In step 4 of the CSMA/CA protocol, a station that successfully transmits a frame begins the CSMA/CA protocol for a second frame at step 2, rather than at step 1. What rationale might the designers of CSMA/CA have had in mind by having such a station not transmit the second frame. Immediately (if the channel is sensed idle)?
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- Protocol 6's frame arrival code includes a part for NAKs.If the incoming frame is a NAK and another condition is satisfied, this section is called.Give an example of a situation in which the existence of this other condition is critical.Use a diagram to illustrate the four-way handshake protocol for channel reservation in IEEE 802.11 (i.e., WiFi). Please indicate the name of each frame, and also the idle period before each frame is transmitted. (Tips: the protocol used is CSMA/CA with RTS/CTS.)NAKs are handled in protocol 6's code for frame arrival.If the incoming frame is a NAK and another condition is satisfied, this section gets activated.Describe a situation in which this other condition is critical.
- Consider a datagram network in which a packet travels through two switches S1 and S2 from source to destination. Packet experiences wait at each switch before it is forwarded. Transmission time, propagation time and wait time at each node is 3 ms, 1.5 ms, and 0.25 ms respectively. Find the total delay.What are the benefits and drawbacks of various access methods based on random access? Why does the slotted ALOHA protocol have twice the channel use of pure Aloha?A sender with window size (W=3) is connected to a selective repeat receiver by a link with bandwidth =106 bytes/second and propagation delay of 0.25 msec. The receiver sends Positive ACK for correctly received packets. For each packet sent, the sender sets a timer with time out value of 5 msec. On receiving an ack for a packet, the timer for that packet is cancelled. If the timer expires, that packet is retransmitted immediately. Assume that the sender sends packets 1, 2, and 3 (at t=0, packet size =1000 Bytes) and packet 2 is lost. NO other packet is lost. Under this scheme , at what time in msec would all the packets been received at the receiver.
- In a byte-oriented link layer protocol, the receiver adds all the bytes between the start and end marker bytes (not including those markers) modulo 239, and expects to get a result of 0 if there have been no errors. The byte immediately before the end marker is a checksum, chosen at the transmitter to make this possible. The probability of an error in any received bit is 0.00075 . The total length of the frame is 205 bytes (including start and end markers). Estimate the probability of errors occurring in the received frame but not being detected. You need to consider what combinations of bit errors could cause the error detection system to fail (i.e, conclude that there are no errors). As you are asked for an estimate, you need only consider the most likely scenarios - it is safe to ignore events that could only occur with much lower probability. Round your answer to three significant figures. As the probability will be very low in some cases, you may have to enter up to 7 digits after…Use a diagram to illustrate the four-way handshake protocol for channel reservation in IEEE 802.11 (i.e., WiFi). Please indicate the name of each frame, and also the idle period before each frame is transmitted. (Tips: the protocol used is CSMA/CA with RTS/CTS.)4. In this problem, you will derive the efficiency of a CSMA/CD-like multiple access protocol. In this protocol, time is slotted and all adapters are synchronized to the slots. Unlike slotted ALOHA, however, the length of a slot (in seconds) is much less than a frame time (the time to transmit a frame). Let S be the length of a slot. Suppose all frames are of
- In a selective repeat sliding window protocol the bandwidth of the link is 10 Mbps. The size of the frame is 10 KB and the one-way delay is 100 msec. If the efficiency of the protocol is 60 % then the minimum number of sequence bits that are required for the protocol is?A circuit-switched TDM-link uses 1024-bit frames over 16 TDM slots'. a. How many 'circuits’ are supported by this configuration? b. If we allowed the TDM-link to be utilized for "TCP connections", how many simultaneous TCP connections could be supported by the TDM link? (Hint: remember TCP connections is a full-duplex bidirectional stream) If the gross bitrate of this link is 8.192Mbit/s, what is the 'frame rate'? C. d. Given the 'gross bitrate', what would the throughput be (in bits/s, Kbits/s, or Mbits/s, as appropriate) for any ONE 'circuit in this configuration?Explain the flow control of TCP protocol when the size of the sliding window will be 500 and the first client sends 201, 301, 402, 455, 330, and once acknowledgments will receive client will send 602. Make a flow control of these packets