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A $1\;\text{Mbps}$ satellite link connects two ground stations. The altitude of the satellite is $36,504\;\text{km}$ and speed of the signal is $3 \times 10^{8}\;\text{m/s}.$ What should be the packet size for a channel utilization of $25\%$ for a satellite link using go-back-$127$ sliding window proto­col? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.

  1. $120\;\text{bytes}$
  2. $60\;\text{bytes}$
  3. $240\;\text{bytes}$
  4. $90\;\text{bytes}$
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6 Answers

Best answer
91 votes
91 votes

Distance from Station $A$ to Satellite $= 36504 \times 10^3$ m

Time to reach satellite $= \dfrac{36504000}{300000000} = 0.12168 s$

RTT $\text{(for a bit)} = 4\times \text{Time to reach satellite}( S1 \to \text{Satellite,
Satellite} \to S2, S2 \to \text{Satellite, Satellite} \to S1)$



Efficiency is the ratio of the amount of data sent to the maximum amount of data that
could be sent. Let $X$ be the packet size.

In Go-Back-N, within RTT we can sent $n$ packets. So, useful data is $n \times X$, where $X$ is the packet size. Now, before we can sent another packet $\text{ACK}$ must reach back. Time for this is transmission time for a packet (other packets are pipelined and we care only for first $\text{ACK}$ ), and $\text{RTT}$ for a bit, $($propagation times for the packet $+$ propagation time for $\text{ACK}\ +$ transmission time for $\text{ACK}\ -$ neglected as per question$)$

$\text{Efficiency} = \dfrac{\text{Transmitted Data Size}}{\text{Packet Size + RTT}_{bit} \times \text{Bandwidth}}$

$\implies 0.25= \dfrac{127\times X}{{X} + 4\times 0.12168 \times B}$

$\implies 0.25 X +  0.25 \times 4 \times 0.12168 \times B = 127X$

$\implies 0.25 X + 0.12168 \times 10^6 = 127 X$

$\implies 121680 = 126.75X$

$\implies X = 9.6 \times 10^{-4} \times 10^6 = 960$

$\therefore$ Packet Size $=960\text{ bits}=120\text{ Bytes}$

So, option (A) is the answer.

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6 votes

Ans B) 60 Bytes

Satellite Bandwidth(B)=1Mbps=$10^{6}$bps

Satellite distance(d)=36504km=$36504\times 10^{3}$m

speed(v)=$3\times 10^{8}$m/s

utilization$\left ( \eta \right )$=0.25

GB-N=GB127

So, we know

$\left ( \eta \right )$=$\frac{N}{1+2a}$    [a=$\frac{t_{p}}{t_{t}}$,$t_{p}=\frac{d}{v}$,$t_{t}=\frac{L}{B}$]

Now, putting all values,

L=60 Bytes

2 votes
2 votes
Normally; $T_{p}$ is directly what's given in the question, and $RTT = 2T_p$

But in satellite communnications, $T_p$ is actually twice the value of what is mentioned in the question; because the distance they give is from one end to the satellite — we've to take into account the distance from the satellite to the other end, as well.
And understandably, $RTT = 4T_p$ for satellite communications.

We know efficiency (or utilization here) = $\frac{W_{s}}{1+2a}$

$=> \frac{W_{s}}{1+2a}=\frac{1}{4}$

Now, $a = \frac{T_p}{T_t}$ $= \frac{{\color{Red} 2}dB}{vL}$ = $= \frac{ 2*36504*10^3*10^6 }{3*10^8L}$ = $244360/L$

So,

$4W_s = 1+2(244360/L)$

$=> 507 = 488720/L$

$=> L = 488720/507 = 963.944 bits = 120 Bytes$
Answer:

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