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+31 votes

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 × 10^{8} m/s. What should be the packet size for a channel utilization of $25$$\text{%}$ 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.

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

+54 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 Satellite,

Satellite \to S2, S2 \to 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$)$

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

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

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

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

$ 121680 = 126.75X$

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

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

**So, option (A) is answer.**

0

it's round trip time, so after reaching B it will come back travelling same distance which is 2*36504 km.

total distance will be 4* 36504.

total distance will be 4* 36504.

+8

Your answer is correct. Just not well explained.Let me reexplain it.

eff=N/(1+2a) =N/(1+2*(d/v)*(B/L))

now N=127, d=2*36504*1000 meters v=3*10^8 meters, B=10^6, Put thesse values in the formula you will get the answer 120 bytes that is the correct answer.

eff=N/(1+2a) =N/(1+2*(d/v)*(B/L))

now N=127, d=2*36504*1000 meters v=3*10^8 meters, B=10^6, Put thesse values in the formula you will get the answer 120 bytes that is the correct answer.

+17

ans is 120 byte and given is altitude of satelite is 36504 km and two ground station A and B linked to this satelite and station A sends the packets to station 2 now total length will be 2*36504 speed not matter everything same

hope you got the point

+1

Isn't it that RTT = transmission time + propagation delay?

So RTT= 2*transmission time ( 1 transmission by station A and other by Satellite) + 4*propagation delay ( which is time to reach satellite).

So RTT= 2*transmission time ( 1 transmission by station A and other by Satellite) + 4*propagation delay ( which is time to reach satellite).

+5

the formula for RTT says RTT = 2 * Tp

where Tp is transmission time , now see how Tp is calculated here

in this question RTT is calculated = 4 * time to reach the satellite one time , just bcoz RTT = time to send the signal + time to get acknowledge the signal

so here from station A to satellite S then S to station B = time to send the signal

and reverse path means from B to S then again S to A = time to get acknowledge

RTT = time to send the signal + time to get the acknowledge = distance from station A to satellite S then S to station B + distance from station B to satellite S then S to station A

this way RTT is calculated without any formula used here.

where Tp is transmission time , now see how Tp is calculated here

in this question RTT is calculated = 4 * time to reach the satellite one time , just bcoz RTT = time to send the signal + time to get acknowledge the signal

so here from station A to satellite S then S to station B = time to send the signal

and reverse path means from B to S then again S to A = time to get acknowledge

RTT = time to send the signal + time to get the acknowledge = distance from station A to satellite S then S to station B + distance from station B to satellite S then S to station A

this way RTT is calculated without any formula used here.

–1

Its quite logical . However in the exam its bit difficult to hit at the first go .I think without figure it can not be understood . Thanks @Bikram Sir :)

0

here packet is transmitted two times once by sender and again by satelite .then why transmission time is not multiplied by 2 ?

+1

In this particular question,

RTT

= time to send the signal + time to get the acknowledge

= distance from station A to satellite S then S to station B + distance from station B to satellite S then S to station A

= 4 * time to reach the satellite one time ,

because RTT = time to send the signal + time to get acknowledge the signal

RTT

= time to send the signal + time to get the acknowledge

= distance from station A to satellite S then S to station B + distance from station B to satellite S then S to station A

= 4 * time to reach the satellite one time ,

because RTT = time to send the signal + time to get acknowledge the signal

0

yeah it is clear to me why we multiply 4 into time to reach the satelite one time . but i am asking about tranmission time. packet is transmitted two time first when sender **A** send it to satelite **S** and satelite again transmit it towards **B **.

0

Transmission from A + Retransmission from B ( get that acknowledgement )

= ( distance from station A to satellite S then S to station B )+ ( distance from station B to satellite S then S to station A )

= ( distance from station A to satellite S then S to station B )+ ( distance from station B to satellite S then S to station A )

0

Why is Time taken for transmission (T_{t}) from satellite to link connecting station B not considered ?

0

Maximum data that can be sent is taken as (PacketSize + Rtt * BandWidth). I did not understand why the packet size is being added. Plz explain.

+2 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

+1

somewhere I read in this cases round trip time is taken as 4Tp [A->satelite,satelite->B,B->satelite,satelite->A] ??

0

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

i think... in this question arjun sir also says the same thing....

0

yes. Tp is distance between sender and receiver. but distance must be doubled. so 160 Bytes should be right

0

@shrestha @gabbar

what will be the final soln?

arjun sir too confirmed on "https://gateoverflow.in/32123/computer-networks-drdo-2008 " that distance must be doubled.

0

@shrestha

visit https://gateoverflow.in/3375/gate2008-it-64

there Tp is taken 4 Tp as I mentioned above. I m confused now

0

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.

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