Sourcing Co-Created Products: Should Your Suppliers Collaborate on Cost Reductions?

Abstract

We study how a multi-product downstream firm should source from the upstream market—single-source versus multi-source—in a situation where the products are co-created with the suppliers. We conceptualize co-creation as investments that are made at the different hierarchical levels and that are aimed at reducing the production costs incurred by the supplies. In the case when the downstream firm sources its products from multiple suppliers, it can foster cost-reduction collaboration among the suppliers. We find that the downstream firm may be worse off when the upstream suppliers collaborate. For a commonly used additively separable cost function, we show that the downstream firm’s optimal strategy is multi-source co-creation without collaboration. Multi-sourcing softens the holdup problem and leads to a positive level of investment by the downstream firm.

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Appendix

Appendix

Proof of Lemma 1

Suppose that v is relatively high, so that in equilibrium all consumers buy (the market is fully covered). The ‘critical’ consumer is located at \(\hat{x}\) that satisfies

$$\begin{aligned} v-t\hat{x}-p_1= & {} v-t(1-\hat{x})-p_2,\\ \hat{x}= & {} \frac{t+p_2-p_1}{2t}. \end{aligned}$$

Given \(p_1^w\) and \(p_2^w\), F faces the following constrained optimization problem:

$$\begin{aligned} \max _{p_1, p_2} \hat{x}(p_1-p_1^w)+(1-\hat{x})(p_2-p_2^w)= & {} \frac{t+p_2-p_1}{2t}(p_1-p_1^w)\\&\quad +\frac{t+p_1-p_2}{2t}(p_2-p_2^w) \end{aligned}$$

subject to

$$\begin{aligned} v-t\hat{x}-p_1\ge 0. \end{aligned}$$

The constraint guarantees that the critical consumer buys the product; it holds with equality at the solution. Thus, we have

$$\begin{aligned} \max _{p_1, p_2} \frac{t+p_2-p_1}{2t}(p_1-p_1^w) +\frac{t+p_1-p_2}{2t}(p_2-p_2^w) \end{aligned}$$

subject to

$$\begin{aligned} v-\frac{t+p_1+p_2}{2}=0. \end{aligned}$$

Substituting the constraint into the objective function yields

$$\begin{aligned}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\max _{p_1} \frac{v-p_1}{t}(p_1-p_1^w)+\frac{t-v+p_1}{t}(2v-t-p_1-p_2^w),\end{aligned}$$
$$\begin{aligned}\text{ F.O.C. } \quad \frac{1}{t}\left( -p_1+p_1^w+v-p_1+2v-t-p_1-p_2^w-t+v-p_1\right) =0,\end{aligned}$$
$$\begin{aligned}\quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, -4p_1+4v-2t+p_1^w-p_2^w=0, \end{aligned}$$

or

$$\begin{aligned} p_1=v-\frac{t}{2}+\frac{p_1^w-p_2^w}{4}, \end{aligned}$$

hence

$$\begin{aligned} p_2=v-\frac{t}{2}+\frac{p_2^w-p_1^w}{4} \end{aligned}$$

and

$$\begin{aligned} \hat{x}= \frac{1}{2}-\frac{p_1^w-p_2^w}{4t}. \end{aligned}$$

Proof of Lemma 2

S1 solves

$$\begin{aligned}&\max _{p_1^w} \hat{x}(p_1^w-c_1)= \left( \frac{1}{2} -\frac{p_1^w-p_2^w}{4t}\right) (p_1^w-c_1),\\&\,\,\,\,\,\,\,\text{ F.O.C. } \quad \frac{1}{2}-\frac{p_1^w-p_2^w}{4t} -\frac{p_1^w-c_1}{4t}=0. \end{aligned}$$

Similarly, S2 solves

$$\begin{aligned}&\max _{p_2^w} (1-\hat{x})(p_2^w-c_2)=\left( \frac{1}{2} -\frac{p_2^w-p_1^w}{4t}\right) (p_2^w-c_2),\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ F.O.C. } \quad \frac{1}{2}-\frac{p_2^w-p_1^w}{4t} -\frac{p_2^w-c_2}{4t}=0. \end{aligned}$$

