Recycling procurement strategies with variable yield suppliers

Abstract

This paper addresses a procurement issue facing a polystyrene packaging manufacturer considering its optimal purchasing strategies between two suppliers—one providing virgin material, the other offering recycled material. We model a single-period scenario where each supplier sells product with a known yield distribution at market pricing. The manufacturer must choose whether to sole-source or dual-source, as well as determine how much material to purchase from each supplier to meet deterministic demand. Our results indicate that there is a range of prices from the recycled material supplier where dual-sourcing will lead to higher manufacturer profits compared to sole-sourcing. We show, based on the procurement strategy, the optimal quantities to purchase to maximize manufacturer’s expected profit.

Appendices

Appendix 1: Proof of property 1

Proof of concavity for the sole -sourcing expected profit functions (second order conditions):

• 1. Sole-Sourcing with Supplier A Given $X_{A}, D, C, R, S, a, b>0; a<b; S<R$ and Eqs. (5) and (6):

  $$\begin{aligned}&\frac{\partial ^{2} E[\Pi _{A}]}{\partial X_{A}^{2}}=-\frac{D^{2}(C+R-S)}{(b-a)X_{A}^{3}} \\&D^2(C+R-S)>0 \text { and } (b-a)X_{A}^3>0 \\&\therefore \frac{\partial ^{2} E[\Pi _{A}]}{\partial X_{A}^{2}}<0, \text { so } E[\Pi _{A}] \text { is strictly concave.} \end{aligned}$$

• 2. Sole-Sourcing with Supplier B Given $X_{B}, D, C, R, S, c, d>0; c<d; S<R$ and Eqs. (9) and (10):

  $$\begin{aligned}&\frac{\partial ^{2} E[\Pi _{B}]}{\partial X_{B}^{2}}=-\frac{D^{2}(C+R-S)}{(d-c)X_{B}^{3}} \\&D^2(C+R-S)>0 \text { and } (d-c)X_{B}^3>0 \\&\therefore \frac{\partial ^{2} E[\Pi _{B}]}{\partial X_{B}^{2}}<0, \text { so } E[\Pi _{B}] \text { is strictly concave.} \end{aligned}$$

Appendix 2: Proof of property 2

Proving the continuity of $E[\Pi _{AB}]$ only requires substituting the boundary conditions into the profit functions of the corresponding cases as follows:

• When $D=aX_A+dX_B$ and $D=bX_A+cX_B$ then $E[\Pi ^{I}_{AB}]$ = $E[\Pi ^{II}_{AB}]$ = $E[\Pi ^{III}_{AB}]$ = $E[\Pi ^{IV}_{AB}]$.

• When $D=aX_A+dX_B$ and $D<bX_A+cX_B$ then $E[\Pi ^{I}_{AB}]$ = $E[\Pi ^{III}_{AB}]$.

• When $D>aX_A+dX_B$ and $D=bX_A+cX_B$ then $E[\Pi ^{I}_{AB}]$ = $E[\Pi ^{IV}_{AB}]$.

• When $D<aX_A+dX_B$ and $D=bX_A+cX_B$ then $E[\Pi ^{II}_{AB}]$ = $E[\Pi ^{III}_{AB}]$.

• When $D=aX_A+dX_B$ and $D>bX_A+cX_B$ then $E[\Pi ^{II}_{AB}]$ = $E[\Pi ^{IV}_{AB}]$.

To prove that the expected profit functions $E[\Pi ^{i}_{AB}]$ for $i=I,II,III,IV$ are strictly concave, the second order conditions and the following optimality conditions will be utilized.

Since $0<S<p_B<p_A<R$, it is never optimal to always overstock. In other words, the following is an optimality condition for this problem:

$$\begin{aligned} D>aX_A+cX_B. \end{aligned}$$

(25)

Also note that it is never optimal to always understock. In other words, the following is another optimality condition for this problem:

$$\begin{aligned} D<bX_A+dX_B. \end{aligned}$$

(26)

Given $X_{A}, X_{B}, D, C, R, S, a, b, c, d>0; c<d<a<b; S<R$, we show that the Hessian is negative definite.

Case I: The second-order partial derivatives of $E\left[ \Pi _{AB}\right]$ under Case I are as follows:

$$\begin{aligned} \frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{A}^2}= & {} -\frac{(C+R-S)\left( 3D^2-3MX_{B}(c+d)+X_{B}^2\left( c^2+cd+d^2\right) \right) }{3X_{A}^3(b-a)}\\ \frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{B}^2}= & {} -\frac{(C+R-S)(c^2+cd+d^2)}{3X_{A}(b-a)} \\ \frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{B} \partial X_{A}}= & {} \frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{A} \partial X_{B}}= \frac{(C+R-S)(2X_{B}(c^2+cd+d^2)-3D(c+d))}{6X_{A}^2(b-a)} \end{aligned}$$

Thus, the determinant of the Hessian = $\frac{D^2(d-c)^2(C+R-S)^{2}}{12X_{A}^4(b-a)^2}$. Note the following observations:

• 1. The second leading principal minor is negative (i.e., $\frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{B}^2}<0)$.

• 2. The determinant of the Hessian is positive (i.e., $\frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{A}^2}\frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{B}^2}-(\frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{B} \partial X_{A}})^{2}>0)$.

• 3. Items (1) and (2) above imply that the first leading principal minor is negative (i.e., $\frac{\partial ^2 E\left[ \Pi ^{I}_{AB}\right] }{\partial X_{A}^2}<0)$.

• 4. From items (1), (2), and (3) we can conclude that $E[\Pi ^{I}_{AB}]$ is strictly concave.

