The organization of project evaluation under competition

Abstract

This article examines the optimal organizational form of project evaluation under competition. The evaluations are carried out by two fallible screening units that sequentially assess projects. Screening can be organized as a hierarchy or a polyarchy. We show that as competitive pressure rises, the polyarchy becomes less attractive. Therefore, different organizational forms might be found in different industries depending on the degree of competition. In addition, we examine endogenous screening rules under competition: For symmetric situations, we show that polyarchies will employ higher decision thresholds compared to hierarchies. Nonetheless, as in the case of exogenous screening rules, the hierarchy becomes more attractive the higher the degree of competition.

Appendix

Proof of Proposition 2

To prove this proposition, let Π _ij be the expected profit of a firm that chooses an organizational structure i ∈ {H,P} whereas its competitor decided for an organizational structure j ∈ {H,P}. Then

$$ \begin{aligned} \Uppi _{HH} &= \alpha p_{1}^{2}\left( p_{1}^{2}z_{D}+(1-p_{1}^{2})z_{M}\right) -(1-\alpha)p_{2}^{2}z_{B} \\ &= (1-\alpha)z_{B}\left( \beta p_{1}^{2}-p_{2}^{2}-\gamma p_{1}^{4}\right) \\ \Uppi _{HP} &= \alpha p_{1}^{2}\left(p_{1}(2-p_{1})z_{D}+(1-p_{1}(2-p_{1}))z_{M}\right) -(1-\alpha)p_{2}^{2}z_{B} \\ &= (1-\alpha)z_{B}\left(\beta p_{1}^{2}-p_{2}^{2}-\gamma p_{1}^{3}(2-p_{1})\right)\\ \Uppi _{PH} &= \alpha p_{1}(2-p_{1})\left(p_{1}^{2}z_{D}+(1-p_{1}^{2})z_{M}\right) -(1-\alpha)p_{2}(2-p_{2})z_{B}\\ &= (1-\alpha)z_{B}\left( \beta p_{1}(2-p_{1})-p_{2}(2-p_{2})-\gamma p_{1}^{3}(2-p_{1})\right)\\ \Uppi _{PP} &= \alpha p_{1}(2-p_{1})\left(p_{1}(2-p_{1})z_{D}+(1-p_{1}(2-p_{1}))z_{M}\right) -(1-\alpha)p_{2}(2-p_{2})z_{B}\\ &= (1-\alpha)z_{B}\left( \beta p_{1}(2-p_{1})-p_{2}(2-p_{2})-\gamma p_{1}^{2}(2-p_{1})^{2}\right)\\ \end{aligned} $$

We have :

$$ \Uppi _{PP}-\Uppi _{HH} = 2(1-\alpha)z_{B}\left( \beta p_{1}(1-p_{1})-p_{2}(1-p_{2})-2\gamma p_{1}^{2}\left( 1-p_{1}\right) \right) $$

(5)

$$ \Uppi _{PP}-\Uppi _{HP} = 2(1-\alpha)z_{B}\left( \beta p_{1}(1-p_{1})-p_{2}(1-p_{2})-\gamma p_{1}^{2}(1-p_{1})(2-p_{1})\right) $$

$$ \Uppi _{PH}-\Uppi _{HH} = 2(1-\alpha)z_{B}\left( \beta p_{1}(1-p_{1})-p_{2}(1-p_{2})-\gamma p_{1}^{3}(1-p_{1})\right) $$

Hence

$$ \Uppi _{PH}-\Uppi _{HH} > \Uppi _{PP}-\Uppi _{HP} > \Uppi _{PP}-\Uppi _{HH}. $$

• 1. Suppose that Π _HH  > Π _PH . Then Π _HP  > Π _PP and Π _HH  > Π _PP . Therefore, H is a best response to H and P. Symmetry implies that (H,H) then is the unique equilibrium and Pareto-dominates (P,P). We have Π _HH  > Π _PH if $( \beta -\gamma p_{1}^{2}) p_{1}(1-p_{1}) < p_{2}(1-p_{2}).$

• 2. Suppose that Π _HH  < Π _PH and Π _HP  > Π _PP . Hence P is a best response to H and H is a best response to P. We have Π _HH  < Π _PH if $(\beta -\gamma p_{1}^{2}) p_{1}(1-p_{1}) > p_{2}(1-p_{2})$ and Π _HP  > Π _PP if $( \beta -\gamma p_{1}(2-p_{1})) p_{1}(1-p_{1}) < p_{2}(1-p_{2}).$ Therefore, we have Π _HP  < Π _PH since p ₂(1−p ₂) < βp ₁(1−p ₁).

