Lot sizing in a deteriorating production system under inspections, imperfect maintenance and reworks

Abstract

The paper studies the joint effect of shift, inspections, imperfect preventive maintenance (PM) and imperfect rework of defective items on optimal decisions for a deteriorating production system. At the time of inspection during a production run, either PM or restoration is done depending on the state of the production process. There is a probability that PM action may shift the process from ‘in-control’ state to ‘out-of-control’ state. During the ‘out-of-control’ phase, the system produces some defective items which go for rework at the end of the production run. A portion of the reworked items may also fail repairing. The model is formulated for the case of general inspections and analyzed under two well known inspection policies—periodic inspection policy and constant cumulative hazard inspection policy. For numerical examples, a comparison of the outcomes of the model with and without reworks under these two inspection policies is made. It is observed that the periodic inspection policy performs better than the constant cumulative hazard policy.

Appendices

Appendix 1

Calculation for the expected number of defective items in the ith inspection interval:

During the regular production time, defective items may be produced in the following two cases:

• Case (I) When the process shifts to the ‘out-of-control’ state in the inspection interval [T _i−1 − T _j ,T _i  − T _j ] subject to the condition that the last restoration was done at t _j . Case (II) When the process is found to be in the ‘in-control’ state at the inspection time T _i−1, PM action is carried out at T _i−1. Since PM action itself is not perfect, so due to imperfectness the process may shift to the ‘out-of-control’ state in the inspection interval [T _i−1 − T _j ,T _i  − T _j ] with probability δ.

Then, the expected number of defective items produced in the ith inspection interval (T _i−1, T _i ], given that the last restoration was done at T _j is given by

$$ \begin{aligned} E(N_i) & = p\left[{\sum_{j=0}^{i-1}} Pr(A_j)\left\{\int\limits_{T_{i-1}-T_j}^{T_i-T_j}(T_i-T_j-\tau)f_{X_j}(\tau) d\tau\right\}\right.\\ & \quad\left. +x_i\left\{\delta\left(1-Pr(A_{i-1})\right)\right\}\right]\int\limits_0^{\infty} y g(y) dy\\ & =p\left[{\sum_{j=0}^{i-1}}Pr(A_j)\left\{\left((1-\delta)^{i-j-1} {\prod_{k=j+1}^{i-1}}R(x_k)\right)\int\limits_{T_{i-1}-T_j}^{T_i-T_j} (T_i-T_j-\tau) f_{X_j}(\tau-T_{i-1}+T_j) d\tau \right\}\right.\\ &\quad\left. +x_i\left\{\delta\left(1-Pr(A_{i-1})\right)\right\}\right]\int\limits_0^{\infty} y g(y) dy\\ & =p\left[I_0\times I_1+x_i\left\{\delta\left(1-Pr(A_{i-1})\right)\right\}\right]\int\limits_0^{\infty} y g(y)dy, \end{aligned} $$

where

$$ \begin{aligned} I_0 & ={\sum_{j=0}^{i-1}}P(A_j)\left[(1-\delta)^{i-j-1} {\prod_{k=j+1}^{i-1}}R(x_k)\right]\\ &= P(A_{i-1})+P(A_{i-2}) (1-\delta)R(x_{i-1})+P(A_{i-3})(1-\delta)^2 R(x_{i-2})R(x_{i-1})\\ & \quad+\cdots + P(A_1)(1-\delta)^{i-2}R(x_2)R(x_3)\ldots R(x_{i-1}) + P(A_0)(1-\delta)^{i-1}R(x_1) R(x_2)\ldots R(x_{i-1}) \\ & = P(A_{i-1})+(1-\delta)\left[R(x_{i-1})Pr(A_{i-2}) +(1-\delta)R(x_{i-1})R(x_{i-2})Pr(A_{i-3})\right.\\ &\quad\left. +\cdots +(1-\delta)^{i-3}R(x_2)R(x_3)\ldots R(x_{i-1})P(A_1)\right.\\ &\quad\left. +(1-\delta)^{i-2}R(x_1)R(x_2)R(x_3)\ldots R(x_{i-1})P(A_0)\right]\\ & = P(A_{i-1})+ (1-\delta)\left[R(x_{i-1})Pr(A_{i-2})+(1-\delta) R(x_{i-1})\left\{1-Pr(A_{i-2})\right\}\right]\\ &= P(A_{i-1})+(1-\delta)\left[1-Pr(A_{i-1})\right], \hbox{using the definition of} \,Pr(A_i)\\ & = 1-\delta\left(1-Pr(A_{i-1})\right) \\ \end{aligned} $$

