CIS 456-102, Spring 2006, Dr Ott
Midterm I, 02/17/2006.

Be clear and concise, and within the constraints of clear and concise, 
be short.

6:00 - 7:00.    Class resumes 7:10.   Name on every sheet!

1. Draw, in detail, the header of an IPv4 packet without options.  For
each field give the size in bits, give a SHORT description of its
meaning and use. Where applicable, mention what the units are it is
expressed in. (e.g., bits, Bytes, meters, miles, coulombs, seconds, etc.)

2. We have an ethernet LAN with NetID 129.129.128.0/19.
   (Don't tell me that LAN has too big an address space for a single 
   ethernet LAN: not by a long way all available adresses are in use.)

a. What is the length of the mask of this LAN? 
b. Write out that mask in dotted decimal.
c. Write out that mask in binary.
d. Write out that mask in hexadecimal.
c. What destination IPv4 address do I have to use to send a broadcast, 
   from 128.235.32.243 to all interfaces on the LAN above?
   What do we call that IP address?

For each of the IP addresses below, can it be a
interface address on this LAN /
source address in a packet on this LAN /
destination address in a packet on this LAN.
(y/y/y, n/n/n. etc, with a BRIEF explanation).

i.    129.129.160.1
ii.   129.129.145.2
iii.  0.0.0.0
iv.   129.130.146.3
v.    127.1.2.3
vi.   129.129.159.252
vii.  129.129.159.255
viii. 224.0.0.1
ix.   255.255.255.255
x.    10.1.2.3

Model answers:

a: 19
b: 255.255.255.224.0
c: 1111 1111 1111 1111 1110 0000 0000 0000
d: FFFFE000
c: 128.129.159.255 (the first 19 bits the same as the network, the last 
                    31 - 19 = 13 bits equal to one: That is the 
                    ``directed broadcast'').
e 
i:    n/y/y  because difference in 19-th bit: can be legal on other LAN.
ii:   y/y/y
iii:  n/y/n  ``I am stupid'' source address.
iv:   n/y/y  because difference in second octet.
v:    n/n/n  loopback address
vi:   y/y/y
vii:  n/n/y  directed broadcast
viii: n/n/y  Multicast address
ix:   n/n/y  limited broadcast
x:    n/n/n  private address, not allowed in public network.

3. Router R1 receives an IPv4 packet with:
   HLEN = 5, TL (Total Length) = 2020, Identifier = 54321 , 
   U = 0 (unused bit in Fragmentation Flags), DF = 0, M = 0,
   Fragmentation Offset = 541, TTL = 19.

a. How many data bytes does this packet have?

b. How many data bytes are there in preceding fragments?
   (R1 may not have seen those preceding fragments, and may never 
   see them.)

   Router R1 determines the next hop for this packet, and finds 
   the next LAN is an ethernet (MTU = 1500 Bytes).

c. How many fragments will result from this packet?

d. For each of these fragments, give HLEN, TL, Identifier, 
   DF, M, FragOffset, and TTL.

Model answers:

a:  Number of Header Bytes = 5 x 4 = 20. Therefore
    number of data bytes = 2020 - 20 = 2000
b.  8 x 541 = 4328 data bytes in preceding fragments of original packet.
c.  With MTU = 1500 and 20 header bytes the number of databytes in 
    fragments in the next LAN can be at most 1500 - 20 = 1480. 
    1480/8 = 185, 1480 is divisible by 8, so fragments CAN indeed have 
    1480 data bytes.
    Fragment 1: 1480 data bytes. Left: 2000 - 1480 = 520.
    Fragment 2:  520 data bytes. Left nothing.
    There will be 2 fragments.

        HLEN       TL            Ident  DF  M  Fragoffset       TTL

Frag 1  5     1480 + 20 = 1500   54321   0  1  541              18

Frag 2  5      520 + 20 +  540   54321   0  0  541 + 185 = 726  18

Grade Distribution:

100 , 92 , 88 , 88 , 82 , 72 , 66 , 54 , 49 , 47 , 46 , 43 , 42 , 20 , 17 , 

  13 , 4 .