Computer Networks - Solved Assignments - Semester Spring 2009

Assignment 1 Question # 1 (Marks: 3+4+3) Answer the following questions: a. Explain what might happen if two stations are accidentally assigned the same hardware address? b. If sharing reduces cost, why are shared networks used only for local communication? c. Why wireless LAN can not use the same CSMA/CD mechanism that Ethernet uses? Solution: a. Depending on the hardware used you generally get an intermittent failure on both devices because the network sees them both as one device. If you have intelligent hubs and network management software they can be identified and locked out. MAC addresses are configured by the hardware manufacturers in the network cards, and no two network cards have the same MAC address around the world.

b. Both technical and economic reasons contribute to the answer; we said that computers attached to a shared network must coordinate use of the network. Because coordination requires communication ad the time required to communicate depends on distance, a large geographical separation between computers introduces longer delays. Thus, shared networks with long delays are inefficient because they spend more time coordinating use of the shared medium and less time sending data. In addition, engineers have learned that providing a high bandwidth communication channel over long distances is significantly more expensive than providing the same bandwidth communication over short distance. c. In contrast with wired LANs, not all participants may be able to reach each other. Because:  It has low signal strength.  In wireless LANs the propagation is blocked by walls etc.  It can’t depend on CD to avoid interference because not all participants may hear. Question # 2 Complete given table and answer yes or no. Characteristic Mutiple access Carrier Sense Collision checking Acknowledgement

CSMA/CD

(Marks: 10) CSMA/CA

Token Ring

Solution: Characteristic Mutiple access Carrier Sense Collision checking Acknowledgement

CSMA/CD yes yes yes yes

CSMA/CA yes yes no yes

Token Ring yes no no yes

Assignment 2 Q#1 Draw an example of ‘Cycle of Bridges’ with the segments below: [5]

Segment a

Segment b

Segment c

Segment d

Segment e

Segment f

Solution: Segment a

Segment b

Segment c

Segment d

Segment e

Segment f Q#2 Answer the following questions:

i. If we can extend the LAN then why we need a WAN? 3 ii. How can a bridge know whether to forward frames? 2 iii. Can the length of an Ethernet be increased to many segments of 500 meter each merely by adding a repeater to connect each additional segment? 2

iv. How can a computer attach to a network that sends and receives bits faster than the computer’s CPU can handle them? 3 Solution: i. A bridge LAN is not considered a wide area technology because bandwidth limitations prevent a bridge LAN from serving arbitrarily many computers at arbitrarily many sites. Performance issue, an in LAN shared medium is used and only two computers are communicate at one time and where as in WAN more than two computers can communicate at one time. Scalability is another issue; a WAN must be able to grow as needed to connect many sites spread across large geographic distance, with many computers at each site. Scalability is very expensive in the extended LAN. ii. When a bridge first boots, it communication with other bridges on the segment to which it connect. The bridges perform a computation know as the distribution spanning tree (DST) algorithm to decide which bridge will not forward frames. DST allows a bridge to determine whether forwarding will introduce a cycle. in essence, a bridge does not forward frames if the bridge finds that each segment to which it attaches already contains a bridge that has agreed to forward frames. after the DST algorithm completes, the bridges that agree to forward frames form a graph that does contain any cycle. iii. The answer is no. Although such an arrangement does guarantee sufficient signal strength, each repeater and segment along the path increases delay. The Ethernet CSMA/CD scheme is designed for low delay. If the delay becomes too large, the scheme fails. In fact, repeaters are a part of current Ethernet standard, which specifies that the network will not operate correctly if more than four repeaters separate any pair of station. iv. The CPU does not handle the transmission or reception of individual bits. Instead, a special purpose hardware component connects a computer to a network, and handles all the details of packet transmission and reception. Physically, the special purpose hardware usually consists of a printed circuit board that contains electronic components. Known as a network adapter card or network interface card (NIC), the printed circuit board plugs into the computer’s bus, and a cable connects it to the network medium.

