Sourcing decision under disruption risk with supply and demand uncertainty: A newsvendor approach

Abstract

Multi-sourcing is considered as a common practice to hedge against supply disruption risk. In this context, this paper proposes two models for optimal order allocation in newsvendor setting, where both supply and demand are uncertain. The first model considers a risk neutral decision maker who maximizes the total expected profit under disruption risk. The second one is for a risk averse decision maker who does so under service level constraints. Analytical closed form solutions for both the models are derived. To overcome the computational complexity of the exact optimal solution, two algorithms are developed to generate optimal order quantity and the corresponding set of suppliers. The solutions with exact optimization algorithms and the proposed ones are illustrated and compared with numerical examples. The results show that the proposed algorithms give the exact optimal solution while being tractable. Finally, a case study is used to illustrate the applicability of the proposed model.

Appendices

Appendix

Proof of Proposition 1

The expected profit function for two supplier case takes the following form:

$$\begin{aligned}&\sum _{b\in B} \bigg [{\prod _{i=1}^2 {(b_i p_i +(1-b_i )(1-p_i ))} }\bigg ] (\int \limits _o^m {\left\{ {sx-<Q_{eff} ,C>+r(m-x)} \right\} } f(x)dx \nonumber \\&\quad +\int \limits _m^\infty {\left\{ {sm-<Q_{eff} ,C>-k(x-m)} \right\} } f(x)dx) \end{aligned}$$

(6)

Applying first order condition with respect to $Q_{1}$ to the above equation, setting to zero and with further algebraic manipulation, we obtain the following equation.

$$\begin{aligned}&(1-p_1 )(1-p_2 )F(Q_1 +Q_2 )+p_1 (1-p_2 )y_1 F(y_1 Q_1 +Q_2 )\nonumber \\&\quad +(1-p_1 )p_2 F(Q_1 +y_2 Q_2 ) +p_1 p_2 y_1 F(y_1 Q_1 +y_2 Q_2 )\nonumber \\&=\left[ {1-p_1 \left( {1-y_1 } \right) } \right] \frac{(s-c_1 +k)}{(s-r+k)} \end{aligned}$$

(7)

This can be rewritten as;

$$\begin{aligned}&\sum _{b\in B} {\left[ \prod _{i=1}^2 (b_i p_i +(1-b_i )(1-p_i ))\right] \left[ \prod _{i=1}^2 (y_i b_i l_i +(1-b_i l_i ) )\right] [F(m(b,Q,y))]} \nonumber \\&=\left[ {1-p_1 \left( {1-y_1 } \right) } \right] \frac{(s-c_1 +k)}{(s-r+k)}\nonumber \\&\hbox {where }l_i =1\hbox { if }i=j,\hbox { otherwise }l_i =0,\hbox { in this case }j=1 \end{aligned}$$

(8)

Similar form will also be obtained, if the partial derivatives with respect to the other variables are considered. In general, for $n$ variable problem, we can have $n$ such equations of the following form:

$$\begin{aligned}&\sum _{b\in B} \left[ \prod _{i=1}^n (b_i p_i +(1-b_i )(1-p_i ))\right] \left[ \prod _{i=1}^n (y_i b_i l_i +(1-b_i l_i ) )\right] [F(m(b,Q,y))] \nonumber \\&=\left[ {1-p_j \left( {1-y_j } \right) } \right] \frac{(s-c_j +k)}{(s-r+k)}\nonumber \\&\hbox {where }l_i =1\hbox { if }i=j,\hbox { otherwise }l_i =0 \end{aligned}$$

(9)

The second order condition for two supplier case leads to a set of four equations each of which can be generalized in the following form:

$$\begin{aligned}&(-(s-r+k)) \Big [\sum _{b\in B} \Big [\prod _{i=1}^n (b_i p_i +(1-b_i )(1-p_i ))\Big ]\nonumber \\&\times \Big [\prod _{i=1}^n (y_i b_i l_i +(1-b_i l_i ) )\prod _{j=1}^n (y_j b_j l_j +(1-b_j l_j ) )\Big ][f(m(b,Q,y))]\Big ] \end{aligned}$$

(10)

where $l_i =1$ if the function is first differentiated with respect to ith variable otherwise $l_i =0;l_j =1$if the function is first differentiated with respect to jth variable otherwise $l_j =0;$

