Computer Networking (COMP2322)

Transcription

1 Computer Networking (COMP2322) Test Two (due at 2:20pm on 28 March 2017) Total: 50 marks Rocky K. C. Chang Name: Student ID: Marks Instructions: 1) Always attach a succinct explanation for your answer. 2) There are five questions. Write down your answers to Q1-Q4 on this test paper. Submit your answer to Q5 to the moodle site ( 3) You can use only your pen and ruler. 4) The slides, answers to A1 and A2, and a copy of the textbook are in Blackboard (under Content). 5) Please get the permission before going to the bathroom. 1) (Sequence number space) [10 MARKS] We consider the sequence number wraparound problem for this scenario. Let SWS = 2 and RWS = 1. Assume that the sender always has frames to send. Consider that the sender sends two data frames at the beginning. Both are received with correct CRCs, and the receiver responds with two corresponding ACKs. The two ACKs, however, are both corrupted, therefore dropped by the sender. Show that the SN space of {0, 1, 2} is not sufficient for guaranteeing the correctness of the protocol for this scenario. Include a time-line diagram to support your answer. Solution: As shown in Figure 1, the retransmitted data frame with SN=0 is interpreted as a new data frame by the receiver, because it can receive SN=2 and SN=0. Therefore, this SN space does not work for this scenario. 2) (Collisions or not) [10 MARKS] This question concerns the CSMA/CD protocol. Assume the followings: The maximum round-trip propagation delay in an Ethernet segment is given by R max seconds. Consider two nodes, A and B, on the Ethernet segment. The transmission time for a frame sent by either A or B is given by R max seconds. The round-trip propagation delay between A and B is given by R R max seconds. Consider that node A has a frame to transmit at time t A, and node B has a frame to send at time t B. Give a condition in terms of t A, t B, and R for no collision between A s frame and B s frame. Solution: The condition is t A t B > 0.5R. If their transmission times are separated by at least 0.5R (whether t A > t B or t B > t A ), then the other node will hear the sender s signal and therefore refrain from sending its frame.

2 LAR=LFS=-1 Server Receiver NFE = 0 LFS=0 LFS=1 NFE = 1 NFE = 2 Timeout Receive it as new data frame Fig. 1. Time-line diagram for SN space of {0, 1, 2}. 3) (Switched LAN) [10 MARKS] Consider hosts W, X, Y, and Z, and bridges B1, B2, and B3 with empty source address tables initially, as shown in Figure 2. Write down the content of the three bridges SATs after each frame transmission in parts (a)-(d). a) [2 MARKS] Y sends a data frame to Z. b) [2 MARKS] W sends an ARP request for Y s MAC address. c) [2 MARKS] Y replies W s ARP request. d) [2 MARKS] Y sends a data frame to X. e) [2 MARKS] How many unnecessary frames are received by hosts other than the destinations in the four scenarios above? Solution: a) B1: Y, B2: Y, B3: Y. Since all the SATs are empty, this data frame is sent to all bridges which will learn Y. b) B1: W, Y, B2: W, Y, B3: W, Y. Since the ARP request is broadcasted to all hosts, all bridges will 2

3 Fig. 2. A switched LAN. learn W. c) Same as (b), since all bridges have already learned Y. d) Same as (b), since all bridges have already learned Y. e) (a): 2 (X and W ), (b): 0 (the request is intended for everyone), (c): 0 (all bridges know W ), (d): 2 (the bridges have not learned X). 4) (Loop or not) [10 MARKS] Consider the layer-two network in Figure 3. Assume bridge B1 already learns the MAC addresses of A and B. However, B2 s SAT is empty (perhaps because it is newly added). When A sends a frame to B, will it create a packet-forwarding loop? Here we do not use the spanning tree protocol to overlay a tree on this network. A B Upper LAN B1 B2 Lower LAN Fig. 3. A simple layer-two network with two bridges B1 and B2, and two hosts A and B. 3

4 Solution: a) B receives the frame. b) When B1 receives the frame, it drops it. c) When B2 receives the frame, it forwards it to the lower LAN and remember A s MAC address. d) When B1 receives the frame from the lower LAN, changes the port for A to the lower one, and forwards a copy to the upper LAN. e) B receives the same frame again. f) When B2 receives the frame and forwards it to the lower LAN. g) Therefore, there is a routing loop for this scenario. 5) (Socket programming) [10 MARKS] Create an application using socket programming where server echo backs the message sent by client along with the current date and time. Client and socket connection should be closed after one exchange of message. The sender-side and client-side execution are given in Figure 4 and Figure 5, respectively. Please submit the codes to the moodle site ( Please name the files to yourstudentid_client.py and yourstudentid_server.py. Fig. 4. The server-side execution. 4

5 Fig. 5. The client-side execution. Solution: #TCPServer - Simple TCP server which receives 16 octets import socket, sys s = socket.socket(socket.af_inet, socket.sock_stream) from datetime import datetime HOST = #localhost PORT = 8888 def recv_all(sock, length): data = while len(data) < length: more = sock.recv(length - len(data)) if not more: raise EOFError( socket closed %d bytes into a %d-byte message % (len(data), length)) data += more return data s.setsockopt(socket.sol_socket, socket.so_reuseaddr, 1) s.bind((host, PORT)) s.listen(1) #listen 1 client only while True: print Listening at, s.getsockname() sc, sockname = s.accept() #wait here until there is a request 5

6 #get the length of the message first msg_length = int(recv_all(sc, 3)) #get the real message with proper length message = recv_all(sc, msg_length) print Client said:, repr(message) msg_to_send = message + + str(datetime.now()) msg_length_in_str = str(len(msg_to_send)) msg_length_in_str = msg_length_in_str.zfill(3) sc.sendall(msg_length_in_str + msg_to_send) break sc.close #TCPClient - Simple TCP client which sends 16 octets import socket, sys s = socket.socket(socket.af_inet, socket.sock_stream) HOST = #localhost PORT = 8888 def recv_all(sock, length): data = while len(data) < length: more = sock.recv(length - len(data)) if not more: raise EOFError( socket closed %d bytes into a %d-byte message % (len(data), length)) data += more return data s.connect((host, PORT)) msg = raw_input( Please enter your message: ) #determine the message length (max 255 characters, i.e. 3 digits), pad with leading zeroes msg_length_in_str = str(len(msg)) msg_length_in_str = msg_length_in_str.zfill(3) s.sendall(msg_length_in_str + msg) #add the length at the beginning of the message msg_length = int(recv_all(s, 3)) message_from_server = recv_all(s, msg_length) print Server said:, repr(message_from_server) s.close 6