Impacts of lead time reduction on fabric sourcing in apparel production with yield and environmental considerations

Abstract

In apparel supply chains, manufacturers usually request a short lead time for fabric supplies. However, a short supply lead time would create environmental problems such as insufficient time for proper control of chemicals and material processing operations, and lead to a lower production yield of good quality supplies. Motivated by this observed industrial practice in fabric sourcing and apparel production, we build a stylized analytical model to investigate how lead time reduction in fabric sourcing affects performances of the fabric supplier and apparel manufacturer as well as the environment. To be specific, we first derive the optimal ordering quantity for the apparel manufacturer and find that it is a production yield scaled newsvendor fractile quantity. We then explore the expected values of lead time reduction, and derive the respective analytical conditions for the apparel manufacturer, fabric supplier and whole supply chain to be benefited by lead time reduction. From the conditions, we reveal that the prior demand mean (which also implies the relative prior demand uncertainty) plays a critical role in determining whether lead time reduction is beneficial. We illustrate how a win–win situation in the supply chain can be achieved by a properly designed deposit payment scheme. For the environment, we show that when the fabric supplier’s profit is improved under lead time reduction, the environment must be hurt. We further investigate how an environment tax can be imposed on the fabric supplier so as to entice it to invest in green technologies.

Appendix (A1): All proofs

Proof of Lemma 4.1

For $i \in (1,\;2)$, we have: $\hat{w}_{i} = \left( {\frac{{w - a(1 - \lambda_{i} )}}{{\lambda_{i} }}} \right)$. Differentiating $\hat{w}_{i}$ with respect to $\lambda_{i}$ yields $\frac{{\partial \hat{w}_{i} }}{{\partial \lambda_{i} }} = - \left( {\frac{w - a}{{\lambda_{i}^{2} }}} \right)$. Since $w > a$, we have: $\frac{{\partial \hat{w}_{i} }}{{\partial \lambda_{i} }} < 0$. This proves Part (a). For Part (b), since $s_{i} = \frac{{r - \hat{w}_{i} }}{r + h - v}$ and $\frac{{\partial \hat{w}_{i} }}{{\partial \lambda_{i} }} < 0$, we have $\frac{{\partial s_{i} }}{{\partial \lambda_{i} }} > 0$. □

Proof of Lemma 4.2

At Time $i,\;i \in (1,\;2)$, it is shown from (4.6) that $\hat{q}_{i}^{*}$ = $\mu_{i} + \sigma_{i} \varPhi^{ - 1} [s_{i} ]$. From (4.3), we have $\hat{q}_{i} = \lambda_{i} q_{i}$. Thus, the apparel manufacturer’s optimal fabric ordering quantity is given by: $q_{i}^{*}$ = $\frac{1}{{\lambda_{i} }}\hat{q}_{i}^{*}$ = $\frac{{\mu_{i} + \sigma_{i} \varPhi^{ - 1} [s_{i} ]}}{{\lambda_{i} }}$. □

Proof of Lemma 5.1

From (5.1), (5.2) and (5.3), we have $EVLR_{M}$ = $E[\pi_{M,2} ]^{*} - E[\pi_{M,1} ]^{*}$, $EVLR_{S}$ = $E[\pi_{S,2} ]^{*} - E[\pi_{S,1} ]^{*}$, $EVLR_{SC}$ = $E[\pi_{SC,2} ]^{*} - E[\pi_{SC,1} ]^{*}$. Substituting the analytical expressions into (5.1), (5.2) and (5.3) yields $EVLR_{M}$ = $T - (\hat{w}_{2} - \hat{w}_{1} )\mu_{1}$, $EVLR_{S}$ = $(w - m)(Y\mu_{1} - L)$, and $EVLR_{SC}$ = $((w - m)Y - \hat{w}_{2} + \hat{w}_{1} )\mu_{1} + (T - (w - m)L)$. □

