Optimal procurement mechanisms: bidding on price and damages for breach

Abstract

We study the optimal procurement mechanism when contract breach and abandoning a project may be efficient, either because of completion costs higher than anticipated or because of new and more lucrative opportunities for the contractor. When contractors have private information about their costs, the procurer finds it optimal to set damages above expectation damages. There is a lock-in effect, or status quo bias; the agent that has won the award will complete the project even in situations when it would be efficient to abandon it. If the cost types of all agents are above a threshold, the optimal bidding procedure assigns the project by lottery. The optimal mechanism cannot be implemented by standard auction formats. However, the larger the number of agents bidding for the project, the closer auctions with a liquidated damage clause approximate the optimal mechanism.

Appendix

In this appendix, first we prove a lemma that deals with the second-order condition of the agents’ reporting problem. Then, we provide a proof of Propositions 4 and 5.

Lemma 1

Consider the mechanism described by the functions $p_{i}(k_{i},k_{-i}),$ $t_{i}(k_{i},k_{-i}),$ $\pi _{i}(k_{i},k_{-i})$ for all $i.$ Suppose that (a) $p_{i}(k_{i},k_{-i})=p_{i}\left( k_{i}\right)$ (i.e., $p_{i}$ does not depend on the types $k_{-i})$; (b) $p_{i}(k_{i})$ is increasing in $k_{i}$ and differentiable; (c) $\Pi _{i}(k_{i})=\int \nolimits _{K_{-i}}\pi _{i}(k_{i},k_{-i})g_{-i}(k_{-i})\text{ d }k_{-i}$ exists and is (weakly) decreasing in $k_{i}.$ If this mechanism satisfies the first-order condition of the agent’s reporting problem, then it also satisfies the second-order condition, and hence, it is incentive compatible.

Proof

Consider the first-order condition of agent $i$ reporting problem:

$$\begin{aligned} \left. \frac{\partial U(z,k_{i})}{\partial z}\right| _{z=k_{i}}=0. \end{aligned}$$

Differentiating it totally yields

$$\begin{aligned} \left. \frac{\partial ^{2}U(z,k_{i})}{\partial z\partial k_{i}}\right| _{z=k_{i}}+\left. \frac{\partial ^{2}U(z,k_{i})}{\partial z^{2}}\right| _{z=k_{i}}=0. \end{aligned}$$

Since

$$\begin{aligned} \frac{\partial U(z,k_{i})}{\partial k_{i}}=-\int \limits _{K_{-i}}\left[ 1-F\left( p_{i}(z,k_{-i})+k_{i}\right) \right] \pi _{i}(z,k_{-i})g_{-i}(k_{-i})\text{ d }k_{-i}, \end{aligned}$$

under the hypotheses of the lemma, we can write the second-order condition as:

$$\begin{aligned}&-\left. \frac{\partial ^{2}U(z,k_{i})}{\partial z^{2}}\right| _{z=k_{i}}=\left. \frac{\partial ^{2}U(z,k_{i})}{\partial z\partial k_{i}}\right| _{z=k_{i}} \\&\quad =-\left[ 1-F\left( p_{i}(k_{i})+k_{i}\right) \right] \frac{\text{ d }\Pi _{i}(k_{i})}{ \text{ d }k_{i}}+f\left( p_{i}(k_{i})+k_{i}\right) \frac{\text{ d }p_{i}(k_{i})}{\text{ d }k_{i}}\Pi _{i}(k_{i})\ge 0. \end{aligned}$$

$\square$

Proof of Proposition 4

Consider first a second-price, bonus auction. Let $\beta (k_{i})$ be the bidding function (for notational simplicity, we drop the dependence on $t$) and assume provisionally that it is strictly increasing everywhere. Let $Q(k)=[1-G(k)]$ . We can write bidder $i$’s problem of determining the optimal value of his bid $b$ as:

$$\begin{aligned}&\max _{b}\int \limits _{\beta ^{-1}(b)}^{k^{+}}\left\{ \int \limits _{c^{-}}^{\beta (k)+k_{i}+t} \left[ \beta (k)+k_{i}+t-c\right] f(c)\text{ d }c\right. \\&\quad \left. -\left( k_{i}+t\right) \right\} \left( N-1\right) g\left( k\right) Q(k)^{N-2}\text{ d }k. \end{aligned}$$

