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A and B are the only two host on a LAN which uses CSMA/CD protocol. A minimum time required by ‘A’ to detect collision is 600 μs. Find the time taken by the packet to travel from host A to host B.
Correct answer is '300'. Can you explain this answer?
Verified Answer
A and B are the only two host on a LAN which uses CSMA/CD protocol. A ...
Concept:
Contention period is minimum time a host must transmit for before it can be sure that no other host's packet has collided with its transmission. It takes minimum of RTT to detect collision.
Calculation:
Contention Period = RTT = 2 × Tp
600 = 2 × Tp
Tp = 300 μs
Most Upvoted Answer
A and B are the only two host on a LAN which uses CSMA/CD protocol. A ...
Understanding CSMA/CD Protocol
In a network using the CSMA/CD (Carrier Sense Multiple Access with Collision Detection) protocol, hosts A and B share the same communication medium. The collision detection mechanism is crucial for ensuring data integrity during transmission.
Minimum Time to Detect Collision
- The minimum time required by host A to detect a collision is given as 600 microseconds (µs).
- This time includes the round-trip time for a signal to travel from A to B and back to A.
Calculating One-Way Propagation Time
- Since the total time for detection (600 µs) involves the time taken for the packet to travel to host B and then the time taken for the collision signal to return to host A, we can deduce that:
- Round-trip time = Time from A to B + Time from B to A
- Given that both paths are equal in a two-host scenario:
- Total round-trip time = 2 * Time from A to B
- Therefore, Time from A to B can be calculated as:
- Time from A to B = Total round-trip time / 2
- Time from A to B = 600 µs / 2 = 300 µs
Conclusion
The time taken by the packet to travel from host A to host B is 300 microseconds. This conclusion is based on the understanding that the total detection time accounts for both the forward and return paths of the signal in the CSMA/CD environment.
In a network using the CSMA/CD (Carrier Sense Multiple Access with Collision Detection) protocol, hosts A and B share the same communication medium. The collision detection mechanism is crucial for ensuring data integrity during transmission.
Minimum Time to Detect Collision
- The minimum time required by host A to detect a collision is given as 600 microseconds (µs).
- This time includes the round-trip time for a signal to travel from A to B and back to A.
Calculating One-Way Propagation Time
- Since the total time for detection (600 µs) involves the time taken for the packet to travel to host B and then the time taken for the collision signal to return to host A, we can deduce that:
- Round-trip time = Time from A to B + Time from B to A
- Given that both paths are equal in a two-host scenario:
- Total round-trip time = 2 * Time from A to B
- Therefore, Time from A to B can be calculated as:
- Time from A to B = Total round-trip time / 2
- Time from A to B = 600 µs / 2 = 300 µs
Conclusion
The time taken by the packet to travel from host A to host B is 300 microseconds. This conclusion is based on the understanding that the total detection time accounts for both the forward and return paths of the signal in the CSMA/CD environment.
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A and B are the only two host on a LAN which uses CSMA/CD protocol. A minimum time required by ‘A’ to detect collision is 600 μs. Find the time taken by the packet to travel from host A to host B.Correct answer is '300'. Can you explain this answer?
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