We put the two first-order conditions into a system of equations and solve for \(p_1^w\) and \(p_2^w\):

$$\begin{aligned} p_1^w=\frac{6t+2c_1+c_2}{4}\quad \text{ and } \quad p_2^w=\frac{6t+2c_2+c_1}{4}. \end{aligned}$$

Proof of Lemma 3

We use the results of Lemmas 1 and 2 to obtain the continuation profits of S1, S2, and F as functions of \(c_1\) and \(c_2\):

$$\begin{aligned} \pi ^{ms}_1(c_1, c_2)&=\left( \frac{1}{2}-\frac{p_1^w-p_2^w}{4t}\right) (p_1^w-c_1)\\&=\left( \frac{1}{2}-\frac{c_1-c_2}{16t}\right) \frac{6t-2c_1+c_2}{4} =\frac{1}{64t}(8t-c_1+c_2)(6t-2c_1+c_2). \end{aligned}$$

Similarly,

$$\begin{aligned} \pi ^{ms}_2(c_1, c_2)=\frac{1}{64t}(8t-c_2+c_1)(6t-2c_2+c_1). \end{aligned}$$

Finally,

$$\begin{aligned} \pi ^{ms}_F(c_1,c_2)&= \hat{x}(p_1-p_1^w)+(1-\hat{x})(p_2-p_2^w)\\&= {} \left( \frac{1}{2}-\frac{p_1^w-p_2^w}{4t}\right) \left( v-\frac{t}{2} +\frac{p_1^w-p_2^w}{4}-p_1^w\right) \\&+\left( \frac{1}{2}+\frac{p_1^w-p_2^w}{4t}\right) \left( v-\frac{t}{2} -\frac{p_1^w-p_2^w}{4}-p_2^w\right) \\&= {} v-\frac{t}{2}-\frac{p_1^w+p_2^w}{2}+\frac{(p_1^w-p_2^w)^2}{8t}\\&= v-2t-\frac{3(c_1+c_2)}{8}+\frac{(c_1-c_2)^2}{128t}. \end{aligned}$$

Proof of Lemma 4

S1 solves

$$\begin{aligned} \max _{I_1} \frac{1}{64t}(8t-c(e,I_1+\gamma I_2)+c(e,I_2+\gamma I_1))(6t-2c(e,I_1+\gamma I_2) +\,c(e,I_2+\gamma I_1))-I_1, \end{aligned}$$
$$\begin{aligned}&\text{ F.O.C. } \quad \frac{1}{64t}\left( \left( -\frac{\partial c(e,I_1+\gamma I_2)}{\partial I}+\gamma \frac{\partial c(e,I_2+\gamma I_1)}{\partial I}\right) (6t-2c(e,I_1+\gamma I_2)\right. +\,c(e,I_2+\gamma I_1))\\&\quad +\,(8t-c(e,I_1+\gamma I_2)\left. +\,c(e,I_2+\gamma I_1)) \left( -2\frac{\partial c(e,I_1+\gamma I_2)}{\partial I} +\gamma \frac{\partial c(e,I_2+\gamma I_1)}{\partial I}\right) \right) =1. \end{aligned}$$

S2 solves

$$\begin{aligned}\max _{I_2} \frac{1}{64t}(8t-c(e,I_2+\gamma I_1) +c(e,I_1+\gamma I_2))(6t-2c(e,I_2+\gamma I_1) +\,c(e,I_1+\gamma I_2))-I_2, \end{aligned}$$
$$\begin{aligned}&\text{ F.O.C. } \quad \frac{1}{64t} \left( \left( -\frac{\partial c(e,I_2+\gamma I_1)}{\partial I}+\gamma \frac{\partial c(e,I_1+\gamma I_2)}{\partial I}\right) (6t-2c(e,I_2+\gamma I_1)\right. \\&\quad +\,c(e,I_1+\gamma I_2)) +\,(8t-c(e,I_2+\gamma I_1)\left. +\,c(e,I_1+\gamma I_2)) \left( -2\frac{\partial c(e,I_2+\gamma I_1)}{\partial I}+\gamma \frac{\partial c(e,I_1+\gamma I_2)}{\partial I}\right) \right) =1. \end{aligned}$$

