Consignment based integrated inventory model for deteriorating goods with price- and green-sensitive demand

Abstract

Now-a-days, investment in greening in the process of manufacturing a product is extremely important for a sustainable supply chain management. Consignment stock policy have been studied for last four decades by many researchers in the development of a supply chain model. In this study, a green supply chain model for deteriorating goods with one seller and one buyer is established. The extent of greening improvement and the selling price have an impact on demand. There are two generalized green supply chain models established, one with and one without consignment stock policy. To compare the outcomes of the two models, a comprehensive computational research is conducted. The sensitivity analysis for key model-parameters is also carried out and several managerial insights are derived from it.

Appendices

Appendix A

The inventory level at time t at the buyer is governed by the differential equation $\frac{dI_b(t)}{dt}=-D-\theta _2I_b(t)=-(a-bp+\alpha \theta )-\theta _2I_b(t)$, the solution of which is $I_b(t)=-\frac{a-bp+\alpha \theta }{\theta _2}+ce^{-\theta _2 t}$. Using initial condition $I_b(t_p)=Q$, we get $c=Qe^{\theta _2 t_p}+\frac{(a-bp+\alpha \theta )e^{\theta _2 t_p}}{\theta _2}$.

$\hbox {Therefore}~~ I_b(t)=\frac{a-bp+\alpha \theta }{\theta _2}\left( e^{\theta _2(t_p-t)}-1\right) ~~~~~~~~~~~~~~~~~~~~~~~+Qe^{\theta _2(t_p-t)}$, $t_p\le t\le 2t_p$.

Since the next shipment is scheduled at time point $2t_p$, the initial inventory level during time interval $(2t_p, 3t_p)$ is $I_b(2t_p)=\frac{a-bp+\alpha \theta }{\theta _2}\left( e^{-\theta _2 t_p}-1 \right) +Qe^{-\theta _2 t_p}+Q$, so that the arbitrary constant of integration c is obtained as $c=\frac{(a-bp+\alpha \theta )e^{\theta _2 t_p}}{\theta _2}+Qe^{\theta _2 t_p}+Qe^{2\theta _2 t_p}$, and inventory level during $(2t_p\le t\le 3t_p)$ is $I_b(t)=\frac{a-bp+\alpha \theta }{\theta _2}\left( e^{(t_p-t)\theta _2}-1\right) ~~~~~~~~~+Qe^{(t_p-t)\theta _2}\left( 1+e^{\theta _2 t_p}\right)$. In a similar manner, the inventory level in the interval $\left[ it_p, (i+1)t_p\right]$ for $i= 1, 2,\dots m-1$ can be obtained as $I_b(t)=\frac{a-bp+\alpha \theta }{\theta _2}\left[ e^{-\theta _2(t-t_p)}-1 \right] ~~~~~~~~+Qe^{-\theta _2(t-t_p)}\sum _{j=0}^{i-1}e^{j\theta _2t_p}$.

Appendix B

From the average profit function

$$\begin{aligned} \begin{aligned} PR_{M1}=&\frac{1}{(m-1)t_p+t_a}\Big [p(a-bp+\alpha \theta )[(m-1)t_p\\ {}&+t_a]-S_v+P_1h_v+m(b_1+b_2Q)\\&+(mt_pP-mQ)c_d+C_pmt_pP+\pi \theta ^2\\ {}&-\big (S_b+h_bB_1+[mQ-(a-bp+\alpha \theta )\\ {}&\{(m-1)t_p+t_a\}]c_d\big )\Big ], \end{aligned} \end{aligned}$$

