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GATE | GATE-CS-2003 | Question 90

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A 2 km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2 × 108 m/s. What is the minimum packet size that can be used on this network?

(A)

50 bytes

(B)

100 bytes

(C)

200 bytes

(D)

None of these


Answer: (D)

Explanation:

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 = Transmission 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 

Source: http://www.btechonline.org/2012/12/gate-computer-networks-ethernet.html


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Last Updated : 25 Nov, 2022
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