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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.

- $120\;\text{bytes}$
- $60\;\text{bytes}$
- $240\;\text{bytes}$
- $90\;\text{bytes}$

0

A $1\ Mbps$ satellite link connects two ground stations.

Since two ground stations are given, total propagation delay is $4\ T_p$.

If nothing is mentioned as in GATE2016-2-55, total propagation delay is $2\ T_p$.

1

88 votes

Best answer

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.**

You are getting 60 because you are not considering the path via which data is transfered. In normal case between any two stations 60 would have been the ansewer, here what is happening is that we have intermediate satellite in between.

So the distance of 36,504 is travelled **4 times**, you are considering only 2 times,

1

**I think transmission is done twice , first one at sender while sending from sender to satellite and second one at satellite while sending from satellite to receiver.**

Tt is Transmission time, Tp is propagation time from sender to satellite

So efficiency = ((2*Tt)*N ) / (2*Tt + 4*Tp)

by solving it we get 60 bytes.

**So why transmission is not taken twice????**

3

edited
Aug 25
by JAINchiNMay

0

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

It should be **120 Bytes**. Since the distance given is from station to satellite (**A--> S**) but to reach signal/data packet into B it has to further travel that is (**S-->B**), basically **twice the distance** it has to cover. So in formula instead of distance Satellite distance(d)=36504km substitute d= **2***36504, then you will get answer as 120 Bytes.

1

https://gateoverflow.in/32123/computer-networks-drdo-2008

this is the example how people forgot the concept in this question you are doing correctly bute hre distance multiply by 2.....

0

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,

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

$=> 507 = 488720/L$

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

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,

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

$=> 507 = 488720/L$

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