diff --git a/src/optimizer/optimizer.py b/src/optimizer/optimizer.py
index 44c4f0bfe..7f0f5b5c3 100644
--- a/src/optimizer/optimizer.py
+++ b/src/optimizer/optimizer.py
@@ -82,6 +82,15 @@ def objective_scale(objective) -> float:
 # those requests but no money, see the fallback in _solve_preferences.
 COST_BOUND_SLACK = 1e-5
 
+# ceiling for the slack ladder in _solve_preferences. CBC declares the pinned tie break LP
+# infeasible on some production splits over a bound its own incumbent satisfies: the model's
+# big M rows put CBC's solve-time perturbation orders above any hundredth-of-a-cent slack, and
+# no algorithm switch helps (primal, dual, barrier and perturbation-off all agree). Measured on
+# two captured requests, feasibility returns at 3e-4 and 1e-3. Each retry multiplies the slack
+# by ten, so the ceiling is reached in three; it equals the default gap_abs, money the cost
+# stage may already have left on the table, and the alternative is no tie break at all.
+COST_BOUND_SLACK_CEILING = 1e-2
+
 # how far below the bound the check in _solve_preferences still accepts a solution. CBC treats
 # that row like any other, so it may miss it by its feasibility tolerance: measured at 1.2e-5 on
 # 024-attenuate-demand-peaks, where rejecting over it threw away a tie break worth four times the
@@ -834,9 +843,20 @@ def keep():
         # clock says and the strategies get something even when the search below never starts.
         pinned = self._pin_integers()
         try:
+            slack = COST_BOUND_SLACK
             with self._timed('tie_break'):
                 self.problem.solve(self._solver(tmpdir, timeLimit=LP_PREFERENCE_TIME_LIMIT))
-            stages.append('LP ' + pulp.LpStatus[self.problem.status] + ('' if keep() else ' unused'))
+                # CBC declares this LP infeasible on ~6% of production splits, over a bound the
+                # incumbent it just returned satisfies, see COST_BOUND_SLACK_CEILING. Walk the
+                # slack up until CBC can hold the row; keep() follows via budget.
+                while (pulp.LpStatus[self.problem.status] == 'Infeasible'
+                       and slack < COST_BOUND_SLACK_CEILING):
+                    slack *= 10
+                    budget = self.settings.preference_budget + slack
+                    self.problem.constraints[COST_BOUND].constant = -(cost - budget)
+                    self.problem.solve(self._solver(tmpdir, timeLimit=LP_PREFERENCE_TIME_LIMIT))
+            label = 'LP' if slack == COST_BOUND_SLACK else f'LP (slack {slack:g})'
+            stages.append(label + ' ' + pulp.LpStatus[self.problem.status] + ('' if keep() else ' unused'))
         finally:
             self._unpin_integers(pinned)
 
diff --git a/tests/test_objective_split.py b/tests/test_objective_split.py
index de11adca6..3db9a098b 100644
--- a/tests/test_objective_split.py
+++ b/tests/test_objective_split.py
@@ -48,6 +48,29 @@ def solve_cost_only(optimizer):
     return optimizer
 
 
+def test_the_lp_floor_relaxes_a_bound_cbc_reports_infeasible(monkeypatch):
+    # production: CBC declares the pinned LP infeasible over a cost bound the incumbent it just
+    # returned satisfies -- the model's big M rows put the solver's perturbation orders above the
+    # slack. The floor must walk COST_BOUND_SLACK up rather than give up the tie break.
+    optimizer = solve_cost_only(build('026-attenuate-grid-peaks'))
+    calls = {'n': 0}
+    real_solve = pulp.LpProblem.solve
+
+    def infeasible_twice(self, solver):
+        result = real_solve(self, solver)
+        calls['n'] += 1
+        if calls['n'] <= 2:
+            self.status = pulp.LpStatusInfeasible
+        return result
+
+    monkeypatch.setattr(pulp.LpProblem, 'solve', infeasible_twice)
+    with TemporaryDirectory() as tmpdir:
+        optimizer._solve_preferences(tmpdir, None)
+
+    # two refusals walk the slack from 1e-5 over 1e-4 to 1e-3, and the tie break still lands
+    assert 'LP (slack 0.001) Optimal' in optimizer.preference_stage
+
+
 @pytest.mark.parametrize('case', CASES)
 def test_objective_split_covers_the_whole_objective(case):
     # the split must stay exhaustive: cost plus preferences has to equal what a single solve would