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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
local 
B
next 
C
higher 
D
lower 
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答案:

C

解析:

题目描述的是在发生碰撞后如何进行随机化的过程。在发生碰撞后,时间被分割成等长的离散槽,每个槽的长度等于以太网最坏情况的往返传播时间。为了适应以太网允许的最长路径,槽时间被设定为512比特时间,即51.2微秒。第一次碰撞后,每个站点在再次尝试之前会等待0或1次随机槽时间。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。第二次碰撞后,每个站点随机选择0、1、2或3个槽时间进行等待。如果发生第三次碰撞(这种情况发生的概率是0.25),那么下一次等待的槽数将从0到2³-1的区间内随机选择。一般来说,经过I次碰撞后,会选择一个介于0和2¹²-1之间的随机数,并跳过相应数量的槽。但是,在达到10次碰撞后,随机化的过程被冻结在最多1023个槽。在发生16次碰撞后,控制器放弃并报告计算机失败。根据题目描述,这个过程发生在网络协议的较低层次,因此选项C “higher” 不符合题意,应选择 “lower” 层次进行进一步的恢复。因此,横线处应选C。

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