Optimal Lot Size with Partial Backlogging Under the Occurrence of Imperfect Quality Items

Abstract

In this study, a continuous review inventory system with deterministic demand, partial backlogging, and imperfect quality items is considered. More precisely, the fraction of imperfect quality items is assumed as a random variable with a known distribution function. The order quantity is subjected to a 100%, error-free, screening process, with finite screening rate. After inspection, the imperfect quality items can be classified into two categories: low quality items and defective items. The demand rate is constant and manifests even during screening period. The demand during the stockout period is satisfied partially as soon as stock is available and before the new demand is met. Perfect and imperfect quality items are charged with different holding cost, giving the chance of different treatment for the two categories of products. The objective is to find the order quantity that maximizes the total profit of the system per unit time. Beyond, the analytical properties are established, the impact of imperfect quality and holding costs differentiation are examined and the behavior of the relative error using the EOQ with partial backlogging solution is displayed graphically. Finally, it is shown that this model can be reduced to other models existing in the literature.

Appendix

Proof of Proposition 1

$$\displaystyle \begin{aligned} \dfrac{\partial TP_{ut}(X,B)}{\partial X}&=\dfrac{\gamma B}{DX^{2}}+\dfrac{K}{X^{2}}-\dfrac{\delta}{2D}+\dfrac{\alpha B^{2}}{2DX^{2}} \end{aligned} $$

(32)

$$\displaystyle \begin{aligned} \dfrac{\partial TP_{ut}(X,B)}{\partial B}&=-\dfrac{\gamma}{DX}+\dfrac{\epsilon}{D}-\dfrac{\alpha B}{DX} \end{aligned} $$

(33)

$$\displaystyle \begin{aligned} \dfrac{\partial^2 TP_{ut}(X,B)}{\partial X^2}&=-\dfrac{2\gamma B}{DX^{3}}-\dfrac{2K}{X^{3}}-\dfrac{\alpha B^{2}}{DX^{3}} \end{aligned} $$

(34)

$$\displaystyle \begin{aligned} \dfrac{\partial^2 TP_{ut}(X,B)}{\partial B\partial X}&=\dfrac{\gamma }{DX^{2}}+\dfrac{\alpha B}{DX^{2}} \end{aligned} $$

(35)

$$\displaystyle \begin{aligned} \dfrac{\partial^2 TP_{ut}(X,B)}{\partial B^2}&=-\dfrac{\alpha}{DX} \end{aligned} $$

(36)

Hence the function TP _ut is concave in X for B constant, concave in B for X constant, and the Hessian matrix is negative definite if

$$\displaystyle \begin{aligned} 2KD\alpha>\gamma^{2} \end{aligned} $$

(37)

□

Proof of Proposition 2

If the inequality (24) holds, then in order to maximize TP _ut(X, B) it is sufficient to solve the following system of equations:

$$\displaystyle \begin{aligned} \alpha B^{2}+2\gamma B-\delta X^{2} + 2KD=0 \end{aligned} $$

(38)

$$\displaystyle \begin{aligned} \alpha B - \epsilon X + \gamma=0 \end{aligned} $$

(39)

Equation (39) always has the solution:

$$\displaystyle \begin{aligned} X^{*}=\dfrac{\alpha}{\epsilon}B+\dfrac{\gamma}{\epsilon} \end{aligned}$$

Replacing X by X ^∗ in (38)

$$\displaystyle \begin{aligned} \dfrac{\partial TP_{ut}(X,B)}{\partial X}\vert_{X=X^{*}}=\alpha(\epsilon^2-\alpha\delta)B^{2}+2\gamma(\epsilon^2-\alpha\delta)B+2KD\epsilon^2-\gamma^2\delta \end{aligned} $$

(40)

is obtained. This quantity has two roots if

$$\displaystyle \begin{aligned} (\epsilon^{2}-\alpha \delta)(\gamma^{2}-2KD\alpha)>0 \end{aligned}$$

Because of the inequality (37), the above inequality holds if

$$\displaystyle \begin{aligned} \epsilon^{2}-\alpha \delta<0 \end{aligned}$$

or, equivalently

$$\displaystyle \begin{aligned} h_{g}^{2}E_{4}^{2}< \left(h_{g}E_{5}+\dfrac{c_{b}E_{2}}{b}\right)[h_{g}z+h_{g}E_{3}+h_{d}zE(p)] \end{aligned}$$

One root is always negative. The other is positive if

$$\displaystyle \begin{aligned} T_{w}>\dfrac{\gamma}{\epsilon} \end{aligned} $$

Taking into account the inequality for the concavity of the objective function and making simplifications, it follows that it is optimal to allow shortages if

$$\displaystyle \begin{aligned} (\epsilon^{2}-\alpha \delta)(\gamma^{2}-2KD\alpha)>0 \end{aligned}$$

and

$$\displaystyle \begin{aligned} T_{w}>\dfrac{\gamma}{\epsilon} \end{aligned} $$

where T _W in (25) is the cycle length in [19] model.

Hence, if the inequalities (26) and (27) hold

$$\displaystyle \begin{aligned} B^{*}&=-\dfrac{\gamma}{\alpha}+ \dfrac{\epsilon}{\alpha}\sqrt{\dfrac{2KD\alpha-\gamma^{2}}{\alpha\delta-\epsilon^{2}}}\\ X^{*}&=\sqrt{\dfrac{2KD\alpha-\gamma^{2}}{\alpha\delta-\epsilon^{2}}} \end{aligned} $$

Using the relation (22) the result is obtained.

Otherwise, B ^∗ = 0 and either there should be an inventory system (i.e. Q ^∗ > 0) or not. If inequality (26) does not hold, then $\dfrac {\partial TP_{ut}(X,B)}{\partial X}\vert _{X=X^{*}}>0$, thus the function TP _ut(X, B) is increasing in X and the maximum value is obtained for X →∞ which gives

$$\displaystyle \begin{aligned} B^{*}&=0\\ Q^{*}&=0 \end{aligned} $$

If inequality (27) does not hold, then

$$\displaystyle \begin{aligned} B^{*}&=0\\ Q^{*}&=\sqrt{\dfrac{2KD}{(h_{g}z+h_{g}E_{3}+h_{d}zE(p))}} \end{aligned} $$

If the inequality (24) does not hold, then either there should not be any inventory system (i.e. Q ^∗ = 0 = B ^∗) or $Q^{*}=\sqrt {\dfrac {2KD}{(h_{g}z+h_{g}E_{3}+h_{d}zE(p))}}$ and B ^∗ = 0. □