The first-order condition is

$$\begin{aligned}&-\frac{\text{ d }\beta ^{-1}(b)}{\text{ d }b}\left\{ \int \limits _{c^{-}}^{b+k_{i}+t}\left[ b+k_{i}+t-c \right] f(c)\text{ d }c\right. \\&\quad \left. -\left( k_{i}+t\right) \right\} \left( N-1\right) g\left( \beta ^{-1}(b)\right) Q(\beta ^{-1}(b))^{N-2}=0, \end{aligned}$$

which yields, using the Nash equilibrium condition $b=\beta (k_{i})$,

$$\begin{aligned} \int \limits _{c^{-}}^{\beta (k_{i})+k_{i}+t}\left[ \beta (k_{i})+k_{i}+t-c\right] f(c)\text{ d }c-k_{i}-t=0 \end{aligned}$$

(5)

or

$$\begin{aligned} \beta (k_{i})=E\left[ c{|}c<\beta (k_{i})+k_{i}+t\right] +\left( k_{i}+t\right) \frac{1-F\left( \beta (k_{i})+k_{i}+t\right) }{F\left( \beta (k_{i})+k_{i}+t\right) }. \end{aligned}$$

(6)

It is immediate to see that the second-order condition is satisfied and that Eq. (6) defines the equilibrium bidding function, as long as $\beta (k_{i})$ is strictly increasing. To see that $\beta \left( k_{i}\right)$ is increasing, note that if we differentiate (5) with respect to $k_{i}$, we obtain

$$\begin{aligned} \int \limits _{c^{-}}^{\beta (k_{i})+k_{i}+t}\left[ \beta ^{\prime }(k_{i})+1\right] f(c)\text{ d }c-1=0 \end{aligned}$$

and hence,

$$\begin{aligned} \beta ^{\prime }(k_{i})=\frac{1-F\left( \beta (k_{i})+k_{i}+t\right) }{ F\left( \beta (k_{i})+k_{i}+t\right) }>0\quad \text{ for } \quad \beta (k_{i})+k_{i}+t<c^{+}. \end{aligned}$$

Using (6), we see that $\beta \left( k_{i}\right) +k_{i}+t<c^{+}$ for types $k_{i}$ such that $k_{i}<c^{+}-E\left[ c\right] -t$. Thus, we have shown that if $k_{i}<c^{+}-E\left[ c\right] -t$, then bidder $k_{i}$ will bid according to $\beta \left( k_{i};t\right) .$

Now suppose $\beta \left( k_{i}\right) +k_{i}+t\ge c^{+}$ or, equivalently, $k_{i}\ge c^{+}-E\left[ c\right] -t$. When winning, bidder $k_{i}$ will complete the contract with certainty and therefore must bid no less than the expected cost; furthermore, he will certainly lose the auction if he asks for more than the expected cost. It follows that the equilibrium bid is $E \left[ c\right] .$ Thus, since $\beta \left( k_{i}\right) =E\left[ c\right]$ for $k_{i}\ge c^{+}-E\left[ c\right]$, bidding according to $\beta \left( k_{i}\right)$ is also an equilibrium for these types.

Now consider a second-price, up-front-payment auction. Let $p$ be the damages and $b(k_{i})$ be the bidding function (for notational simplicity, we drop the dependence on $p$) and assume provisionally that it is strictly increasing everywhere. We can write bidder $i$’s problem of determining the optimal value of his bid $\widetilde{b}$ (the up-front payment) as:

$$\begin{aligned}&\max _{\widetilde{b}}\int \limits _{b^{-1}(\widetilde{b})}^{k^{+}}\left\{ b(k)-\int \limits _{c^{-}}^{p+k_{i}}cf(c)\text{ d }c\right. \\&\qquad \qquad \qquad \left. -\int \limits _{p+k_{i}}^{c^{+}}\left[ k_{i}+p \right] f(c)\text{ d }c\right\} \left( N-1\right) g\left( k\right) Q(k)^{N-2}\text{ d }k. \end{aligned}$$