F solves

$$\begin{aligned}\max _{e} v-2t-\frac{3(c(e, I_1+\gamma I_2)+c(e,I_2+\gamma I_1))}{8} +\frac{(c(e, I_1+\gamma I_2)-c(e, I_2+\gamma I_1))^2}{128t}-e, \end{aligned}$$
$$\begin{aligned} &\text{ F.O.C. } \quad -\frac{3}{8} \left( \frac{\partial c(e, I_1+\gamma I_2)}{\partial e} +\frac{\partial c(e, I_2+\gamma I_1)}{\partial e}\right) \\&\quad +\,\frac{(c(e, I_1+\gamma I_2)-c(e, I_2+\gamma I_1))}{64t} \left( \frac{\partial c(e, I_1+\gamma I_2)}{\partial e} -\frac{\partial c(e, I_2+\gamma I_1)}{\partial e}\right) =1. \end{aligned}$$

We put the three first-order conditions into a system of equations to solve for the equilibrium values of \(I_1\), \(I_2\), and e. In equilibrium S1 and S2 will choose the same level of investment; we denote it by \(I^C\). Let \(e^C\) denote the equilibrium level of F’s investment. \(I^C\) and \(e^C\) must satisfy

$$\left\{ \begin{array}{l} -\dfrac{3}{4}\dfrac{\partial c(e^C, I^C+\gamma I^C)}{\partial e} = 1 \\ -\dfrac{1}{64t}(22t-14\gamma t-(1-\gamma )c(e^C, I^C+\gamma I^C)) \dfrac{\partial c(e^C, I^C+\gamma I^C)}{\partial I} = 1. \end{array}\right.$$

Proof of Proposition 1

It follows from Lemma 3 that \(\tilde{I}^C\) and \(e^C\) are implicitly defined by

$$\left\{ \begin{array}{l} -\dfrac{3}{4}\dfrac{\partial c(e^C, \tilde{I}^C)}{\partial e} = 1 \\ -\dfrac{1}{64t}(22t-14\gamma t-(1-\gamma )c(e^C, \tilde{I}^C)) \dfrac{\partial c(e^C,\tilde{I}^C)}{\partial I} = 1. \end{array}\right.$$

Taking the total derivative with respect to \(\gamma \) yields

$$\left\{ \begin{array}{l} -\dfrac{3}{4}\left( \dfrac{\partial ^2 c}{\partial e^2} \dfrac{\mathrm{d} e^C}{\mathrm{d} \gamma } +\dfrac{\partial ^2 c}{\partial e \partial I} \dfrac{\mathrm{d} \tilde{I}^C}{\mathrm{d} \gamma }\right) = 0 \\ -\dfrac{1}{64t}\left( \left( -14t+c+\gamma \left( \dfrac{\partial c}{\partial e}\dfrac{\mathrm{d} e^C}{\mathrm{d} \gamma } +\dfrac{\partial c}{\partial I}\dfrac{\mathrm{d} \tilde{I}^C}{\mathrm{d} \gamma }\right) \right) \dfrac{\partial c}{\partial I}\right. \\ \quad \quad \left. +(22t-14\gamma t-(1-\gamma )c) \left( \dfrac{\partial ^2 c}{\partial e \partial I} \dfrac{\mathrm{d} e^C}{\mathrm{d} \gamma }+\dfrac{\partial ^2 c}{\partial I^2} \dfrac{\mathrm{d} \tilde{I}^C}{\mathrm{d} \gamma }\right) \right) =0. \end{array}\right.$$

