An assignment of tasks to servers can resist every small improvement and still make tasks wait longer than necessary. We determine exactly how inefficient such an assignment can be when each task requires one unit of service and the eligibility constraints permit all tasks to use distinct servers. For every move size $r$ and maximum current server load $K$, we give a closed formula for the worst ratio between locally optimal and globally optimal total completion time. Local optimality here allows every feasible reassignment changing at most $r$ tasks. Every finite-cap bound is attained on a tree where each task has at most two eligible servers. Thus the worst behavior already occurs under simple eligibility constraints. At load cap two, the exact ratio is $1+1/(r+2)$, attained on a path with $r+2$ tasks. Without a load cap, the worst-case supremum is $3/2$ for single-task moves and approximately $1.294503159$ for two-task moves; its excess above one is $1/(r+2)+O(2^{-r}/r)$ as $r$ grows. The proof uses an explicit rational potential on a comparison graph and matching extremal constructions. These results give sharp guarantees for bounded-size local search on matchable semi-matchings, including exact guarantees under degree bounds.

翻译：暂无翻译