The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+37 votes
6.2k views

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 × 108 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.

  1. $120$ $\text{bytes}$
  2. $60$ $\text{bytes}$
  3. $240$ $\text{bytes}$
  4. $90$ $\text{bytes}$
asked in Computer Networks by Boss (19.1k points)
edited by | 6.2k views
0
here they asked for  channel utilization, why every one is calculating for sender utilization?
+1
Both are same !!

3 Answers

+62 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.

answered by Active (3.9k points)
edited by
+2
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
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 )
0

Why is Time taken for transmission (Tt) 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
efficeancy = $\frac{TT}{TT+ RTT}$

TT = packet size / bandwidth

now put in place of TT in above formula you will get same
0
Thanks!
0

How to attempt sums with this heavy calculation?

0
Acc to this it gives 60 not 120.
0
Because A sends the message whose size is significant so its transmission time is considered. When B receives the message, it will send acknowledgement (ACK packet) not the message and in the question it is clearly stated that acknowledgment packets are of negligible size so its transmission time is not considered. Hope your doubt is clear now.
+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

answered by Veteran (104k points)
+1
somewhere I read in this cases round trip time is taken as 4Tp  [A->satelite,satelite->B,B->satelite,satelite->A]  ??
0
120 should be ans
0
+1
same way as explained srestha ... only distance doubled... you will get 120B
0
but why u take A->satelite as tp?

means we can take A->B as tp.rt?
0

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

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

+1
yes. Tp is distance between sender and receiver. but distance must be doubled. so 160  Bytes should be right
0
@Gabbar

I am sorry

but in selected ans it is not considered, rt?
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

+1

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 vote

Ans: A

answered by (151 points)
Answer:

Related questions



Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

44,337 questions
49,834 answers
164,735 comments
65,874 users