Question

Suppose an 802.11b station is configured to always reserve the channel with the RTS/CTS sequence. We assume a frame without data is 35 bytes long, and the transmission rate is 10 Mbps. Suppose this station suddenly wants to transmit 500 bytes of data, and all other stations are idle at this time. As a function of SIFS and DIFS, and ignoring propagation delay and assuming no bit errors, calculate the time required to transmit the frame and receive the acknowledgement. (Proof is needed)

Answer 1

Solution:

Given,

=>Frame without data = 35 bytes

=>Transmission rate(B) = 10 Mbps

=>Size of data = 500 bytes

=>Propagation delay = 0

=>No bit errors are there.

Explanation:

Calculating size of frame:

=>Frame size = frame without data size + data size

=>Frame size = 35 bytes + 500 bytes

=>Frame size = 535 bytes

Calculating time for control frame:

=>Generally size of control frame = 256 bits

=>Transmission time = frame size/transmission rate

=>Transmission time = 256 bits/10 Mbps

=>Transmission time = 256 bits/ 10*10^6 bps as 1 Mbps = 10^6 bps

=>Transmission time = 25.6 microseconds as 1 sec = 10^6 microseconds

Calcualting transmission time for frame:

=>Transmission time = 535 bytes/10 Mbps

=>Transmission time = 535*8 bits/10*10^6 bps as 1 byte = 8 bits and 1 Mbps = 10^6 bps

=>Transmission time = 428 microseconds as 1 sec = 10^6 microseconds

Calculating total transmission time:

=>Total transmission time = DIFS + RTS + SIFS + CTS + SIFS + Frame transmission + SIFS + ACK

=>As we know RTS, CTS and ACK are control frames so transmission time for RTS = CTS = ACK = 25.6 microseconds

=>Transmission time = DIFS + 25.6 microseconds + SIFS + 25.6 microseconds + SIFS + 428 microseconds + SIFS + 25.6 microseconds

=>Transmission time = DIFS + 3*SIFS + 3*25.6 microseconds + 428 microseconds

=>Transmission time = DIFS + 3*SIFS + 76.8 microseconds + 428 microseconds

=>Transmission time = DIFS + 3*SIFS + 504.8 microseconds

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