Our last example is a window of four segments. These segments are sent at :math:`t_{0}`, :math:`t_{0}+1`, :math:`t_{0}+2` and :math:`t_{0}+3`. The first segment reaches host `D` at :math:`t_{0}+4`. Host `D` replies to this segment by sending an acknowledgment that enables host `A` to transmit its fifth segment. This segment reaches router `R1` at :math:`t_{0}+5`. At that time, router `R1` is transmitting the third segment to router `R2` and the fourth segment is still in its buffers. At time :math:`t_{0}+6`, host `D` receives the second segment and returns the corresponding acknowledgment. This acknowledgment enables host `A` to send its sixth segment. This segment reaches router `R1` at roughly :math:`t_{0}+7`. At that time, the router starts to transmit the fourth segment to router `R2`. Since link `R1-R2` can only sustain 500 Kbps, packets will accumulate in the buffers of `R1`. On average, there will be two packets waiting in the buffers of `R1`. The presence of these two packets will induce an increase of the round-trip-time as measured by the transport protocol. While the first segment was acknowledged within 4 msec, the fifth segment (`data(4)`) that was transmitted at time :math:`t_{0}+4` is only acknowledged at time :math:`t_{0}+11`. On average, the sender transmits at 500 Kbps, but the utilization of a large window induces a longer delay through the network.