Ethernet requires a minimum frame size. For example, 10Base2 Ethernet imposes a

Question

Ethernet requires a minimum frame size. For example, 10Base2 Ethernet imposes a minimum of 64 bytes. This is necessary because when a node transmits in the network it can only detect a collision during its transmission. Therefore, the time of transmission needs to be sufficiently long for a node to find out if another node is also transmitting. From this time, the size of the frame can be derived using the node's transmission speed.

Taking into consideration the distance (d) between nodes in a network, the propagation speed (S) of a signal (measured in meters/second) and the transmission speed (R) for sending bits (measured in bits/second), derive a formula that determines the minimum size in bytes of a frame. Use S, R, and d in your formula to compute length L of the frame size expressed in bytes.

Possible choices:

1) L = 2 * R * d/S
2) L = 0.125 * R * d/S
3) L = R * d/S
4) L = 0.25 * R * d/S
5) None of these

1) L = 2 * R * d/S
2) L = 0.125 * R * d/S
3) L = R * d/S
4) L = 0.25 * R * d/S
5) None of these

Explanation / Answer

ANS;

1) L = 2 * R * d/S

JUSTIFICATION;

Let X and Y be the two nodes here.

At t=0, X sends a frame.

At t= (Propagation Delay-1), Y sends frame just before it identifies X's first bit.

Propagation Delay = (d)Distance between X and Y / (s)Propagation Speed * (R)Transmission Speed

If transmission of Last bit of X is complete before the arrival of Y, then there wont be any collision. To have a collision, The frame size should be greater than or equal to 2* (Propagation Delay-1)

So, L= 2* d/s *R -1

If R= 1 Gbps or 8*10 9 bits per sec, s=1.8*108 m per sec, d= 0.2 km

L= 2 * 0.2/1.8*108 * (8*10 9 )-1

=16.77777

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