1. Data Link Layer Protocols always put CRC in there trailer rather than in the header. Why?
Ans-The CRC is computed while the packet is being transmitted and then incorporated in a trailer. Similarly, the receiver computes the CRC and compares it with the transmitted one , It is better to have the CRC in a trailer.
Because then we can compute the CRC as we transmit the packet, and then splat out the final CRC at the end. If we put it in the header, we will have to make *one* pass over the packet computing the CRC, then send the CRC, and then a *second* pass actually stuffing each byte out the interface.The CRC is computed during transmission and appended
to the output stream as soon as the last bit goes out onto the wire. If the CRC were in the
header, it would be necessary to make a pass over the frame to compute the CRC before transmitting. This would require each byte being handled twice-once for check summing and once for transmitting.
Using the trailer cuts the work in half.
2. Slotted aloha is a improved version on aloha protocol. On what factors the improvement is implemented in slotted aloha.
Ans -Slotted Aloha improves contention management , in which a receiver transmits signals at exact intervals, indicating to each source when the channel is clear to send a frame of data. Slotted Aloha essentially advertises the availability of a time slot in a channel. If more than one station transmits packets in the same slot, there is a collision, and the receivers cannot receive the packets correctly. Successful transmission happens only when there is exactly one packet transmitted in a slot. If no packet is transmitted in a slot, the slot is called idle.
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-Loss insertion and modification can be detected by checksome because by change of one bit, the sum and the checksum can never be equal and thus error can be detected.Checksum works by comparing the sum and the checksum at the sender and the receiver side.If the sum and the checksum are equal then there is no error.
4. Give two reasons why network might use an error correcting code instead of error detection and retransmission?
Ans-Error detectionis the detection of errors caused by noise or other impairments during transmission from the transmitter to the receiver.
Error correctionis the detection of errors and reconstruction of the original, error-free data.
Two reasons for using error correcting code instead of error detection and retransmission :-
1)Time taken to error correcting is less than doing detection and retransmission.Bandwidth use will be less.
2) In detection and retransmission if back messaging occours the bandwidth will be more
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- Detection is not possible in wireless transmission.After the collision the energy is decreased and hence can't detect.Moreever in wiresless listen and transmission occours simultaneously.
6.Bluetooth supports two types of links between a slave and master .What are they and why is each one used for?
Ans-The two types of links between a slave and master are
i)SCO(synchronous Connection Oriented)-It used where the proper transmission is more important than the speed.
ii)ACL(Asynchornous connectionless LInk)-It is used where speed is more important than accuracy of transmission.
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
b. Go-Back-N ARQ
c. Selective-Repeat ARQ
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- If an Ethernet destination address is 07:01:02:03:04:05 then it is the MULTICASTING cozright most bit of the least singnificant bytein the MAC address is odd(7=0111)