Question

TCP provides congestion control through slow start and AIMD (additive) increase and multiplicative decrease). The receive window is 24 KB and congestion window is set to 18 KB. How big will the window be if the next three transmission bursts are all successful? Assume that the maximum segment size is 2 KB - If the round-trip time is 12 -msee and no congestion, how long does it take to send message of size 30 KB ? How big will be the window if time out occurs after sending the fourth transmission burst?

Answer 1

TCP window size and round-trip time calculations, and the effects of timeouts on congestion window.

Step-by-step explanation:

The TCP window size is determined by the receive window and the congestion window. In this case, the receive window is 24 KB and the congestion window is 18 KB. To find the size of the window after three successful transmission bursts, we need to add the maximum segment size (2 KB) to the congestion window for each successful transmission burst. So, the window size would be 18 KB + 2 KB + 2 KB + 2 KB = 24 KB.

When calculating the time taken to send a message of size 30 KB, we need to consider the round-trip time and the window size. The time taken to send one window-full of data is equal to the round-trip time multiplied by the window size. In this case, the window size is 24 KB, so the time taken to send one window-full would be 12 ms * 24 KB = 288 ms. Since the message size is 30 KB, it would take multiple window-fulls to send the entire message. Therefore, the total time taken would be the time taken for one window-full multiplied by the number of window-fulls needed, which is 30 KB / 24 KB = 1.25. So, the total time taken would be 1.25 * 288 ms = 360 ms.

If a timeout occurs after sending the fourth transmission burst, TCP will assume that there is congestion and perform a multiplicative decrease. This means that the congestion window will be reduced by half. In this case, the congestion window would be reduced to 18 KB / 2 = 9 KB.