For example when $n=2$,

$$\begin{aligned}&\frac{\partial G^{2}\left( {Q_1 ,Q_2 ,...Q_n } \right) }{\partial Q_1 ^{2}}=(-\left( {s-r+k} \right) ) \nonumber \\&\quad \times \left[ {\begin{array}{l} \left( {1-p_1 } \right) \left( {1-p_2 } \right) f\left( {Q_1 +Q_2 } \right) +p_1 \left( {1-p_2 } \right) y_1 ^{2}f\left( {y_1 Q_1 +Q_2 } \right) \\ \quad +p_2 \left( {1-p_1 } \right) f\left( {Q_1 +y_2 Q_2 } \right) +p_1 p_2 y_1 ^{2}f\left( {y_1 Q_1 +y_2 Q_2 } \right) \\ \end{array}} \right] \end{aligned}$$

(11)

Please note that the above derivative is the product of $(-(s-r+k))$ which is always negative, and a positive term. All the elements of the Hessian matrix will be of this form. Therefore, the first order principal determinant of this matrix is in the following form.

$$\begin{aligned} \left| {D_1 } \right| =\frac{\partial G^{2}\left( {Q_1 ,Q_2 ,...Q_n } \right) }{\partial Q_1 ^{2}}\le 0 \end{aligned}$$

The second order principal determinant is given by,

$$\begin{aligned} \left| {D_2 } \right|&= \frac{\partial G^{2}\left( {Q_1 ,Q_2 ,...,Q_n } \right) }{\partial Q_1^2 }\frac{\partial G^{2}\left( {Q_1 ,Q_2 ,...,Q_n } \right) }{\partial Q_2^2 }-\left[ {\frac{\partial G^{2}\left( {Q_1 ,Q_2 ,...,Q_n } \right) }{\partial Q_1 \partial Q_2 }} \right] ^{2}\nonumber \\&= (-(s-r+k))^{2}\nonumber \\&\times \left[ {\begin{array}{l} p_1 \left( {1-p_1 } \right) \left( {1-p_2 } \right) ^{2}f\left( {Q_1 +Q_2 } \right) f\left( {y_1 Q_1 +Q_2 } \right) \left( {1-y_1 } \right) ^{2} \\ +p_2 \left( {1-p_1 } \right) ^{2}\left( {1-p_2 } \right) f\left( {Q_1 +Q_2 } \right) f\left( {Q_1 +y_2 Q_2 } \right) \left( {1-y_2 } \right) ^{2} \\ +p_1 p_2 \left( {1-p_1 } \right) \left( {1-p_2 } \right) f\left( {Q_1 +Q_2 } \right) f\left( {y_1 Q_1 +y_2 Q_2 } \right) \left( {y_1 -y_2 } \right) ^{2} \\ +p_1 p_2 \left( {1-p_1 } \right) \left( {1-p_2 } \right) f\left( {y_1 Q_1 +Q_2 } \right) f\left( {Q_1 +y_2 Q_2 } \right) \left( {1-y_1 y_2 } \right) ^{2} \\ +p_1^2 p_2 \left( {1-p_2 } \right) f\left( {y_1 Q_1 +Q_2 } \right) f\left( {y_1 Q_1 +y_2 Q_2 } \right) y_1^2 \left( {1-y_2^2 } \right) \\ +p_2^2 p_1 \left( {1-p_1 } \right) f\left( {Q_1 +y_2 Q_2 } \right) f\left( {y_1 Q_1 +y_2 Q_2 } \right) y_2^2 \left( {1-y_1^2 } \right) \\ \end{array}} \right] >0\nonumber \\ \end{aligned}$$

(12)

It may be noted that the second order determinant is the product of the square of a negative number (positive), and a positive term. Proceeding in this manner for specific cases of $n$, it can be shown that the $i{th}$ order principal determinant will have the form $\left| {D_i } \right| =(-(s-r+k))^{i}\left[ {\hbox {a positive term}} \right]$, the sign of which will depend on the power ‘$i$’. Hence all the odd principal determinants are negative and the even ones are positive. Hence the function is proved to be negative definite and thus concave to the optimal order lot sizes.

