英语翻译When A comes up,the other routers learn about it via the vector exchanges.For simplicity we will assume that there is a gigantic gong somewhere that is struck periodically to initiate a vector exchange at all routers simultaneously.At the

英语翻译When A comes up,the other routers learn about it via the vector exchanges.For simplicity we will assume that there is a gigantic gong somewhere that is struck periodically to initiate a vector exchange at all routers simultaneously.At the
英语翻译
When A comes up,the other routers learn about it via the vector exchanges.For simplicity we will assume that there is a gigantic gong somewhere that is struck periodically to initiate a vector exchange at all routers simultaneously.At the time of the first exchange,B learns that its left neighbor has zero delay to A.B now makes an entry in its routing table that A is one hop away to the left.All the other routers still think that A is down.At this point,the routing table entries for A are as shown in the second row of Fig.5-10(a).On the next exchange,C learns that B has a path of length 1 to A,so it updates its routing table to indicate a path of length 2,but D and E do not hear the good news until later.Clearly,the good news is spreading at the rate of one hop per exchange.In a subnet whose longest path is of length N hops,within N exchanges everyone will know about newly-revived lines and routers.
Now let us consider the situation of Fig.5-10(b),in which all the lines and routers are initially up.Routers B,C,D,and E have distances to A of 1,2,3,and 4,respectively.Suddenly A goes down,or alternatively,the line between A and B is cut,which is effectively the same thing from B's point of view.
At the first packet exchange,B does not hear anything from A.Fortunately,C says:Do not worry; I have a path to A of length 2.Little does B know that C's path runs through B itself.For all B knows,C might have ten lines all with separate paths to A of length 2.As a result,B thinks it can reach A via C,with a path length of 3.D and E do not update their entries for A on the first exchange.
On the second exchange,C notices that each of its neighbors claims to have a path to A of length 3.It picks one of the them at random and makes its new distance to A 4,as shown in the third row of Fig.5-10(b).Subsequent exchanges produce the history shown in the rest of Fig.5-10(b).
From this figure,it should be clear why bad news travels slowly:no router ever has a value more than one higher than the minimum of all its neighbors.Gradually,all routers work their way up to infinity,but the number of exchanges required depends on the numerical value used for infinity.For this reason,it is wise to set infinity to the longest path plus 1.If the metric is time delay,there is no well-defined upper bound,so a high value is needed to prevent a path with a long delay from being treated as down.Not entirely surprisingly,this problem is known as the count-to-infinity problem.There have been a few attempts to solve it (such as split horizon with poisoned reverse in RFC 1058),but none of these work well in general.The core of the problem is that when X tells Y that it has a path somewhere,Y has no way of knowing whether it itself is on the path.

英语翻译When A comes up,the other routers learn about it via the vector exchanges.For simplicity we will assume that there is a gigantic gong somewhere that is struck periodically to initiate a vector exchange at all routers simultaneously.At the
当A出现时,其他路由器了解它通过媒介交流.为了简单起见我们假设有一个巨大的功是平衡的地方开始定期在所有的路由器矢量交流同时进行.在第一次交流时间,乙得知其左邻居零延迟到A
B现在使得在它的路由表项A是一跳距离到左边.所有其他路由器仍然认为,一个已关闭.在这一点上,对阿现图中的第二行显示的路由表项.5-10（1）.在接下来的交流,ç得知B的长度为1的路径,因此更新其路由表来表示
路径长度为2,但D和E不直到后来听到这个好消息.显然,好消息是散布在每汇率跳率.在一个子网的最长路径长度为N啤酒花在n交流,每个人,会知道新恢复系和路由器.
现在让我们考虑图的情况.5月10日（二）,所有的线条和路由器初步建立.路由器乙,丙,丁,E有距离的1,2,3 A和4分别.突然出现故障,或者,在A和B之间的线路被切断,这实际上是从B的角度来看,同样的事情.
在第一次分组交换,B没有听到答：幸运的是什么,C说：别担心,我有一个路径的长度为2.小乙不知道C的道路,通过B本身运行.对于所有B知道,C可能有10条线所有的长度为2一个单独的路径.因此,乙认为它可以达到通过C一个有3路径的长度.
D和E不更新阿的第一次交流作品.
在第二次交换,ç通知书,其每一个邻国,声称路径长度为3甲.它选取其中一个随机他们并提出新的路程,前往4,显示在图第三排.5-10（b）项.随后的交流,产生其余的图所示的历史.5-10（b）项.
从这个数字,应该很清楚,为什么坏消息传缓慢：没有路由器都具有一个以上的价值高于其所有邻国的最低.渐渐地,所有路由器的工作,跑的无穷大,但所需的交流人数为无穷使用的数值而定.为此,明智的做法是设置无穷最长的路径加1.
如果度量时间延迟,没有明确的上限,因此需要高价值以防止长时间的延迟路径中被当作了处理.不完全不足为奇,这个问题被称为计数到无穷大的问题.有一些试图解决这个问题（如与中毒水平分割在RFC 1058相反）,但没有Ø
f这些工作良好.问题的核心是,当X告诉辎它有一个路径下,Y已无法知道自己是否在路径上的方式.