Technical Change and the Decentralization Penalty

Abstract

We consider the organizer of a firm who compares a decentralized arrangement where divisions are granted total autonomy with a centralized arrangement where perfect monitoring and policing guarantee that all divisions make the choices the organizer wants them to make. We ask: when does improvement in the divisions’ technology strengthen the case for decentralization and when does it weaken it? In a simple one-division model with complete information and linear contracts we obtain conditions under which the welfare loss due to decentralizing rises (falls) when technology improves.

Appendix

1.1 Proofs of Parts (a), (g), and (h) of Theorem 1

1.1.1 Proof of Part (a)

The function r ⋅ R(x) − tC(x), where t is fixed, displays strictly increasing differences in r, x if r ⋅ R(x) displays strictly increasing differences in r, x. But that is the case, since R is nondecreasing. Since, for fixed t, the effort $\hat {x}(r,t)$ maximizes r ⋅ R(x) − tC(x) on the effort set Σ, it is indeed the case that $\hat {x}(r_H,t) \geq \hat {x}(r_L,t)$, as (a) asserts. Part (a) also asserts that the Agent strictly prefers the higher share. That is the case since $\hat {x}(r_H,t)$ is a maximizer of r _H ⋅ R(x) − t ⋅ C(x), so we have

$$\displaystyle \begin{aligned} r_H \cdot R(\hat{x}(r_H,t))-t \cdot C (\hat{x}(r_H,t))&\geq r_H \cdot R(\hat{x}(r_L,t))-t \cdot C (\hat{x}(r_L,t))\\ &> r_H \cdot R(\hat{x}(r_L,t))-t \cdot C (\hat{x}(r_L,t)). \end{aligned} $$

1.1.2 Proof of Part (g)

Part (g) says:

$$\displaystyle \begin{aligned}W(r,t_L) > W(r,t_H) \mbox{ whenever }t_L,t_H \in \tilde{\Gamma} \mbox{and }0<t_L<t_H.\end{aligned}$$

The effort $\hat {x}(r,t_L)$ is a maximizer of rR(x) − t _H ⋅ C(x). Hence,

$$\displaystyle \begin{aligned}r \cdot R(\hat{x}(r,t_L))-t_L \cdot C(\hat{x}(r,t_L)) \geq r \cdot R(\hat{x}(r,t_H))-t_L \cdot C(\hat{x}(r,t_H))\end{aligned}$$

or

$$\displaystyle \begin{aligned}r \cdot [R(\hat{x}(r,t_L))-R(\hat{x}(r,t_H))] \geq t_L \cdot [C(\hat{x}(r,t_L))-C(\hat{x}(r,t_H))].\end{aligned}$$

That implies—since 0 < r < 1—that

$$\displaystyle \begin{aligned}R(\hat{x}(r,t_L))-R(\hat{x}(r,t_H)) > t_L \cdot [C(\hat{x}(r,t_L))-C(\hat{x}(r,t_H))]\end{aligned}$$

or

$$\displaystyle \begin{aligned}R(\hat{x}(r,t_L))-t_L \cdot C(\hat{x}(r,t_L)) >R(\hat{x}(r,t_H))-t_L \cdot C(\hat{x}(r,t_H))\end{aligned}$$

and hence (since t _H > t _L)

$$\displaystyle \begin{aligned}R(\hat{x}(r,t_L))-t_L \cdot C(\hat{x}(r,t_L)) >R(\hat{x}(r(t_H),t_H))-t_H \cdot C(\hat{x}(r,t_H))\end{aligned}$$

The term on the left of the inequality is W(r, t _L) and the term on the right is W(r, t _H). That completes the proof of Part (g).