This is a linear system of equations with respect to \(\mathrm{d} e^C/\mathrm{d} \gamma \) and \(\mathrm{d}\tilde{I}^C/\mathrm{d} \gamma \):

$$\left\{ \begin{array}{l} \dfrac{\partial ^2 c}{\partial e^2}\dfrac{\mathrm{d} e^C}{\mathrm{d} \gamma } +\dfrac{\partial ^2 c}{\partial e \partial I}\dfrac{\mathrm{d} \tilde{I}^C}{\mathrm{d} \gamma }= 0 \\ \left( \gamma \dfrac{\partial c}{\partial e}\dfrac{\partial c}{\partial I} +(22t-14\gamma t-(1-\gamma )c)\dfrac{\partial ^2 c}{\partial e \partial I}\right) \dfrac{\mathrm{d} e^C}{\mathrm{d} \gamma }\\ \quad \quad +\left( \gamma \left( \dfrac{\partial c}{\partial I}\right) ^2 +(22t-14\gamma t-(1-\gamma )c)\dfrac{\partial ^2 c}{\partial I^2}\right) \dfrac{\mathrm{d} \tilde{I}^C}{\mathrm{d} \gamma } =(14t-c)\dfrac{\partial c}{\partial I}. \end{array}\right. $$

Recall that \(\partial c/\partial e<0\), \(\partial c/\partial I<0\), \(\partial ^2 c/\partial e^2>0\), \(\partial ^2 c/\partial I^2>0\), and \(\partial ^2 c/\partial e \partial I>0\). We can rewrite our system of equations as

$$\begin{aligned} \left\{ \begin{array}{l} C_1\dfrac{\mathrm{d} e^C}{\mathrm{d} \gamma }+C_2\dfrac{\mathrm{d} \tilde{I}^C}{\mathrm{d} \gamma }= 0 \\ C_3\dfrac{\mathrm{d} e^C}{\mathrm{d} \gamma }+C_4\dfrac{\mathrm{d} \tilde{I}^C}{\mathrm{d} \gamma } =-C_5, \end{array}\right. \end{aligned}$$

where

$$\begin{aligned} C_1= & {} \frac{\partial ^2 c}{\partial e^2},\\ C_2= & {} \frac{\partial ^2 c}{\partial e \partial I}, \end{aligned}$$
$$\begin{aligned}C_3= & {} \gamma \frac{\partial c}{\partial e}\frac{\partial c}{\partial I} +(22t-14\gamma t-(1-\gamma )c)\frac{\partial ^2 c}{\partial e \partial I},\\ C_4= & {} \gamma \left( \frac{\partial c}{\partial I}\right) ^2 +(22t-14\gamma t-(1-\gamma )c)\frac{\partial ^2 c}{\partial I^2}, \end{aligned}$$

and

$$\begin{aligned} C_5=-(14t-c)\frac{\partial c}{\partial I} \end{aligned}$$

are all positive. Solving the system of equations for \(\mathrm{d} e^C/\mathrm{d} \gamma \) and \(\mathrm{d} \tilde{I}^C/\mathrm{d} \gamma \) yields

$$\begin{aligned} \dfrac{\mathrm{d} e^C}{\mathrm{d} \gamma }=\frac{C_2 C_5}{C_1 C_4-C_2 C_3} \end{aligned}$$

and

$$\begin{aligned} \dfrac{\mathrm{d} \tilde{I}^C}{\mathrm{d} \gamma }=-\frac{C_1 C_5}{C_1 C_4-C_2 C_3}. \end{aligned}$$

Therefore, when \(C_1 C_4-C_2 C_3>0\), \(\mathrm{d} e^C/\mathrm{d} \gamma \) is positive, and \(\mathrm{d} \tilde{I}^C/\mathrm{d} \gamma \) is negative. Otherwise, \(\mathrm{d} e^C/\mathrm{d} \gamma \) is negative, and \(\mathrm{d} \tilde{I}^C/\mathrm{d} \gamma \) is positive.

