Outsourcing management under various demand Information Sharing scenarios

Abstract

Outsourcing has been praised as a cost effective solution for many companies. One of the major potential issues of outsourcing is the poor quality of outsourcing work. In this research, Stackelberg game is used to design an outsourcing contract for a supplier and a buyer to decide the optimal outsourcing price, retail price, and outsourcing quality with demand uncertainty. The supplier and buyer’s demand forecasting are private information to each part and cause asymmetric information in the outsourcing contract design. Three different forecast scenarios are studied, including Non-Information Sharing, Information Sharing, and Buyer Forecasting case. Optimal values of outsourcing price, retail price, and outsourcing quality level are derived. We then compare the optimal policies of three cases and derive several managerial insights. Results of extensive numerical experimentation are also presented.

Appendix

Proof of Proposition 1

The buyer (follower, decision variable is p) and the supplier (leader, decision variables are c and Q) maximize her and his profit, respectively.

We solve this game backward.

• The leader supplier calculates the follower buyer’s best response function. Please note that the follower buyer can infer the leader supplier’s forecast, $f_{\mathrm{S}}$, after the supplier announce c and Q (or $f_{\mathrm{S}}$ can be inferred by solving linear function of c and Q).

  $$\begin{aligned} E(\pi _B |f_B ,f_S )= & {} (p-c)(E(a|f_B ,f_S )-bp+\theta Q) \\ \frac{\partial E(\pi _B |f_B ,f_S )}{\partial p}= & {} E(a|f_B ,f_S )-2bp+cb+\theta Q=0 \\ p^{BR}= & {} \frac{E(a|f_B ,f_S )+cb+\theta Q}{2b} \\ \end{aligned}$$

• The leader supplier plugs buyer’s best response function into his profit function to get c and Q.

  $$\begin{aligned} E(\pi _S |f_S )= & {} c\left[ {E(a|f_S )-bE(p^{BR}|f_S )+\theta Q} \right] -\frac{L}{2}Q^{2} \\= & {} c\left[ {E(a|f_S )-bE\left( \frac{E(a|f_B ,f_S )+cb+\theta Q}{2b}|f_S \right) +\theta Q} \right] -\frac{L}{2}Q^{2} \\= & {} c\left[ {E(a|f_S )-b\frac{E(a|f_S )+cb+\theta Q}{2b}+\theta Q} \right] -\frac{L}{2}Q^{2} \\= & {} c\left[ {\frac{E(a|f_S )}{2}-\frac{cb}{2}+\frac{\theta Q}{2}} \right] -\frac{L}{2}Q^{2} \\= & {} \frac{cE(a|f_S )}{2}-\frac{c^{2}b}{2}+\frac{c\theta Q}{2}-\frac{L}{2}Q^{2} \\ \end{aligned}$$

  $\frac{\partial E\left( {\pi _S \hbox {|}f_S } \right) }{\partial c}=\frac{E\left( {a|f_m } \right) }{2}-cb+ \frac{\theta Q}{2}=0$ or $\frac{A_S}{2}-cb +\frac{\theta Q}{2}=0$ and $\frac{\partial E\left( {\pi _S \hbox {|}f_S } \right) }{\partial Q}=\frac{c \theta }{2}-LQ=0$ We can get the following from the above equations.

  $$\begin{aligned} \left\{ {{\begin{array}{l} {c=\frac{A_S +\theta Q}{2b}} \\ {Q=\frac{c\theta }{2L}} \\ \end{array} }} \right. \end{aligned}$$

  Solving the above two equations simultaneously, we get the following. $c^{N*}=\frac{2LA_S }{4Lb-\theta ^{2}}$ and $Q^{N*}=\frac{\theta A_S }{4Lb-\theta ^{2}}$

• Optimal retail price can get by plug $c^{N*}$ and $Q^{N*}$into the best response function.

