1.2 The first Internet (i.e., one of the first packet switched networks) was referred to as the ARPANET.

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1 CPSC 360 Spring 2011 Exam 1 Solutions This exam is closed book, closed notes, closed laptops. You are allowed to have one 8.5x11 sheets of paper with whatever you like written on the front and back. You may also use a calculator. Name: Section 1 True/False (20 points) 1.1 A host can be a router but a router is not considered a host. False. A host can be considered a router if it forwards packets from one interface to another. A router is ALWAYS considered a host. However, the router might have a different set of network applications active than a general purpose computer. 1.2 The first Internet (i.e., one of the first packet switched networks) was referred to as the ARPANET. True (19.4.1) 1.3 Shannon s Theorem tells us that cleverly encoding data bits to signal state changes allows more bits to be transmitted per second. False this is Nyquist s Theorem. Shannon s theorem says there is an absolute limit of the number of bits per second that can be transmitted in a real system. 1.4 The Internet s best effort datagram service is not the best network model to support telephony. True - the current circuit switched telephone network is much better suited. 1.5 An application such as ssh is best suited for the UDP transport protocol rather than TCP. False 1.6 An IP packet header is usually 20 bits large. False 1

2 1.7 The process of forwarding at a router means that a frame that arrives on one interface is transferred and sent out another interface. False 1.8 TCP/IP needs to know if it s operating over a wireless link so that it can correctly perform error correction. So False 1.9 An application needs to know if it will use UDP or TCP. True 1.10 The main difference between a UDP socket recv (recvfrom) and a TCP socket receive (recv) is that the latter requires the application to loop forever until all expected data is received OR until an error occurs. True 2. Short answer questions (50 points) 2.1 Television channels use a bandwidth of 6 Mhz. What is the maximum possible data rate if four-level digital signals are used in the transmission? Assume ideal, noiseless channels. Make sure your answer is in units of bits per second. Solution: Nyquist says: D=B log2(4) = 2 * *2 = 24,000,000 bps 2.2 What is the largest size IP datagram that is allowed? Solution: the total length field is 16 bits. So the largest datagram is 2EXP16 or bytes. 2.3 Given the following IP address: Write the address in binary Solution:

3 -What is the network directed broadcast address? Please put your answer in dotted decimal format. Solution: What percentage of the IPV4 IP address space do all class C addresses consume? Solution: There are 2EXP21 class C networks, each offers 256 addresses. So the fraction of address consumed by class C is 2EXP29/2EXP32 = 2EXP-3 or 12.5% 2.5 The UDPEcho program sends a UDP message of 1472 bytes. If we add 8 bytes of overhead for UDP and 20 bytes of overhead for IP, the IP datagram is size 1500 bytes. Host A sends this datagram to Host B, and Host B echoes back the data. This is performed exactly one time. How many frames in total are sent (by all hosts/routers)? You should account for any ARP messages. Assume that ALL ARP caches are empty prior to the first transmission. Assume no packets are dropped. And assume there is no other traffic generated over the network. To receive partial credit you must show me your thinking draw arrows representing frames that are sent. For 5 points extra credit, identify exactly how large every IP packet is. There s no partial credit for this you need to correctly identify the size of EVERY IP packet. Host A LAN (MTU1500) R Network (MTU 1000)------Host B Solution Host A LAN (MTU1500) R Network (MTU 1000)------Host B ARP who is R ARP I am R IP Datagram (1500 bytes) ARP who is Host B < ARP I am Host B IP Packet (1000 bytes) IP Packet (520 bytes) < IP Packet (1000 bytes) < IP Packet (520 bytes) < IP Packet (1000 bytes) < IP Packet (520 bytes) So 11 frames are sent all together. 3

