Study Guide for Midterm 1 CSC/ECE , Fall, 2012

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1 Study Guide for Midterm 1 CSC/ECE , Fall, 2012 The focus of this midterm will be on the IP and transport layer protocols. Together with IP, the functions which are required for lower layer interfacing, i.e. ARP and fragmentation, will also be covered. At the transport layer, protocol operation details of UDP and TCP will be the focus. Questions may make reference to general protocol operation, or specific issues by making reference, explicitly or implicitly, to specific RFCs, as in the sample questions discussed in class and presented in the slides entitled Samples of descriptive problems. There will also be reflective questions on Internet architecture in general. Questions may be posed on their own, or refer to specific papers (see below). Details of the type of question to be expected for the various types are below. IP/ARP Details of the mechanism of the IP forwarding engine. IP header structure, significance of various fields. The IP forwarding table address classes, CIDR, matching forwarding rules. ARP and RARP, fragmentation. Representative questions: What class is the IP address ? Examining the binary form of the first octet, , we see that it starts with 10, so it falls into Class B. What class is the IP address ? Such an IP address is impossible, of course none of the octets can have value more than 255. (Adapted from Tanenbaum, 5.37) Suppose that instead of using 24 bits for the network part (and remaining 8 bits for host part) of a Class C address originally, 20 bits had been used (and remaining 12 for host part). (a) How many Class C networks would there have been? (b) How large could a Class C network be? (c) What would be the lowest IP address and what would be the highest IP address in Class C? (a) With a 20 bit network part, of which the first 3 bits are fixed (to indicate Class C),

2 the number of networks would have been 2 17, or (b) The number of hosts each network could contain would be or just over 4000 possibly a much more attractive size than 254, which might have led to better utilization of the address space. (c) The lowest and highest Class C address would not change they would still be, respectively, the two 32- bits numbers beginning with 110, followed by all zeros and all ones. That is, they would be and , respectively. (If you took into account the fact that all zeros and all ones are not valid as host addresses, and said and , this is perfectly acceptable also.) (Tanenbaum, 5.38) Convert the IP address whose hexadecimal representation is C22F1582 to dotted decimal notation. C2 2F can be converted to binary number, but much easier is to first separate the four octets, then convert C216 = 12(16)+2 = 19410, etc. The IP address in dotted decimal format is A block of CIDR addresses start with What is the maximum size of this block? What is the minimum size? Clearly, the last 16 bits are zeros in the starting address, but the 17 th LSB (LSB of second octet) is clearly a 1 (this can also be verified in general by writing this number as a 32- bit binary number). Thus this address is aligned on a 2- boundary, 4- boundary, boundary, but not a boundary. Thus the largest such block would be a block of 2 16, or addresses. The minimum block could conceptually be as small as one (a 32 bit mask). Because CIDR blocks are meant to be non- overlapping, very small block sizes tend to fragment the address space, and in practice some address allocation authorities may refuse to grant very small blocks 8 is sometimes cited as the minimum block size. Some authors also state that /31 and /32 masks are impossible because such networks, after excluding the all zeros and all ones addresses, have no valid IP address but the relevant RFC actually allows /31 for point- to- point links, and top- level address allocation authorities appear to define address blocks all the way to /32. Any of these answers is okay, especially if you also cite the source where you found your information. Consider the IP address whose decimal representation is 3,257,865,600. What is the largest block of CIDR addresses that might start with this address? Writing this number as a 32- bit binary number shows that the least significant 7 bits are zeros, but the 8 th is a 1. Thus this address is aligned on a 2- boundary, 4- boundary, 128- boundary, but not a 256- boundary. Thus the largest such block would be a block of 128 addresses.

3 A router R needs to forward packets destined to IP addresses between and (inclusive) to router A, packets destined to IP addresses between to to router B, and packets destined to IP addresses between to to router A. What is the minimum number of routes that can be included in the forwarding table of R in order to realize this, and what are they? Writing the various addresses in binary helps us see the boundary alignments again, we only do this for the third octet only, since the first two do not vary over this range, and the last octet is either all zeros or all ones in the addresses given above. Then the desired routing is: A B A It is clear that the biggest block that can start from is a /19 block, which ends at This gives us the first rule in Table 1 below. The next starting address is the one immediately after the end of this block, that is , which is aligned on a /18 block. But we only need to go up to , so we use a /21 block, giving us the second rule of Table 1. From the next immediate address after the end of this block, , we need to create a block till , which is exactly a /21 block, which gives us the next rule. Finally, from , which is aligned on /20, the largest block we can create is the /20 block, which ends at , which is exactly what we need, hence the fourth rule. In a particular physical network using IP subnetting, a specific host is seen to have an IP address of The physical network has 100 hosts, and the subnet mask used for this network is the tightest, or most specific possible (i.e., has the largest number of 1 bits), given this number of hosts. (a) Write down the subnet mask being used, in dotted decimal form. (b) How many bits of the IP address are being used for the subnet part? (a) For this part we only need the number of hosts in the network. The smallest

