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A production-inventory model with imperfect production process and partial backlogging under learning considerations in fuzzy random environments

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Abstract

In this paper, we investigates the learning effect of the unit production time on optimal lot size for the imperfect production process with partial backlogging of shortage quantity in fuzzy random environments. It is assumed that the setup cost, the average holding cost, the backorder cost, the raw material cost and the labour cost are characterized as fuzzy variables and the elapsed time until the machine shifts from “in-control” state to “out-of-control” state is characterized as a fuzzy random variable. As a function of these parameters, the average total cost is also a random fuzzy variable. Based on the credibility measure of fuzzy event, the fuzzy random total cost function is transformed into an equivalent crisp function. We propose an algorithm to determine the optimal solution. Furthermore, the model is illustrated with the help of numerical example. Finally, sensitivity analysis of the optimal solution with respect to major parameters is carried out.

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Acknowledgments

The authors wish to thank the anonymous referees for their valuable comments and suggestions that improved the quality of the paper to a great extent.

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Authors and Affiliations

  1. Department of Mathematics, Sitananda College, Purba Medinipur District, Nandigram, 721631, West Bengal, India

    Gour Chandra Mahata

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  1. Gour Chandra Mahata
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Correspondence to Gour Chandra Mahata.

Appendices

Appendix 1

Proof of inference 1

For the outer integral, integral interval \((-\infty , \infty )\) with respect to \(x\) can be divided into the following five intervals \((-\infty , t(q_i)-\Delta -\Delta _2]\), \([t(q_i)-\Delta -\Delta _2, t(q_i)-\Delta ]\), \([t(q_i)-\Delta , t(q_i)]\), \([t(q_i), t(q_i)+\Delta _1]\) and \([t(q_i)+\Delta _1, \infty )\). Then we have

$$\begin{aligned}&E[N(t(q_i))]\nonumber \\&\quad =\tau \hat{p}(q_i) \Big [\int _{-\infty }^{t(q_i)- \Delta -\Delta _2} \int _0^{t(q_i)} \Phi (y;x)dy f(x)dx \nonumber \\&\quad \quad +\, \int _{t(q_i) -\Delta -\Delta _2}^{t(q_i) -\Delta } \int _0^{t(q_i)}\Phi (y;x)dy f(x)dx\nonumber \\&\quad \quad +\, \int _{t(q_i)-\Delta }^{t(q_i)} \int _0^{t(q_i)}\Phi (y;x)dy f(x)dx\nonumber \\&\quad \quad +\, \int _{t(q_i)}^{t(q_i)+\Delta _1}\int _0^{t(q_i)} \Phi (y;x)dy f(x)dx \nonumber \\&\quad \quad +\, \int _{t(q_i)+\Delta _1}^{\infty } \int _0^{t(q_i)}\Phi (y;x)dy f(x)dx\Big ]. \end{aligned}$$
(28)

For the first item in the square bracket, since \(x\in (-\infty , t(q_i)-\Delta -\Delta _2]\) which implies \(x+\Delta +\Delta _2\le t(q_i)\), then the inner integral interval \([0,t(q_i)]\) with respect to \(y\) can be divided into the following five intervals \([0, x-\Delta _1]\), \([x-\Delta _1, x]\), \([x, x+\Delta ]\), \([x+\Delta , x+\Delta +\Delta _2]\) and \([x+\Delta +\Delta _2, t(q_i)]\) (see Fig. 3). The rest items in the square bracket can be similarly discussed. Hence we have

