Computer Networking

1. Data Link Layer Protocols always put CRC in there trailer rather than in the header. Why?

Ans.

The CRC is computed during transmission and appended to the output stream as soon as the last bit goes out onto the wire. But If we put CRC in the header, before transmitting it is necessary to compute CRC bits and then add it. This approach will result in handling of each byte twice once for check summing and once for transmitting. While addition of CRC in trailer results is the easy way and less effort. Thats the reason why we add CRC at the trailer part of the frame.

2. Slotted aloha is a improved version on aloha protocol. On what factors the improvement is implemented in slotted aloha.

Ans.

In case of slotted ALOHA concept of time slot has been introduced. During the transmission we make use of this time slot to send a frame. A frame will be send only at the beginning of the time slot. As a result of this method the chances of collision decreases considerably.

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Vulnerable time in case of slotted ALOHA is also less than the vulnerable time in pure aloha. Or in other words we can say that the vulnerable time is just half in the slotted ALOHA as compared to that of ALOHA.

As the probability of collisions has been decreased this results in the increased efficiency

of the transmission.

3.When bit stuffing is used, is it possible for loss, insertion or modification of a single bit to cause an error not detected by checksum? If not why not? If so how? Does checksum play a role there?

Ans.

Yes. There is a possibility of modification a single bit which will result in an error.

But this error can easily be detected by the checksum error detection method.

This is so because checksum is the sum of the data elements to be transmitted.

If a bit in frame get modified then it will result in the change of sum of the data elements.

Which in turn will change the checksum. And if the chechsum dose not match at both the sides then error will be detected by the receiver.

Also there is chance of loss of a bits.

Lets suppose if there change occurs in the bit which was stuffed in the data part means stuffed 0 bit becomes 1. in this case receiver will take it as flag and conseder it as the end of the frame which will finally result in the loss of some data bits. This loss of data bits can easily be detected by the checksum method. As due to the loss of data bits the sum will change and it will result in change in checksum number. And the error will be detected easily.

Insertion of a bit is not possible because flags are used to indicate the start and end of a frame. And also the size of the frame is fixed. So we can not add one extra bit to the frame.

4. Give two reasons why network might use an error correcting code instead of error detection and retransmission?

Ans.

Sending data with error correcting code and detecting error then ask for retransmission of data are two different methods used for data transmission.

The 1st option have some advantages over the 2nd one.

1> Fast transmission of data takes place.

If any frame is received with error then receiver can regenerate the correct frame using the error control information. While in 2nd case receiver will 1st send request for retransmission of that particular frame then sender will again send that frame, which is a very time consuming process.

2> Efficiency of transmission increases.

If we use 1st option then the whole bandwidth of the channel will be used to send the data only in one direction which will automatically increases the data transmission efficiency. While in case of 2nd option bandwidth of the channel is divided in two parts which results in loss of efficiency.

5. Wireless transmission and wired transmission use different set of multiple channel allocation strategies. Why there was a need of avoidance when detection was already available?

Ans.

In case of wired transmission we use wires for the transmission. If data have to send over a small distance it is easily transferred but if data is to be send over a long distance then we makes use of repeaters. Because of the repeaters the energy of the frames are maintained. So if any collision occurs in wired connection then it is easy to recover the data.

While in case of wireless transmission data travels through the air which results in loss of energy of the packets. During transmission energy of the data packets decreases. So if collision occurs then the packets destroy easily and completely. So we need to avoid collisions in case of wireless transmission.

6. Blue tooth supports two types of links between a slave and master .What are they and why is each one used for?

Ans.

There are two types of link that can be created between a primary ( Master ) and a secondary ( Slave ) stations.

1> SCO, Synchronous connection oriented.

This connection is used when it is more important to deliver data in time ( to avoid latency) than integrity ( i. e. error-free delivery ).

The basic units of connection is two slots, one for each direction. At regular intervals specific slots are reserved for primary and secondary stations to establish the connection. If any error occurs ( such as packet lost) then it is never retransmitted.

This type of connection is used in real-time applications.

2> ACL, Asynchronous connectionless link.

This type of connection is used when data integrity is more important than avoiding delay in data delivery.

If any error occurs, then the damaged frame is retransmited.

The number of slots are not fixed for ACL, it can use one, two or more number of slots.

After the arrival of the data frame secondary station sends ACL frame if and only if the previous slot has been addressed to it.

7. Using 5-bit sequence numbers, what is the maximum size of the send and receive windows for each of the following protocols?

a. Stop-and-Wait ARQ

Ans. In this protocol sequence number is based on modulo-2 arithmeetic.

Send window Size :- 1 ( always )

Receive Window Size :- 1 ( always )

b. Go-Back-N ARQ

Ans.

Send window size :- 32 ( frames numbering from 0 to 31 )

receive window size:- 1 ( always )

c. Selective-Repeat ARQ

Ans. In this protocol the size of send window is equal to receive window.

Send Window size :- 16

Receive Window size :- 16

8. If an Ethernet destination address is 07:01:02:03:04:05, what is the type of the

address (unicast, multicast, or broadcast)?

Ans.

A source address is always a unicast address as frame comes only from one station.

Now destination address can be unicast, multicast or broadcast.

To identify a address wheather it is a unicast or a multicast we conseder least significant bit of the 1st byte.

If this bit is 0 then address is Unicast.

If this bit is 1 then address is Multicast.

While broadcast address is a special case of the multicast address.

If all the bits in the this 6 byte address are 1 then its a broadcast address.