  $$\begin{aligned} p^{BR}= & {} \frac{E(a|f_B ,f_S )+bc^{*}+\theta Q^{*}}{2b}=\frac{A+bc^{*}+\theta Q^{*}}{2b}\\ p^{N*}= & {} \frac{A}{2b}+\frac{\left( {\theta ^{2}+2Lb} \right) A_S }{2b\left( {4Lb-\theta ^{2}} \right) }\\ E\left( {\pi _B \hbox {|}f_B ,f_S } \right) ^{N*}= & {} E(\left( {p^{*}-c^{*}} \right) \left( {A-bp^{*}+\theta Q^{*})\hbox {|}f_B ,f_S } \right) ^{N*}\\= & {} \frac{\left( {4ALb-A\theta ^{2}+A_S \theta ^{2}-2A_S Lb} \right) ^{2}}{4b\left( {4Lb-\theta ^{2}} \right) ^{2}} \end{aligned}$$

  Since $E\left( {A^{2}} \right) =\bar{a}^{2}+J^{2}\left( {U+V_S } \right) +K^{2}\left( {U+V_B } \right)$, $E\left( {A_S ^{2}} \right) =\bar{a}^{2}+t_S U$, and

  $$\begin{aligned} E\left( {A*A_S } \right) =\bar{a}^{2}+JU, \end{aligned}$$

  we can get unconditional buyer’s profit

  $$\begin{aligned} E\left( {\pi _B \hbox {|}f_B ,f_S } \right) ^{N*}= & {} \frac{\bar{a}^{2}+J^{2}\left( {U+V_S } \right) +K^{2}\left( {U+V_B } \right) }{4b}+\frac{\left( {\theta ^{2}-2Lb} \right) \left( {\bar{a}^{2}+JU} \right) }{2b\left( {4Lb-\theta ^{2}} \right) }\\&+\,\frac{\left( {\theta ^{2}-2Lb} \right) ^{2}\left( {\bar{a}^{2}+t_S U} \right) }{4b\left( {4Lb-\theta ^{2}} \right) ^{2}} \end{aligned}$$

  Similar, we can get

  $$\begin{aligned} E(\pi _S |f_S )^{N*}= & {} E(c^{*}\left( {A-bp^{*}+\theta Q^{*})\hbox {|}f_S } \right) -\frac{LQ^{{*}^{2}}}{2}=\frac{A_S ^{2}L}{2\left( {4Lb-\theta ^{2}} \right) }\\ E\left( {\pi _S } \right) ^{N*}= & {} \frac{\left( {\bar{a}^{2}+t_S U} \right) L}{2\left( {4Lb-\theta ^{2}} \right) }=\frac{(\bar{a}^{2}+Ut_S )L}{2\left( {4Lb-\theta ^{2}} \right) } \end{aligned}$$

Proof of Proposition 2

The proof of Proposition 6 is similar to Proposition 1. We will briefly show the steps as below.

• The leader supplier calculates the follower buyer’s best response function. Please note that the follower buyer has both information $f_{\mathrm{S}}$ and $f_{\mathrm{B}}$ in Information Sharing case.

  $$\begin{aligned} \begin{array}{l} E(\pi _B |f_B ,f_S )=(p-c)(E(a|f_B ,f_S )-bp+\theta Q) \\ \frac{\partial E(\pi _B |f_B ,f_S )}{\partial p}=E(a|f_B ,f_S )-2bp+cb+\theta Q=0 \\ p^{BR}=\frac{E(a|f_B ,f_S )+cb+\theta Q}{2b} \\ \end{array} \end{aligned}$$

• The leader supplier plugs buyer’s best response function into his profit function to get c and Q. Please note that the leader supplier also has both information $f_{\mathrm{S}}$ and $f_{\mathrm{B}}$ in Information Sharing case.

