forouzan exercises
-------- Chapter 12 Multiple Access -------
After reading the text about contention window at page 378 carefully and repeatedly, I'm sure that I have mistaked something.
"The number of slots in the window changes according to the binary exponential back-off strategy. This means that it's set to one slot the 1st time and then doubles each time the station cannot detect an idle channel after the IFS time."
According the text listed above, I assumed a station which suffered 3 times back-off, then listed all of the time slots it has occupied below:
At 1st back-off, the station taked 1 time slot of contetion window.
At 2nd back-off, the station taked 2 time slots of contention window.
At 3rd back-off, the station taked 4 time slots of contention window.
But after reading the text flowing the previous one, I found it says: "This is very similar to (the original CSMA's) p-persistent method except that a random outcome defines the number of slots taken by the waiting station."
So in figure 12.17 the block of "choice a random number R between 0 and (2^k)-1" is undoubted exact.
Once again, the example I prefered must be changed below:
At 1st back-off, the station taked a random number of time slots of contetion window with size 0, which is derived from the expression (2^k)-1 where K=0. In other words, the station's 1st sending didn't wait. As the flow in the figure 12.17 showed, it didn't receive ACK and increased k by 1, where k = 1 and less than 15.
At 2nd back-off, the station taked a random number of time slots of contetion window with size 1, which is derived from the express (2^k)-1 where K=1. It's obviously the number of time slots is neither 0 nor 1. No more one another to "be randomly choose". The station still failed to sending data and k was increased to 2.
At 3rd back-off, the station taked a random number of time slots of contetion window with size 3, which is derived from the express (2^k)-1 where K=2.
The station would have 0 time slot to wait and send immediately.
Or It would try to send after waiting for 1 time slot.
Maybe it tried to send after waiting for 2 or 3 time slots.
The number of time slots to wait before try to sending data is between 0 and 3 where the size of contention window is 3. I think that the binary exponential back-off strategy means the growth of the size of contention window, not the "random" waiting time.
I should read this good book more carefully.