Friday, April 01, 2011

Computer Network Questions and Answers

Q-1 What are the goals in mind of IEEE 802 committee?
Ans: IEEE 802 committee has few goals in mind, namely
–To promote compatibility
–Implementation with minimum efforts
–Accommodate diverse applications

Q-2. List the functions performed by the physical layer of 802.3 standard?
Ans. Functions of physical layer are:
i) Data encoding/decoding (To facilitate synchronization and efficient transfer of signal through the medium).
ii) Collision detection (It detects at the transmit side)
iii) Carrier sensing (Channel access senses a carrier on the channel at both the transmit and receive sides)
iv) Transmit/receive the packets (Frame transmitted to all stations connected to the channel)
v) Topology and medium used (Mediums are co-axial cable, twisted pair and fiber optic cable)

Q-3. Why do you require a limit on the minimum size of Ethernet frame?
Ans. To detect collision, it is essential that a sender continue sending a frame and at the same time receives another frame sent by another station. Considering maximum delay with five Ethernet segments in cascade, the size of frame has been found to be 64 bytes such that the above condition is satisfied.

Q-4. What are the different types of cabling supported by Ethernet standard?
Ans. Types of cabling are:
i) 10 BASE 5 - Maximum cable length is 500 meters using 4” diameter coaxial cable.
ii) 10 BASE 2 - Maximum cable length is 185 meters using 0.25” diameter CATV cable.
iii) 10 BASE T - Maximum cable length is 100 meters using twisted-pair cable (CAT-3 UTP).
iv) 10 BASE FL - Maximum cable length is 2 Km using multimode fiber optic cable (125/62.5 micrometer).

Q-5. Explain the basic difference between IEEE 802.3 and switched Ethernet, as far as implementation is concerned.
Ans: In Ethernet (IEEE 802.3) the topology, though physically is start but logically is BUS. i.e. the collision domain of all the nodes in a LAN is common. In this situation only one frame can send the frame, if more than one station sends the frame, there is a collision.
In Switched Ethernet, this collision domain is separated. Hub is replaced by a switch, a device that can recognize the destination address and can route the frame to the port to which the destination station is connected, the rest of the media is not involved in the transmission process. The switch can receive another frame from another station at the same time and can route this frame to its own final destination.

Q-6. Explain the two techniques for implementing Ethernet switches.
Ans: There are two techniques used in the implementation of Ethernet switches: store-and-forward and cut-through. In the first case, the entire frame is captured at the incoming port, stored in the switch’s memory, and after an address lookup to determine the LAN destination port, forwarded to the appropriate port. The lookup table is automatically built up. On the other hand, a cut-through switch begins to transmit the frame to the destination port as soon as it decodes the destination address from the frame header.
Store-and-forward approach provides a greater level of error detection because damaged frames are not forwarded to the destination port. But, it introduces longer delay of about 1.2 msec for forwarding a frame and suffers from the chance of loosing data due to reliance on buffer memory. The cut-through switches, on the other hand, has reduced latency but has higher switch cost.

Q-7. What are the different categories of Fast Ethernet?
Ans: IEEE has designed two categories of Fast Ethernet: 100Base-X and 100Base-T4. 100Base-X uses two cables between hub and the station while 100Base-T4 uses four. 100-Base-X itself is divided into two: 100Base-TX and 100base-FX.
* 100 BASE-T4: This option is designed to avoid overwriting. It is used for half-duplex communication using four wire-pairs of the existing category 3 UTP cable, which is already available for telephone services in homes/offices. Two of four pairs are bi-directional; other two are unidirectional. This means that there are 3 pairs to be used for carrying data, in each direction (2 bi-directional and 1 uni-directional). Because 100Mbps data cannot be handled by voice-grade UTP, this specification splits the 100 Mbps flow into three 33.66Mbps flow.
* 100 BASE TX: This option uses two category 5 UTP or two shielded (STP) cable to connect a station to hub. One pair is used to carry frames from the hub to the station and other to carry frames from station to hub. Encoding is 4B/5B to handle 100 Mbps; signaling is NRZ-I. The distance between station and hub should be less than 100 meters.
* 100 BASE FX: This option uses two Fiber optic cables, one carry frames from station to hub and other from hub to station. The encoding is 4B/5B and signaling in NRZ-I. the distance between station and hub should be less than 2000 meters.

