Optimization in the Development of Target Contracts

Abstract

Target contracts are commonly used in construction and related project industries. However, research to date has largely been qualitative, and there is no universal agreement on how any sharing of project outcomes should be allocated between contracting parties. This chapter demonstrates that by formulating the sharing problem in optimization terms, specific quantitative results can be obtained for all the various combinations of the main variables that exist in the contractual arrangements and project delivery. Such variables include the risk attitudes of the parties (risk-neutral, risk-averse), single or multiple outcomes (cost, duration, quality), single or multiple agents (contractors, consultants), and cooperative or non-cooperative behaviour. The chapter gives original, newly derived results for optimal outcome sharing arrangements. The chapter will be of interest to academics and practitioners interested in the design of target contracts and project delivery. It provides an understanding of optimal sharing arrangements within projects, broader than currently available.

Appendices

Appendix A. Cooperative Owner-Contractor

Derivation of the solution to the maximization problem presented in expression (8)

Consider where both the owner and contractor are risk-averse, though the level of aversion may range from very large to being risk-neutral. Risk aversion is characterized by a concave utility function. Exponential, power and linear-exponential are candidate functions [48]. Here, the exponential utility function, because it has been popularly adopted [40, 48], is used, and for the owner and the contractor, respectively, have the form,

$$\begin{aligned} \mathrm{{U}}_\mathrm{{o}} (\mathrm{{x}}-\mathrm{{Fee}})=1-\exp \left[ {-\mathrm{{r}}_\mathrm{{o}} \left( {\mathrm{{x}}-\mathrm{{Fee}}} \right) } \right] \end{aligned}$$

(A1)

$$\begin{aligned} \mathrm{{U}}_\mathrm{{c}} (\mathrm{{Fee}})=1-\exp [-\mathrm{{r}}_\mathrm{{c}} \mathrm{{Fee}}] \end{aligned}$$

(A2)

where $\mathrm{{r}}_\mathrm{{o}}$ and $\mathrm{{r}}_\mathrm{{c}}$ are the owner’s and the contractor’s level of risk aversion, respectively. The shapes of the owner and contractor utility functions change with $\mathrm{{r}}_\mathrm{{o}}$ and $\mathrm{{r}}_\mathrm{{c}}$.

Substituting Eqs. (A1) and (A2) into expression (8), differentiating the result with respect to Fee, setting to zero, and simplifying,

$$\begin{aligned} -\mathrm{{r}}_\mathrm{{o}} \exp \left[ {-\mathrm{{r}}_\mathrm{{o}} \left( {\mathrm{{x}}-\mathrm{{Fee}}} \right) } \right] +\lambda \mathrm{{r}}_\mathrm{{c}} \exp \left[ {-\mathrm{{r}}_\mathrm{{c}} \mathrm{{Fee}}} \right] =0 \end{aligned}$$

(A3)

from which an expression for $\lambda$ can be obtained.

Taking the derivative of Eq. (A3) with respect to x gives,

$$\begin{aligned} \mathrm{{r}}_\mathrm{{o}}^2 \left[ {1-\frac{\mathrm{{dFee}}}{\mathrm{{dx}}}} \right] \exp \left[ {-\mathrm{{r}}_\mathrm{{o}} \left( {\mathrm{{x}}-\mathrm{{Fee}}} \right) } \right] -\lambda \mathrm{{r}}_\mathrm{{c}}^2 \frac{\mathrm{{dFee}}}{\mathrm{{dx}}}\exp \left[ {-\mathrm{{r}}_\mathrm{{c}} \mathrm{{Fee}}} \right] =0 \end{aligned}$$

(A4)

Substituting $\lambda$ from Eqs. (A3) into (A4),

$$\begin{aligned} \frac{\mathrm{{dFee}}}{\mathrm{{dx}}}=\frac{\mathrm{{r}}_\mathrm{{o}} }{\mathrm{{r}}_\mathrm{{o}} +\mathrm{{r}}_\mathrm{{c}} } \end{aligned}$$

(A5)

Integrating Eq. (A5) with respect to x gives the optimal outcome sharing arrangement of Eqs. (44) and (45).

