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+19 votes

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 Veteran (52.2k points)
edited by | 3.2k views

7 Answers

+30 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 Veteran (424k points)
edited by
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 ....

bps means bits per second not Bytes per second

(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?


50B is valid but it is not minimum.

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

well that's quite dependent on how we interpret question in one go during Exam, anyway thanks for clearing it
+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 Active (5.1k points)
+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 Active (1.3k points)
edited by
RTT = 2* Propagation Time
0 votes
L/B=2D/V ( we are using 2 because we want minimum packet size)


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

=>25 bytes

so the answer is D
by (11 points)
–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 (177 points)
The answer will be 200 bits, not bytes.

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



Put  the data in above quetion and u will find answer like below :: 200Byte.
by Boss (10k points)
The answer will be 200 bits, not bytes.

hence option D).
–6 votes
ans 200 bit
by Loyal (5.2k points)
Should be100 bits.

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