1.1.3 Proof of Part (h)

When the Agent’s share is r _H, he chooses an effort $\hat {x}(r_H,t)$ which satisfies

$$\displaystyle \begin{aligned}r_H R(\hat{x}(r_H,t))-tC(\hat{x}(r_H,t))\geq r_H R(\hat{x}(r_L,t))-tC(\hat{x}(r_L,t)),\end{aligned}$$

or equivalently

$$\displaystyle \begin{aligned} r_H \cdot \left[R(\hat{x}(r_H,t))-R(\hat{x}(r_L,t))\right]\geq t \cdot \left[C(\hat{x}(r_H,t))-C(\hat{x}(r_L,t))\right]. \end{aligned} $$

(1)

Part (a) of Theorem 1 tells us that $\hat {x}(r_H,t) \geq \hat {x}(r_L,t)$. Since R is strictly increasing, that means that the left side of (1) is either positive or zero. First, suppose that it is positive. Then, since r _H < 1, (1) implies that

$$\displaystyle \begin{aligned} R(\hat{x}(r_H,t))-R(\hat{x}(r_L,t)) > t \cdot \left[C(\hat{x}(r_H,t))-C(\hat{x}(r_L,t))\right], \end{aligned} $$

(2)

or equivalently

$$\displaystyle \begin{aligned} R(\hat{x}(r_H,t))- t \cdot C(\hat{x}(r_H,t))> R(\hat{x}(r_L,t))- t \cdot C(\hat{x}(r_L,t)),\end{aligned} $$

(3)

i.e.,

$$\displaystyle \begin{aligned} W(r_H,t) > W(r_L,t).\end{aligned} $$

(4)

If $\hat {x}(r_H,t) \neq \hat {x}(r_L,t)$, then, since R is strictly increasing, the left side of (1) is indeed positive, so (4) holds. If, on the other hand, $\hat {x}(r_H,t) = \hat {x}(r_L,t)$, then both sides of (1) equal zero and (2),(3),(4) become equalities. So, as claimed, W(r _H, t) ≥ W(r _L, t) and the inequality is strict if and only if $\hat {x}(r_H,t) \neq \hat {x}(r_L,t)$.

1.2 Proofs of Parts (a) and (b) of Theorem 2

1.2.1 Proof of Part (a)

We note first that the Agent’s chosen effort $\hat {x}(r,t)$ depends only on the ratio ${r \over t}$, which we shall call ρ. The set of possible values of ρ is $\left (0, {1 \over t} \right ]$. The Agent’s effort is a value of x which maximizes t ⋅ (ρR(x) − C(x)) on the effort set Σ and is therefore a maximizer of ρR(x) − C(x). We shall use a new symbol, namely ϕ(ρ) to denote the Agent’s chosen effort when the ratio is ρ. So, $\phi (\rho )= \hat {x}(r,t)$. The function ρR(x) − C(x) displays strictly increasing differences with respect to ρ, x. Hence, the maximizer ϕ(ρ) is nondecreasing in ρ, so we have

$$\displaystyle \begin{aligned} \phi(\rho_H) \geq \phi(\rho_L) \mbox{ whenever }0<\rho_L<\rho_H. \end{aligned} $$

(+)

We can now reinterpret the Principal as the chooser of a ratio. For a given t, he chooses the ratio $\rho ^{*}(t)={r^{*}(t) \over t}$, where

$$\displaystyle \begin{aligned}\rho^{*}(t)= \min \{\mbox{argmax}_{\rho \in (0,1/t)}~ M(\rho, -t)~ \},\end{aligned} $$

and

$$\displaystyle \begin{aligned}M(\rho, -t) = (1-t\rho) \cdot R(\phi (\rho)) = R(\phi(\rho)) -t \cdot \rho \cdot R(\phi(\rho)).\end{aligned} $$

The function M has strictly increasing differences in ρ, −t if the function − t ⋅ ρ ⋅ R(ϕ(ρ)) has strictly increasing differences in ρ, −t. But that is the case, since R is nondecreasing, which implies (using (+)) that R(ϕ( ⋅ )) is also nondecreasing. Since ρ ^∗(t) is a maximizer of M(ρ, −t), we conclude that

$$\displaystyle \begin{aligned}{r^{*}(t_L) \over t_L} =\rho^{*}(t_L) \geq \rho^{*}(t_H) ={r^{*}(t_H) \over t_H} \mbox{ whenever }0<t_L<t_H,\end{aligned} $$

as Part (a) asserts.

1.2.2 Proof of Part (b)

We use the terminology just used in the proof of Part (a). Since $\phi \left ({r^{*}(t) \over t} \right )= \hat {x}(r^{*}(t),t)$, we have, using (+) in the proof of part (a), $\hat {x}(r^{*}(t_L),t_L) \geq \hat {x}(r^{*}(t_H),t_H)$, as (b) asserts.