Appendix B. Cooperative Owner-Contractor-Design Consultant (Traditional Delivery)

Derivation of the solution to the maximization problem presented in expression (12)

Consider a multiple-contract arrangement with traditional delivery, in which all parties are assumed to be risk-averse. Consistent with the above development, the parties’ utilities are described by an exponential form and for the owner and agents, respectively, are,

$$\begin{aligned} \mathrm{{U}}_\mathrm{{o}} (\mathrm{{x}}-\sum _{\mathrm{{j}}=1}^2 {\mathrm{{Fee}}_\mathrm{{j}} } )=1-\exp \left[ {-\mathrm{{r}}_\mathrm{{o}} \left( {\mathrm{{x}}-\sum _{\mathrm{{j}}=1}^2 {\mathrm{{Fee}}_\mathrm{{j}} } } \right) } \right] \end{aligned}$$

(B1)

$$\begin{aligned} \mathrm{{U}}_\mathrm{{i}} (\mathrm{{Fee}}_\mathrm{{i}} )=1-\exp [-\mathrm{{r}}_\mathrm{{i}} \mathrm{{Fee}}_\mathrm{{i}} ] \qquad \qquad \mathrm{{i}} = 1, 2 \end{aligned}$$

(B2)

where $\mathrm{{r}}_{\mathrm{{o}}}, \mathrm{{r}}_{1}$ and $\mathrm{{r}}_{2}$ are the levels of risk aversion of the owner, the contractor and the design consultant, respectively.

Substituting Eq. (B1) and (B2) into expression (12), differentiating the result with respect to $\mathrm{{Fee}}_{\mathrm{{i}}}$, i $=$ 1, 2, setting to zero, and simplifying,

$$\begin{aligned} -\mathrm{{r}}_\mathrm{{o}} \exp \left[ {-\mathrm{{r}}_\mathrm{{o}} \left( {\mathrm{{x}}-\sum _{\mathrm{{j}}=1}^2 {\mathrm{{Fee}}_\mathrm{{j}} } } \right) } \right] +\lambda _\mathrm{{i}} \mathrm{{r}}_\mathrm{{i}} \exp \left[ {-\mathrm{{r}}_{\mathrm{i}}\mathrm{{Fee}}_\mathrm{{i}} } \right] =0 \qquad \mathrm{{i}} = 1, 2 \end{aligned}$$

(B3)

from which expressions for $\lambda _\mathrm{{i}}$, i $=$ 1, 2 are obtained.

Differentiating Eq. (B3) with respect to x and substituting $\lambda _\mathrm{{i}}$, i $=$ 1, 2 from Eq. (B3),

$$\begin{aligned} \mathrm{{r}}_\mathrm{{o}} \left( {1-\frac{\mathrm{{d}} \sum _{\mathrm{{j}}=1}^2 {\mathrm{{Fee}}_\mathrm{{j}} } }{\mathrm{{dx}}}} \right) -\mathrm{{r}}_1 \left( {\frac{\mathrm{{dFee}}_1 }{\mathrm{{dx}}}} \right) =0 \end{aligned}$$

(B4)

$$\begin{aligned} \mathrm{{r}}_\mathrm{{o}} \left( {1-\frac{\mathrm{{d}} \sum _{\mathrm{{j}}=1}^2 {\mathrm{{Fee}}_\mathrm{{j}} } }{\mathrm{{dx}}}} \right) -\mathrm{{r}}_2 \left( {\frac{\mathrm{{dFee}}_2 }{\mathrm{{dx}}}} \right) =0 \end{aligned}$$

(B5)

The first terms in Eqs. (B4) and (B5) are the same. This then requires the second terms in Eqs. (B4) and (B5) to be the same. That is,

$$\begin{aligned} \mathrm{{r}}_1 \left( {\frac{\mathrm{{dFee}}_1 }{\mathrm{{dx}}}} \right) =\mathrm{{r}}_2 \left( {\frac{\mathrm{{dFee}}_2 }{\mathrm{{dx}}}} \right) \end{aligned}$$

(B6)

Substituting $\left( {\frac{\mathrm{{dFee}}_2 }{\mathrm{{dx}}}} \right)$ from Eqs. (B6) into (B4), and integrating with respect to x provides the contractor’s optimal fee, given in Eq. (48), i $=$ 1, and Eq. (49). Similarly substituting $\left( {\frac{\mathrm{{dFee}}_1 }{\mathrm{{dx}}}} \right)$ from Eqs. (B6) into (B5), and integrating with respect to x provides the design consultant’s optimal fee, given in Eqs. (48), i = 2, and (50).

