GATE IT 2008 | Question: 64

Question

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.

• A. $120\;\text{bytes}$
• B. $60\;\text{bytes}$
• C. $240\;\text{bytes}$
• D. $90\;\text{bytes}$

Comments

0.25  =  127 /  [1 + 4* Tp/ Tt]
Tt = size / 10^6
Tp = dist / speed

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Another approach is
 

No. of links does the DATA packet traverse in one direction = 2 links or 2 ways to reach there

hence, we are multiplying by 2.

Find Tp and then 2 x Tp
 

                             W
                U =   ----------
                          1 + 2a

From here, u will get value of a

                          Tp
               a  =    ------
                          Tt

From here, u will get value of Tt

                          L
              Tt  =  -------
                        BW

From here, u will get value of L      

Lastly, u will get value in bits  divide by 8  to convert into bytes     and done.

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Here Satellite acts as a transparent/bent-pipe relay: it receives and simultaneously forwards the signal; no extra processing or store-and-forward delay is considered.

Answer 1

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.

Comments

Transmission time won't calculate twice because once packets are transmitted it will go through in an order just like pipeline so again no need to transmit but according to propagation we are moving 4 times to reach destination with transmitted packets so 4*tp + tt

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we treat the satellite as a "Bent Pipe" (repeater). A satellite usually doesn't wait to receive the entire packet before it starts broadcasting it back down to Earth. It works at the physical layer: as soon as the first bit of the packet hits the satellite's receiver, it is amplified and re-transmitted toward Ground Station B.

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 Assume that the acknowledgment packets are negligible in size

this is given is question. and we knw transmission time= length / bandwidth
so essentially for ack transmission time becomes zero.
hence our rtt is = T_t + 4* T_p

only transmission time will take from station A. 
4*T_p because distance between station A to satellite and satellite to station B is same and vice versa . Also speed of same.
so t_p is same for all.
$\ 0.25 = \frac{127  * T_t}{T_t + 4 \cdot T_p}$

then open up t_t as L/bw . and find L

Answer 2

Ans: A

Answer 3

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

Comments

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.....

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papesh sir, srestha ma’am, Can we calculate the same as this is calculate just by Taking the total distance between both the stations? i.e 2∗36,504=73,008Km. And taking the rest of values same as @srestha ma’am took? 
 

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@srestha

consider the satellite as transmition device so, we have to take 4*Tp..

Answer 4

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,

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

$=> 507 = 488720/L$

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

Comments

Here packet is transmitted two time first when sender A send it to satellite S and satellite again transmit it towards B, so why transmission time is not doubled?

Answer 5

Given altitude of the satellite is 36504 km. Let $A$ and $B$ be the two ground stations. Let $d$ be the distance between the two ground stations. 

By Trigonometry, we can say, $tan$ $45^{\circ}$ = $perpendicular / base$ $\implies$

$tan$ $45^{\circ}$ = $\frac{36504}{(d/2)}$ $\implies$ $d = 2*36504 = 73008 km$

Now, In $GBN$, Utilization = $\dfrac{n}{1+2a}$, where n is the window size and $\large a = \dfrac{\text{Propagation Time}}{\text{Transmission Time}}$

Given utilization = $25\%$ or $1/4$ and $n = 127$, $v = 3*10^8m/s$, $ packet size = L$, $Bandwidth = 1Mbps$

So, $1/4$ = $\frac{1}{1+ \frac{2*73008*10^3*10^6}{3*10^8*L}}$ $\implies$ $L = 960 bits$ = $120 B$ 

 

Comments

Very good explanation 👍

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can u clarify say how the angles are at 45° ?

The altitude being 36,504 km does not imply the base angles are 45°

Answer 6

EVERY ONE CONFUSED WHY WE MULTIPLY 2 IN DISTANCE BECAUSE  SENDER AND RECIEVER ARE BOTH ON ERTH

BUT IN GATE 2022 FLOW CONTROL QUESTION SATELLITE IS THE RECIVER THATS WHY IN THIS CASE WE DO NOT MULTIPLY THE 2