Q-8. What are the Objectives of The Gigabit Ethernet Alliance?
Ans: The objectives of the alliance are:
• supporting extension of existing Ethernet and Fast Ethernet technology in response to demand for higher network bandwidth.
• developing technical proposals for the inclusion in the standard
• establishment of inter-operability test procedures and processes

Q-9. Explain GMII (Gigabit Media Independent Interface) in brief.
Ans: The GMII is the interface between the MAC layer and the Physical layer. It allows any physical layer to be used with the MAC layer. It is an extension of the MII (Media Independent Interface) used in Fast Ethernet. It uses the same management interface as MII. It supports 10, 100 and 1000 Mbps data rates. It provides separate 8-bit wide receive and transmit data paths, so it can support both full-duplex as well as half-duplex operation.
The GMII provides 2 media status signals: one indicates presence of the carrier, and the other indicates absence of collision. With the GMII, it is possible to connect various
media types such as shielded and unshielded twisted pair, and single-mode and multi mode optical fiber, while using the same MAC controller. It has three sub-layers namely: PCS (Physical Coding Sublayer), PMA (Physical Medium Attachment) and PMD (Physical Medium Dependent)

Wednesday, March 23, 2011

Computer Network Questions and Answers

1. The IP datagram for a TCP ACK message is 40 bytes long: It contains 20 bytes of TCP header and 20 bytes of IP header. Assume that this ACK is traversing an ATM network that uses AAL5 to encapsulate IP packets. How many ATM layer packets will it take to carry the ACK for each of AAL layer?
a. AAL5
b. AAL3/4
ANS
a)) The length of the AAL5 CS-PDU into which the ACK is encapsulated is exactly 48 bytes, and this fits into a single ATM cell.
b)) When AAL3/4 is used the CS-PDU is again 48 bytes, but now the per-cell payload is only 44 bytes and two cells are necessary.

2. Suppose there are N stations on a LAN that has capacity (transmission rate) C. All packets have a fixed length L and the end-to-end propagation delay of the channel is P. If only one station ever has a message to send (i.e., the other N-1 stations generate no traffic). What is the maximum possible throughput seen by this single node under each of the protocols:
a. CSMA
Assuming that the station senses the channel immediately after finishing its transmission, and finding it idle, sends another packet, the throughput is C. Note that this ignores the overhead time needed to do channel sensing.

b. Slotted Aloha
Assuming the station always has data to send, it will always to do under slotted Aloha, so the throughput would be C.

c. Token Passing
After sending a message, a station gives up the token. The token circulates around the ring (taking time P). Assuming that L is the amount of time needed to transmit a message, the throughput is thus (L/(L+PC))*C.


3. Consider sending a 1500-byte datagram into a link that has an MTU of 500 bytes. Suppose the original datagram is stamped with the identification number 1. Assume that IPv4 is used.
a. Where does fragmentation happen? Where are the fragments reassembled?
Fragmentation happens in the router preceding the link with the small MTU. The fragments are reassembled in the end system.

b. How many fragments are generated?
The maximum size of the data field in each fragment = 480 (because there are 20 bytes IP header). Thus the number of required fragments= 1500-20/480 = 4

c. What are the values of the fragmentation-related fields in the generated IP datagram(s)?
Each fragment will have an identical identification number. Each fragment except the last one will be of size 500 bytes (including the IP header). The last datagram will be of size 60 bytes (including the IP header). The offsets of the 4 fragments will be 0, 60, 120, 180. Each of the first 3 fragments will have flag=1; the last fragment will have flag=0.

d. What changes if IPv6 were used?
The router preceding the link with the small MTU will drop the packet and send an ICMP error message “Packet Too Big” back to the source. The source is responsible for adjusting the packet size.

4. Consider building a CSMA/CD network running at 1 Gbps over a 1-km cable with no repeaters. The signal speed in the cable is 200,000 km/sec. What is the minimum frame size?
ANS
For a 1-km cable, the one-way propagation time is 5 μsec, so 2τ = 10 μsec. To make CSMA/CD work, it must be impossible to transmit an entire frame in this interval. At 1 Gbps, all frames shorter than 10,000 bits can be completely transmitted in under 10 μsec, so the minimum frame is 10,000 bits or 1250 bytes

5. In the CSMA/CD protocol, what condition on the transmission delay Ttrans and the propagation delay Tprop has to be satisfied to guarantee that a node always detects a collision?
ANS Ttrans>2Tprop


6. At low load, consider the delay of pure ALOHA versus slotted ALOHA. Which one is less? Explain your answer.
ANSWith pure ALOHA, transmission can start instantly. At low load, no collisions are expected so the transmission is likely to be successful. With slotted ALOHA, it has to wait for the next slot. This introduces half a slot time of delay.

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