Appendix C. Cooperative Owner-Contractor-Design Consultant (Managing Contractor Delivery)

Derivation of the solution to the maximization problem presented in expression (18)

Using exponential utility functions, the utilities of the owner and contractor are described respectively by,

$$\begin{aligned} \mathrm{{U}}_\mathrm{{o}} (\mathrm{{x-Fee}}_1 )=1-\exp \left[ {-\mathrm{{r}}_\mathrm{{o}} \left( {\mathrm{{x-Fee}}_1 } \right) } \right] \end{aligned}$$

(C1)

$$\begin{aligned} \text {U}_\text {1} \left( {\text {Fee}_\text {1} -\mathrm{{Fee}}_2 } \right) =1-\exp \left[ {-\mathrm{{r}}_1 \left( {\text {Fee}_\text {1} -\mathrm{{Fee}}_2 } \right) } \right] \end{aligned}$$

(C2)

The design consultant’s utility is obtained from Eq. (B2), with i = 2.

Substituting Eqs. (C2) and (B2), i $=$ 2, into expression (18), differentiating the result with respect to $\mathrm{{Fee}}_{2}$, setting to zero, and simplifying,

$$\begin{aligned} -\lambda _1 \mathrm{{r}}_1 \exp \left[ {-\mathrm{{r}}_1 \left( {\text {Fee}_\text {1} -\mathrm{{Fee}}_2 } \right) } \right] +\lambda _2 \mathrm{{r}}_2 \exp \left[ {-\mathrm{{r}}_2 \text {Fee}_\text {2} } \right] =0 \end{aligned}$$

(C3)

from which an expression for $\lambda _2$ can be obtained.

Taking the derivative of Eq. (C3) with respect to x gives,

$$\begin{aligned} \lambda _1 \mathrm{{r}}_1 ^{2}\left( {\frac{\text {dFee}_\text {1} }{\mathrm{{dx}}}-\frac{\text {dFee}_\text {2} }{\mathrm{{dx}}}} \right) \exp \left[ {-\mathrm{{r}}_1 \left( {\text {Fee}_\text {1} -\mathrm{{Fee}}_2 } \right) } \right] -\lambda _2 \mathrm{{r}}_2 ^{2}\frac{\text {dFee}_\text {2} }{\mathrm{{dx}}}\exp \left[ {-\mathrm{{r}}_2 \text {Fee}_\text {2} {}} \right] =0 \end{aligned}$$

(C4)

Substituting $\lambda _2$ from Eqs. (C3) into (C4) gives,

$$\begin{aligned} \frac{\mathrm{{dFee}}_2 }{\mathrm{{dx}}}=\left( {\frac{\mathrm{{r}}_1 }{\mathrm{{r}}_1 +\mathrm{{r}}_2 }} \right) \frac{\mathrm{{dFee}}_1 }{\mathrm{{dx}}} \end{aligned}$$

(C5)

Substituting Eqs. (C1), (C2) and (B2), i $=$ 2, into expression (18), and differentiating the result with respect to $\mathrm{{Fee}}_{1}$, setting to zero, and simplifying,

$$\begin{aligned} \text {-r}_\text {o} \exp \left[ {-\mathrm{{r}}_\mathrm{{o}} \left( {\mathrm{{x-Fee}}_1 } \right) } \right] +\lambda _1 \mathrm{{r}}_1 \exp \left[ {-\mathrm{{r}}_1 \left( {\text {Fee}_\text {1} -\mathrm{{Fee}}_2 } \right) {}} \right] +\lambda _2 \frac{\mathrm{{dFee}}_2 }{\mathrm{{dFee}}_1 }\mathrm {r}_2 \exp \left[ {-\mathrm {r}_2 \text {Fee}_\text {2} } \right] =0 \end{aligned}$$

(C6)

