Fuzzy mixture two warehouse inventory model involving fuzzy random variable lead time demand and fuzzy total demand

Abstract

This paper considers a two-warehouse fuzzy-stochastic mixture inventory model involving variable lead time with backorders fully backlogged. The model is considered for two cases—without and with budget constraint. Here, lead-time demand is considered as a fuzzy random variable and the total cost is obtained in the fuzzy sense. The total demand is again represented by a triangular fuzzy number and the fuzzy total cost is derived. By using the centroid method of defuzzification, the total cost is estimated. For the case with fuzzy-stochastic budget constraint, surprise function is used to convert the constrained problem to a corresponding unconstrained problem in pessimistic sense. The crisp optimization problem is solved using Generalized Reduced Gradient method. The optimal solutions for order quantity and lead time are found in both cases for the models with fuzzy-stochastic/stochastic lead time and the corresponding minimum value of the total cost in all cases are obtained. Numerical examples are provided to illustrate the models and results in both cases are compared.

Appendices

Appendix 1

Remark 1

If per unit holding cost at $RW$ and $OW$ are same (i.e. $h_1 = h_2 = h$, say) then the total average holding cost expression is same with the holding cost expression of single warehouse EOQ model for the same order quantity.

Proof

$$\begin{aligned}&\text{ AHCRW+} \text{ AHCOW}\\&\quad = \frac{h_2}{2Q}\left({Q-W+k\sigma \sqrt{L}} \right)^{2}+\frac{h_1 }{Q}\left[{WQ-\frac{1}{2}\left({W-k\sigma \sqrt{L}} \right)^{2}} \right]\\&\quad = \frac{h}{Q}\left[{\frac{1}{2}\left\{ {(Q\!-\!W\!+\!k\sigma \sqrt{L}) ^{2}\!-\!\frac{1}{2}(W\!-\!k\sigma \sqrt{L})^{2}} \right\} \!+\!WQ} \right] [\text{ Putting}\,\,h_1 = h_2 = h]\\&\quad =\frac{h}{Q}\left[{Q\left({\frac{Q}{2}-W+k\sigma \sqrt{L}} \right)+WQ} \right] =h\left({\frac{Q}{2}+k\sigma \sqrt{L}} \right)\\&\quad =h\left({\frac{Q}{2}+r-\mu L} \right) [\text{ Proved}]. \end{aligned}$$

Appendix 2

$$\begin{aligned}&\int \limits _r^\infty {(x-r)f(x)dx} = \int \limits _r^\infty {(x-r)\frac{1}{\sqrt{2\pi } \sigma \sqrt{L}}e^{-\frac{1}{2}\left({\frac{x-\mu L}{\sigma \sqrt{L}}} \right)^{2}}dx}\\&\quad =\int \limits _r^\infty {\left\{ {(x-\mu L)-(r-\mu L)} \right\} \frac{1}{\sqrt{2\pi } \sigma \sqrt{L}}e^{-\frac{1}{2}\left({\frac{x-\mu L}{\sigma \sqrt{L}}} \right)^{2}}dx}\\&\quad = \int \limits _r^\infty {(x-\mu L)\frac{1}{\sqrt{2\pi } \sigma \sqrt{L}}e^{-\frac{1}{2}\left({\frac{x-\mu L}{\sigma \sqrt{L}}} \right)^{2}}dx} -\int \limits _r^\infty {(r-\mu L)\frac{1}{\sqrt{2\pi } \sigma \sqrt{L}}e^{-\frac{1}{2}\left({\frac{x-\mu L}{\sigma \sqrt{L}}} \right)^{2}}dx}\\&\quad =\int \limits _k^\infty {\frac{1}{\sqrt{2\pi } }t} e^{-\frac{1}{2}t^{2}}dt-\sigma \sqrt{L} \int \limits _k^\infty {\frac{1}{\sqrt{2\pi } }} e^{-\frac{1}{2}t^{2}}dt\\&\qquad \quad [\text{ Putting} \frac{x-\mu L}{\sigma \sqrt{L}}=t\,\,\text{ in} \text{ both} \text{ integrations} \text{ and} \text{ using}\,\,r = \mu L + k\sigma \sqrt{L}]\\&\quad = -\int \limits _k^\infty {\frac{1}{\sqrt{2\pi } }\frac{d}{dt}\left({e^{-\frac{1}{2}t^{2}}} \right)} dt-\sigma \sqrt{L} \int \limits _k^\infty {\frac{1}{\sqrt{2\pi } }} e^{-\frac{1}{2}t^{2}}dt\\&\quad = \sigma \sqrt{L}\Psi (k),\text{ where}, \Psi (k)\equiv \phi (k)-k[1-\Phi (k)][\text{ Proved}]. \end{aligned}$$

Appendix 3 (Markov inequality)

$$\begin{aligned} \Pr ob [B+pX\ge p(Q+r)]\ge \lambda \Rightarrow \int \limits _{B+pX\ge p(Q+r)} {f(x)} dx\ge \lambda \end{aligned}$$

(34)

Again, $\int \nolimits _{B+pX\ge p(Q+r)} {\left({\frac{B+pX}{p(Q+r)}} \right)f(x)} dx\ge \int \nolimits _{B+pX\ge p(Q+r)} {f(x)} dx$

We have

$$\begin{aligned} \int \limits _{-\infty } ^\infty {\left({\frac{B+pX}{p(Q+r)}} \right)f(x)dx}&\ge \int \limits _{B+pX\ge p(Q+r)} {\left({\frac{B+pX}{p(Q+r)}} \quad \right)f(x)} dx\nonumber \\&\ge \int \limits _{B+pX\ge p(Q+r)} {f(x)} dx\nonumber \\&\Rightarrow \frac{E(B+pX)}{p(Q+r)}\ge \lambda [\text{ By} \text{(A1)}]\nonumber \\&\Rightarrow B+pE(X)\ge \lambda (pQ+r) [\text{ Proved}]. \end{aligned}$$

(35)

Appendix 4

$$\begin{aligned}&\frac{1}{\sqrt{2\pi } \Delta _2} \int \limits _k^{\frac{\Delta _2}{\sigma _L}} {(\Delta _2 -\sigma _L w)e^{-\frac{w^{2}}{2}}dw}\\&\quad = \frac{1}{\sqrt{2\pi } }\int \limits _k^{\frac{\Delta _2}{\sigma _L }} {e^{-\frac{w^{2}}{2}}dw} -\frac{\sigma _L}{\sqrt{2\pi } \Delta _2} \int \limits _k^{\frac{\Delta _2}{\sigma _L} } {we^{-\frac{w^{2}}{2}}dw}\\&\quad = \Phi \left({\frac{\Delta _2}{\sigma _L} } \right)-\Phi \left(k \right) + \frac{\sigma _L}{\Delta _2} \frac{1}{\sqrt{2\pi }}\int \limits _k^{\frac{\Delta _2}{\sigma _L} } {\frac{d}{dw}\left({e^{-\frac{w^{2}}{2}}} \right)dw}\\&\quad = \Phi \left({\frac{\Delta _2}{\sigma _L} } \right)-\Phi \left(k \right) + \frac{\sigma _L}{\Delta _2} \left| {\phi (w)} \right|_k^{\frac{\Delta _2}{\sigma _L} }\\&\quad = \Phi \left({\frac{\Delta _2}{\sigma _L} } \right)-\Phi \left(k \right) + \frac{\sigma _L}{\Delta _2} \left({\phi \left({\frac{\Delta _2}{\sigma _L}} \right)-\phi (k)} \right) [\text{ Proved}] \end{aligned}$$