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GATE | GATE-IT-2004 | Question 83

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  • Difficulty Level : Basic
  • Last Updated : 28 Jun, 2021

A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the “go back n ARQ” scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?
(A) 5Kbps
(B) 10Kbps
(C) 15Kbps
(D) 20Kbps


Answer: (B)

Explanation:

It uses the sliding window protocol for transmission of data. 
The question takes into consideration the variant of sliding window protocol 
namely GO BACK N ARQ. In this protocol the sender can have up to N packets 
unacknowledged that are still remaining in the pipeline. The receiver only 
sends cumulative acknowledgements. In case of encountering an error the sender 
has to resend all the data frames following the error.
According to the question:
The data rate of the link is 20 Kbps and the propagation delay = 400 ms
So, the time required to transmit 100 bytes long data will be given by

Transmission Time t = Number of bits to be transmitted / data rate of the link 
                    = (100* 8 bits) /20 Kbps =  40 ms

Now, the propagation delay is given as d = 400 ms

So the efficiency of the link is given by:

Efficiency E = N * t / ( t+ 2 * d )

Where N = window size

E = 10 * 40 / (40+2*400) = 0.476

So, the maximum data rate attainable = 0.476 * 20 Kbps = 9.52 Kbps

This is close to 10.
So, the answer will be 10Kbps.

This explanation has been contributed by Namita Singh.

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