$$\begin{aligned}&E[N(t(q_i))]\nonumber \\&\quad =\tau \hat{p}(q_i) \int _{-\infty }^{t(q_i)-\Delta -\Delta _2}\Big [\int _0^{x-\Delta _1} \Phi (y;x)dy \nonumber \\&\quad \quad +\, \int _{x-\Delta _1}^x \Phi (y;x)dy+ \int _x^{x+\Delta } \Phi (y;x)dy \nonumber \\&\quad \quad +\, \int _{x+\Delta }^{x+\Delta +\Delta _2}\Phi (y;x)dy \!+\! \int _{x+\Delta +\Delta _2}^{t(q_i)} \Phi (y;x)dy \Big ]f(x)dx \nonumber \\&\quad \quad +\, \tau \hat{p}(q_i) \int _{t(q_i)-\Delta -\Delta _2}^{t(q_i)-\Delta } \Big [\int _0^{x-\Delta _1} \Phi (y;x)dy \nonumber \\&\quad \quad +\, \int _{x-\Delta _1}^x \Phi (y;x)dy + \int _x^{x+\Delta } \Phi (y;x)dy \nonumber \\&\quad \quad +\, \int _{x+\Delta }^{t(q_i)} \Phi (y;x)dy\Big ]f(x)dx \nonumber \\&\quad \quad +\, \tau \hat{p}(q_i) \int _{t(q_i)-\Delta }^{t(q_i)} \Big [\int _0^{x-\Delta _1} \Phi (y;x)dy \nonumber \\&\quad \quad +\, \int _{x-\Delta _1}^x \Phi (y;x)dy + \int _x^{t(q_i)}\Phi (y;x)dy \Big ]f(x)dx\nonumber \\&\quad \quad +\,\tau \hat{p}(q_i) \int _{t(q_i)}^{t(q_i)+\Delta _1} \Big [\int _0^{x-\Delta _1} \Phi (y;x)dy\nonumber \\&\quad \quad +\, \int _{x-\Delta _1}^{t(q_i)} \Phi (y;x)dy\Big ]f(x)dx\nonumber \\&\quad \quad +\, \tau \hat{p}(q_i) \int _{t(q_i)+\Delta _1}^\infty \int _0^{t(q_i)} \Phi (y;x)dy f(x)dx. \end{aligned}$$
(29)

From the credibility distribution \(\Phi (y;x)\) of \(\xi (x)\), we can obtain

$$\begin{aligned}&E[N(t(q_i))]= \tau \hat{p}(q_i)\int _{-\infty }^{t(q_i)-\Delta -\Delta _2} \Big [\int _0^{x-\Delta _1} 0 dy\\&\quad +\, \int _{x-\Delta _1}^x \frac{L(y;x)}{2}dy + \int _x^{x+\Delta } \frac{1}{2}dy \\&\quad +\, \int _{x+\Delta }^{x+\Delta +\Delta _2}\Big [1\!-\!\frac{R(y;x)}{2} \Big ]dy \!+\! \int _{x+\Delta +\Delta _2}^{t(q_i)} 1dy \Big ]f(x)dx \\&\quad +\, \tau \hat{p}(q_i) \int _{t(q_i)-\Delta -\Delta _2}^{t(q_i)-\Delta } \Big [\int _0^{x-\Delta _1} 0dy \\&\quad +\, \int _{x-\Delta _1}^x \frac{L(y;x)}{2}dy + \int _x^{x+\Delta } \frac{1}{2}dy\\&\quad +\, \int _{x+\Delta }^{t(q_i)} \Big [1-\frac{R(y;x)}{2}\Big ]dy\Big ]f(x)dx \\&\quad +\, \tau \hat{p}(q_i) \int _{t(q_i)-\Delta }^{t(q_i)} \Big [\int _0^{x-\Delta _1} 0dy + \int _{x-\Delta _1}^x \frac{L(y;x)}{2}dy\\&\quad +\, \int _x^{t(q_i)}\frac{1}{2}dy \Big ]f(x)dx \! +\!\tau \hat{p}(q_i)\! \int _{t(q_i)}^{t(q_i)+\Delta _1} \Big [\int _0^{x-\Delta _1} 0 dy\\&\quad +\, \int _{x-\Delta _1}^{t(q_i)} \frac{L(y;x)}{2}dy\Big ]f(x)dx \\&\quad +\, \tau \hat{p}(q_i) \int _{t(q_i)+\Delta _1}^\infty \int _0^{t(q_i)} 0 dy f(x)dx. \end{aligned}$$
$$\begin{aligned}&= \tau \hat{p}(q_i) \int _{-\infty }^{t(q_i)-\Delta -\Delta _2} \Big [\int _{x-\Delta _1}^x \frac{L(y;x)}{2}dy\\&\quad +\,\frac{\Delta }{2}dy + \int _{x+\Delta }^{x+\Delta +\Delta _2}\Big [1-\frac{R(y;x)}{2} \Big ]dy\\&\quad +\, (t(q_i)-x-\Delta -\Delta _2) \Big ]f(x)dx \\&\quad +\, \tau \hat{p}(q_i) \int _{t(q_i)-\Delta -\Delta _2}^{t(q_i)-\Delta } \Big [\int _{x-\Delta _1}^x \frac{L(y;x)}{2}dy + \frac{\Delta }{2}dy\\&\quad +\, \int _{x+\Delta }^{t(q_i)} \Big [1-\frac{R(y;x)}{2}\Big ]dy\Big ]f(x)dx\\&\quad +\, \tau \hat{p}(q_i) \int _{x-\Delta _1}^x \frac{L(y;x)}{2}dy + \frac{1}{2}(t(q_i)-x) \Big ]f(x)dx \\&\quad +\,\tau \hat{p}(q_i) \int _{t(q_i)}^{t(q_i)+\Delta _1} \Big [\int _{x-\Delta _1}^{t(q_i)} \frac{L(y;x)}{2}dy\Big ]f(x)dx \end{aligned}$$
$$\begin{aligned}&\quad = \frac{\tau \hat{p}(q_i)}{2} \int _{-\infty }^{t(q_i)-\Delta -\Delta _2} \Big [\int _{x-\Delta _1}^x L(y;x)dy\nonumber \\&\quad \quad -\, \int _{x+\Delta }^{x+\Delta +\Delta _2} R(y;x)dy (2t(q_i)-2x-\Delta ) \Big ]f(x)dx\nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i)-\Delta -\Delta _2}^{t(q_i)-\Delta }\Big [\int _{x-\Delta _1}^x L(y;x)dy \nonumber \\&\quad \quad -\, \int _{x+\Delta }^{t(q_i)}R(y;x)dy + (2t(q_i)-2x-\Delta )\Big ]f(x)dx \nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i)- \Delta }^{t(q_i)}\! \Big [\int _{x-\Delta _1}^x\! L(y;x)dy +\,(t(q_i)\!-\!x)\Big ]f(x)dx \nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i)}^{t(q_i)+\Delta _1} \int _{x-\Delta _1}^{t(q_i)} L(y;x)dy f(x)dx. \end{aligned}$$
(30)

