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  3. Let us now see how randomization is done

Let us now see how randomization is done

考试题库2022-08-02  11
问题 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 the ether(2τ). To accommodate the longest path allowed by Ethernet, the slot time has been set t0 512 bit times, or 51.2μsec.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 recoveryis up to （　　） layers.问题1选项A.datagramB.collisionC.connectionD.service问题2选项A.slotB.switchC.processD.fire问题3选项A.restB.randomC.onceD.odds问题4选项A.unicastB.multicastC.broadcastD.interval问题5选项A.localB.nextC.higherD.lower
选项
答案 BABDC
解析 现在让我们看看当发生冲突时，随机性操作是如何体现的。出现冲突时，时间被划分为离散的时槽，其长度等于最坏情况下以太网的周转传播时间（2τ），为了适应以太网中的最长通路，时槽被设为512比特的发送时间，即15.2微秒。
第一次冲突后，每个站在再次试图发送前等待0或1个时槽。如果两个站出现冲突，并且每个站都选用了同样的随机数，那么就会再一次发生冲突。第二次发生冲突后，每个站随机地选取数字0、1、2或者3，并等待相应的时槽数。如果发生了第三次冲突（这种情况出现的概率为0.25），则下一次等待的时槽数目就随机的在0~23-1中选取。
一般情况下，第i次冲突后，随机数在0到2i-1之间选取，相应的时槽数被跳过。然而，达到10次冲突后，随机数被固定在最大1023个时槽之内。16次冲突后，控制器放弃发送，向计算机发出故障报告。近一步的恢复措施由上层协议实施。
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本试题收录于： 中级 网络工程师题库软件水平考试初中高级分类

中级 网络工程师

软件水平考试初中高级

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