Comp 443 Midterm Exam Information, Fall 2008

The midterm, October 15, will cover the following sections (as numbered in the third and fourth edition of the text). Sections in bold are the most important.

1.2  Requirements
1.3  Network architecture (eg layers)
1.5  Performance
2.1  Hardware Building Blocks (Nodes and Links)
2.2  Encoding: Manchester, 4B/5B, etc
2.3  Framing: byte-oriented, bit-oriented, Sonet
2.4  Error Detection
2.5  Sliding windows
2.6   Ethernet (see also my ethernet notes)
3.1  Switching and Forwarding (but not virtual circuits)
3.2  Bridges and LAN switching
4.1  Basic IP, ARP
4.2.1 Routing-table basics
4.2.2  Distance Vector/RIP
4.3.1  Subnets
5.1 UDP
5.2 TCP (through 5.2.4)

The following exercises from the book should give you a fair idea of what to expect on the exam; you shouldn't necessarily complete every detail of these, but you should study the text until you're reasonably confident you know how to approach them. The following exercise numbers are the same for both the 4th edition and the 3rd edition (I'm not sure this is always the case, but it's true for these).

Solutions are here.

Chapter 1:
  #22    Network delivery times 
Chapter 2:
  #36    Sliding windows and intermediate routers
  #37    Sliding windows and intermediate router queues
  #39    Ethernet minimum packet size
  #47    Ethernet collision timeline
Chapter 3:
#3    datagram forwarding
  #4     datagram forwarding with DEFAULT entry
  #15   Ethernet learning bridges.  Some problem like this will be on the exam.
Chapter 4:
  #1     IP addresses are per host, not per machine
  #2     Fragmentation, header layout issues
  #15   IP distance-vector routing, without link breaks.
  #18   IP forwarding, sort of backwards
  #20   IP distance-vector routing, with link breaks. Some problem like this will be on the exam.
  #40   Subnets
Chapter 5:
   #2   TFTP-style connection failure
   #8  sequence number rollover
   #14  Duplicate SYN

Here are two further study problems:

1. The following problem deals with IP routing tablesas maintained by routers; machines A, B, C, D, etc. are routers. No hostmachines are shown. The special machine DEFAULT is also a router, but youneed not give its routing table (it represents a default destination).Nets involved are all class C, with addresses 200.0.5, 200.0.6, 200.0.7,.... The following convention is used: machine A always has host portionof its address equal to 1; e.g. 200.0.5.1, 200.0.6.1, etc. Similarly, Bhas host portion 2, C has 3, D has 4. The special machine DEFAULT has hostportion 100.

(a). Give the routing tables for the following connections. You may use symbolic names (A, B, C...) for routers instead of IP addresses. Use default routes whenever possible, but be sure that a packet destinedfor some net other than 200.0.x gets routed to machine DEFAULT.

  net               net               net                 net
200.0.5____A_____200.0.6_____B______200.0.7_______D____200.0.8______DEFAULT
                             |                                                                                                                   200.0.8.100
                        net 200.0.9
                             |
                             C
                             |
                        net 200.0.10

b. Do the same for the following configuration.
                               B
                         /            \
        net 200.0.5             net 200.0.6
     /                                           \
    A                                             D------------200.0.9
     \                                            /                              \
       net 200.0.7                net 200.0.8                        DEFAULT
                        \             /
                               C

c. Suppose two routers, A and B, have tables as below. What will happen to an IP packet sent from A to address 147.126.4.9?

     200.0.5----A----------------------B----200.0.6

A: ___________________              B: ___________________
   200.0.5  |  direct                  200.0.6  |  direct
   default  |   B                      default  |   A


2. In real implementations of ARP, hosts are allowed to extract address mapping info from any broadcast ARP query packet: every machine sending such a packet includes its own IP-to-physical address binding info, and every machine receiving such a broadcast (whether or not intended for that machine) adds the source IP-to-physical address info to its ARP cache. Thus, in the example above, not only would A get B's address info but also every machine on the net would get A's address info.

(a). Explain why this means that if
   1. A broadcasts an ARP query "where is B?"
   2. A sends B a regular IP packet
   3. B wants to send an IP packet in reply to A
then A's physical address will already be in B's ARPcache.

(b). Suppose A broadcasts a request "where is B", but inadvertently lists the physical address of another machine C instead of its own (i.e. the ARP packet has IP src=A, phys src = ethernet addressof C).

What will happen? Specifically, will A get a reply? What entries will be made in the ARP caches on A, B, C, and a 4th machine D?

Suppose D uses its newly updated cache to send to A. Will the packet arrive at A? What if C tries to send to A?


3. Suppose A and B create a TCP connection with ISNA=2000 and ISNB=5000. A sends three 1000-byte packets, and B ACKs each. Then B sends a 1000-byte packet to A and terminates the connection. List all packets, together with SEQ and ACK fields. The table below does not provide space for all ACKs at the end.

A sends   

A sends B Sends
SYN, ISNA=2000
SYN, ISNB=5000, ACK=_____________
ACK, SEQ=___________, ACK=_____________
Data1, SEQ=___________, ACK=_____________
ACK, SEQ=____________, ACK=____________
Data2, SEQ=___________, ACK=_____________
ACK, SEQ=____________, ACK=____________
Data3, SEQ=___________, ACK=_____________
ACK, SEQ=____________, ACK=____________

DataB, SEQ=___________, ACK=_____________
FIN, SEQ=____________, ACK=____________