3.2.3 Layering

1. Data Transmission Time = Packet size/Bandwidth = 1000×8/2×10⁶ = 4ms

   Propagation delay = Distance/Signal speed = ${\frac{{3000}}{{2 \times 10^{5}}}}$ = 15ms

   Ack Transmission Time = Packet size/Bandwidth = 10×8/2×10⁶ = 0.04ms

   Round Trip Time = 2× propagation delay + data transmission time + ack transmission time = 2×15 + 4 + 0.04 = 34.04ms

   For 100% utilisation, window size × data transmission time > rtt $\rightarrow$ window size = $\lceil$34.04/4$\rceil$ = 9.

   We ignore processing delay at sender and receiver.

2. Assume single channel, ie one in which packets arrive in order but can be lost. Then maximum window size is 2^N - 1 otherwise, the window may advance into space which hasn't been acknowledged, assuming go-back-n.

   If the network can delay, duplicate and reorder packets arbitrarily, then the maximum window size depends upon the maximum packet lifetime.

3. There are 2³² labelled bytes. These can be transmitted in ${\frac{{2^{32}}}{{12.5 \times 10^{6}}}}$ = 343s. Maximum lifetime of a packet must be less than 343 seconds.
4. TCP breaks. Basically a lot of the TCP flows throttle back due to congestion control till they are no longer sending, and those packets sent are often lost, leading to high numbers of retransmissions.