The Gateway to Computer Science Excellence
+20 votes
4.4k views

A $2$ $km$ long broadcast LAN has $10^7$ bps bandwidth and uses CSMA/CD. The signal travels along the wire at $2 \times 10^8$ m/s. What is the minimum packet size that can be used on this network?

  1. $50$ $\text{bytes}$
  2. $100$ $\text{bytes}$
  3. $200$ $\text{bytes}$
  4. None of the above
in Computer Networks by
edited by | 4.4k views

7 Answers

+32 votes
Best answer
In CSMA/CD, to detect a collision the transmission time (which depends on the
packet size) must be greater than twice the propagation delay.

Propagation delay here = $\dfrac{2 km}{  2 \times 10^8 m/s}=10 \text{ microseconds}$

Now, transmission time for $x$ bytes = $\dfrac{x \times 8}{10^7} = 0.8x\text{ microseconds}$

So, $0.8x > 2 \times 10 \implies x > 25 \text{ bytes}$

So, None of these.

Correct Answer: $D$
by
edited by
+3
Total distance for RTT = 4 Km

Transfer rate =2 *10^8   m/s  

Time to transfer =(4*10^3)/(2*10^8)=2*10^-5 sec

Data rate  =  10^7 bps

Packet size  = 2*10^-5*10^7 bits = 200 bits

so its option D ....
+8

bps means bits per second not Bytes per second

0
(doubt) yes the packet size ahould be 25 bytes, but question says what could be minimum size .

a packet should be equal to or more than 25bytes so can't we say 50 bytes(option a) is fulfilling the question criteria?
0

@skyby

50B is valid but it is not minimum.

but if it is asking minimum from the options then 50B is correct.

0
well that's quite dependent on how we interpret question in one go during Exam, anyway thanks for clearing it
0

@Arjun   @Bikram  

Sir,Why we are not adding jamming signal 32 bits long  when collision occurs .i.e

Tx of frame = 2*propogation time+ Tx of Jamming Signal  in this question it is not mention about jamming signal but whenever collision occurs jammming signal is generated  

Similar type of question is asked in Gate 2005 https://gateoverflow.in/1397/gate2005-74 but in that jamming signal was mentioned in the question

 

 

 

+7 votes

For CSMA/CD Protocol formula of minimum packet size 

L >= 2*Tp*B 

where B is the bandwidth and Tp is the propagation delay which is D/V D is the distance and V is the velocity that signal travels through the wire so by putting all these values we get that L>= 200 bits which is L>= 25 bytes 

Ans D

by
+6 votes
In CSMA/CD, the transmitting node is listening for collisions while it transmits it's frame. Once it has finished transmitting the final bit without hearing a collision, it assumes that the transmission was successful. In this worst-case collision scenario, the time that it takes for a Node to detect that its frame has been collided with is twice the propagation delay. Hence to confirm that the collision has not occurred the condition for the minimum size of the packet is:
 
RTT = 2 * Propagation Time
Transmission Time = Length of packet / Bandwidth
RTT = 2 (d/v) = 2(2000/2×108
Therefore to find minimum size of the packet, 
RTT <= Length of packet / Bandwidth
Length of packet >= RTT x Bandwidth
                         = 2(2000/2×108) x 107 = 200bits = 25bytes
 
Therefore, minimum size of the packet = 25bytes
by
edited by
+2
RTT = 2* Propagation Time
0 votes
L/B=2D/V ( we are using 2 because we want minimum packet size)

L=(2*B*D)/V

2*(10^7)/8 * 2000/2*10^8

=>25 bytes

so the answer is D
by
–5 votes

Lets explore the answer 

As we know:

TT = framesize/Bandwidth;

and minimum time taken to acquire the line is = 2*pt

TT = 2*pt

min frame size = 2*pt*b.w;

=200 bits(C)

by
+3
The answer will be 200 bits, not bytes.

hence option D).
–5 votes
In CSMA/CD length of Packet is suppose is L

L>=2*Tp*Bw

L>=2*d/v*Bw

Put  the data in above quetion and u will find answer like below :: 200Byte.
by
0
The answer will be 200 bits, not bytes.

hence option D).
–6 votes
ans 200 bit
by
–2
Should be100 bits.
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
52,375 questions
60,615 answers
202,053 comments
95,435 users