Homework 6 Solution

Problem 1 (65+10): Addressing

(5) problem 39 on page 478
The binary representation of the mask 255.255.240.0 is 11111111.11111111.11110000.00000000. Therefore, the host portion is 12 bits. 2^12=4098. Subtract from it the all-zero for itself and all-one for broadcasting, this mask can support 4098-2=4096 hosts.
(10) A server has an IP address of 160.36.30.110, network mask of 255.255.254.0, derive the broadcast address, network ID, and the number of hosts that be supported on this network. Show details.
The binary representation of the IP address is 160.36.00011110.110 and t he binary representation of the mask is 255.255.11111110.0, indicating the network portion of the mask is 23 bits. Therefore,
the broadcast address is 160.36.31.255 (all host bits are zero)
the network ID is 160.36.30.0 (all host bits are zero)
the number of hosts supported is 2^9-2=510
(15) Page 478, problem 43
First, represent the IP addresses in binary format:
135.46.56.0/22 ==> 135.46.00111000.0/22 for interface 0
135.46.60.0/22 ==> 135.46.00111100.0/22 for interface 1
192.53.40.0/23 ==> 192.53.00101000.0/23 for Router 1
For the IP addresses asked for:
(a) 135.46.63.10 ==> 135.46.00111111.10, has to go through interface 1
(b) 135.46.57.14 ==> 135.46.00111001.14, has to go through interface 0
(c) 135.46.52.2 ==> 135.46.00110100.2, doesn't have common network ID for any interfaces above, therefore, go through Router 2
(d) 192.53.40.7 ==> 192.53.00101000.7, has to go through Router 1
(e) 192.53.56.7 ==> 192.53.00111000.7, doesn't have common network ID with any interfaces above, therefore, take the default which is Router 2
(15) problem 41 on page 478
This question is basically asking what are the common network ID of the four IP addresses given.
57.6.96.0/21 ==> 57.6.01100000.0/21
57.6.104.0/21 ==> 57.6.01101000.0/21
57.6.112.0/21 ==> 57.6.01110000.0/21
57.6.120.0/21 ==> 57.6.01111000.0/21
Therefore, the aggregate address is 57.6.01100000/19
(20) Page 478, problem 40 (pay attention to the "in that order" requirement)
For organization A, which requested 4000 IP addresses, the number of bits for host portion would be at least 12, i.e., 2^12=4096. Therefore,
For organization B, which requested 2000 IP addresses, the number of bits for host portion would be at least 11, i.e., 2^11=2048. Therefore,
- its network mask should be 198.16.00010000/21, or 198.16.16.0/21
- its first IP address assigned is 198.16.16.1
- its last IP address assigned is 198.16.00010111.11100001 or 198.16.23.225
For organization C, which requested 4000 IP addresses, the number of bits for host portion would be at least 12, i.e., 2^12=4096. Therefore,
- its network mask should be 198.16.00110000/20, or 198.16.48.0/20
- its first IP address assigned is 198.16.48.1
- its last IP address assigned is 198.16.00111111.10100001 or 198.16.63.161
For organization D, which requested 8000 IP addresses, the number of bits for host portion would be at least 12, i.e., 2^13=8192. Therefore,
- its network mask should be 198.16.01100000/19, or 198.16.96.0/19
- its first IP address assigned is 198.16.96.1
- its last IP address assigned is 198.16.01111111.01000001 or 198.16.127.65
(+10)AICIP lab uses LDAP (a directory management software) to manage the user accounts. User can login from any computer in the lab and get to his/her home directory maintained at the server. LDAP manages host computers using hostname. All the lab computers use DHCP The server, aicip.ece.utk.edu sits in Ferris Hall There needs to be a "hosts.allow" file and an "exports" file to tell LDAP which computers are managed from which domain. Question: when the lab is moved from ferris hall to SERF (only user computers), none of the users can login. Discuss the problem and provide solutions to the problem.
```
/etc/exports
/home   160.36.30.0/255.255.254.0(rw)
```
```
/etc/hosts.allow
ALL: 127.0.0.1
ALL: 160.36.30.
sshd:ALL
sendmail:ALL
```
Because of the subnet changes, two steps need to be carried out:
```
/etc/exports
/home 160.36.30.0/255.255.254.0(rw) 160.36.40.0/255.255.252.0(rw)
```
```
/etc/hosts.allow
ALL: 127.0.0.1
ALL: 160.36.30. 160.36.40.
ssh: ALL
sendmail:ALL
```

Problem 2 (20): Fragmentation

Problem 34 on page 477
We need to understand the definition of fields first:
Total length: total IP packet length, not including Ethernet header
DF: if 1, indicating fragment is not allowed
MF: if 1, indicating more fragments are coming
offset: has to be multiple of 8 bytes
Between A and R1
Between R1 and R2
- The link can support 512-8=504 bytes. So fragmentation is needed.
  Total Length: 464 (msg len) + 20 + 20 = 504 bytes
  ID: x
  DF: 0
  MF: 1
  Offset: 0
- Total Length: 900-464 (msg len) + 20 + 20 = 476 bytes
  DF: 0
  MF: 0
  Offset: 504 bytes
Between R2 and B
- The link can support 512-12=500 bytes. So fragmentation is needed. In addition, we have to keep in mind that the offset has to be multiples of 8 bytes
  Total Length: 436 (msg len) + 20 + 20 = 496 bytes
  ID: x
  DF: 0
  MF: 1
  Offset: 0
- Total Length: 436 (msg len) + 20 + 20 = 496 bytes
  DF: 0
  MF: 1
  Offset: 496 bytes
- Total Length: 900-436*2 (msg len) + 20 + 20 = 68 bytes
  DF: 0
  MF: 0
  Offset: 992 bytes

Problem 3 (15): tcpdump

Requirement: turn in the dump result of ping. Identify each field of the IP header and ICMP header. Note that there are many options provided by 'tcpdump'. Use the best combinations of options so that your dump result is easy to read and not redundant.

See lecture slides