patternMinor
Why is the vulnerable time in pure aloha twice the frame time?
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whythepuretwicetimealohavulnerableframe
Problem
The time required to send a frame is called frame time.
Vulnerable time is the time during which no transmission should be done to avoid any collision.
My question is what kind of problem will be created if vulnerable time was equal to frame time?
Vulnerable time is the time during which no transmission should be done to avoid any collision.
My question is what kind of problem will be created if vulnerable time was equal to frame time?
Solution
This is because in pure ALOHA, even if a bit of a frame collides with a bit of another frame, both the frames get discarded. Also, in pure ALOHA, a station doesn't listen to the medium before transmitting. So, it has no way of knowing that another frame was already underway. If another frame was indeed underway already, then the newly transmitted frame becomes vulnerable. That is why it is called "vulnerable time". It equals twice of the frame time because it counts the time in which the transmission of another frame should start so as to make the current frame vulnerable. This time interval includes:
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the frame time because if transmission of another frame were started in the frame time of the current frame, collision would occur.
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a time interval (equal to the frame time) before the frame time because if transmission of another frame were started in this time interval, collision would still occur.
My question is what kind of problem will be created if vulnerable time was equal to frame time?
If the vulnerable time were equal to frame time, any frame (say A) transmitted prior to the considered frame (say B) could be transmitted within the frame time of the considered frame (B), resulting in a collision.
Suppose a sender X starts transmitting at time instant '0' and finishes at instant 't'. Then any other sender Y should start at or after instant 't' and continue till instant '2t' (because if Y start its transmission even a tiny amount of time before 't', there will be collision). Here, the vulnarable time for the frame transmitted by Y is the sum of its own frame time (i.e. 't' to '2t') and another time duration (i.e. '0' to 't'), effectively being '0' to '2t'. This is the time interval during which no sender should send on the channel to ensure that the frame sent by Y doesn't collide.
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the frame time because if transmission of another frame were started in the frame time of the current frame, collision would occur.
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a time interval (equal to the frame time) before the frame time because if transmission of another frame were started in this time interval, collision would still occur.
My question is what kind of problem will be created if vulnerable time was equal to frame time?
If the vulnerable time were equal to frame time, any frame (say A) transmitted prior to the considered frame (say B) could be transmitted within the frame time of the considered frame (B), resulting in a collision.
Suppose a sender X starts transmitting at time instant '0' and finishes at instant 't'. Then any other sender Y should start at or after instant 't' and continue till instant '2t' (because if Y start its transmission even a tiny amount of time before 't', there will be collision). Here, the vulnarable time for the frame transmitted by Y is the sum of its own frame time (i.e. 't' to '2t') and another time duration (i.e. '0' to 't'), effectively being '0' to '2t'. This is the time interval during which no sender should send on the channel to ensure that the frame sent by Y doesn't collide.
Context
StackExchange Computer Science Q#82542, answer score: 4
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