• 3. Suppose that Π _HP  < Π _PP . Then Π _HH  < Π _PH . Therefore, P is a best response to P and H. Π _HP  < Π _PP if (β −γ p ₁(2−p ₁)) p ₁(1−p ₁) > p ₂(1−p ₂). However, we have Π _PP  > Π _HH only for (β −2γ p ₁) p ₁(1−p ₁) > p ₂(1−p ₂). If the reverse is true, coordination offers an improvement potential since Π _PP  < Π _HH . $\square$

Proof of Proposition 3

We examine the effect of a change in the competitive pressure on the decision threshold. For example, consider the symmetric case where both firms organize as a hierarchy. From (4) we have $f( z_{D},R^{HH}) ={\frac{\alpha } {(1-\alpha)z_{B}}}( z_{M}-(z_{M}-z_{D})p_{1}(R^{HH})^{2}) p_{1}^{\prime}(R^{HH})p_{1}(R^{HH})-p_{2}^{\prime}(R^{HH})p_{2}(R^{HH})=0$ and for R ^HH to be a maximum we must have ${\frac{\partial f} {\partial R^{HH}}} < 0.$ We have

$$ \frac{\partial R^{HH}}{\partial z_{D}}=-\frac{\frac{\partial f}{\partial z_{D}}}{\frac{\partial f}{\partial R^{HH}}}=-\frac{\alpha }{(1-\alpha)z_{B}} \frac{p_{1}(R^{HH})^{3}p_{1}^{\prime }(R^{HH})}{\frac{\partial f}{\partial R^{HH}}} <\ 0. $$

since p′(R) < 0. The same reasoning applies to the other competitive organizational forms in (4).

Therefore, the stronger the competitive pressure (small z _D ), the higher will be the optimal decision threshold.$\square$

Proof of Proposition 4

To see that R ^PP > R ^HH, define $c(z):={\frac{1-p(z)} {p(z)}}$ (i.e. $c(z_{M})={\frac{( 1-p_{1}) } {p_{1}}}$ and $c(z_{B})={\frac{( 1-p_{2}) } {p_{2}}}.$) Calculate

$$ \begin{aligned} c(0){\frac{\partial \Uppi _{HH}} {\partial R}}-{\frac{\partial \Uppi _{PP}} {\partial R}} &= 2\alpha p_{1}^{\prime }(R)p_{1}\left( c(0)-c(z_{M})\right) \left( z_{M}-p_{1}^{2}\left( z_{M}-z_{D}\right) \right)\\ & \quad +2\alpha p_{1}^{\prime }(R)p_{1}2p_{1}\left( z_{M}-z_{D}\right) c(z_{M})\left( 1-p_{1}\right)\\ & \quad -2\left( 1-\alpha \right) p_{2}p_{2}^{\prime }(R)z_{B}\left( c(0)-c(z_{B})\right). \\ \end{aligned} $$

We have c′(z) < 0 and therefore c(0) − c(z _M ) > 0 since z _M  > 0. We also have p′(R) < 0, therefore the first and the second term are negative. We have c(0) − c(z _B ) < 0 since z _B  < 0. We also have p′(R) < 0, therefore the third term is also negative. Assume to the contrary that R ^HH > R ^PP. Then from ${\frac{\partial \Uppi _{PP}} {\partial R}} > c(0){\frac{\partial \Uppi _{HH}} {\partial R}}$ it follows that for R ^HH > R ^PP:

$$ \Uppi _{PP}\left(R^{HH}\right)-\Uppi _{PP}\left(R^{PP}\right) > c\left(0,R^{HH}\right)\Uppi _{HH}\left(R^{HH}\right)-c\left(0,R^{PP}\right)\Uppi _{HH}\left(R^{PP}\right). $$

Since $c(0,R)={\frac{M(R)} {1-M(R)}},$ we have ${\frac{\partial } {\partial R}}c(0,R)={\frac{m(R)} {(1-M(R)) ^{2}}} > 0$ and thus c(0,R ^HH) > c(0,R ^PP). Therefore,

$$ c\left(0,R^{HH}\right)\Uppi _{HH}\left(R^{HH}\right)-c\left(0,R^{PP}\right)\Uppi _{HH}\left(R^{PP}\right) > c\left(0,R^{PP}\right)\left( \Uppi _{HH}\left(R^{HH}\right)-\Uppi _{HH}\left(R^{PP}\right)\right).$$

Hence, for R ^HH > R ^PP we must have

$$ \Uppi _{PP}\left(R^{HH}\right)-\Uppi _{PP}\left(R^{PP}\right) > c\left(0,R^{PP}\right)\left(\Uppi _{HH}\left(R^{HH}\right)-\Uppi _{HH}\left(R^{PP}\right)\right).$$

The LHS is negative since the optimal R is R ^PP. The RHS is positive since the optimal R is R ^HH and c(0,R ^PP) > 0. This contradicts R ^HH > R ^PP. $\square$