Graphic design python turtle 🐢

from turtle import * import colorsys bgcolor('black') pensize(0) tracer(50) h=0 for i in range(300): c=colorsys.hsv_to_rgb(h,1,1) h+=0.9 color(c) forward(300) left(100) fd(i) goto(0,0) down() rt(90) begin_fill() circle(0) end_fill() rt(10) for j in range(5): rt(30) done() Please follow my blog and subscribe my channel for more videos and newly updates 👍👍👍👍👍 import turtle as t import colorsys t.bgcolor('black') t.tracer(100) h=0.4 def draw(ang,n): t.circle(5+n,60) t.left(ang) t.circle(5+n,60) for i in range(200): c=colorsys.hsv_to_rgb(h,1,1) h+=0.005 t.color(c) t.pensize(2) draw(90,i*2) draw(120,i*2.5) draw()

Explain binary exponential back-off algorithm.

Explain binary exponential back-off algorithm.

                                  Or 

Discuss any one algorithm for allocating a multiple access channel.

To ensure that random back-off maintains stability, CSMA/CD uses a technique known as binary exponential back-off. Binary exponential back-off algorithm varies the limit on number of slot times. It works in the following manner-

1) if a stations frame collides for the first time, wait 0 or 1 time slot before trying again.

2) In case of colliding a second time, wait 0, 2 or 3 slot times.

3) If it collides third time, wait anywhere from 0 to 7 slots.

4) After 16 collisions, give up and report an error.

Thus, as congestion increase, stations back-off by larger and larger amounts to reduce the probability of collision. This method clearly attempts to reduce excessive waits by keeping the number of possible time slots small. After all if two stations collide, there is a 50% opportunity they will succeed on the next attempt.

If several stations collide, however, the opportunity are very small that even one will be successful on next attempt. The successful one would have to select either 0 or 1 slot with all the others making the other selections. By raising the number of possible slots after every collision, the opportunity of colliding again decrease exponentially.

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