4 3 Socket program (30 points). The questions are based on the UDPEcho2.tar.gz package. 3.1 This is a snippet of code from UDPEchoClient2.c that fills in the server address structure. //BEGIN CODE SNIPPET /* Construct the server address structure */ memset(&echoservaddr, 0, sizeof(echoservaddr)); /* Zero out structure */ echoservaddr.sin_family = AF_INET; /* Internet addr family */ echoservaddr.sin_addr.s_addr = inet_addr(servip); /* Server IP address */ if (echoservaddr.sin_addr.s_addr == -1) thehost = gethostbyname(servip); echoservaddr.sin_addr.s_addr = *((unsigned long *) thehost->h_addr_list[0]); echoservaddr.sin_port = htons(echoservport); /* Server port */ //END CODE SNIPPET Look at the very last line in the code snippet. echoservaddr.sin_port = htons(echoservport); /* Server port */ Correction to this question: Assume the server port that is specified is 0x0102. When the frame containing the first IP packet (that contains the UDP message) eventually gets sent, which octet gets transmitted first over the channel, the 0x01 or the 0x02? Solution: The htons puts the port number in network byte order (big endian). This means the higher order byte is sent first. So 0x01 gets sent first followed by the 0x I run the client on LinuxVM1 as follows:./client LinuxVM The 4 parameters are: Server name Port number Delay between iterations Size of message. I allow the client to run for 10 iterations and then issue a CNT-C to terminate the program. We see the following displayed by the client: Ping(1): 2860 microseconds 4

5 Ping(2): 353 microseconds Ping(3): 361 microseconds Ping(4): 556 microseconds Ping(5): 355 microseconds Ping(6): 349 microseconds Ping(7): 399 microseconds Ping(8): 332 microseconds Ping(9): 347 microseconds Ping(10): 364 microseconds Avg Ping: 627 microseconds Loss: 0 Percent Question: Why is the first RTT time so much larger than the other RTT times? Solution: The first iteration requires ARP which adds hundreds of microseconds to the observed application RTT. 3.3 Add to the UDPEchoServer code (shown below) a printf after the recvfrom showing the client s IP address in dotted decimal format AND the client s port number in decimal format. The UDPEchoServer.c code is shown below. Add the printf in this code. HINT: Look at the code snippet in question 3.1 for help. int main(int argc, char *argv[]) int sock; /* Socket */ struct sockaddr_in echoservaddr; /* Local address */ struct sockaddr_in echoclntaddr; /* Client address */ unsigned int cliaddrlen; /* Length of incoming message */ char echobuffer[echomax]; /* Buffer for echo string */ unsigned short echoservport; /* Server port */ int recvmsgsize; /* Size of received message */ if (argc!= 2) /* Test for correct number of parameters */ fprintf(stderr,"usage: %s <UDP SERVER PORT>\n", argv[0]); exit(1); echoservport = atoi(argv[1]); /* First arg: local port */ /* Create socket for sending/receiving datagrams */ if ((sock = socket(pf_inet, SOCK_DGRAM, IPPROTO_UDP)) < 0) printf("failure on socket call, errno:%d\n",errno); 5

6 /* Construct local address structure */ memset(&echoservaddr, 0, sizeof(echoservaddr)); /* Zero out structure */ echoservaddr.sin_family = AF_INET; /* Internet address family */ echoservaddr.sin_addr.s_addr = htonl(inaddr_any); /* Any incoming interface */ echoservaddr.sin_port = htons(echoservport); /* Local port */ /* Bind to the local address */ if (bind(sock, (struct sockaddr *) &echoservaddr, sizeof(echoservaddr)) < 0) printf("failure on bind, errno:%d\n",errno); for (;;) /* Run forever */ /* Set the size of the in-out parameter */ cliaddrlen = sizeof(echoclntaddr); /* Block until receive message from a client */ if ((recvmsgsize = recvfrom(sock, echobuffer, ECHOMAX, 0, (struct sockaddr *) &echoclntaddr, &cliaddrlen)) < 0) DieWithError("recvfrom() failed"); /* Send received datagram back to the client */ if (sendto(sock, echobuffer, recvmsgsize, 0, (struct sockaddr *) &echoclntaddr, sizeof(echoclntaddr))!= recvmsgsize) DieWithError("sendto() sent a different number of bytes than expected"); Solution: Correction to this question: It turns out the hint is not in question 3.1. printf("handling client with an IP address of %s and a port number of: %d\n", inet_ntoa(echoclntaddr.sin_addr), ntohs(echoclntaddr.sin_port)); 6