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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 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

Tt >= 2*Tp

or     Tt 2*Tp

which is correct

@JAINchiNMay It is $Tt >=2*Tp$.

In order to detect the collision, in worst case keep in transmitting the data→ Until collision is detected (got back). Now suppose Stations $A$ and $B$ are at extreme  ends, and $A$ start transmitting it’s packet at $t=0$. Assume exactly at $Tp-1 s$ (say units are in sec), B also starts, then at $t = Tp$ there will be collision. So it will take one more $Tp$ for the collision to reaach at $A$. So $A$ must transmit atleast for $2*Tp$.

thanks :)

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
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

### 1 comment

RTT = 2* Propagation Time
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