Go back n ARQ

Question

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 efficiency of channel? (with 2 decimal places)

Comments

is answer b)?

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yes.

please give your solution

Answer 1

Transmission time= $\large \frac{L}{B}$ = 40ms

Propagation time = 400ms

a=$\large \frac{T_p}{T_t}$ = 10

max window size for 100% efficiency = $1+2a$ =  21

efficiency =$\large \frac{10\times 100}{21 } = 47.6$%

Comments

i agree with you that means no option is correct in this question?

https://gateoverflow.in/147466/computer-networks

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I saw the answer and actually in that answer RTT = 2*Tp , but this is true when we neglect Tt. Here , the exact answer will be 9.52Kbps, but round off we can say it is 10 Kbps.

That is in 800+ 40 ms we are transferring 10*800 bits,not in 800 ms with this we get 10*800/840 = 9.52 Kbps.

I think this is a made easy question. I remember seeing it somewhere.

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It's GATE 2004 - IT question

Answer 2

Given Data
Bandwidth = 20 Kbps
Propogation dealy = 400 ms, so RTT = 2 * PD = 800 ms
window size = 10
frame size = 100 bytes or 800 bits

Even though bandwidth is 20kbps, we are restricted with window size here

In 1 RTT we are sending 10 * 800 bits
In 800 ms we are sending 10 * 800 bits
In 1 ms we are sending 10 bits
In 1 sec we are sending 10 * 1000 bits
In 1 sec we are sending 10 Kb

So effective bandwidth = 10 Kb/sec, which is option B

Answer 3

We use the formula Throughput =Data Size/(Transmission time +2Propagation Delay)

or data size =Windows size* packet Size (here windows n=10)

Bandwidth =20 Kbps =20*10^bps

Propagation Delay=400ms =400*10^-3 seconds 

transmission time (Data )=100/20*10^-3 seconds =0.4 seconds 

And using  formula  of Throughput = (10*100*8)/(0.4+2*400*10^-3)= 10000 bps =10Kbps 

 

Answer 4

Here, efficiency = N / (1 + 2 a), where N is the sender window size (given as 10).

a = T_p / T_t                         

T_p = 400 ms (given) and T_t = (length of each frame) / (bandwidth of the satellite link) = (100 * 8) / (20 * 210).

Evaluating the above, we find a = 10.24.

Substituting the values in the above formula, we get efficiency = 0.4655 = 46.55 %.