  $$\begin{aligned} E(\pi _S |f_B ,f_S )= & {} c\left[ {E(a|f_B ,f_S )-bE(p^{BR}|f_B ,f_S )+\theta Q} \right] -\frac{L}{2}Q^{2} \\= & {} c\left[ {E(a|f_B ,f_S )-b\frac{E(a|f_B ,f_S )+cb+\theta Q}{2b}+\theta Q} \right] -\frac{L}{2}Q^{2} \\= & {} \frac{cE(a|f_B ,f_S )}{2}-\frac{c^{2}b}{2}+\frac{c\theta Q}{2}-\frac{L}{2}Q^{2} \\&\quad \left\{ {{\begin{array}{l} {\frac{\partial E\left( {\pi _S \hbox {|}f_B ,f_S} \right) }{\partial c}=\frac{E\left( {a|f_B ,f_S} \right) }{2}-cb+\frac{\theta Q}{2}=\frac{A}{2}-cb+\frac{\theta Q}{2}=0} \\ {\frac{\partial E\left( {\pi _S \hbox {|}f_B ,f_S} \right) }{\partial Q}=\frac{c \theta }{2}-LQ=0} \\ \end{array} }} \right. \end{aligned}$$

  Solving the above two equations simultaneously, we get the following. $c^{I*}=\frac{2LA}{4Lb-\theta ^{2}}$ and $Q^{I*}=\frac{\theta A}{4Lb-\theta ^{2}}$

• Optimal retail price can get by plug $c^{I*}$ and $Q^{I*}$ into the best response function.

  $$\begin{aligned} p^{I*}=\frac{3LA}{4Lb-\theta ^{2}} \end{aligned}$$

  Then we can get the expected profits and unconditional expected profits for the buyer and the supplier, as in Proposition 2.

Proof of Proposition 3

The proof of Proposition 6 is similar to Proposition 2. We will briefly show the steps as below.

• The leader supplier calculates the follower buyer’s best response function. Please note that only the buyer forecasts the demand. All decision making is based on the buyer’s forecast $f_{\mathrm{B}}$ in Buyer Forecast case.

  $$\begin{aligned} E(\pi _B |f_B )= & {} (p-c)(E(a|f_B )-bp+\theta Q) \\ \frac{\partial E(\pi _B |f_B )}{\partial p}= & {} E(a|f_B )-2bp+cb+\theta Q=0 \\ p^{BR}= & {} \frac{E(a|f_B )+cb+\theta Q}{2b} \\ \end{aligned}$$

• The leader supplier plugs buyer’s best response function into his profit function to get c and Q. Please note that the leader supplier use buyer’s forecast $f_{\mathrm{B}}$ in his decision making under Buyer Forecast case.

  $$\begin{aligned} E(\pi _S |f_B )= & {} c\left[ {E(a|f_B )-bE(p^{BR}|f_B )+\theta Q} \right] -\frac{L}{2}Q^{2} \\= & {} \frac{cE(a|f_B )}{2}-\frac{c^{2}b}{2}+\frac{c\theta Q}{2}-\frac{L}{2}Q^{2} \\&\quad \left\{ {{\begin{array}{l} {\frac{\partial E\left( {\pi _S \hbox {|}f_B } \right) }{\partial c}=\frac{A_{B}}{2}-cb+\frac{\theta Q}{2}=0} \\ {\frac{\partial E\left( {\pi _S \hbox {|}f_B } \right) }{\partial Q}=\frac{c \theta }{2}-LQ=0} \\ \end{array} }} \right. \end{aligned}$$

  Solving the above two equations simultaneously, we get the following. $c^{B*}=\frac{2LA_B }{4Lb-\theta ^{2}}$ and $Q^{B*}=\frac{\theta A_B }{4Lb-\theta ^{2}}$

• Optimal retail price can get by plug $c^{B*}$ and $Q^{B*}$ into the best response function.

  $$\begin{aligned} p^{B*}=\frac{3LA_B }{4Lb-\theta ^{2}} \end{aligned}$$

  Then we can get the expected profits and unconditional expected profits for the buyer and the supplier, as in Proposition 3.

