●Let us now see how randomization is done when a collision occurs.After a(71),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 l (72) 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 recoveryis up to (75 )layers.(71) A.datagramB.collisionc.connectionD. service(72) A.slotB.switchC.processD.fire(73) A.restB.randomC.onceD.odds(74) A.unicastB.multicastC.broadcastD.interval(75) A.localB.nextC.higherD.lower

●Let us now see how randomization is done when a collision occurs.After a(71),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 l (72) 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 recoveryis up to (75 )layers.

(71) A.datagram

B.collision

c.connection

D. service

(72) A.slot

B.switch

C.process

D.fire

(73) A.rest

B.random

C.once

D.odds

(74) A.unicast

B.multicast

C.broadcast

D.interval

(75) A.local

B.next

C.higher

D.lower


相关考题:

38.—__________ do you want _________for a party?—Let's ask Peter for advice. He knows all the parks in this city.A. What,to buyB. Where,to goC. How,to startD.When,to begin

计划的主要内容包括(),计划必须清楚地确定和描述这些内容。A.What,Why,Who,Where,When,Howmany,HowmuchB.What,Why,We,Where,When,Howto,HowmuchC.What,Why,We,Where,When,How,HowmuchD.What,Why,Who,Where,When,How,Howmuch

消费者消费行为“5W”原则是指:何时(When)、何地(Where)、如何(how)、何人(Who)、()。

在职业生涯决策的过程中,我们需要考虑的因素包括() A、whoB、whatC、when和whereD、whyE、how

5W2H法要求人们从以下7个方面进行设问:What、Who、、When、Where、How、Howmuch。

计划的主要内容包括______计划必须清楚的确定和描述这些内容。A.What,Why,Who,Where,When,How many,How muchB.What,Why,We,Where,When,How to,How muchC.What,Why,We,Where,When,How,How muchD.What,Why,Who,Where,When,How,How much

使用哪个函数可插入当前日期?A.today()B.now()C.time()D.new()

计划的主要内容包括______,计划必须清楚地确定和描述这些内容。A.What, Why, Who, Where, When, How many, How muchB.What, Why, We, Where, When, How to, How muchC.What, Why, We, Where, When, How, How muchD.What, Why, Who, Where, When, How, How much

5W2H分析法,包括7个要素:why, what, who, when, where, how, how much.

3、5W2H分析法,包括7个要素:why, what, who, when, where, how, how much.