Proof of Lemma 6

When \(p_1^w\) and \(p_2^w\) are high, F solves

$$\begin{aligned} \max _{p_1, p_2} \hat{x}_1(p_1-p_1^w)+ (1-\hat{x}_2)(p_2-p_2^w), \end{aligned}$$

where

$$\begin{aligned} v-p_1-t\hat{x}_1=0 \quad \Rightarrow \quad \hat{x}_1=\frac{v-p_1}{t} \end{aligned}$$

and

$$\begin{aligned} v-p_2-t(1-\hat{x}_2)=0 \quad \Rightarrow \quad 1-\hat{x}_2=\frac{v-p_2}{t}. \end{aligned}$$

Thus, we have

$$\begin{aligned} \max _{p_1, p_2} \frac{v-p_1}{t}(p_1-p_1^w)+ \frac{v-p_2}{t}(p_2-p_2^w). \end{aligned}$$

Differentiating with respect to \(p_1\) and setting the derivative to zero yield

$$\begin{aligned} p_1=\frac{v+p_1^w}{2}. \end{aligned}$$

Differentiating with respect to \(p_2\) and setting the derivative to zero yield

$$\begin{aligned} p_2=\frac{v+p_2^w}{2}. \end{aligned}$$

Hence,

$$\begin{aligned} \hat{x}_1=\frac{v-p_1^w}{2t} \quad \text{ and } \quad 1-\hat{x}_2=\frac{v-p_2^w}{2t}. \end{aligned}$$

In equilibrium, S sets wholesale prices \(p_1^w\) and \(p_2^w\) just low enough so that all consumers purchase in equilibrium. (Recall that v has been assumed to be sufficiently high, implying high margins.) That is, \(\hat{x}_1\) and \(1-\hat{x}_2\) must sum up to one:

$$\begin{aligned} \frac{v-p_1^w}{2t}+\frac{v-p_2^w}{2t}=1, \end{aligned}$$

or

$$\begin{aligned} p_1^w+p_2^w=2v-2t \quad \Rightarrow \quad p_2^w=2v-2t-p_1^w. \end{aligned}$$

Next, given \(c_1\) and \(c_2\), S solves

$$\begin{aligned} \max _{p_1^w, p_2^w} \frac{v-p_1^w}{2t}(p_1^w-c_1)+\frac{v-p_2^w}{2t}(p_2^w-c_2) \end{aligned}$$

subject to

$$\begin{aligned} p_2^w=2v-2t-p_1^w. \end{aligned}$$

Substituting the constraint into the objective function yields

$$\begin{aligned}\,\,\,\,\,\,\,\,\,\,\,\,\max _{p_1^w} \frac{v-p_1^w}{2t}(p_1^w-c_1)+\frac{2t-v+p_1^w}{2t}(2v-2t-p_1^w-c_2), \end{aligned}$$
$$\begin{aligned}\mathrm{F.O.C.} \quad \frac{1}{2t}(-p_1^w+c_1+v-p_1^w+2v-2t-p_1^w-c_2-2t+v-p_1^w)=0, \end{aligned}$$
$$\begin{aligned}\quad \,\,\,\,\,\,\,\,\,\,\,\,-4p_1^w+4v-4t+c_1-c_2=0, \end{aligned}$$
$$\begin{aligned}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p_1^w=v-t+\frac{c_1-c_2}{4}, \end{aligned}$$

hence

$$\begin{aligned} p_2^w=v-t+\frac{c_2-c_1}{4}. \end{aligned}$$

Finally, we derive S’s continuation profit as a function of \(c_1\) and \(c_2\):

$$\begin{aligned} \pi ^{ss}_S(c_1,c_2)= & {} \frac{v-p_1^w}{2t}(p_1^w-c_1)+\frac{v-p_2^w}{2t}(p_2^w-c_2)\\= & {} \left( \frac{1}{2}-\frac{c_1-c_2}{8t}\right) \left( v-t+\frac{c_1-c_2}{4}-c_1\right) +\left( \frac{1}{2}+\frac{c_1-c_2}{8t}\right) \left( v-t-\frac{c_1-c_2}{4}-c_2\right) \\= & {} v-t-\frac{c_1+c_2}{2}+\frac{(c_1-c_2)^2}{16t}. \end{aligned}$$