Appendix 2

Proof of Proposition 2

For the model $(P_\beta )$ to be a convex programming problem in $Q_1,Q_2 ,...,Q_n$, the objective function to be maximized should be concave and service constraint function should be convex. The objective function $G_\beta \left( {Q_1 ,Q_2 ,...Q_n } \right)$ is proved to be concave in $Q_1 ,Q_2 ,...,Q_n$ as above. For the convexity of the service level constraint, hessian matrix $\nabla ^{2}G_\beta \left( {Q_1 ,Q_2 ,...,Q_n } \right)$ should be calculated.

$$\begin{aligned}&G_\beta \left( {Q_1 ,Q_2 ,...Q_n } \right) \nonumber \\&\quad =\beta _0-1+\frac{1}{\mu }\left[ {\sum _{b\in B} \bigg [{\prod _{i=1}^n {(b_i p_i +(1-b_i )(1-p_i )} } )\bigg ]\left( \int \limits _m^\infty {(x-m)} f(x)dx\right) } \right] \\&\hbox {where, }m(b,Q,y)=\sum _{i=1}^n {y_i Q_i b_i +\sum _{i=1}^n {Q_i \left( {1-b_i } \right) } }\nonumber \end{aligned}$$

(13)

The first order principal determinant is given by,

$$\begin{aligned}&\left| {D_1 } \right| =\frac{\partial G_\beta ^{2}\left( {Q_1 ,Q_2 ,...Q_n } \right) }{\partial Q_1 ^{2}}\nonumber \\&=\left( {\frac{1}{\mu }} \right) \times \left[ {\begin{array}{l} \left( {1-p_1 } \right) \left( {1-p_2 } \right) f\left( {Q_1 +Q_2 } \right) +p_1 \left( {1-p_2 } \right) y_1 ^{2}f\left( {y_1 Q_1 +Q_2 } \right) \\ +p_2 \left( {1-p_1 } \right) f\left( {Q_1 +y_2 Q_2 } \right) +p_1 p_2y_1 ^{2}f\left( {y_1 Q_1 +y_2 Q_2 } \right) \\ \end{array}} \right] \quad \end{aligned}$$

(14)

Continuing the arguments as the proof of Proposition 1 and with simple algebraic manipulation for a specific ‘$n$’ it can be shown that the higher order principal determinants will have the form which is always positive.

$$\begin{aligned} \left| {D_i } \right| =\left( {\frac{1}{\mu }} \right) ^{i}\left[ {\hbox {a positive term}} \right] \end{aligned}$$

Hence, $H$ is positive definite as its leading principal minors are positive. Therefore, function $G_\beta$ is proved to positive definite and thus strictly convex. Thus, the model $\left( {P_\beta } \right)$ is a convex programming problem. Corresponding Lagrangian relaxation problem as follows:

$$\begin{aligned} LP_\beta :MaxL_\beta \left( {Q_1 ,Q_2 ,...\lambda _\beta } \right) \end{aligned}$$

With

$$\begin{aligned}&L_\beta (Q_1 ,Q_2 ,\ldots \lambda _3 )=G(Q_1 ,Q_2 ,\ldots Q_n )-\lambda _\beta (\beta _o -\beta ) \nonumber \\&=\sum _{b\in B} \bigg [{\prod _{i=1}^n {(b_i p_i +(1-b_i )(1-p_i ))\bigg ]\bigg [\int \limits _o^m {\{sx-<Q_{eff} ,C>+r(m-x)\}f(x)dx} } } \nonumber \\&\quad +\int \limits _m^\infty {\{sm-<Q_{eff} ,C>-k(x-m)\}f(x)}\bigg ] \nonumber \\&\quad -\lambda _\beta \left\{ {\beta _0 -1+\frac{1}{\mu }\left[ {\sum _{b\in B} \bigg [{\prod _{i=1}^n {(b_i p_i +(1-b_i )(1-p_i ))\bigg ](\int \limits _m^\infty {(x-m)f(x)dx)} } } } \right] } \right\} \nonumber \\ \end{aligned}$$

(15)

The solution of the first order conditions of the Lagrangian relaxation problem gives the global maximum value for problem $LP_\beta$ and subsequently to the problem $P_\beta$. The first order conditions of the above equation leads to Eqns 4 and 5.