2. 计算机
  3. Let us now see how randomization is done when a collision occurs．After a (1) ，time is divided into discrete slots whose length

Let us now see how randomization is done when a collision occurs．After a (1) ，time is divided into discrete slots whose length

admin2019-05-15  27
问题 Let us now see how randomization is done when a collision occurs．After a  (1)  ，time is divided into discrete slots whose length is equal to the worst-case round-trip propagation time on the ether(2t)．To accommo-date the longest path allowed by Ethernet，the slot tome has been set to 512 bit times，or 512 μsec．
    After the first collision，each station waits either 0 or 1  (2)  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，or3 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  (3)  from the interval 0 to 23～1．
    In general，after i collisions，a random number between 0 and 2i一1 is chosen，and that number of slots is skipped．However，after ten collisions have been reached，the randomization  (4)  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 recovery iS up to  (5)  layers．
(5)
选项 A、local
B、next
C、higher
D、lower
答案C
解析 现在让我们观察冲突发生时如何做随机处理。冲突发生后，时间被划分成离散的长度等于最坏的往返传播时间的时槽。为了容纳以太网允许的最长路径，冲突时槽缩小为5．12μs。
    第一次冲突后，每个站再次尝试前需要等待0或1时槽。如果每一站发生冲突，且每一个挑选相同的随机数，它们将再次发生冲突。笫2次冲突之后，每一站随机选择0、1、2或3，等待时槽的个数。如果第三次冲突发生(发生的概率为0．25)，则下次时槽的等待数量都是在0～23一1之间随机挑选的。
    总的来说，第i次冲突后，将在O～2i一1之间挑选随机数，而且那个时槽数是被略过的。然而，当冲突次数达到10次，随机间隔被锁定在最高1023时槽。当16次冲突后，控制器丢弃报告而没有返回计算机。进一步恢复已经达到更高的层次。
转载请注明原文地址:https://jikaoti.com/ti/Xbi7FFFM
本试题收录于： 网络工程师上午基础知识考试题库软考中级分类
0

网络工程师上午基础知识考试

软考中级

相关试题推荐
随机试题
最新回复(0)