GATE Overflow Test Series | Computer Networks | Test 1 | Question: 22

Question

A $2\;\text{Mbps}$ satellite link connects two ground stations. The altitude of the satellite is $32,604\;\text{km}$ and speed of the signal is $3\times 10^8\;\text{ m/s}$. Assuming that the acknowledgment packets are negligible in size and that there are no errors during communication, the minimum packet size (in bytes rounded to one decimal point) required for a channel utilization of $35\%$ for a satellite link using $\text{GoBack-63}$ sliding window protocol is _______

Answer 1

In satellite communication, $RTT_{bit} = 4\; PD$ as both the frame and ACK must travel to Satellite and then back.
    
So, $RTT_{bit} = 4\; PD = 4 \times \dfrac{32604\times 10^3}{3 \times 10^8}=0.43472\;\text{sec}$
    
With $\text{GoBack-63},\; 63$ packets are sent. Transmission time for this, $T_t=\dfrac{\text{Frame Size}}{2 \times 10^6}$
    
For $35\%$ channel utilization we need,

$\qquad \dfrac{63 \times T_t}{T_t+0.43472} \geq 0.35$

$\qquad \implies 62.65 \times T_t \geq 0.152152$

$\qquad \implies \dfrac{62.65 \times \text{Frame Size}}{2 \times 10^6} \geq 0.152152$

$\qquad \implies \text{Frame Size} \geq 4.85\;\text{kb} = 607.15 \; \text{bytes}.$

Comments

Can anyone explain why RTT=4*PD here?

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ok guys, even i made this mistake of yours   the reason why its going to be 4rtt is that dont confuse satellite with router  if it would be router the router could have boosted the signal hence the frames wont fell to error but satellite is just carrier which doesnt boost the signal hence you have to consider it just like its the part of link

if it would be a router then we would have taken only the maximum distance path that is →  propagation time = 32604km/velocity

 

please notify me if this way thinking is wrong @Dheeraj Varma @ijnuhb @gatecse

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@gatecse Sir, why is the total time not – $(2*T_t+4*T_p)$ ? Since the satellite should also need to transmit the frame once, don’t we need two times the transmission delay ($T_t$)?