TDC 375-701                   Student Name:
Quiz 2
2009-09-23


1. Classful addressing

   ___  resulted in an efficent allocation of addresses

   ___  is a useful technique for mitigating DDoS attacks

   _x_  has been made obsolete with the advent of CIDR

   ___  has made classless addressing obsolete



2. What is the directed broadcast address for 192.0.2.5/26

   ___ 192.0.2.255

   _x_ 192.0.2.63

   ___ 255.255.255.255

   ___ 192.0.2.31



Q. Convert 10101011.01100010.00101101.01110110 to dotted decimal
   notation.

   171.98.45.118


Q. From Steve Deering's IETF 51 Hourglass presentation, which,
   according to him, has clearly been the most damaging to the
   the Internet architecture/model?

   ___ IP in IP tunneling

   ___ IPv6

   _x_ Network address translation (NAT)

   ___ A thin network layer (waist)


Q. A network adminstrator has been assigned an address block of
   192.78.0.0/22.  The administrator wants to divide this space
   into three networks that can support at least 100 hosts, two
   networks that support at least 32 hosts and one network that
   supports 5 hosts.  Perform the following:

   a. Specify each of the sub netblocks in slash (/) notation.
   b. Specify the range of addresses for each sub netblock.
   c. Specify the number of usable hosts for each sub netblock.
   d. Show any remaining address space that is left unused.

   NOTE: I screwed up the netblock originally on the quiz in class.
         Technically you could do this if by "networks" you don't
         mean separate router subnets, but just by way of dividing
         up the address space.  That way you wouldn't lose two
         addresses due to subnetting each time (but it probably
         means you're supporting lots more networks and hosts on
         a single router interface than you may otherwise want, so
         its a trade-of).  I meant to have you create separate
         router networks so while I typed it up as /23 originally,
         I should have said /22 (which is how its written now).  Its
         only fair I give you some leeway in grading right?  So if it
         was clear you were on the right track and knew how to do
         subnetting (which was pretty obvious in my estimation) I
         gave full credit even if all parts weren't perfectly correct.

   There are potentially multiple ways to solve this, but this is
   one possible way:

   part a:
     Three nets that support at least 100 hosts:
       192.78.0.0/25, 192.78.0.128/25, 192.78.1.0/25
     Two nets that support at least 32 hosts:
       Note, /27 is too small if we are using routed subnets, we will
       lose two addresses, one for the network and one for the directed
       broadcast address. 
       192.78.1.128/26, 192.78.1.192/26 
     One net that supports at least 5 hosts:
       192.78.2.0/29
   part b:
     192.78.0.0/25   = 192.78.0.0   - 192.78.0.127,
     192.78.0.128/25 = 192.78.0.128 - 192.78.0.255,
     192.78.1.0/25   = 192.78.1.0   - 192.78.1.127,
     192.78.1.128/26 = 192.78.1.128 - 192.78.1.191,
     192.78.1.192/26 = 192.78.1.192 - 192.78.1.255,
     192.78.2.0/29   = 192.78.2.0   - 192.78.2.7
   part c (assuming each netblock is its own routed subnet, subtract
           two addresses, one for "the network" address and one for the
           directed broadcast address):
     /25 = 128 - 2 = 126
     /26 =  64 - 2 = 62
     /29 =   8 - 2 = 6
   part d:
     192.78.2.8 - 192.78.2.255,
     192.78.3.0/24
     (not making any further assumptions about routed subnet sizes):
       (256-8) + 256 = 504 addressses
    
$Id: quiz2.txt,v 1.3 2009/09/28 22:55:27 jtk Exp $