we obtain

$$\begin{aligned} \frac{\partial (PR_{M1})}{\partial p}= & {} \frac{1}{(m-1)t_p+t_a}\left[ (a-bp+\alpha \theta )\right. \\ {}{} & {} \left. ((m-1)t_p+t_a)-pb((m-1)t_p\right. \\ {}{} & {} \left. +t_a) -h_b\frac{\partial B_1}{\partial p}+c_d\frac{\partial S_1}{\partial p}\right] \\= & {} \frac{1}{(m-1)t_p+t_a}\left[ \left( a-2bp+\alpha \theta \right) \right. \\ {}{} & {} \left. \left( (m-1)t_p+t_a \right) -h_b\left\{ \frac{(m-1)bt_p}{\theta _2}\right. \right. \\ {}{} & {} \left. \left. -\frac{b(1-e^{-\theta _2 t_p})}{\theta _2^2}\sum _{i=0}^{m-2}e^{-i\theta _2t_p}\right. \right. \\ {}{} & {} \left. \left. +\frac{bt_a}{\theta _2}+\frac{be^{-(m-1)\theta _2 t_p}}{\theta _2^2}(e^{-\theta _2t_a}\right. \right. \\ {}{} & {} \left. \left. -1) \right\} -bc_d((m-1)t_p+t_a) \right] ,\\ \hbox {and~}\\ \frac{\partial (PR_{M1})}{\partial \theta }= & {} \frac{1}{T}\left[ p\alpha \left[ (m-1)t_p +t_a\right] -2\pi \theta \right. \\ {}{} & {} \left. -h_b\frac{\partial B_1}{\partial \theta }+c_d\frac{\partial S_1}{\partial \theta } \right] \\= & {} \frac{1}{T}\left[ p\alpha \left[ (m-1)t_p +t_a\right] -2\pi \theta \right. \\ {}{} & {} \left. -h_b\left\{ -\frac{(m-1)\alpha t_p}{\theta _2}+\right. \right. \\ {}{} & {} \left. \left. \frac{\alpha (1-e^{-\theta _2t_p})}{\theta _2^2}\sum _{i=0}^{m-2}e^{-i\theta _2t_p}-\frac{\alpha t_a}{\theta _2}\right. \right. \\ {}{} & {} \left. \left. -\frac{\alpha e^{-(m-1)\theta _2t_p}}{\theta _2^2}(e^{-\theta _2t_a}-1)\right\} \right. \\ {}{} & {} \left. +c_d\alpha \left[ (m-1)t_p +t_a\right] \right] . \end{aligned}$$

The optimal values of p and $\theta$, i.e. $p^*$ and $\theta ^*$ are obtained by setting $\frac{\partial PR_{M1}}{\partial p}=0$ and $\frac{\partial PR_{M1}}{\partial \theta }=0$, which give

$$\begin{aligned} p^*= & {} \frac{1}{2b\left[ \left( m-1\right) t_p+t_a\right] }\left[ \left( a+\alpha \theta \right) \left[ (m-1)t_p\right. \right. \\ {}{} & {} \left. \left. +t_a\right] -h_b\left\{ \frac{(m-1)bt_p}{\theta _2}-\frac{b(1-e^{-\theta _2t_p})}{\theta _2^2}\right. \right. \\ {}{} & {} \left. \left. \sum _{i=0}^{m-2}e^{-i\theta _2t_p}+\frac{bt_a}{\theta _2}+\frac{be^{-(m-1)\theta _2t_p}}{\theta _2^2}(e^{-\theta _2t_a}\right. \right. \\ {}{} & {} \left. \left. -1) \right\} -bc_d\left[ \left( m-1\right) t_p+t_a\right] \right] ,\\ \hbox {and~}\\\theta ^{*}= & {} \frac{1}{2\pi }\left[ p\alpha \left[ \left( m-1\right) t_p+t_a\right] -h_b\right. \\ {}{} & {} \left. \left\{ -\frac{(m-1)\alpha t_p}{\theta _2}+\frac{\alpha (1-e^{-\theta _2t_p})}{\theta _2^2}\right. \right. \\ {}{} & {} \left. \left. \sum _{i=0}^{m-2}e^{-i\theta _2t_p}-\frac{\alpha t_a}{\theta _2}-\frac{\alpha e^{-(m-1)\theta _2t_p}}{\theta _2^2}(e^{-\theta _2t_a}\right. \right. \\ {}{} & {} \left. \left. -1) \right\} +c_d\alpha \left[ (m-1)t_p+t_a\right] \right] . \end{aligned}$$

