Solution to Homework #2. IP address and End Host Routing due 9/17/2003

• Learn about IP address.
  Learn how end host routing works.
• Learn how to interpret the IP classless notation and routing table entries.
• Learn how to partition a subnet and assign gateway address.

Assignment Date: 9/10/2003
Due Day: 9/17/2003
Related Material: Section 5.6.0-5.6.3, and http://cs.uccs.edu/~cs522/intro/pcnetsetup/pcnetsetup.htm
Description:
Exercise 1. Classless Address Notation

• Question 5.43. A router has the following (CIDR) entries in its routing table:
  
  Address/mask    Next hop
  135.46.56.0/22   Interface 0
  135.46.60.0/22   Interface 1
  192.53.40.0/23   Router 1
  default                  Router 2
  
  For each of the following IP addresses, what does the router do if a packet with that address arrives?
  (a) 135.46.63.10 (b) 135.46.57.14 (c) 135.46.52.2 (d) 192.53.40.7 (e) 192.53.56.7
  
  Ans: (a) 135.46.63.10
  Taking the first 22 bits of 135.46.63.10 as network address, we have 135.46.60.0.
  The bit pattern of 135.46.63.10 is 10000111.00101110.00111111.00001010
  When we perform the bit and operation with 22 leading bit 1s and 10 bit 0s, it is equivalent of making the last 10 bit zero. We get the following network address bit pattern: 10000111.00101110.00111100.00000000. The first two bytes are not changed. The 3rd type changes from 63 to 60 while the 4th byte become zero.
  Match with network address in the routing table. The 2rd row matches. The router will forward the packet to Interface 1.
  
  (b) 135.46.57.14
  Taking the first 22 bits of the above IP address as network address, we have 135.45.56.0. It matches the network address of the first row. The packet will be forwarded to Interface 0.
  
  (c) 135.46.52.2
  Taking the first 22 bits of the above IP address as network address, we have 135.45.52.0. It dose not matche the network addresses of the first three rows. The packet will be forwarded to default gateway which is Router 2.
  
  (d) 192.53.40.7
  Taking the first 23 bits of the above IP address as network address, we have 192.53.40.0. It matches the network address of the third row. The packet will be forwarded to Router 1.
  
  (e) 192.53.56.7
  Taking the first 23 bits of the above IP address as network address, we have 192.53.56.0. It does not matche the network addresses of the first three rows. The packet will be forwarded to default gateway which is Router 2.

  
  

• Question 5.51. lPv6 uses 16-byte addresses. If a block of 1 million addresses is allocated every picosecond, how long will the addresses last?
  Ans: With 16 bytes, there are 2^128 or ~3.4E38 addresses. (3.4E38/(1000000))*1E-12~=3.4E20 seconds~=1E13 years.

• Problem 3. IP Address and Routing.
  Given the following routing table on blanca.uccs.edu.
  blanca.uccs.edu> netstat -rn
  Kernel IP routing table
  Destination   Gateway       Genmask       Flags MSS Window irtt Iface
  128.198.160.0 0.0.0.0       255.255.224.0 U     0   0      0    eth0
  169.254.0.0   0.0.0.0       255.255.0.0   U     0   0      0    eth0
  127.0.0.0     0.0.0.0       255.0.0.0     U     0   0      0    lo
  0.0.0.0       128.198.160.1 0.0.0.0       UG    0   0      0    eth0
  

  From the first entry of the routing table, we have destination IP address 128.198.160.0 and genmask (or called netmask) 255.255.224.0 that identifies the subnet which blanca is in.