大门|大门-CS-2014-（第1组）|问题65

当超时发生时，让TCP连接的拥塞窗口大小为32 KB。连接的往返时间为100毫秒，使用的最大段大小为2 KB。TCP连接返回32 KB拥塞窗口所用的时间（以毫秒为单位）为________。 （A） 1100至1300 （B） 800到1000 （C） 1400至1600 （D） 1500到1700 答复: （A） 说明：

Current size of congestion window in terms of number of segments
                            = (Size in Bytes)/(Maximum Segment Size)
                            = 32KB / 2KB 
                            = 16 MSS
          
When timeout occurs, in TCP's Slow Start algorithm, threshold is 
reduced to half which is 16KB or 8MSS. Also, slow start phase begins 
where congestion window is increased twice. 
So from 1MSS to 8 MSS window size will grow exponentially. 
Congestion window becomes 2MSS after one RTT and becomes 4MSS after
2 RTTs and 8MSS after 3 RTTs.  At 8MSS, threshold is reached and
congestion avoidance phase begins.  In congestion avoidance phase,
window is increased linearly. So to cover from 8MSS to 16MSS, it needs
8 RTTs

Together, 11RTTs are needed (3 in slow start phase and 8 in congestion
avoidance phase).

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