Question 4 (P2P Tit-for-Tat, 16%). In this question, we consider a slightly modified Tit-for-Tat algorithm in a P2P system. As shown in the figure below, A and B are communicating with their top-6 partners in a P2P system. In this scenario, each peer sends chunks to those six (instead of four) peers currently sending her chunks at highest rate. A's uploading and downloading data rates of the ith partner are u and di respectively; B's uploading and downloading data rates of the ith partner are u and di respective. For i = 1, 2, ..., 6, ui, u', di, d' are randomly distributed. They are independent random variables, following uniform distribution in [0, 1] Mbps. Now A optimistically unchoked B, with a sending data rate of rab. Tab is a random variable, independent of ui, u, di, dr. It follows uniform distribution in [0,1] Mbps. If A becomes a top-6 sender of B, B will start to serve A with a sending data rate of Tba = Tab. What is the probability that both A and B find each other a top-6 sender? Show your mathematical derivations. ud 42 46 B