The proof is completed.

Appendix 2

Proof of inference 2 Since \(L(y;x)=\frac{y-(x-\Delta _1)}{\Delta _1}\) for \(y\in [x-\Delta _1, x)\) and \(R(y;x)=\frac{(x+\Delta +\Delta )-y}{\Delta _2}\) for \(y\in [x+\Delta , x+\Delta +\Delta _2)\), then by substituting them into \(E[N(t(q_i))]\) in Inference 1, we can get

$$\begin{aligned}&E[N(t(q_i))]\nonumber \\&\quad =\frac{\tau \hat{p}(q_i) }{2}\int _{-\infty }^{t(q_i) -\Delta -\Delta _2}\Big [ \int _{t(q_i) -\Delta _1}^{t(q_i)} \frac{y-(x-\Delta _1)}{\Delta _1}dy \nonumber \\&\quad \quad -\, \int _{x+\Delta }^{x+\Delta +\Delta _2} \frac{(x+\Delta +\Delta _2)-y}{\Delta _2}dy \nonumber \\&\quad \quad +\,(2t(q_i)-2x-\Delta )\Big ]f(x)dx\nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i)-\Delta -\Delta _2}^{t(q_i) -\Delta }\Big [ \int _{x-\Delta _1}^x \frac{y-(x-\Delta _1)}{\Delta _1}dy \nonumber \\&\quad \quad -\, \int _{x+\Delta }^{t(q_i)} \frac{(x+\Delta +\Delta _2) -y}{\Delta _2}dy \nonumber \\&\quad \quad +\,(2t(q_i)-2x-\Delta )\Big ]f(x)dx\nonumber \\&\quad \quad + \,\frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i) -\Delta }^{t(q_i)}\Big [ \int _{x-\Delta _1}^x \frac{y-(x-\Delta _1)}{\Delta _1}dy\nonumber \\&\quad \quad +\,(t(q_i)-x)\Big ]f(x)dx \nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i)}^{t(q_i) +\Delta _1} \int _{x-\Delta _1}^{t(q_i)} \frac{y-(x-\Delta _1)}{\Delta _1}dy f(x)dx.\nonumber \\ \end{aligned}$$
(31)