Case II: The second-order partial derivatives of $E\left[ \Pi _{AB}\right]$ under Case II are as follows:

$$\begin{aligned} \frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{A}^2}= & {} -\frac{(C+R-S)(a^2+ab+b^2)}{3X_{B}(d-c)} \\ \frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{B}^2}= & {} -\frac{(C+R-S)(3D^2-3MX_{A}(a+b)+X_{A}^2(a^2+ab+b^2))}{3X_{B}^3(d-c)}\\ \frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{B} \partial X_{A}}= & {} \frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{A} \partial X_{B}}\\= & {} \frac{(C+R-S)(2X_{A}(a^2+ab+b^2)-3D(a+b))}{6X_{B}^2(d-c)} \end{aligned}$$

Thus, the determinant of the Hessian $=\frac{D^2(b-a)^2(C+R-S)^{2}}{12X_{B}^4(d-c)^2}$. Note the following observations:

• 1. The first leading principal minor is negative (i.e., $\frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{A}^2}<0)$.

• 2. The determinant of the Hessian is positive (i.e., $\frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{A}^2}\frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{B}^2}-(\frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{B} \partial X_{A}})^{2}>0)$.

• 3. Items (1) and (2) above imply that the second leading principal minor is negative (i.e., $\frac{\partial ^2 E\left[ \Pi ^{II}_{AB}\right] }{\partial X_{B}^2}<0)$.

• 4. From items (1), (2), and (3) we can conclude that $E[\Pi ^{II}_{AB}]$ is strictly concave.

Case III: The second-order partial derivatives of $\Pi _{AB}$ under Case III are as follows:

$$\begin{aligned} \frac{\partial ^2 E\left[ \Pi ^{III}_{AB}\right] }{\partial X_{A}^2}= & {} \frac{(C+R-S)\left( a^3X_{A}^3-(D-cX_{B})^3\right) }{3X_{A}^3X_{B}(b-a)(d-c)} \\ \frac{\partial ^2 E\left[ \Pi ^{III}_{AB}\right] }{\partial X_{B}^2}= & {} \frac{(C+R-S)\left( c^3X_{B}^3-(D-aX_{A})^3\right) }{3X_{A}X_{B}^3(b-a)(d-c)} \\ \frac{\partial ^2 E\left[ \Pi ^{III}_{AB}\right] }{\partial X_{B} \partial X_{A}}= & {} \frac{\partial ^2 E\left[ \Pi ^{III}_{AB}\right] }{\partial X_{A} \partial X_{B}}\\= & {} \frac{(C+R-S)\left( -D^3+3D\left( a^2X_{A}^2+c^2X_{B}^2\right) -2\left( a^3X_{A}^3+c^3X_{B}^3\right) \right) }{6X_{A}^2X_{B}^2(b-a)(d-c)} \end{aligned}$$

Thus, the determinant of the Hessian $=\frac{D^2(C+R-S)^2(aX_{A}+cX_{B}-D)^4}{12X_{A}^4X_{B}^4(b-a)^2(d-c)^2}$. Note the following observations:

• 1. Under Eq. (25) the first leading principal minor is negative (i.e., $\frac{\partial ^2 E\left[ \Pi ^{III}_{AB}\right] }{\partial X_{A}^2}<0)$.

• 2. Using (25) it is also easy to show that second leading principal minor is negative (i.e., $\frac{\partial ^2 E\left[ \Pi ^{III}_{AB}\right] }{\partial X_{B}^2}<0)$.

• 3. The determinant of the Hessian is positive.

• 4. From items (1), (2), and (3) we can conclude that $E[\Pi ^{III}_{AB}]$ is strictly concave.

Case IV: The second-order partial derivatives of $\Pi _{AB}$ under Case IV are as follows:

$$\begin{aligned} \frac{\partial ^2 E\left[ \Pi ^{IV}_{AB}\right] }{\partial X_{A}^2}= & {} \frac{(C+R-S)\left( (D-dX_{B})^3-b^3X_{A}^3\right) }{3X_{A}^3X_{B}(b-a)(d-c)} \\ \frac{\partial ^2 E\left[ \Pi ^{IV}_{AB}\right] }{\partial X_{B}^2}= & {} \frac{(C+R-S)\left( (D-bX_{A})^3-d^3X_{B}^3\right) }{3X_{A}X_{B}^3(b-a)(d-c)} \\ \frac{\partial ^2 E\left[ \Pi ^{IV}_{AB}\right] }{\partial X_{B} \partial X_{A}}\!= & {} \! \frac{\partial ^2 E\left[ \Pi ^{IV}_{AB}\right] }{\partial X_{A} \partial X_{B}}\!=\! \frac{(C\!+\!R\!-\!S)\left( D^3\!-\!3D(b^2X_{A}^2\!+\!d^2X_{B}^2)\!+\!2(b^3X_{A}^3\!+\!d^3X_{B}^3)\right) }{6X_{A}^2X_{B}^2(b-a)(d-c)} \end{aligned}$$

Thus, the determinant of the Hessian = $\frac{D^2(C+R-S)^2(bX_{A}+dX_{B}-D)^4}{12X_{A}^4X_{B}^4(b-a)^2(d-c)^2}$. Note the following observations:

• 1. Under Eq. (26) the first leading principal minor is negative (i.e., $\frac{\partial ^2 E\left[ \Pi ^{IV}_{AB}\right] }{\partial X_{A}^2}<0)$.

• 2. Using (26) it is also easy to show that second leading principal minor is negative (i.e., $\frac{\partial ^2 E\left[ \Pi ^{IV}_{AB}\right] }{\partial X_{B}^2}<0)$.

• 3. The determinant of the Hessian is positive.

• 4. From items (1), (2), and (3) we can conclude that $E[\Pi ^{IV}_{AB}]$ is strictly concave.