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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: $D$

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: $D$

@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 $2*Tp$.

0

9 votes

For CSMA/CD Protocol formula of minimum packet size

L >= 2*T_{p}*B

where B is the bandwidth and T_{p} 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

6 votes

RTT = 2 * Propagation Time

Transmission Time = Length of packet / Bandwidth

RTT = 2 (d/v) = 2(2000/2×10^{8})

Therefore to find minimum size of the packet,

RTT <= Length of packet / Bandwidth

Length of packet >= RTT x Bandwidth

= 2(2000/2×10^{8}) x 10^{7} = 200bits = 25bytes

Therefore, minimum size of the packet = 25bytes