By solving the inner integral, we have

$$\begin{aligned}&E[N(t(q_i))]\nonumber \\&\quad =\frac{\tau \hat{p}(q_i)}{2} \int _{-\infty }^{t(q_i) -\Delta -\Delta _2} \Big [\frac{y^2}{2\Delta _1}\Big |_{x-\Delta _1}^x\nonumber \\&\quad \quad -\,\frac{(x- \Delta _1)y}{\Delta _1} \Big |_{x-\Delta _1}^x -\frac{(x+\Delta + \Delta _2)y}{\Delta _2} \Big |_{x+\Delta }^{x+ \Delta + \Delta _2} \nonumber \\&\quad \quad +\,\frac{y^2}{2\Delta _2}\Big |_{x+\Delta }^{x+ \Delta + \Delta _2} + (2t(q_i)-2x-\Delta )\Big ]f(x)dx\nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i) -\Delta -\Delta _2}^{t(q_i) -\Delta } \Big [\frac{y^2}{2\Delta _1} \Big |_{x-\Delta _1}^x \nonumber \\&\quad \quad -\,\frac{(x-\Delta _1)y}{\Delta _1}\Big |_{x-\Delta _1}^x -\frac{(x+\Delta +\Delta _2)y}{\Delta _2}\Big |_{x+\Delta }^{t(q_i)}\nonumber \\&\quad \quad +\,\frac{y^2}{2\Delta _2}\Big |_{x+\Delta }^{t(q_i)}+ (2t(q_i) -2x- \Delta ) \Big ]f(x)dx \nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2}\int _{t(q_i)- \Delta }^{t(q_i)} \Big [\frac{y^2}{2\Delta _1} \Big |_{x-\Delta _1}^x\nonumber \\&\quad \quad -\,\frac{(x-\Delta _1)y}{\Delta _1} \Big |_{x-\Delta _1}^x + (t(q_i) -x)\Big ]f(x)dx \nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2}\int _{t(q_i)}^{t(q_i)+\Delta _1}\Big [\frac{y^2}{2\Delta _1} \Big |_{x-\Delta _1}^{t(q_i)} \nonumber \\&\quad \quad -\,\frac{(x-\Delta _1)y}{\Delta _1} \Big |_{x-\Delta _1}^{t(q_i)} \Big ]f(x)dx\nonumber \\ \end{aligned}$$
$$\begin{aligned}&\quad = \frac{\tau \hat{p}(q_i)}{2} \int _{-\infty }^{t(q_i)-\Delta -\Delta _2} \Big [\frac{x^2-(x- \Delta _1)^2}{2\Delta _1}\nonumber \\&\quad \quad -\,\frac{(x-\Delta _1) \Delta _1}{\Delta _1}- \frac{(x+ \Delta + \Delta _2) \Delta _2}{\Delta _2} \nonumber \\&\quad \quad +\,\frac{(x+\Delta + \Delta _2)^2 -(x+\Delta )^2}{2\Delta _2}\nonumber \\&\quad \quad +\, (2t(q_i) -2x-\Delta )\Big ]f(x)dx \nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i)-\Delta -\Delta _2}^{t(q_i)-\Delta } \Big [\frac{x^2-(x-\Delta _1)^2}{2\Delta _1}\nonumber \\&\quad \quad -\,\frac{(x-\Delta _1) \Delta _1}{\Delta _1}-\frac{(x+\Delta +\Delta _2)(t(q_i)-x- \Delta )}{\Delta _2} \nonumber \\&\quad \quad +\,\frac{t(q_i)^2 -(x+\Delta )^2}{2\Delta _2} + (2t(q_i) -2x- \Delta )\Big ] f(x)dx\nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2}\int _{t(q_i)-\Delta }^{t(q_i)} \Big [\frac{x^2-(x-\Delta _1)^2}{2 \Delta _1} \nonumber \\&\quad \quad -\,\frac{(x-\Delta _1) \Delta _1}{\Delta _1} +(t(q_i)-x) \Big ]f(x)dx \nonumber \\&\quad \quad +\, \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i)}^{t(q_i) +\Delta _1} \Big [\frac{t(q_i)^2 -(x-\Delta _1)^2}{2 \Delta _1}\nonumber \\&\quad \quad -\,\frac{(x-\Delta _1)(t(q_i)-x+ \Delta _1)}{\Delta _1}\Big ]f(x)dx. \end{aligned}$$
(32)