Proof of Proposition 4

Part 4(a)

$c^{I}-c^{B}=\frac{(A-A_B )}{4Lb-\theta ^{2}}<0$ if $A<A_B$ or $f_S <E(f_S |f_B )$.

We next show that $A<A_B$ is equivalent to $f_S <E(f_S |f_B )$.

• Proof of $E\left( {A\hbox {|}f_B } \right) =A_B$

$$\begin{aligned} E(A|f_B)= & {} E[(1-J-K)\bar{{a}}+Jf_S |f_B +Kf_B ] \\= & {} (1-J-K)\bar{{a}}+JE(f_S |f_B )+Kf_B \\= & {} (1-J-K)\bar{{a}}+J(\bar{{a}}+t_B (f_B -\bar{{a}}))+Kf_B \\= & {} \bar{{a}}+(f_B -\bar{{a}})t_B =A_B \\ \end{aligned}$$

From the above, we can get $A_B =E(A|f_B )=(1-J-K)\bar{{a}}+JE(f_S |f_B )+Kf_B$

Recall A can be written as

$$\begin{aligned} A\equiv E\left[ {a\hbox {|}f_B ,f_S } \right] =\bar{a}+J\left( {f_S -\bar{a}} \right) +K\left( {f_B -\bar{a}} \right) =\left( {1-J-K} \right) \bar{a}+Jf_S +Kf_B \end{aligned}$$

Comparing the above $A_B$ and A, we can say that $A<A_B$ is equivalent to $f_S <E(f_S |f_B )$.

Demand difference between I and B case can be shown as $d^{B*}-d^{I*}=\frac{bL\left( {A_B -A} \right) }{4Lb-\theta ^{2}}>$0 if $A_B >A$

Part 4(b), Similar to Proposition 4(a) and is omitted.

Proof of Proposition 5

5(a1) and 5(a2), proofs are Straightforward and are omitted.

5(b1)

$$\begin{aligned}&E\left( {\pi _S \hbox {|}f_S } \right) ^{N*}-E\left( {\pi _S \hbox {|}f_B ,f_S } \right) ^{I*}\\&\quad c=\frac{1}{4}\frac{1}{b(4Lb-{\uptheta } ^{2})^{2}}(12L^{2}A^{2}b^{2}\\&\qquad \quad \, -\,8A^{2}Lb{\uptheta } ^{2}+12ALbAS{\uptheta } ^{2}-16AL^{2}b^{2}AS+A^{2}{\uptheta } ^{4} \\&\qquad \quad \, -\,2A{\uptheta } ^{4}AS-AS^{2}{\uptheta } ^{4}-4AS^{2}{\uptheta } ^{2}Lb+4L^{2}AS^{2}b^{2}) \\&\quad ={\uptheta } (12A^{2}b^{2}-16Ab^{2}AS+4AS^{2}b^{2})^{2}L^{2}+{\uptheta } (-4AS^{2}{\uptheta } ^{2}b \\&\qquad \, -\,8A^{2}b{\uptheta } ^{2}+12AbAS{\uptheta } ^{2})^{2}L+{\uptheta } (A^{2}{\uptheta } ^{4}-2A{\uptheta } ^{4}AS+AS^{2}{\uptheta } ^{4})^{2} \\&\qquad >0 \\ \end{aligned}$$

5(b2)

$$\begin{aligned} \pi _B ^{N*}-\pi _B ^{I*}=\frac{1}{2}\frac{LU^{2}VS(UVSVB-UVS^{2}+2U^{2}VB-VS^{2}VB)}{(U+VS)(4Lb-{\uptheta } ^{2})(UVS+VBU+VBVS)^{2}} \end{aligned}$$

We can see that $\pi _B ^{N*}>\pi _B ^{I*}$ if $\frac{UV_S V_B -UV_S ^{2}+2U^{2}V_B -V_S ^{2}V_B }{\left( {4Lb-\theta ^{2}} \right) }>0$.

All other proofs are omitted due to straightforward reasoning.