This continues through iteration i. Since the user is saturated at I, its throughput y satisfies y = x(r(Z)). Also, since all users of I are saturated, ysat(I, i + 1) = y(I) and y(l, i + 1) = x(&) - y(2)) = y. Thus, due to the saturation allocation at I, (1976)

by iteration
Venue:in Proc. 3rd Int. Conf, Comput. Commun