= Solution
\b[No: arbitrary ties can produce an assignment that is not <Pareto optimal>.] A two-student <counterexample> suffices. Give the first student value $1$ for both rooms; give the second student value $1$ for $r_1$ and $0$ for $r_2$. If the first student breaks the tie by taking $r_1$, the second must take $r_2$, and the <utility> <vector> is $(1,0)$.
Swapping the rooms gives <utility> <vector> $(1,1)$, so it is a <Pareto improvement>. Both choices in the original run maximized the corresponding student's available room value. The strict-ranking argument from (a) fails because the first student's <utility> need not decrease when that student's room changes.
Back to article page