Assignment 3 Question: For the packet switched network below:

1

5

3

2

4

a. Determine the routing table? Solution: Node 1 Dest. Next hop

Node 2 Dest. Next hop

Node 3 Dest. Next hop

Node 4 Dest. Next hop

Node 5 Dest. Next hop

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

(1,2) (1,5) (1,5) (1,5)

(2,1) (2,1) (2,1) (2,1)

(3,5) (3,5) (3,4) (3,5)

(4,5) (4,5) (4,3) (4,5)

(5,1) (5,1) (5,3) (5,4) -

b. Determine the collapsed routing table using default route? Solution: Node 1 Dest. Next hop

Node 2 Dest. Next hop

Node 3 Dest. Next hop

Node 4 Dest. Next hop

Node 5 Dest. Next hop

1 2 *

2 *

3 4 *

4 3 *

5 3 4 *

(1,2) (1,5)

(2,1)

(3,4) (3,5)

(4,3) (4,5)

(5,3) (5,4) (5,1)

Assignment 4 Q.1 What is gratuitous ARP? [3] When a host sends an ARP request to resolve its own IP address, it is called gratuitous ARP. In the ARP request packet, the source IP address and destination IP address are filled with the same source IP address itself. The destination MAC address is the Ethernet broadcast address (FF:FF:FF:FF:FF:FF). Q.2 Find the net ID and host ID of the following IP Addresses. [8] a) 117.34.4.8 b) 29.34.41.5 c) 23.67.12.1 d) 139.33.4.5 Solution: a) b) c) d)

Net id: 117; Net id: 29; Net id: 23; Net id: 139.33;

host id: 34.4.8 host id: 34.41.5 host id: 67.12.1 host id: 4.5

Q.3 Find the class of the following IP Addresses. [4] a) 218.34.54.12 b) 238.34.2.1 c) 124.34.2.8 d) 139.14.6.8 Solution: a) b) c) d)

Class C Class D Class A Class B

Assignment 5 a. Show the shortest form of the following IPV6 address: 2340:1ABC:119A:A000:0000:0000:0000:0000 2340:1ABC:119A:A000::0

b. Show the original (unabbreviated) form of the following IPV6 address: 0:AA::0 0000:00AA:0000:0000:0000:0000:0000:0000

c. In IP datagram header format, what is the value of the data field given H.LEN value of 12 and TOTAL LENGTH value of 40,000? Total number of bytes in the header = 12 x 4 = 48 bytes Total length = 40,000 bytes Therefore data field = 40,000 – 48 = 39952 bytes

d. What are the main advantages of IPv6 over IPv4? The main advantages of IPv6 over IPv4 are as follows: Larger Address Space: An IPv6 address is 128 bits long. Compared with the 32-bit address of IPV4, this is a huge increase in the address space, and this space is large enough to accommodate continued growth of the worldwide Internet for many decades. Header Format: The IPV6 datagram header is completely different from IPv4 header. IPV6 uses a new header format in which options are separated from the base header and inserted, when needed, between the base header and upper-layer data. This simplifies and speeds up the routing process because most of the options do not need to be checked by routers. Allowance for Extension: IPv6 is designed to allow the extension of the protocol if required by new technologies or applications. IPv6 has new options to allow additional functionalities. Support for more Security: The encryption and authentication options in IPV6 provide confidentiality and integrity of the packet. Extension Headers: Unlike IPV4, which uses a single header format for all datagrams, IPV6 encodes information into separate headers. A datagram consists of the base IPV6 header followed by zero or more extension headers, followed by data. Support for Audio and Video: In IPV6, the type-of-service field has been removed, but a mechanism has been added that allows a sender and receiver to establish a high-quality path through the underlying network and to associate datagrams with that path. Although the mechanism is intended for use with audio and video applications that require high performance guarantees, the mechanism an also be used to associate datagrams with lowcost paths.

Extension Protocol: Unlike IPv4, IPv6 does not specify all possible protocol features. Instead, the designers have provided a scheme that allows a sender to add additional information to the datagram. The extension scheme makes IPv6 more flexible than IPv4, and means that new features can be added to the design as needed.