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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 (73) 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 (74) 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
slot
B
switch
C
process 
D
fire 
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答案:

A

解析:

根据题目描述,在发生碰撞后,时间被分割成离散的时隙,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,时隙时间被设定为512比特时间,即51.2微秒。在第一次碰撞后,每个站点在再次尝试之前会等待0或1个时隙时间。因此,根据上下文,“slot”(时隙)是合适的选项。

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