Combining the similar terms, we get

$$\begin{aligned}&E[N(t(q_i))]\nonumber \\&\quad = \frac{\tau \hat{p}(q_i)}{2} \int _{-\infty }^{t(q_i)-\Delta -\Delta _2} \Bigg [ \frac{2x-\Delta _1}{2}\nonumber \\&\qquad -\,(x-\Delta _1)-(x+\Delta +\Delta _2)\nonumber \\&\qquad +\, \frac{2x+2\Delta +\Delta _2}{2} +(2t(q_i)-2x-\Delta )\Bigg ]f(x)dx \nonumber \\&\qquad +\,\frac{\tau \hat{p}(q_i)}{2}\int _{t(q_i) -\Delta -\Delta _2}^{t(q_i) -\Delta }\Bigg [ \frac{2x-\Delta _1}{2}-(x-\Delta _1)\nonumber \\&\qquad -\,\frac{(x+\Delta +\Delta _2) (t(q_i) -x-\Delta )}{\Delta _2}\nonumber \\&\qquad +\, \frac{(t(q_i) +x+\Delta )(t(q_i) -x-\Delta )}{2\Delta _2} \nonumber \\&\qquad +\,(2t(q_i) -2x-\Delta )\Bigg ]f(x)dx\nonumber \\&\qquad +\,\frac{\tau \hat{p}(q_i)}{2}\int _{t(q_i) -\Delta }^{t(q_i)} \Bigg [\frac{(2x\!-\! \Delta _1)}{2} \!-\!(x\!-\!\Delta _1) \nonumber \\&\qquad +\,(t(q_i) \!-\!x)\Bigg ]f(x)dx \!+\! \frac{\tau \hat{p}(q_i)}{2} \int _{t(q_i)}^{t(q_i) +\Delta _1}\nonumber \\&\qquad \, \Bigg [\frac{(t(q_i) -x+\Delta _1)(t(q_i) +x-\Delta _1)}{2\Delta _1}\nonumber \\&\qquad -\,\frac{(x\!-\!\Delta _1)(t(q_i) \!-\!x\!+\! \Delta _1)}{\Delta _1} \Bigg ]f(x)dx. \end{aligned}$$
(33)

By further simplification, we can derive

$$\begin{aligned}&E[N(t(q_i))]\nonumber \\&\quad =\frac{\tau P}{2}\int _{-\infty }^{t(q_i) -\Delta -\Delta _2} \Big (\frac{\Delta _1}{2} -\frac{\Delta _2}{2}\nonumber \\&\qquad +\,2t(q_i)-2x-\Delta \Big )f(x)dx \nonumber \\&\qquad +\,\frac{\tau \hat{p}(q_i)}{2}\int _{t(q_i) -\Delta -\Delta _2}^{t(q_i) -\Delta }\Big [\frac{\Delta _1}{2}\nonumber \\&\qquad +\,\frac{(t(q_i) -x-\Delta )^2}{2\Delta _2} +t(q_i) -x \Big ]f(x)dx\nonumber \\&\qquad +\,\frac{\tau \hat{p}(q_i)}{2}\int _{t(q_i) -\Delta }^{t(q_i)} \Big [\frac{\Delta _1}{2}+(t(q_i) -x)\Big ]f(x)dx \nonumber \\&\qquad +\, \frac{\tau \hat{p}(q_i)}{2}\int _{t(q_i)}^{t(q_i) +\Delta _1}\nonumber \\&\qquad \,\times \frac{(t(q_i)-x +\Delta _1)(t(q_i) -x+\Delta _1)}{2\Delta _1}f(x)dx.\nonumber \\&\quad =\tau \hat{p}(q_i)\int _{-\infty }^{t(q_i)\!-\!\Delta \!-\!\Delta _2}\Big (\frac{\Delta _1\!-\!2\Delta \!-\!\Delta _2}{4} \!+\!t(q_i) \!-\!x\Big )f(x)dx \nonumber \\&\qquad +\, \tau \hat{p}(q_i)\int _{t(q_i) -\Delta -\Delta _2}^{t-\Delta }\Big [ \frac{\Delta _1}{4} +\frac{(t(q_i) -x-\Delta )^2}{4\Delta _2}\nonumber \\&\qquad +\,\frac{t(q_i) -x}{2} \Big ]f(x)dx\nonumber \\&\qquad +\, \tau \hat{p}(q_i) \int _{t(q_i)-\Delta }^{t(q_i)} \Big [\frac{\Delta _1}{4} +\frac{(t(q_i)-x)}{2}\Big ]f(x)dx \nonumber \\&\qquad +\, \tau \hat{p}(q_i) \int _{t(q_i)}^{t(q_i) +\Delta _1}\frac{(t(q_i)-x+\Delta _1)^2}{4 \Delta _1}f(x)dx. \end{aligned}$$
(34)

The proof is completed.

