Midterm I, CIS 656, 02/19/2005 Dr Ott Use only the paper provided. Name on every sheet. Save a margin for Mr Jain and Dr Ott! Some of the information I give you will not need. MTU sizes: PPP 296, X.25 576, ethernet 1,500, FDDI 4,534, Hyperchannel 65,535. In an ICMP Echo-request, the type is 8 and the code is 0. In an ICMP Echo-response, the type is 0 and the code is 0. In an ICMP Time-exceeded error reporting message, the type is 11. In this message, the code is 0 if a router reports that the TTL decreased to zero. The code is 1 if a (destination) host reports that some but not all fragments arrived, and the packet timed out. Whenever you discuss checksums or CRC fields, only tell me what fields it is computed over. No details of how the computation is done. Inside NJIT, assume all LANs are ethernet (most or all are 10/100Base-T) Remember: 127.0.0.0/8 are all loop-back addresses. 10.0.0.0/8, 172.16.0.0/12, and 192.168.0.0/16 are set aside for addresses in private networks. 1. Draw the header of an IP packet without options. For each field, give the size and a BRIEF explanation of its use, role, or function. Where appropriate, give the units in which the content is expressed. (Like liters, gallons, barrels, msec, etc.) 2. NJIT has a subnetwork with address 128.235.32.0/22 . For each of the following addresses, tell whether it can be: interface address on this subnet / source address of packet on this network / destination address of packet on this network. (e.g. ``yes / no / yes''). Give a BRIEF explanation, just enough to convince us you did not just guess. Feel free to use the exam sheet for your answer. (Next sheet) (i) 128.235.32.1 / / / / (ii) 128.235.31.2 / / / / (iii) 128.235.33.255 / / / / (iv) 128.235.35.255 / / / / (v) 128.235.36.255 / / / / (vi) 128.236.32.3 / / / / (vii) 0.0.0.0 / / / / (viii) 255.255.255.255 / / / / (ix) 224.0.0.1 / / / / (x) 128.235.33.34 / / / / (xi) 127.0.0.127 / / / / (xii) 0.0.3.3 / / / / (xiii) 128.235.32.0 / / / / (xiv) 10.10.10.10 / / / / (xv) 11.11.11.11 / / / / 3. An IP packet with Total Length TL = 4532, without options, with DF = 0, M = 1 , FragOffset = 564, TTL = 11, Identification 12345, arrives (from the outside) on a router on the subnet in problem 2 above. It has destination address 128.235.32.8 . The router will do a ``direct delivery'' but first has to fragment the packet. a. How many databytes does the incoming IP packet have? b. How many fragments will there be? (make the ``most reasonable'' assumption about sizes). c. For each of these fragments, give: TL, Identification, DF bit, More bit, Fragmentation Offset, TTL, Number of databytes. 4. In problem 3, explain how the fragment arriving at the router (no options, TL = 4532, Identification 12345, DF = 0, M = 1, FragOffset = 564, TTL = 11) probably came into existence. Keep it clear and concise and short! ---- Problems 1, 2, 3 are straight from notes or book, or have been seen in homework. Model response to problem 4: The packet has fragmentation offset 564. Hence, it is the result of previous fragmentation. (This also follows from M=1 ). The fragmentation offset shows there are 564*8 = 4512 data bytes in previous fragments. That is the same as in the current fragment! (4512 + 20 = 4532). That can hardly be a coincidence. Look in the list of MTUs: FDDI has 4534. The previous fragmentation must have occurred when entering an FDDI WAN. (If you say LAN it will not be counted as wrong. FDDI used to be used as LAN as well as WAN.) That packet or fragment must have had more than 9026 data bytes. In the fragmentation there: The first fragment could get at most 4534 - 20 = 4514 data bytes. But since there was more data following, it had to get a multiple of 8 data bytes. The highest multiple of 8 that is .LEQ. 4514 is 4512. So, the first fragment there got 4512 data bytes, Total Size 4512 + 20 = 4532, M = 1, frag_offset 0. The second fragment there is the fragment we now look at. It got frag offset 4512/8 = 564, and also 4512 data bytes, total size 4532, and M=1. We know that after this fragment there are at least 3 more data bytes. Because, if there were only two more data bytes this fragment would have received them, with M=0 (no more fragments), 4514 data bytes (if there is no next fragment the number of databytes need not be a multiple of 8), and Total Length 4534. (Similar if there were only 1 more data byte.)