当超时发生时，设 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 连接恢复到 32 KB 拥塞窗口所用的时间(以毫秒为单位)是_________。
(A) 1100 至 1300
(B) 800 至 1000
(C) 1400 至 1600
(D) 1500 至 1700答案：(一)
解释：
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).这个问题的测验