Csma/Cd

Question

We know that transmission time Tx=L/B where L is the frame size and B is the bandwidth.

In Csma/Cd, in order to find out the min frame size L, we equate it with the bandwidth delay product and obtain the formula L=BR(R is the round trip time).

R=L/B

R=2Tp (Tp is the one way propagation delay) and Tp is the function of distance between sender and receiver and propagation speed.

Hence Tp=L/(2B)

How is this possible that the Tx and R has the same formula and Tp becomes the function of L and B?

Please clear me this thing .

Answer 1

In CSMA/CD in order to detect the collision we need to transmit the frame twice of the propagational delay i.e Tp time to reach from sender to receiver and another Tp time of collision signal from receiver to sender. So from this $Tt \geq 2*Tp$

Answer 2

u said R = L/B & Tp = L/2B... ie. Tp = 1/2(L/B)..... This means propagation time is half of the round trip. this is what the basic definition says.

Comments

to detect collision in csma/cd Tt >= 2*Tp.

L/B >= 2*Tp

Tp <= L/2B

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to detect collision in csma/cd Tt $\geq$ 2*Tp.

L/B >= 2*Tp

Tp <= L/2B