Appendix C

The delay shipment starts at time $t_p$. We consider the time point $t_p$ as $t=0$. The inventory level at t is governed by the differential equation $\frac{dI_v(t)}{dt}=P-\theta _1I_v(t)$ with $I_v(0)=0.$ The solution of the differential equation is $I_v(t)=\frac{P}{\theta _1}+ce^{-\theta _1t}$. This gives the inventory level in the time interval $0\le t \le t_q$ as $I_v(t)=\frac{P}{\theta _1}\left( 1-e^{-\theta _1t}\right)$. At $t=t_q$, $I_v(t_q)=\frac{P}{\theta _1}\left( 1-e^{-\theta _1t_q}\right) -Q$. Now $\frac{P}{\theta _1}+ce^{-\theta _1t_q}=\frac{P}{\theta _1}\left( 1-e^{-\theta _1t_q}\right) -Q$ gives $c=-\frac{P}{\theta _1}-Qe^{\theta _1t_q}$, so that $I_v(t)=\frac{P}{\theta _1}\left( 1-e^{-\theta _1t}\right) -Qe^{\theta _1(t_q-t)}$, $t_q\le t\le 2t_q$. $I_v(2t_q)=\frac{P}{\theta _1}\left( 1-e^{-2\theta _1t_q}\right) -Qe^{-\theta _1t_q}-Q.$ Again, $\frac{P}{\theta _1}+ce^{-2\theta _1t_q}=\frac{P}{\theta _1}\left( 1-e^{-2\theta _1t_q}\right) -Qe^{-\theta _1t_q}-Q$, so that $c=-\frac{P}{\theta _1}-Qe^{\theta _1t_q}-Qe^{2\theta _1t_q}$, giving $I_v(t)=\frac{P}{\theta _1}\left( 1-e^{-\theta _1t}\right) -Qe^{\theta _1(t_q-t)} (1+e^{\theta _1t_q})$ in $2t_q\le t \le 3t_q$.

Proceeding in the same way, we can infer that the inventory level in $(i-1)t_q \le t \le it_q ~\hbox {for}~i=1,2,\dots ,m$ is $I_v(t)=\frac{P}{\theta _1}\left( 1-e^{-\theta _1t}\right) -Qe^{\theta _1(t_q-t)}\sum _{j=0}^{i-2}e^{j\theta _1t_q}$.

Appendix D

The total inventory $(P_2)$ can be obtained as

$P_2=\int _{0}^{t_p}I_v(t)dt+\int _{0}^{t_q}I_v(t)dt+\int _{t_q}^{2t_q}I_v(t)dt+ \int _{2t_q}^{3t_q}I_v(t)dt+\dots +\int _{(m-1)t_q}^{mt_q}I_v(t)dt +\int _{0}^{t_q}I_v(t)dt+\int _{t_q}^{2t_q}I_v(t)dt+\int _{2t_q}^{3t_q}I_v(t)dt+\dots +\int _{(k-1)t_q}^{kt_q}I_v(t)dt$

