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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 (75) layers.

横线处应选(   )。 

A
rest
B
random
C
once  
D
odds 
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答案:

B

解析:

根据题目描述,当发生冲突时,时间被分成离散的时隙,每个时隙的长度等于以太网最坏情况的往返传播时间。为了容纳以太网允许的最长路径,时隙时间被设置为512比特时间,即51.2微秒。第一次冲突后,每个站点在再次尝试之前会等待0或1次随机数。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。因此,可以推断出第二次冲突后,每个站点随机选择等待的时隙数是在0到某个值之间。第三次冲突发生时,等待的插槽数量从0到某个更大的值范围内随机选择。因此,可以推断出空格处应填入与随机数相关的词语。选项B中的“random”符合语境,表示随机选择等待的时隙数。因此,正确答案是B。

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