Proof of Proposition 5.1

(a) Since $EVLR_{M}$ = $T - (\hat{w}_{2} - \hat{w}_{1} )\mu_{1}$ and $(\hat{w}_{2} - \hat{w}_{1} > 0$, simple algebra with rearranging terms easily gives the following relationship: $EVLR_{M} > 0 \Leftrightarrow$$\mu_{1} < \bar{\mu }_{1} \equiv \frac{T}{{\hat{w}_{2} - \hat{w}_{1} }}$. (b) As $EVLR_{S}$ = $(w - m)(Y\mu_{1} - L)$,$(w - m) > 0$ and $Y > 0$, we can easily see that $EVLR_{S} > 0 \Leftrightarrow$$\mu_{1} > \underline{\mu }_{1} \equiv \frac{L}{Y}$. □

Proof of Proposition 5.2

• (a) If $\underline{\mu }_{1} < \mu_{1} < \bar{\mu }_{1}$, we have $EVLR_{M} > 0$ and $EVLR_{S} > 0$. Since $EVLR_{SC} = EVLR_{M} + EVLR_{S}$, we have $EVLR_{SC} > 0$ if $\underline{\mu }_{1} < \mu_{1} < \bar{\mu }_{1}$.

• (b) Directly checking the closed-form analytical expression of $EVLR_{SC}$, we can see that $EVLR_{SC} > 0 \Leftrightarrow$$((w - m)Y - \hat{w}_{2} + \hat{w}_{1} )\mu_{1} + (T - (w - m)L) > 0$. □

Proof of Lemma 5.2

As $((w - m)Y - \hat{w}_{2} + \hat{w}_{1} )\mu_{1} + (T - (w - m)L) > 0$$\Leftrightarrow EVLR_{SC} > 0$ and “$\underline{\mu }_{1} < \mu_{1} < \bar{\mu }_{1}$ does not hold” implies that the scenario with “$EVLR_{M} > 0$ and $EVLR_{S} > 0$” does not occur (P.S.: Proposition 5.2), we must have either one of the following situations: (i) The apparel manufacturer is benefited but the fabric supplier suffers. (ii) The apparel manufacturer suffers but the fabric supplier is benefited. □

Proof of Proposition 5.3

(a) DP Case 1: In this case, the apparel manufacturer is benefited and the fabric supplier suffers. Thus, the apparel manufacturer should contribute $DP_{M \to S}$ and the respective interest generated amounts to $\xi DP_{M \to S}$. For the apparel manufacturer, after contributing this deposit payment, its benefit after adopting the lead time reduction scenario is: $\Delta_{SC} - l_{S} - \xi DP$ and we require it to be positive. Thus, we have:

$$ \Delta_{SC} - l_{S} - \xi DP_{M \to S} > 0 . $$

(A1)

For the fabric supplier, to ensure it does not suffer a loss after lead time reduction, we require:

$$ \xi DP > l_{S}. $$

(A2)

Combining (A1) and (A2) yields $\frac{{l_{S} }}{\xi } < DP_{M \to S} < \frac{{\Delta_{SC} - l_{S} }}{\xi }$, which is the analytical condition for the establishment of the win–win situation. Part (b), similarly, in DP Case 2, we can prove that the condition for win–win is $\frac{{l_{M} }}{\xi } < DP_{S \to M} < \frac{{\Delta_{SC} - l_{M} }}{\xi }$.□

Proof of Proposition 6.1

• (a) From (6.4), we have $EHTE^{(LTR)}$ = $TP_{2} - TP_{1}$. From (6.1) and (6.3), we have $TP_{1}$ = $\left( {\frac{{\mu_{1} + \sigma_{1} \varPhi^{ - 1} [s_{1} ]}}{{\lambda_{1} }}} \right)Q_{1}$ and $TP_{2}$ = $\left( {\frac{{\mu_{1} + \sigma_{2} \varPhi^{ - 1} [s_{2} ]}}{{\lambda_{2} }}} \right)Q_{2}$. Putting (6.1) and (6.3) into (6.4) immediately gives $EHTE^{(LTR)}$$= \left( {\frac{{Q_{2} }}{{\lambda_{2} }}} \right)(\mu_{1} + \sigma_{2} \varPhi^{ - 1} [s_{2} ]) - \left( {\frac{{Q_{1} }}{{\lambda_{1} }}} \right)(\mu_{1} + \sigma_{1} \varPhi^{ - 1} [s_{1} ])$.