Now,

$$\begin{aligned} \int _{0}^{t_p}I_v(t)dt{} & {} =\int _{0}^{t_p}\left( \frac{P}{\theta _1}(1-e^{-\theta _1t})\right) dt =\frac{Pt_p}{\theta _1}+\frac{P(e^{-\theta _1t_p}-1)}{\theta _1^2} ,\\{} & {} \quad \int _{t_q}^{2t_q}\left[ \frac{P}{\theta _1}+(-\frac{P}{\theta _1}-Qe^{\theta _1t_q})e^{-\theta _1t} \right] dt\\{} & {} =\frac{Pt_q}{\theta _1}+\frac{Pe^{-\theta _1t_q}}{\theta _1^2}(e^{-\theta _1t_q}-1)+\frac{Q(e^{-\theta _1t_q}-1)}{\theta _1}, \int _{2t_q}^{3t_q}\left[ \frac{P}{\theta _1}+\left( -\frac{P}{\theta _1}-Qe^{\theta _1t_q}-Qe^{2\theta _1t_q} \right) e^{-\theta _1t} \right] dt\\{} & {} =\frac{Pt_q}{\theta _1}+\frac{Pe^{-2\theta _1t_q}}{\theta _1^2}(e^{-\theta _1t_q}-1)+\frac{Q}{\theta _1}(e^{-2\theta _1t_q}-1). \end{aligned}$$

Proceeding in the same way, we can deduce that $\int _{(m-1)t_q}^{mt_q}I_v(t)dt=\frac{Pt_q}{\theta _1}+\frac{Pe^{-(m-1)\theta _1t_q}}{\theta _1^2}(e^{-\theta _1t_q}-1)+\frac{Q}{\theta _1}(e^{-(m-1)\theta _1t_q}-1)$. After production stops, we have

$$\begin{aligned}{} & {} \int _{0}^{t_q}\left[ \frac{P(1-e^{-\theta _1t})}{\theta _1}-Qe^{\theta _1(t_q-t)}\sum _{i=0}^{m-1}e^{i\theta _1t_q} \right] dt\\{} & {} =\frac{Pt_q}{\theta _1}+\frac{P(e^{-\theta _1t_q}-1)}{\theta _1^2}+\frac{Q}{\theta _1}(1-e^{m\theta _1t_q}), \int _{t_q}^{2t_q}\left[ \frac{P}{\theta _1}+\left\{ -\frac{P}{\theta _1}-Qe^{\theta _1t_q}\left( 1+\sum _{i=0}^{m-1}e^{i\theta _1t_q} \right) \right\} \right. \\{} & {} \quad \left. e^{-\theta _1t} \right] dt\\{} & {} =\frac{Pt_q}{\theta _1}+\frac{Pe^{-\theta _1t_q}}{\theta _1^2}(e^{-\theta _1t_q}-1)+\frac{Q}{\theta _1}(e^{-\theta _1t_q}-1)\\{} & {} \quad +\frac{Qe^{-\theta _1t_q}}{\theta _1}(1-e^{m\theta _1t_q}), \int _{2t_q}^{3t_q}\left[ \frac{P}{\theta _1}+\left\{ -\frac{P}{\theta _1}-Qe^{\theta _1t_q}\left( 1+\sum _{i=0}^{m-1}e^{i\theta _1t_q} \right) \right. \right. \\{} & {} \quad \left. \left. -Qe^{2\theta _1t_q} \right\} e^{-\theta _1t} \right] dt\\{} & {} =\frac{Pt_q}{\theta _1}+\frac{Pe^{-2\theta _1t_q}}{\theta _1^2}(e^{-\theta _1t_q}-1)+\frac{Q}{\theta _1}(e^{-2\theta _1t_q}-1)\\{} & {} \quad +\frac{Qe^{-2\theta _1t_q}}{\theta _1}(1-e^{m\theta _1t_q}), \end{aligned}$$

so that we can ultimately have

$$\begin{aligned} \int _{(k-1)t_q}^{kt_q}I_v(t)dt{} & {} =\frac{Pt_q}{\theta _1}+\frac{Pe^{-(k-1)\theta _1t_q}}{\theta _1^2}(e^{-\theta _1t_q}-1)\\{} & {} \quad +\frac{Q}{\theta _1}(e^{-(k-1)\theta _1t_q}-1)+\frac{Qe^{-(k-1)\theta _1t_q}}{\theta _1}(1-e^{m\theta _1t_q}). \end{aligned}$$

