This work studies the deadlock and blockage control problem of an automated manufacturing system (AMS) with a single unreliable resource. It aims to develop a robust control policy to ensure that AMS can produce all parts in the absence of resource failures, and when the unreliable resource fails, the system can continuously produce all parts that do not require the failed resource. To this end, we divide the system into two regions, continuous and non-continuous, based on whether all parts in them can be produced continuously or not. For the non-continuous region, dominating region constraints are established to ensure that all parts in it do not block the production of parts in the continuous region, and an optimal deadlock avoidance policy based on a Petri net model is introduced to guarantee its deadlock-free operation. For the continuous region, we configure a resource order policy to ensure the smooth productions of AMS. By integrating the dominating region constraints and deadlock avoidance policy with the configured resource order policy, we propose a novel robust control policy. It is proven to be of polynomial complexity and more permissive than the existing one with the same resource order policy. Also, it is tested to be more permissive than other existing policies. φ(r s )∩φ(r t ) = ∅, ∀r s , r t ∈R F such that s ≠ t.Proof. Suppose that p jk ∈φ(r s )∩φ(r t ) for s ≠ t. Since p jk ∈φ(r s ), for some c ≥ 0, ρ(p j(k+c) ) = r s and p jk ∈φ(p j(k+c) ). Also, for some d ≥ 0, ρ(p j(k+d) ) = r t and p jk ∈φ(p j(k+d) ). Since r s , r t ∈R F , for some e ≥ c, d, ρ(p j(k+e) ) = r u . Without loss of generality, suppose that 0 ≤ c < d ≤ e. p jk ∈φ(p j(k+d) ) implies that ∀v∈[k, k + d), ρ(p jv )∉R F . This is a contradiction, since (k + c)∈[k, k + d) and ρ(p j(k+c) ) = r s ∈R F . ■We can easily prove the following properties. Property 3. Movements of parts within a dominating set do not violate Δ 1 . Property 4. Movements of parts out of N f do not violate Δ 1 . Theorem 5. Let S be an AMS, (N, M 0 ) be its buffer net, and (N f , M f 0 ) = (P f ∪R F , T f , F f , M f 0 ) be its FBS. Given M∈R(N, M 0 , Δ 1 ), let M f = ϒ (M) and ς(M) = {t | t ∈T f , M f [t>M f1 ∈R(N f , M f 0 , ς)}. If λ = 1 and ∀p∉P F , M(p) = 0, then ς(M) ≠ ∅and ∀t∈ς(M), t ∈ D M ð Þ = {t | t ∈T f , M[t>M 1 ∈R(N, M 0 , D)}. Proof. By Theorems 2 and 4, ς is a DAP for (N f , M f 0 ), and thus ς(M) ≠ ∅. Suppose that ∃t∈ς(M) but t ∉D(M). We consider two cases a) t is not enabled at M in N and b) firing t at M violates D. Case a) t is not enabled at M in N. Since M f [t>M f1 , M f ( (a) t) > 0. Since M f =ϒ (M), M(p) = M f (p) holds for p ∈P f . Hence, M( (a) t) > 0. Thus, t is not resource-enabled at M in N, that is, M(r 1 ) = 0, where r 1 = (r) t in N. If r 1 ∈R F and (r 1 , t)∈N f , then M(r 1 ) > 0 since M f (r 1 ) > 0 (M f [t>)and by Lemma 1, M(r 1 ) ≥ M f (r 1 ). Thus r 1 ∈R F and (r 1 , t) ∉N f . In this situation, firing t means moving a part within a dominating set. According to Property 3, M f1 (φ(r 1 )) = M f (φ(r 1 )). Since M∈R(N, M 0 , Δ ...