and

$$ \begin{aligned} I_1&= \int\limits_{T_{i-1}-T_j}^{T_i-T_j}(T_i-T_j-\tau) f_X(\tau-T_{i-1}+T_j) d\tau\\ & =\int\limits_0^{x_i}(x_i-u) d F_X(u), \hbox{by taking}\,\tau-T_{i-1}+T_j=u\\ &= \int\limits_0^{x_i}F_X(u)du. \end{aligned} $$

Substituting the values of I ₀ and I ₁ and then using (1), we get the expected number of defective items produced in the interval (T _i−1, T _i ] as

$$ \begin{aligned} E(N_i)&= p\left[\left\{1-\delta {\sum_{k=0}^{i-2}} (-\delta)^k {\prod_{j=i-k-1}^{i-1}}R(x_j)\right\}\int\limits_0^{x_i} F_X(u) du +x_i\left\{\delta {\sum_{k=0}^{i-2}} (-\delta)^k \prod_{j=i-k-1}^{i-1} R(x_j)\right\}\right]\\ &\quad\times \int\limits_0^{\infty} y g_Y(y) dy \end{aligned} $$

(23)

which is the expression given in (3).

Appendix 2

Derivation of expected restoration cost given in (5):

Substituting z = T _i  − T _j  − τ and assuming the restoration cost ϕ as a linear function of detection delay i.e. ϕ(z) = r ₀ + r ₁ z, we get from (4)

$$ \begin{aligned} &\sum_{i=1}^{n}\left[{\sum_{j=0}^{i-1}} Pr(A_j)\int\limits_{T_{i-1}-T_j}^{T_i-T_j}\left\{r_0+r_1(T_i-T_j-\tau)\right\} \left((1-\delta)^{i-j-1}{\prod_{k=j+1}^{i-1}}R(x_k)\right)\right.\\ &\left.\times f_X(\tau-T_{i-1}+T_j) d\tau\right]+{\sum_{i=1}^{n}}\delta \left\{1-Pr(A_{i-1})\right\}\left\{r_0+r_1(T_i-T_{i-1})\right\}\\ & ={\sum_{i=1}^{n}}\left[{\sum_{j=0}^{i-1}} Pr(A_j)\int\limits_0^{x_i} \left\{r_0+r_1(x_i-u)\right\} \left((1-\delta)^{i-j-1}{\prod_{k=j+1}^{i-1}}R(x_k)\right) f_X(u) du\right]\\ &+{\sum_{i=1}^{n}}\delta \left\{1-Pr(A_{i-1})\right\} \left\{r_0+r_1x_i\right\}\\ & ={\sum_{i=1}^{n}}\left[{\sum_{j=0}^{i-1}}Pr(A_j) \left((1-\delta)^{i-j-1}{\prod_{k=j+1}^{i-1}}R(x_k)\right) \left(r_0 F_X(x_i)+r_1\int\limits_0^{x_i}F_X(u) du \right)\right]\\ & +{\sum_{i=1}^{n}}\delta \left\{1-Pr(A_{i-1})\right\} \left\{r_0+r_1x_i\right\}\\ &= {\sum_{i=1}^{n}}\left[\left\{1-\delta {\sum_{k=0}^{i-2}}(-\delta)^k {\prod_{j=i-k-1}^{i-1}} R(x_j)\right\}\left(r_0F_X(x_i)+r_1\int\limits_0^{x_i}F_X(u) du\right)\right]\\ &+{\sum_{i=1}^{n}}\left\{\delta {\sum_{k=0}^{i-2}} \left[(-\delta)^k {\prod_{j=i-k-1}^{i-1}}R(x_j)\right]\right\} \left\{r_0+r_1x_i\right\}, \hbox{using the Eq.} (1)\\ & ={\sum_{i=1}^{n}}\left[\left(r_0F_X(x_i)+r_1\int\limits_0^{x_i} F_X(u) du\right)+\delta \left(r_0+r_1x_i-r_0F_X(x_i)-r_1\int\limits_0^{x_i}F_X(u) du\right)\right.\\ &\left. \times \left\{{\sum_{k=0}^{i-2}} (-\delta)^k {\prod_{j=i-k-1}^{i-1}}R(x_j)\right\}\right]\\ &={\sum_{i=1}^{n}}\left[\psi(x_i) +\delta(r_0+r_1x_i-\psi(x_i)){\sum_{k=0}^{i-2}}(-\delta)^k {\prod_{j=i-k-1}^{i-1}}R(x_j)\right], \end{aligned} $$

