Optimal production time and number of maintenance actions for an imperfect production system under equal-interval maintenance policy

Abstract

This paper deals with the optimal production/maintenance (PM) policy for a deteriorating production system which may shift from the in-control state to the out-of-control state while producing items. The process is assumed to have a general shift distribution. Under the commonly used maintenance policy, equal-interval maintenance, the joint optimizations of the PM policy are derived such that the expected total cost per unit time is minimized. Different conditions for optimality, lower and upper bounds and uniqueness properties on the optimal PM policy are provided. The implications of another commonly used policy, to perform a maintenance action only at the end of the production run, are also discussed. Structural properties for the optimal policy are established so that an efficient solution procedure is obtained. In the exponential case, some extensions of the results obtained previously in the literature are presented. A numerical example is provided to illustrate the solution procedure for the optimal production and maintenance policy.

Appendix

Proof of Lemma 1

• Equation (3) and T ^*=T ₁ are obtained by setting the partial derivatives of with respect to T and z equal to zero. To verify that a solution (z*, T*) to Equation (3) and T ^*=T ₁ gives a local minimum, it is sufficient to show that

  

  and the determinant ω of the matrix of the second partial derivatives of at (z*,T*) is positive. It is easy to see that the first partial derivative of on z is given by

  

  From (A.1), we have

  

  Furthermore, the second partial derivative of on z is given by

  

  Thus,

  

  has a unique solution, z*. Furthermore, form (A.1), it is easy to verify that

  

  Taking the first partial derivative of with respect to T results in

  

  Since (z ^*, T ^*) is a solution of

  

  and

  

  it follows that from Equations (A.2) and (A.3), we obtain T ^*=T ₁. Moreover, we have

  

  Furthermore, it is easy to see that

  

  Therefore, the determinant ω of the matrix of the second partial derivatives of at (z ^*, T ^*) is positive, since □

Proof of Lemma 2-(i)

• From (2), we have

  

  Since

  

  as described in the proof of Lemma 1, the result follows from (A.4) and from the fact that is increasing in n. □

Proof of Lemma 2-(ii)

• From (4), if we let

  

  for n=1, 2, …, then the first derivative of L _n (T) on T is given by

  

  This implies that for a larger T, we need a larger n*(T) to satisfy Equation (4) since is increasing in n as shown before. This completes the proof. □

Proof of Lemma 3-(i)

•  From Equation (A.5), it is easy to verify that

  

In addition, L _n (T) is decreasing in T from Equation (A.6). Therefore, for any T>0, if r<(sαP+a)E(X)⩽r+ν, then the smallest natural number which satisfies Equation (4) is one. Now, we will show that has a unique minimizer, T _U ^*. The first derivative of is given by where

Furthermore, it is not hard to see that

Equations (A.9), (A.10), (A.11) and (A.12) collectively imply that has a unique minimizer, T _U ^*. From (A.8), the first derivative of M _U (T) on T is given by M _U ′(T)=T[(sαP+a)f(T)+rf′(T)]⩾0, ∀T⩾0. Also, we have M _U (0)=0. Therefore, M _U (T)⩾0, ∀T⩾0. Hence, U(T ₂ )⩾0. On the other hand, since

we have

Therefore, we have U(T ₁)⩽0. That is, for the general distribution case, the finite interval [T ₁, T ₂] can be searched to obtain T _U ^* using any numerical procedure.

Proof of Lemma 3-(ii)

• It is obvious from Lemmas 2–3.

Proof of Lemma 4

• If (sαP+a)f(x)+rf′(x)⩽0, ∀x>0, then from Equation (A.6), we have L _n (T) is increasing in T for n=1, 2, … . Besides, since lim_{T → 0} L _n (T)=−r, we have L _n (T)⩾−r, for T⩾0.

  This implies that lim_{T → ∞} L _n (T)=−(sαP+a)E(X)⩾−r, and hence (sαP+a)E(X)⩽r, for n=1, 2, … .

  Next, we show the properties of n* and T ^*. When (sαP+a)f(x)+rf′(x)⩽0, we have a cost function as given in (2) (see Tseng⁷). Now, let G _n (x)=∫₀ ^nx[(sαP+a)F(y)+rf(y)] dy−n∫₀ ^x[(sαP+a)F(y)+rf(y)] dy, ∀n=1, 2, 3, …. Taking the first derivative of G _n (x) with respect to x results in G _n ′(x)=n[(sαP+a)F(nx)+rf(nx)]−n[(sαP+a)F(x)+rf(x)]⩽0 since (sαP+a)F(x)+rf(x) is a decreasing function in x and G _n (0)=0.

  Using G _n (x)⩽0, we have

  

  Thus, we have for n=2, 3, 4, …. This implies that n*=1. Next, we show that has a unique minimizer, T _V *. The first derivative of is given by where

  

  and

  

  It is easy to see that

  
  

  which implies that

  

  In addition,

  
  

  Equations (A.14), (A.15), (A.16), (A.17) and (A.18) collectively imply that has a unique minimizer, T _V ^*. From (A.13), the first derivative of M _V (T) on T is given by M _V ′(T)=T[(sαP+a)f(T)+rf′(T)⩽0, ∀T⩾0. Also, we have M _V (0)=0. Therefore, M _V (T)⩽0 ∀T⩾0. Hence, V(T ₂)⩽0. On the other hand, from (A.15), we have V(T ₃)⩾0. Thus, the finite interval [T ₂, T ₃] can be searched to obtain T _V ^* using any numerical procedure. □