The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
0 votes
Consider two Station communicating via 1 Mbps Satellite link with propagation delay of 270 msec. The Satellite merely serves to retransmit data received from one station to another,with negligible switch delay. If Ethernet frame size is 1024 bits with 3 bit sequence number,then max. possible data throughput is.................
asked in Computer Networks by Loyal (6.9k points) | 80 views
Please clear my doubt here:-

From an uplink station when the signal will reach Satellite and then from satelite to receiver i.e. Down link so total 2*Tpropagation. But say collision occurs at the receiver place ( It may be when say receiver wants to send data back to sender it sends and collision happens) Now it generates jamming signal to reach that jamming signal to sender it will require 2*Tpropagation again so wont the time will be = 4*Tpropagation.


Approach i used : -

Total time = 4 *270 + 1.024 = 1081.024 ms

So Utilization = (7 * 1.024)/1081.024

Throughput = Utilization * Bandwidth = 6.630Kbps

Why its wrong ?
It's Ethernet. I have Already mentioned Above There is no Ack.
It ain't the Acknowledgement.

In an ethernet Why we say Transmission is atleast 2*Tpropagation ? Becasue

When say sender is transmitting the data it listens for collision and say when the frame reaches near to destination the destination just a while before sensed the medium to be free and start transmitting resulting in collision. Now my transmission should be taken in such a way that the worst case collision can also be heard so now receiver sends a jamming signal for alerting everyone that collision had occured to reach that jamming signal to sender total time is 2*Tpropagation.

Likewise here mode of transportation is by satellite thats why it should be 4*Tpropagation


can you explain me this line 1024 -144(why ?)

Truly speaking i also didnt get this line they are subtracting the overheads like checksum and all that and taking into consideration Pure Payload i guess :/
@Magma can u see my doubt above written here :)

Na462  I had also  same doubt  !!

Dharmendra Lodhi   can you please help  ???

transmission time is 1 ms is right. why taking 1Mbps as 10^6 bps instead of 2^20 bps? 2^20 bps is more logical and intuitive, as we generally deal with bits and bytes.
what will be efficiency in this case???
correct me if i am wrong for efficiency (for GBN)

sender window size=7

maximum possible window size=1+2*$T_{P}/T_{t}$

now putting this value


now for throughput=efficiency*bandwidth

2 Answers

0 votes

In the Above question, the maximum windows Size is 1+2a

if we are using the max sender windows size(stop and wait ARQ, GBN ARQ, Selective Repeat ARG)

windows Size of GBN is (using 3 sequence bit ) 7

than efficiency = 7/(1+2a)        (a= tp/tt )

tt =1024/1*10^6 = 1024 micro second 

a =270/1024=0.26

1+2a =0.26*2+1=1.52

efficiency =7/1.52  =4.6




answered by Active (1k points)
–1 vote

Is the solution correct??

answered by (9 points)
Hey what kind of answer is this ??

Related questions

0 votes
0 answers
asked Sep 15 in Computer Networks by Na462 Loyal (6.9k points) | 53 views

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

42,626 questions
48,616 answers
63,876 users