Appendix 3

Proof of inference 3

Substituting \(\Delta =0\) into \(E[N(t(q_i))]\) in Inference 2, we can get

$$\begin{aligned}&E[N(t(q_i))]\nonumber \\&\quad = \tau \hat{p}(q_i)\int _{-\infty }^{t(q_i)-\Delta _2}\Big (\frac{\Delta _1-\Delta _2}{4} +t(q_i)-x\Big )f(x)dx \nonumber \\&\qquad +\, \tau \hat{p}(q_i)\int _{t(q_i)-\Delta _2}^{t(q_i)}\Big [ \frac{\Delta _1}{4}+\frac{(t(q_i)-x)^2}{4\Delta _2} \nonumber \\&\qquad +\,\frac{t(q_i)-x}{2} \Big ]f(x)dx + \tau \hat{p}(q_i)\int _{t(q_i)}^{t(q_i)} \Big [\frac{\Delta _1}{4}\nonumber \\&\qquad +\,\frac{(t(q_i)-x)}{2}\Big ]f(x)dx \nonumber \\&\qquad +\, \tau \hat{p}(q_i)\int _{t(q_i)}^{t(q_i)+\Delta _1}\frac{(t(q_i)-x+\Delta _1)^2}{4 \Delta _1}f(x)dx.\nonumber \\&\quad =\tau \hat{p}(q_i)\int _{-\infty }^{t(q_i)-\Delta _2}(t(q_i)-x)f(x)dx\nonumber \\&\qquad +\, \tau \hat{p}(q_i)\int _{-\infty }^{t(q_i)-\Delta _2}\frac{\Delta _1-\Delta _2}{4} f(x)dx \nonumber \\&\qquad +\, \tau \hat{p}(q_i)\int _{t(q_i)-\Delta _2}^{t(q_i)} \frac{\Delta _1}{4}f(x)dx\nonumber \\&\qquad +\,\tau \hat{p}(q_i) \int _{t(q_i)\!-\!\Delta _2}^{t(q_i)} \frac{(t(q_i)\!-\!x)(t(q_i)\!-\!x\!+\!2\Delta _2)}{4\Delta _2} f(x)dx \nonumber \\&\qquad +\, \tau \hat{p}(q_i)\times 0\nonumber \\&\qquad +\, \tau \hat{p}(q_i) \int _{t(q_i)}^{t(q_i)+\Delta _1}\frac{(t(q_i)-x+\Delta _1)^2}{4 \Delta _1}f(x)dx \nonumber \\&\quad =\tau \hat{p}(q_i)\Big [\int _{-\infty }^{t(q_i)- \Delta _2}(t(q_i)-x)f(x)dx\nonumber \\&\qquad +\, \frac{\Delta _1-\Delta _2}{4} \int _{-\infty }^{t(q_i)-\Delta _2} f(x)dx \nonumber \\&\qquad +\, \frac{\Delta _1}{4} \int _{t(q_i)-\Delta _2}^{t(q_i)}f(x)dx\nonumber \\&\quad + \frac{1}{4\Delta _2} \int _{t(q_i)-\Delta _2}^{t(q_i)} (t(q_i)-x)(t(q_i)-x+2\Delta _2)f(x)dx \nonumber \\&\qquad +\,\frac{1}{4 \Delta _1} \int _{t(q_i)}^{t(q_i)+ \Delta _1}(t(q_i)-x+\Delta _1)^2f(x)dx. \end{aligned}$$
(35)

The proof is completed.

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Mahata, G.C. A production-inventory model with imperfect production process and partial backlogging under learning considerations in fuzzy random environments. J Intell Manuf 28, 883–897 (2017). https://doi.org/10.1007/s10845-014-1024-2

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