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单选题

Let us now see how randomization is done when a collision occurs . After a (   ), time is divided into discrete slots whose length is equal to the worst-case round-trip propagation time on ether. To accommodate the longest path allowed by Ethernet, the slot time has been set to 512 bit times, or 51.2us.

 After the first collision, each station waits either 0 or 1 (  ) times before trying again. If two stations collide and each one picks the same random number, they will collide again. After the second collision, each one picks either 0,1,2 or 3 at random and waits that number of slot times. If a third collision occurs (the probability of this happening is 0.25), then the next time the number of slots to wait is chosen at () from the interval 0 to 23-1.  In general, after I collisions, a random number between 0 and 2i -1 is chosen, and that number of slots is skipped. However, after ten collisions have been reached, the randomization ( _____ ) is frozen at a maximum of 1023 slots. After 16 collisions, the controller throws in the towel and reports failure back to the computer. Further recoveries up to (     ) layers.

横线处应选()。

A
unicast
B
multicast
C
broadcast
D
interval 
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答案:

D

解析:

根据题目描述,当发生碰撞后,时间被分割成离散的时隙,每个时隙的长度等于以太网最坏情况的往返传播时间。在第一次碰撞后,每个站点在再次尝试之前会等待0或1个时隙的时间。如果两个站点发生碰撞并且都选择了相同的随机数,它们将再次发生碰撞。在第二次碰撞后,每个站点随机选择0、1、2或3,并等待相应数量的时隙时间。如果在第三次碰撞中(这种情况发生的概率是0.25),那么下一次要等待的时隙数是从0到2^3-1的间隔中随机选择的。一般来说,在I次碰撞后,会从一个介于0和2^i-1的随机数中做出选择,并跳过相应数量的时隙。但在达到十次碰撞后,随机化的间隔被冻结在最大值为1023个时隙。因此,横线上应填入的词是“interval”,表示间隔。选项D为正确答案。

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