So,

$$\begin{aligned} P_2{} & {} = \frac{Pt_p}{\theta _1}+\frac{P}{\theta _1^2}(e^{-\theta _1t_p}-1)+(m+k-1)\frac{Pt_q}{\theta _1}\\{} & {} \quad +\frac{P(e^{-\theta _1t_q}-1)}{\theta _1^2}\left[ \sum _{i=0}^{m-1}e^{-i\theta _1t_q}+\sum _{i=0}^{k-1}e^{-i\theta _1t_q}\right] \\{} & {} \quad +\frac{Q}{\theta _1}\left[ \sum _{i=1}^{m-1}e^{-i\theta _1t_q}+\sum _{i=1}^{k-1}e^{-i\theta _1t_q}-(m+k\right. \\{} & {} \quad \left. -2) \right] +\frac{Q}{\theta _1}(1-e^{m\theta _1t_q})\sum _{i=0}^{k-1}e^{-i\theta _1t_q}. \end{aligned}$$

Appendix E

The profit function $(PR_{M2})$ is

$$\begin{aligned} PR_{M2}{} & {} =\frac{1}{T}\left[ p(m+k+1)(a-bp+\alpha \theta )t_q\right. \\{} & {} \quad \left. -TCV_2-TCB_2\right] \\{} & {} =\frac{1}{T}\left[ p(m+k+1)(a-bp+\alpha \theta )t_q-(S_v\right. \\{} & {} \quad \left. + c_d[P(t_p+mt_q)-(m+k+1)Q]+C_pP(t_p\right. \\{} & {} \quad \left. +mt_q)+(m+k+1)(b_1+b_2Q)+h_vP_2+\pi \theta ^2)\right. \\{} & {} \quad \left. -(S_b+c_d[(m+k-1)Q-S_2]+h_bB_2)\right] . \end{aligned}$$

Now,

$$\begin{aligned} \frac{\partial (PR{M2})}{\partial p}{} & {} =\frac{1}{T}\left[ (m+k+1)(a-bp+\alpha \theta )t_q\right. \\{} & {} \quad \left. -bp(m+k+1)t_q-h_b\frac{\partial B_2}{\partial p}+c_d\frac{\partial S_2}{\partial p}\right] \\{} & {} =\frac{1}{T}\left[ (m+k+1)(a-2bp+\alpha \theta )t_q-(m+k\right. \\{} & {} \quad \left. +1)h_b\left\{ \frac{bt_q}{\theta _2}-\frac{b(1-e^{-\theta _2t_q})}{\theta _2^2}\right\} -(m+k+1)bc_dt_q \right] , \end{aligned}$$

and $\frac{\partial ^2 (PR_{M2})}{\partial p^2}=-2b<0$, so that the profit is maximized at $p^*=\frac{1}{2bt_q(m+k+1)}\left[ (a+\alpha \theta )(m+k+1)t_q-h_b(m\right. \left. +k+1)\left\{ \frac{bt_q}{\theta _2}-\frac{b(1-e^{-\theta _2t_q})}{\theta _2^2} \right\} +(m+k+1)bc_dt_q \right]$. Again,

$$\begin{aligned}{} & {} \frac{\partial (PR_{M2})}{\partial \theta }=\frac{1}{T}\left[ p\alpha (m+k+1)t_q-2\pi \theta \right. \\{} & {} \quad \left. ~~~~~~~~~~~~~~~~~~~~~~~-h_b\frac{\partial B_2}{\partial \theta }+c_d\frac{\partial S_2}{\partial \theta } \right] \\{} & {} \quad =\frac{1}{T}\left[ p\alpha (m+k+1)t_q-2\pi \theta -(m+k+1)h_b\right. \\{} & {} \quad \left. \left\{ \frac{\alpha (1-e^{-\theta _2 t_q})}{\theta _2^2}-\frac{\alpha t_q}{\theta _2}\right\} +(m+k+1)c_d\alpha t_q \right] , \end{aligned}$$

