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试卷年份2012年下半年
试题题型【单选题】
试题内容

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 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 (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 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 (3) 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 (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 recoveries up to (5) layers.

(1)A.datagram
 B.collision
 C.connection
D.service
(2)A.slot
 B.switch
 C.process
 D.fire
(3)A.rest
 B.random
 C.once
 D.odds
(4)A.unicast
 B.multicast
 C.broadcast
 D.interval
(5)A.local
 B.next
 C.higher
 D.lower

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