376 0, , Problem 5 We have said that an application may choose UDP for a transport protocol because UDP offers finer application control (than TCP) of what data is sent in a segment and when. a) Why does an application have more control of what data is sent in a segment? b) Why does an application have more control on when the segment is sent? Solution proposed by Group 2: a) The info flows not controlled, TCP decides how many data can carry in T_UDP. b) TCP has own mechanism to control each T_PDU (buffer and segmentation). Is more control over the moment when data is sent. UDP just passes directly the data. Problem 6 Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes. a) What is the maximum value of L such that TCP sequence numbers are not exhausted? Recall that the TCP sequence number field has 4 bytes. Note: MSS is the maximum size of user data carried by any segment in a TCP connection. In SYN segment, using TCP options header, each TCP entity informs to the other about which MSS wants to use. The smaller will be used. b) For the L you obtain in (a), find how long it takes to transmit the file. Assume that a total of 66 bytes of transport, network, and data link header are added to each segment before the resulting packet is sent out over a 155 Mbps link. Ignore flow control and congestion control so A can pump out the segments back to back and continuously. Solution proposed by Group 1: a) To calculate it, we must know that 1byte is 8 bits. So, if there is 4 bytes, we have 32 bits. The maximum value of L must be: 2^32= bits Teacher note: MSS doesn t care because TCP sequence numbers are to enumerate data bytes not segments. b) We have 536 bytes for every packet sent. Of those 536, 66 bytes are needed for transport, network and data link header. This means we have 470 bytes for data in every packet. We would need to sent /470 = packets to send the file So, packets of 536 bytes, make a total of bits, megas in total mb/155mbs = 31,6 seconds Teacher note: There is an error in this solution, which is it? 3/8

4 Problem 7 Host A and B are communicating over a TCP connection, and Host B has already received from A all bytes up through byte 126. Suppose Host A then sends two segments to Host B back to back. The first and second segments contain 70 and 50 bytes of data, respectively. In the first segment, the sequence number is 127, the source port number is 302, and the destination port number is 80. Host B sends an acknowledgement whenever it receives a segment from Host A. a) In the second segment sent from Host A to B, what are the sequence number, source port number, and destination port number? b) If the first segment arrives before the second segment, in the acknowledgement of the first arriving segment, what is the acknowledgment number, the source port number, and the destination port number? Solution proposed by Group 3: a) The sequence number will be first byte of the next sequence of bytes transmitted. As the first segment sent had 70 bytes, the next sequence number will be 197. The source port will be the same used in the connection: 302. The destination port will be 80. b) The ACK number is the next byte expected, in this case the ACK number of the acknowledgement of the first segment will be 197. The source port: 80 (now, we are in the server side). The destination port: 302. c) In this case the server has received the segment with the seq. number 198, but it hasn t received the segment with the seq. number 127,which is, actually, the segment it has expecting to receive. In this case, the server indicates in the ACK number of the response that it has the first 126 bytes, but it still expects to receive the one with the seq. number 127. So the ACK # would be 127 again. Note that the server always sent as ACK number the number of bytes that has received all the bytes until the ACK# 1. In this case, although the sender retransmits the segment with seq# 127, the server tells the sender that it has everything OK until byte /8

7 Solution proposed by Group 5: TCP has a mechanism called flow control which prohibits sending too much data to the receiver. In all ACKs the receiver tells the sender how much free space there still is in the buffer (the RcvWindow). The sender never sends more unacknowledged data to the receiver than the RcvWindow. Therefore it cannot overflow. Solution proposed by Group 7: In this problem, there is no danger in overflowing the receiver since the receiver s receive buffer can hold the entire file. 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. Once the send buffer becomes full, the process will pass data at an average rate or R << S. Teacher note: Flow control doesn t slow A down because in B there is a buffer (in TCP entity) that is able to store the whole file, then reception window always allows A to send. The problem is that TCP entity in A is not able to take out data from its transmission buffer at the same rate that its application layer introduces it, because R is less than S. So, application layer has to go down its data sending rate to TCP socket to not overflow its transmission buffer, where the whole file doesn t fit. Groups and grades (out of 5 points): Surnames Name Group Problems Grade ARCINIEGA FERNANDEZ ALEJANDRO 1 1, 6 3 ARJONILLA COBREROS PABLO MANUEL 1 1, 6 3 DIAZ CASAL DARIO 2 2, 5 3,5 GARCIA MORENO FRANCISCO MANUEL 2 2, 5 3,5 MERCHAN CACHINERO BORJA MANUEL 2 2, 5 3,5 JARANA PEREZ JOSE ANDRES 3 3, 7 5 CASTRO MATEOS HECTOR 3 3, 7 5 GUERRERO MONTORO ALEJANDRO 4 4, 8 5 RODRIGUEZ DE PRA AVILES MARIA JOSE 4 4, 8 5 SANJUAN SEGOVIA FELIX MANUEL 4 4, 8 5 SANCHEZ GUAJARDO FAJARDO CARLOS 5 9, 13 3 SCHUETZ ROLAND 5 9, 13 3 VARO HERRERO PEDRO ANTONIO 5 9, 13 3 VAZQUEZ SANCHEZ RAFAEL 6 10, 11, 12 1,5 VERA HERNANDEZ PABLO 6 10, 11, 12 1,5 YAÑEZ HERRERA ALFONSO 6 10, 11, 12 1,5 ELVIRA PEREZ SERGIO 7 9, 13 5 RINCON BORREGUERO ALBERTO 7 9, /8

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