The first-order condition is

$$\begin{aligned}&-\frac{db^{-1}(\widetilde{b})}{d\widetilde{b}}\left\{ \widetilde{b} -\int \limits _{c^{-}}^{p+k_{i}}cf(c)\text{ d }c\right. \\&\qquad \qquad \qquad \quad \left. -\int \limits _{p+k_{i}}^{c^{+}}\left[ k_{i}+p\right] f(c)\text{ d }c\right\} \left( N-1\right) g\left( b^{-1}(\widetilde{b})\right) Q(b^{-1}(\widetilde{b}))^{N-2}=0, \end{aligned}$$

which yields, using the Nash equilibrium condition $\widetilde{b}=b(k_{i};p):$

$$\begin{aligned} b(k_{i};p)=\int \limits _{c^{-}}^{p+k_{i}}cf(c)\text{ d }c+\int \limits _{p+k_{i}}^{c^{+}}\left[ k_{i}+p \right] f(c)\text{ d }c. \end{aligned}$$

It remains to check that $b$ is increasing in $k_{i}$. This follows immediately:

$$\begin{aligned} \frac{\partial b(k_{i};p)}{\partial k_{i}}=1-F\left( p+k_{i}\right) . \end{aligned}$$

Note that the fact that $b$ is increasing also implies that the second-order condition holds. $\square$

Proof of Proposition 5

Observe from (4) that

$$\begin{aligned} \frac{\partial b(k_{i};p)}{\partial p}=1-F\left( p+k_{i}\right) . \end{aligned}$$

Let $k^{(1)}$ be the lowest cost type, $k^{(2)}$ be the second lowest cost type, and $g_{k^{\left( 1\right) },k^{\left( 2\right) }}(x,y)$ be their joint density. The principal chooses $p$ to maximize

$$\begin{aligned} \int \limits _{k^{-}}^{k^{+}}\int \limits _{k^{-}}^{k^{+}}\left\{ \int \limits _{c^{-}}^{p+x}Vf(c)\text{ d }c+\int \limits _{p+x}^{c^{+}}pf\left( c\right) \text{ d }c-b(y;p)\right\} g_{k^{\left( 1\right) },k^{\left( 2\right) }}(x,y)\text{ d }x\text{ d }y \end{aligned}$$

which yields the first-order condition

$$\begin{aligned}&\int \limits _{k^{-}}^{k^{+}}\int \limits _{k^{-}}^{k^{+}}\left\{ \left[ V-p\right] f(p+x)+ \left[ 1-F\left( p+x\right) \right] \right. \\&\quad \left. -\left[ 1-F\left( p+y\right) \right] \right\} g_{k^{\left( 1\right) },k^{\left( 2\right) }}(x,y)\text{ d }x\text{ d }y=0 \end{aligned}$$

or

$$\begin{aligned} \left[ V-p\right]&\int \limits _{k^{-}}^{k^{+}}\int \limits _{k^{-}}^{k^{+}}f(p+x)g_{k^{\left( 1\right) },k^{\left( 2\right) }}(x,y)\text{ d }x\text{ d }y\\&\quad -\int \limits _{k^{-}}^{k^{+}}\int \limits _{k^{-}}^{k^{+}}\left[ F\left( p+x\right) -F\left( p+y\right) \right] g_{k^{\left( 1\right) },k^{\left( 2\right) }}(x,y)\text{ d }x\text{ d }y=0 \end{aligned}$$

from which we obtain

$$\begin{aligned} p=V+\frac{E\left[ F\left( p+k^{\left( 2\right) }\right) \right] -E\left[ F\left( p+k^{\left( 1\right) }\right) \right] }{E\left[ f\left( p+k^{\left( 1\right) }\right) \right] }. \end{aligned}$$

Since $k^{(1)}<k^{(2)}$, it is immediate to see that the optimal damage clause is $p>V$ as long as $V+k^{-}<c^{+}.$ If $V+k^{-}\ge c^{+}$, then $F\left( V+k_{1}\right) =F\left( V+k_{2}\right) =1$ for all $k_{1},k_{2}$ and $p=V.$ $\square$