F’s continuation profit as a function of \(c_1\) and \(c_2\) is:

$$\begin{aligned} \pi ^{ss}_F(c_1,c_2)&=\frac{v-p_1}{t}(p_1-p_1^w)+ \frac{v-p_2}{t}(p_2-p_2^w) =\frac{(v-p_1^w)^2}{4t}+\frac{(v-p_2^w)^2}{4t}\\&=\frac{1}{4t}\left( \left( t-\frac{c_1-c_2}{4}\right) ^2 +\left( t+\frac{c_1-c_2}{4}\right) ^2\right) \\&=\frac{1}{4t}\left( 2t^2+2\left( \frac{c_1-c_2}{4}\right) ^2\right) =\frac{t}{2}+\frac{(c_1-c_2)^2}{32t}. \end{aligned}$$

Proof of Proposition 3

F solves

$$\begin{aligned} \max _{e} \frac{t}{2}+\frac{(c(e,I_1+\delta I_2) -c(e,I_2+\delta I_1))^2}{32t}-e. \end{aligned}$$

The higher is the difference in the production costs, the higher is F’s profit. Suppose for definitiveness \(I_2>I_1\). Then

$$\begin{aligned} |c(e,I_1+\delta I_2)-c(e,I_2+\delta I_1)| =c(e,I_1+\delta I_2)-c(e,I_2+\delta I_1) \end{aligned}$$

and

$$\begin{aligned} \frac{\partial }{\partial e}(c(e,I_1+\delta I_2)-c(e,I_2+\delta I_1)) =-\int _{I_1}^{I_2} \frac{\partial ^2 c(e,I)}{\partial e \partial I} \mathrm{d}I<0. \end{aligned}$$

It follows that zero investment is F’s dominant strategy.

Proof of Lemma 7

S solves

$$\begin{aligned}&\max _{I_1, I_2} \frac{v}{2}-\frac{t}{2}-\frac{c(0,I_1+\delta I_2) +c(0,I_2+\delta I_1)}{2} +\frac{(c(0,I_1+\delta I_2) -c(0,I_2+\delta I_1))^2}{16t}-I_1-I_2,\\&\,\,\,\,\,\,\,\left\{ \begin{array}{l} \dfrac{1}{2}\left( \dfrac{\partial c(0,I_1+\delta I_2)}{\partial I} +\delta \dfrac{\partial c(0,I_2+\delta I_1)}{\partial I}\right) \\ \quad \quad +\dfrac{(c(0,I_1+\delta I_2)-c(0,I_2+\delta I_1))}{8t} \left( \dfrac{\partial c(0,I_1+\delta I_2)}{\partial I}-\delta \dfrac{\partial c(0,I_2+\delta I_1)}{\partial I}\right) =1\\ \dfrac{1}{2}\left( \dfrac{\delta \partial c(0,I_1+\delta I_2)}{\partial I} +\dfrac{\partial c(0,I_2+\delta I_1)}{\partial I}\right) \\ \quad \quad +\dfrac{(c(0,I_1+\delta I_2)-c(0,I_2+\delta I_1))}{8t} \left( \delta \dfrac{\partial c(0,I_1+\delta I_2)}{\partial I} -\dfrac{\partial c(0,I_2+\delta I_1)}{\partial I}\right) =1. \end{array}\right. \end{aligned}$$

Depending on the parameters t and \(\delta \), as well as the functional from of the cost function, we can obtain a symmetric solution, or two asymmetric ones. We assume that \(\delta \) is sufficiently close to one, so that the solution is symmetric: \(I_1=I_2=I^*\). \(I^*\) must satisfy

$$\begin{aligned} -\frac{1}{2}(1+\delta )\frac{\partial c(0,I^*+\delta I^*)}{\partial I}=1. \end{aligned}$$

