Use the nonnegative potential F=Phi/[M(b+w)]+Xi/(Qn). Its initial average is at most two. Fix one whole schedule a_v or d_f at a time. For each candidate, compute the average F over the still-unfixed independent schedules, conditional on all choices already made. The current average is the average of these candidate averages, so at least one candidate is no larger. Choose one such candidate. Once every schedule is fixed, no randomness remains and the actual F is at most two. This is a deterministic selection argument, not a claim that every random assignment succeeds or that either imbalance must become zero.