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A $20$ $\text{Kbps}$ satellite link has a propagation delay of $400$ $\text{ms}$. The transmitter employs the "go back $n$ $ARQ$" scheme with $n$ set to $10$. Assuming that each frame is $100$ $\text{byte}$ long, what is the maximum data rate possible?

1. $5$ $\text{Kbps}$
2. $10$ $\text{Kbps}$​​​​​​​
3. $15$ $\text{Kbps}$​​​​​​​
4. $20$ $\text{Kbps}$​​​​​​​

SWS= 10*100*8 bits

capacity= RTT*BW = 2 * 400 * 10^-3 * 20 * 10^3 bits.

therfore efficiency= sws / capacity = 0.5

Thus, maximum data rate possible = efficiency * BW = 0.5 * 20 Kbps = 10Kbps
Tt = 40ms,  Pt = 400ms

Maximum packets that can be sent in GBN ARQ = 1 + 2a, where a = Pt/Tt.

Max packets = 1 + 2*(400/40)=21 packets.

However we are sending only 10 packtes  (sender window size is 10).

Efficiency = 10/21

= 0.476   (efficiency is just a ratio)

Max Data rate = Efficiency *  Actual Bandwidth

= 0.476 * 20Kbps

= 9.523Kbps = 10Kbps approx
why $t_{t}$= 40 ms?

Transmission TIme $=\dfrac{100\times 8\text{bits}}{20\text{ Kbps}}=40\ ms$

Propagation Time $=400\ ms$

Efficiency $=\dfrac{\text{Window Size$\times$Transmission Time}}{\text{(Transmission Time + 2$\times$Propagation Time)}}$

$=\dfrac{10\times 40}{(40+2\times 400)}=0.476$

Maximum Data Rate$=0.476\times 20\text{ Kbps}=9.52\text{ Kbps}$

which is close to option B.

@Ajit_Singh

Propagation delay is usually specified for links. If sender and receiver are directly connected, then you can directly calculate Tp between them as the Tp of the link, but if they are connected via multiple links, then you have to consider the Tp of each link separately. Although sometimes the total Tp between S and R is given in question so we don’t have to calculate it, but here the Tp is given for satellite link and so the total Tp between S and R will be 4Tp.
In this question they are talking about the communication between satellite and host (which will take 2*Tp) and not host to host (which will take 4*Tp). I am saying this because the term “transmitter’ is used instead of “sender” and “receiver”. But it can still be really tricky to answer in gate exam because 1.33 or 2.66 marks will be at stake based on an assumption. (PS: I am no expert)

https://gateoverflow.in/3375/gate2008-it-64 This question should clear up any further confusion regarding this topic

Option(B) is correct!

There are 3 ways to do  this question.
1.
Tp = 400 ms,  R.T.T = 2*400 ms = 800 ms
Tt = 40 ms
In one round trip, we only  send 10 packets = 10*100 = 1000 Bytes = 8000 bits.

In 800 ms we send 8000 bits, throughput = In 1 second, how many bits can be sent?
1 ms = $\frac{8000}{800}$ = 10 bits, and in 1 second = 10 k bits
2.
$$\text{Throughput = efficiency*B.W}$$
efficiency = $\frac{n}{(1+2*a)}$ where $a=\frac{tp}{tt}$
efficiency = $\frac{10}{1+2*400/40}$ = $\frac{10}{21}$
Throughout = (10/21)*20Kbps = 10 Kbps (almost)

3.
R.T.T = 2*TP = 800 ms
Tt = 40 ms
in 1 R.T.T we can send maximum $\frac{800ms}{40ms}$ = 20 packets

But as mentioned in the question we are sending only 10 packets, it means efficiency is 50% (or) we are utilizing the bandwidth only 50%, so effective bandwidth or throughout will be 50% of total bandwidth i.e. 10 kbps.

@Manu Thakur @Puja Mishra

What puts in in trouble is why we did not taken RTT = 4 x PT. It is a satellite link right so 2 x PT will be the time needs to be taken from source to destination. Same from destination to source as well.

@Manu Thakur  in your first approach, why haven't you included transmission time for sending 8000 bits?

Tt+2Tp=840 ms----------------can send 8000 bits,

1 ms------------------------------8000/840 =9.52 bits

1 s----------------------------------9.52 K bits =10 K bits(almost)

calculate transmission time = (100*8)/20 =40ms
throughput is 100*8*10/(40+2*400)= 9.52 Kbps = 10Kbps

but why 8 ??
to convert byte into bit

Thanks to Shiva Sagar Rao

by