where $\psi(x_i)=r_0 F(x_i)+r_1 \int_0^{x_i}F(u) du$.

Appendix 3

Calculation of holding cost

We calculate the expected holding cost using Figs. 1 and 2. Since $t_2=\frac{E(N)}{p_1}$, therefore, we have

$$ \begin{aligned} \Upomega &= p_1-d-\frac{\theta E(N)}{t_2}\\ &= p_1-d-\theta p_1\\ &=(1-\theta)p_1-d,\hbox{ which is a constant}. \end{aligned} $$

Let I(t) denote the inventory level at any time t. Then we have the differential equation

$$ \frac{d I(t)}{dt}=p-d-\frac{E(N)}{t_1}, \hbox{ for } 0\le t \le t_1, $$

(24)

with I(0) = 0. Solving (24), we get $I(t)=(p-d-\frac{E(N)}{t_1})t, 0\le t \le t_1$. Suppose $\Updelta_i$’s denote the areas as indicated in Figs. 1 and 2. Then

$$ \begin{aligned} \Updelta_1 &= \int\limits_0^{t_1} I(t)dt = \frac{1}{2}\left\{(p-d)t_1-E(N)\right\} t_1,\\ \Updelta_2 &=I(t_1)t_2=\left\{(p-d)t_1-E(N)\right\}t_2. \end{aligned} $$

If I(t ₂) denotes the inventory level at the end of time t ₂, then

$$ I(t_2)=I(t_1)+\left\{p_1-d-\frac{\theta E(N)}{t_2}\right\}t_2. $$

Hence,

$$ \begin{aligned} \Updelta_3 &= \frac{1}{2}\left\{I(t_2)-I(t_1)\right\}t_2= \frac{1}{2}\left\{(p_1-d)t_2-\theta E(N)\right\}t_2,\\ \Updelta_4 &= \frac{1}{2}I(t_2)t_3=\frac{1}{2} \left[\left\{(p-d)t_1-E(N)\right\}+\left\{(p_1-d)t_2-\theta E(N)\right\}\right]t_3,\\ \Updelta_5=&\frac{1}{2}E(N)t_1,\\ \Updelta_6=&\frac{1}{2}E(N)t_2=\frac{1}{2}p_1t_2^2, \hbox{ since } t_2=\frac{E(N)}{p_1}. \end{aligned} $$

Now, the expected holding cost of both defective and non-defective items in each production cycle is

$$ \begin{aligned} &c_h\left(\Updelta_1+\Updelta_2+\Updelta_3+\Updelta_4\right)+{c_{h}^{\prime}} \left(\Updelta_5+\Updelta_6\right)\\ & = \frac{c_h}{2}\left[\left\{(p-d)t_1-E(N)\right\}\left(t_1+2 t_2+t_3\right) +\left\{(p_1-d)t_2-\theta E(N)\right\}\left(t_2+t_3\right)\right]\\ &+\frac{c_h'}{2}\left[E(N)t_1+p_1t_2^2\right], \end{aligned} $$

which is the expression given in (6).