• (b) From the analytical expression of $EHTE^{(LTR)}$, we can easily find that if $\mu_{1} > \underline{\mu }_{1}$, we have $EHTE^{(LTR)} > 0$. Furthermore, when $\mu_{1} \le \underline{\mu }_{1}$, we can prove by rearranging terms that (i)$EHTE^{(LTR)} > 0$ if and only if $\frac{{Q_{2} }}{{Q_{1} }} > \left( {\frac{{\lambda_{2} (\mu_{1} + \sigma_{1} \varPhi^{ - 1} [s_{1} ])}}{{\lambda_{1} (\mu_{1} + \sigma_{2} \varPhi^{ - 1} [s_{2} ])}}} \right)$, and (ii) $EHTE^{(LTR)} \le 0$ if and only if $\frac{{Q_{2} }}{{Q_{1} }} \le \left( {\frac{{\lambda_{2} (\mu_{1} + \sigma_{1} \varPhi^{ - 1} [s_{1} ])}}{{\lambda_{1} (\mu_{1} + \sigma_{2} \varPhi^{ - 1} [s_{2} ])}}} \right)$.□

Proof of Proposition 6.2

• (a) Under the lead time reduction ordering scenario (i.e. ordering at Time 2), in the presence of the environment taxation scheme as well as the ability to reduce $Q_{2}$ from the original $Q_{2}$ to $\hat{Q}_{2}$, we denote the EVLR for the fabric supplier by $EVLR_{S}^{(ET)} (\hat{Q}_{2} )$, and we can easily see that it is given as follows.

  $$ EVLR_{S}^{(ET)} \left( {\hat{Q}_{2} } \right) = (w - m)(Y\mu_{1} - L) - \left( {\left( {\frac{{\hat{Q}_{2} }}{{\lambda_{2} }}} \right)\left( {\mu_{1} + \sigma_{2} \varPhi^{ - 1} [s_{2} ]} \right) - \left( {\frac{{Q_{1} }}{{\lambda_{1} }}} \right)\left( {\mu_{1} + \sigma_{1} \varPhi^{ - 1} [s_{1} ]} \right)} \right) - \frac{{\varepsilon (Q_{2} - \hat{Q}_{2} )^{2} }}{2}. $$

  Checking the 2nd order derivative shows that $\frac{{\partial^{2} EVLR_{S}^{(ET)} (\hat{Q}_{2} )}}{{\partial \hat{Q}_{2}^{2} }} < 0$ which means that it is a concave function. Solving $\frac{{\partial EVLR_{S}^{(ET)} (\hat{Q}_{2} )}}{{\partial \hat{Q}_{2} }} = 0$ yields $\hat{Q}_{2}^{*} = Q_{2} - \frac{t}{\varepsilon }\left( {\frac{{\mu_{1} + \sigma_{2} \varPhi^{ - 1} [s_{2} ]}}{{\lambda_{2} }}} \right).$

• (b) In the presence of the environment tax, it is optimal to reduce $Q_{2}$ to $\hat{Q}_{2}^{*}$. Thus, we have $EHTE^{(LTR)} (\hat{Q}_{2}^{*} )$$= \left( {\frac{{\hat{Q}_{2}^{*} }}{{\lambda_{2} }}} \right)(\mu_{1} + \sigma_{2} \varPhi^{ - 1} [s_{2} ]) - \left( {\frac{{Q_{1} }}{{\lambda_{1} }}} \right)(\mu_{1} + \sigma_{1} \varPhi^{ - 1} [s_{1} ])$.

  It is easy to find that $EHTE^{(LTR)} (\hat{Q}_{2}^{*} )$ = 0 if and only if $t = \varepsilon (Q_{2} - \beta Q_{1} ) = \hat{t}.$ A further checking reveals that $EHTE^{(LTR)} (\hat{Q}_{2}^{*} )\left\{ {\begin{array}{*{20}l} { > 0, \quad if\;t < \hat{t}} \\ { = 0 \qquad if\;t = \hat{t}} \\ { < 0, \quad if\;t > \hat{t}} \\ \end{array} } \right.$.□