and $\frac{\partial ^2 (PR_{M2})}{\partial \theta ^2}=-2\pi <0$, so that the optimum green investment level is obtained as $\theta ^* = \frac{1}{2\pi }\left[ (m+k+1)p\alpha t_q-(m+k+1)h_b\left\{ \frac{\alpha }{\theta _2^2}(1\right. \right. \left. \left. -e^{-\theta _2t_q})-\frac{\alpha t_q}{\theta _2} \right\} +(m+k+1)c_d\alpha t_q \right]$.

Appendix F

• (i) Straightforward to derive from (7), hence omitted.

• (ii) From (8), we have $\frac{\partial t_q}{\partial p}=-\frac{b\theta _1}{\theta _2(\theta _1(a-bp+\alpha \theta )+P\theta _2(1-e^{-\theta _1 t_p}))} +\frac{b\theta _1}{\theta _2\theta _1(a-bp+\alpha \theta )}>0$, and $\frac{\partial t_q}{\partial \theta }=\frac{\theta _1}{\theta _2(\theta _1(a-bp+\alpha \theta )+P\theta _2(1-e^{-\theta _1 t_p}))} -\frac{\theta _1}{\theta _2\theta _1(a-bp+\alpha \theta )}<0$.

• (iii) From (9), we have

$$\begin{aligned} \frac{\partial k}{\partial t_q}= & {} \frac{\ln [D/N]}{\theta _1t_q^2}+\frac{1}{N}\left[ 2Q\theta _1^2e^{2\theta _1t_q}-\theta _1Pe^{\theta _1t_q}\right. \\ {}{} & {} \left. -Q\theta _1^2e^{\theta _1t_q}(e^{m\theta _1t_q}-1)-mQ\theta _1^2e^{(m+1)\theta _1t_q}\right] \\{} & {} +\frac{1}{D}(P-\theta _1)Qe^{\theta _1t_q} > 0, \end{aligned}$$

Where $N=\left[ Q\theta _1e^{2\theta _1 t_q}-P(e^{\theta _1t_q}-1)\right. \left. -Q\theta _1e^{\theta _1t_q}(e^{m\theta _1t_q}-1)\right]$ and $D=(Q\theta _1-P)(e^{\theta _1t_q}-1)+Q\theta _1$. (iv) Straightforward, hence omitted. (v) From (3), we get

$$\begin{aligned} \frac{\partial t_a}{\partial p}=&\frac{1}{\theta _2}\left[ \frac{-be^{-(m-1)\theta _2t_p}(1-e^{-\theta _2 t_p})}{(a-bp+\alpha \theta )(1-e^{-\theta _2 t_p})e^{-(m-1)\theta _2 t_p}+Q\theta _2e^{-\theta _2 t_p}(1-e^{-(m-1)\theta _2t_p})+Q\theta _2(1-e^{-\theta _2t_p})}\right. \\ {}&\left. +\frac{b(1-e^{-\theta _2 t_p})}{(a-bp+\alpha \theta )(1-e^{-\theta _2 t_p})}\right] >0,\\ \hbox {and} \\ \frac{\partial t_a}{\partial \theta }=&\frac{1}{\theta _2}\left[ \frac{\alpha (1-e^{-\theta _2 t_p})e^{-(m-1)\theta _2t_p}}{(a-bp+\alpha \theta )(1-e^{-\theta _2 t_p})e^{-(m-1)\theta _2 t_p}+Q\theta _2e^{-\theta _2 t_p}(1-e^{-(m-1)\theta _2t_p})+Q\theta _2(1-e^{-\theta _2t_p})}\right. \\ {}&\left. -\frac{\alpha (1-e^{-\theta _2t_p})}{(a-bp+\alpha \theta )(1-e^{-\theta _2 t_p})} \right] <0. \end{aligned}$$