4 power of 2 which is greater than 100 is 2 7 = 128. So the mask will have only 7 zeros at the end for the host part, since we are told it is the tightest possible mask. (Incidentally, this is usually considered good design.) A detail is that the two special cases (all zeros and all ones) need to be excluded, but = 126 is still larger than 100, so this works. In slash notation, this mask is /25 (since 32-7= 25), in dotted decimal notation it is (b) We know where the subnet part ends (7 bits from the LSB), but to answer this part we need to know where it begins also. Since nothing is mentioned about CIDR, we assume that this network only uses subnetting on top of classful addressing. Examining the first few bits of the first octet (which is ), this address is in Class B space. Hence the net ID consists of the first two octets and the subnet part must begin after that. Therefore it is 9 bits long. Consider the following rule: Destination Address Netmask Metric Interface Next- Hop Router /22 10 eth0 B What is the range of destination IP addresses that match this rule (i.e., what is the lowest IP address that matches this rule, and what is the highest)? The lowest IP address matching any rule can be found by taking the destination network address part of the rule and masking it with the netmask, i.e. setting all bits of the host part to zero and leaving all other bits unchanged. (For a valid rule, this is always the same as the destination network address, otherwise nothing would ever match this rule!) The highest IP address is found by setting all bits of the host part to one and leaving all other bits unchanged in the destination network address. Accordingly, the lowest and highest addresses matching the given rule are and , respectively. UDP/TCP General transport functions, protocol details of UDP and TCP, header structure, significances of header fields. The concept of sockets, typical socket functions, clientserver socket operation. TCP three-way handshake, congestion control mechanism. Representative Questions: (Adapted from Tanenbaum, 6.14) Imagine a case where we do not want any of the value added features of TCP reliability, ordered delivery, or congestion control. Why do we need something like UDP? Is it not sufficient to let user applications have direct access to IP, and allow applications to use IP to inject raw datagrams?

5 The basic function of the transport layer would still be needed that is endpoint abstraction (ports; multiplexing multiple applications). IP only distinguishes between hosts. A TCP endpoint can get into the SYN_RCVD state by receiving a SYN segment. There is another way a TCP endpoint can get into that state what is it? Through the simultaneous open feature of TCP. If an endpoint sends the SYN, and is in the SYN_SENT state, it might receive a SYN segment from its intended destination endpoint (not a SYN- ACK) if the other end was simultaneously trying to open a socket. Then this endpoint would transit to the SYN_RCVD state. (Adapted from Tanenbaum, 6.29) Suppose that a TCP sender s congestion window is set to 18 KB and a timeout occurs. Assume that the maximum segment size for this connection is 1 KB. The next four transmission bursts are all successful. What will be the congestion window after this? The congestion window after the timeout will be 1 MSS, i.e. 1 KB, and the sender will be in slow- start: one segment will be transmitted. If this is successful, after one RTT, the ACK will be received, and the window will become 2 KB, now two segments will be transmitted. Similarly after the next two successful transmissions, and receiving corresponding ACKs, the congestion window will be 4 and 8 KB respectively. It will stay at 8 KB until the ACKs for the fourth burst of successful transmissions are received. One thing to observe here is that TCP did not get into the congestion avoidance regime since the last congestion window before a timeout was 18 KB. Assume TCP is sending segments using a maximum window size (64 KB) on a channel that has infinite bandwidth and an average roundtrip time of 20 ms. What is the maximum throughput? How does the throughput change if the roundtrip time increases to 40 ms? The key insight in this question is that after probing the bandwidth, if congestion does not limit the throughput, TCP will settle down to the throughput allowed by flow control. However, in flow control the data transfer is always advanced by the ACKs (for TCP, non- zero window advertisements in conjuction with ACKs), so this will end up determining the final throughput. Since the maximum window size is 64KB, no more than 64KB can be sent over the wire until an ACK is received for the first byte within this window, so that the window can be advanced by one byte. Hence, only 64KB can be sent every 20 ms. Hence, the throughput is found to be 25.6Mbps. The throughput will decrease when the RTT increases to 40 ms since 64KB will be sent every 40 ms in this case. Hence, the throughput in this case will be: 12.8Mbps. Without knowing the values of cwnd or window advertisement, by only observing

6 the sequence numbers of successive segments transmitted, is it possible to determine whether flow control or congestion control is operative at any given time in a connection? No, without knowing the window advertisement it is not possible to determine if flow control is in use or not. Similarly, cwnd, the sequence numbers of segments transmitted and the window advertisements all need to be known to determine if congestion control is operative or not. (Tanenbaum 6.28) Consider the effect of using slow start on a line with a 10 ms round- trip time and no congestion. The receive window is 24 KB and the maximum segment size is 2 KB. How long does it take before the first full window can be sent? We know that in slow start, the congestion window increases by one every time an ACK is received, and thus doubles in every RTT. Initially the window is just one segment. Thus the first transmission will be of one 2 KB segment. After one RTT, the ACK for this segment is received, and the sender sends two segments back- to- back (because the congestion window has been increased to 1+1 = 2 segments, i.e. 4 KB). After one more RTT, two acknowledgments come back, and the congestion window increases to 2+2 = 4 segments = 8 KB. After one more RTT, the window is at 16 KB. After one more, the congestion window is 32 KB, but now flow control dominates and the actual transmission window becomes 24 KB. Counting the RTTs, this took 40 ms. Architecture and Retrospective Questions similar to those asked in Homework 2. The papers questions may be asked out of are [CerfKahn74, Clark84, Clark88, Barabasi01].

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