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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?

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  • $\begingroup$ This is an excellent question for you to ponder. $\endgroup$ Oct 16 '17 at 7:54
  • $\begingroup$ @YuvalFilmus The stations participating in aloha don't need to transmit frame starting at the initial time t=0. But the vulnerable time(T) is added to initial time to create some kind of check points (t, t+T, t+2T,...). So if a station decides to send a frame somewhere between (t, t+T), the transmission time will cross the t+T point and use some of the time between t+2T. In order to avoid collision we have to include the next duration (t+T, t+2T) also into vulnerable time because another station might decide to transfer between time period (t+T, t+2T) causing collision. $\endgroup$
    – Sahil
    Oct 16 '17 at 8:29
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    $\begingroup$ Great! So you can write an answer now. $\endgroup$ Oct 16 '17 at 8:30
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    $\begingroup$ @YuvalFilmus thanks sir for confirming that my reasoning was correct. I'll write the answer. $\endgroup$
    – Sahil
    Oct 16 '17 at 8:33
  • $\begingroup$ @YuvalFilmus, but why is twice the frame time the optimal checkpoint? (Frame time is $F_t$) Why not $\frac{3F_t}{2}$? A sender could initiate at $t=0$ and finish sending by $t=F_t$, so how is time$F_t, 2F_t$ vulnerable to the sender? Clearly we are not including propagation delays in the model so I assume the receiver has the data at $t=F_t$. $\endgroup$ Jan 27 '21 at 11:13
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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:

  • the frame time because if transmission of another frame were started in the frame time of the current frame, collision would occur.

  • 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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  • $\begingroup$ Could you expand on the last paragraph? If a sender $A$ has finished sending at a time instant, why can't a new sender start sending immediately afterwards? Does this have anything to do with propagation delays? $\endgroup$ Jan 27 '21 at 11:19
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    $\begingroup$ Hi @sprajagopal, If a sender A has finished sending at a time instant, a new sender (say B) can indeed start sending immediately afterwards. But its transmission should start strictly after A has finished sending. If B starts its transmission even a tiny amount of time before the transmission of A ends, it can lead to collision. (continued...) $\endgroup$ Jan 28 '21 at 4:53
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    $\begingroup$ Suppose A starts transmitting at time instant '0' and finishes at instant 't'. Then B should start at or after instant 't' and continue till instant '2t'. Here, the vulnarable time for the frame transmitted by B 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 to ensure that the frame sent by B doesn't collide. $\endgroup$ Jan 28 '21 at 4:53
  • $\begingroup$ so it's basically a frame time behind that's vulnerable... yeah I think I get it now. Maybe you could add this example as part of the answer. It definitely helps understand better. $\endgroup$ Jan 28 '21 at 5:16
  • $\begingroup$ @sprajagopal, I am glad to know that you are satisfied with my answer. Thanks for your advice. I am working on updating my answer. $\endgroup$ Jan 28 '21 at 5:24
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If vulnerable time was equal to frame time, there is a chance to collide with a frame as there is no gap. Let me give an example :

If there is a car in road and it maintains a gap of 2 meter from front and back there is less chance to hit with another car on the road, or if each and every car in a road maintains 2 meter gap there will be less hit if some one push brake.

But if vulnerable time was equal to frame time that is, if each car on the road maintains zero distance there is maximum possibility to get hit. I think it's clear now.

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    $\begingroup$ But why is it exactly twice the frame time? $\endgroup$ Jun 1 '18 at 9:35
  • $\begingroup$ @Gouranga, could you back up these with some math? Using the Poisson distribution? $\endgroup$ Jan 27 '21 at 11:17
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The stations participating in aloha don't need to transmit frame starting at the initial time t=0.

But the vulnerable time(T) is added to initial time to create some kind of check points (t, t+T, t+2T,...).

So if a station decides to send a frame somewhere between (t, t+T), the transmission time will cross the t+T point and use some of the time between t+2T.

In order to avoid collision we have to include the next duration (t+T, t+2T) also into vulnerable time because another station might decide to transfer between time period (t+T, t+2T) causing collision

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  • $\begingroup$ In the Poisson distribution, I understand that we allow only one transmission during the first frame time, but why restrict any transmission from starting after this frame time is complete? Can I allow a new transmission after $\frac{1}{1000}$-th of a frame time after a sender? Why is it a full frame time of no initiations? $\endgroup$ Jan 27 '21 at 11:23
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Suppose at time t=0 station A starts sending the frames, then station A takes T(i.e Transmission Delay) unit time to transmit the frame during this interval if any other station sends the frame then the collision will occur. Then the idea comes if after the time T if any station wants to send the frame then there should not be any collision but wait what if any other sender sends the frame at time T-t'(where t'->0 i.e very small) then it will collide with the frame which is sent by station A and after the collision that useless data transmitted till T-t'+T i.e 2T-t' and up to this time if any station sends frame then that will also collide. So the vulnerable time for the transmission of the frame is taken 2*T because t' is very small it tends to zero so we can neglect that.

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In aloha vulnerable time is twice frame time. To see why this is, suppose you transmit the msg bit in 10sec(FT) to the receiver. So when you start transmission of msg bit and before completing 10 sec if anyone is transmit another msg bit than there will be collision occur same as 10sec before you transmit the msg bit than less collision occur. So we can say total vulnerable time is twice of frame time.

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  • $\begingroup$ I tried to improve the answer a bit but it's still hard to understand. Can you edit it to write in complete sentences? $\endgroup$
    – 6005
    Apr 9 '20 at 15:01

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