Proof of Lemma 8

S1 solves

$$\begin{aligned}&\max _{I_1} \frac{1}{64t}(8t-c(0, I_1)+c(0, I_2))(6t-2c(0, I_1)+c(0, I_2))-I_1,\\&\,\,\,\,\text{ F.O.C. } \quad \frac{1}{64t}\left( -\frac{\partial c(0,I_1)}{\partial I}(6t-2c(0,I_1)+c(0,I_2))\right. \\&\,\,\,\,\,\,\,\,\,\,\,\quad + \left. (8t-c(0,I_1)+c(0,I_2)) \left( -2\frac{\partial c(0,I_1)}{\partial I}\right) \right) =1. \end{aligned}$$

S2 solves

$$\begin{aligned}&\max _{I_2} \frac{1}{64t}(8t-c(0,I_2)+c(0,I_1))(6t-2c(0,I_2)+c(0,I_1))-I_2,\\&\,\,\,\,\text{ F.O.C. } \quad \frac{1}{64t}\left( -\frac{\partial c(0,I_2)}{\partial I}(6t-2c(0,I_2)+c(0,I_1))\right. \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\quad + \left. (8t-c(0,I_2)+c(0,I_1)) \left( -2\frac{\partial c(0,I_2)}{\partial I}\right) \right) =1. \end{aligned}$$

In equilibrium S1 and S2 will choose the same level of investment; we denote it by \(I^{\dagger }\). \(I^{\dagger }\) must satisfy

$$\begin{aligned} -\frac{1}{64t}(22t-c(0, I^{\dagger }))\frac{\partial c(0, I^{\dagger })}{\partial I} = 1. \end{aligned}$$

Proof of Proposition 4

With \(c(e,I)=\bar{c}-a\sqrt{e}-b\sqrt{I}\), the system of equations in Lemma 5 becomes

$$\begin{aligned}&\left\{ \begin{array}{l} \dfrac{3}{4}\times \dfrac{a}{2\sqrt{e^{NC}}} = 1 \\ \dfrac{1}{64t}\left( 22t-\left( \bar{c}-a\sqrt{e^{NC}}-b\sqrt{I^{NC}}\right) \right) \dfrac{b}{2\sqrt{I^{NC}}}= 1, \end{array}\right. \\ &\,\,\,\,\,\left \{ \begin{array}{l} \sqrt{e^{NC}}=\dfrac{3a}{8}\\ \left. 22t-\bar{c}+a\sqrt{e^{NC}}+b\sqrt{I^{NC}}\right) =\dfrac{128t}{b}\sqrt{I^{NC}}, \end{array}\right.\end{aligned}$$

so

$$\begin{aligned} e^{NC}=\left( \frac{3a}{8}\right) ^2 \end{aligned}$$

and

$$\begin{aligned} I^{NC}=\left( \frac{22t-\bar{c}+a\sqrt{e^{NC}}}{128t/b-b}\right) ^2. \end{aligned}$$

Setting \(e^{NC}\) to zero, we obtain

$$\begin{aligned} I^{\dagger }=\left( \frac{22t-\bar{c}}{128t/b-b}\right) ^2. \end{aligned}$$

Observe that \(I^{NC}>I^{\dagger }\). It immediately follows that strategy A without collaboration dominates strategy C. Indeed, the downstream firm F can choose \(e=0\) under strategy A without collaboration and get a higher profit than under strategy C. Since \(e=e^{NC}\) is F’s best response to \(I_1=I_2=I^{NC}\), F’s equilibrium profit under strategy A without collaboration is even higher.

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Loginova, O., Syam, N.B. Sourcing Co-Created Products: Should Your Suppliers Collaborate on Cost Reductions?. Rev Ind Organ 56, 329–355 (2020). https://doi.org/10.1007/s11151-019-09700-x

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Keywords

  • Co-creation
  • Collaboration
  • Holdup
  • Product sourcing

JEL Classification

  • C72
  • D4
  • L1
  • M31