Consider now a window of two segments. Host `A` can send two segments within 2 msec on its 1 Mbps link. If the first segment is sent at time :math:`t_{0}`, it reaches host `D` at :math:`t_{0}+4`. Host `D` replies with an acknowledgment that opens the sending window on host `A` and enables it to transmit a new segment. In the meantime, the second segment was buffered by router `R1`. It reaches host `D` at :math:`t_{0}+6` and an acknowledgment is returned. With a window of two segments, host `A` transmits at roughly 500 Kbps, i.e. the transmission rate of the bottleneck link.