Taking the derivative of Eq. (C6) with respect to x gives,

$$\begin{aligned} \begin{array}{l} \text {r}_\text {o} ^{2}\left( {\text {1-}\frac{\text {dFee}_\text {1} }{\mathrm{{dx}}}} \right) \exp \left[ {-\mathrm{{r}}_\mathrm{{o}} \left( {\mathrm{{x-Fee}}_1 } \right) } \right] -\lambda _1 \mathrm{{r}}_1 ^{2}\left( {\frac{\text {dFee}_\text {1} }{\mathrm{{dx}}}-\frac{\text {dFee}_\text {2} }{\mathrm{{dx}}}} \right) \exp \left[ {-\mathrm{{r}}_1 \left( {\text {Fee}_\text {1} -\mathrm{{Fee}}_2 } \right) } \right] {} \\ -\lambda _2 \frac{\mathrm{{dFee}}_2 }{\mathrm{{dFee}}_1 }\frac{\text {dFee}_\text {2} }{\mathrm{{dx}}}\mathrm{{r}}_2 ^{2}\exp \left[ {-\mathrm{{r}}_2 \text {Fee}_\text {2} } \right] =0 \\ \end{array} \end{aligned}$$

(C7)

Substituting $\text {r}_\text {o} \exp \left[ {-\mathrm {r}_\text {o} \left( \mathrm{{x}-\mathrm {Fee}_1 } \right) } \right]$ from Eqs. (C6) into (C7), using Eq. (C5) and simplifying gives,

$$\begin{aligned} \text {r}_\text {o} \left( {1-\frac{\text {dFee}_\text {1} }{\mathrm{{dx}}}} \right) -\left( {\frac{\mathrm{{r}}_1 }{\mathrm{{r}}_1 +\mathrm{{r}}_2 }} \right) \frac{\text {dFee}_\text {1} }{\mathrm{{dx}}}\mathrm{{r}}_2 =0 \end{aligned}$$

(C8)

Integrating Eq. (C8) with respect to x provides the contractor’s optimal fee, given in Eq. (48), i $=$ 1, and Eq. (51).

Substituting Eq. (48), i $=$ 1, and Eqs. (51) into (C5), and integrating with respect to x, provides the design consultant’s optimal fee, given in Eq. (48), i $=$ 2 and Eq. (52).

Appendix D. Non Cooperative Contracting, Single-Agent, Single-Outcome Case

Derivation of the solution to the maximization problem presented in expressions (27), (28) and (29)

Maximizing expression (29) with respect to e yields the optimal level of effort,

$$\begin{aligned} \mathrm{{e}}=\frac{\mathrm{{k}}}{\mathrm{{b}}}\mathrm{{n}} \end{aligned}$$

(D1)

The optimal value of F is such that expression (28) holds as an equality, that is,

$$\begin{aligned} \mathrm{{F}}=\mathrm{{MinFee-nke}}+\frac{\mathrm{{b}}}{2}\mathrm{{e}}^{2}+\frac{1}{2}\mathrm{{n}}^{2}\mathrm{{r}}_\mathrm{{c}} \sigma ^{2} \end{aligned}$$

(D2)

Substituting Eqs. (D1) and (D2) into (27), the owner’s problem can be restated as,

$$\begin{aligned} \mathop {\mathrm{{Max}}}\limits _\mathrm{{n}} {}\left\{ {\frac{\mathrm{{k}}^{2}}{\mathrm{{b}}}\mathrm{{n-MinFee}}-\frac{\mathrm{{k}}^{2}}{2\mathrm{{b}}}\mathrm{{n}}^{2}-\frac{\mathrm{{n}}^{2}\mathrm{{r}}\sigma ^{2}}{2}-\frac{1}{2}\left( {1-\mathrm{{n}}} \right) ^{2}\mathrm{{r}}_\mathrm{{o}} \sigma ^{2}} \right\} {}{} \end{aligned}$$

(D3)

Differentiating expression (D3) with respect to n and setting to zero, leads to the optimal sharing ratio,

$$\begin{aligned} \mathrm{{n}}=\frac{1}{1+{\mathrm{{r}}_\mathrm{{c}} }/{(\mathrm{{r}}_\mathrm{{o}} +{\mathrm{{k}}^{2}}/{\sigma ^{2}\mathrm{{b}}})}} \end{aligned}$$

(D4)

Substituting Eq. (D1) into (D2) leads to the optimal fixed fee,

$$\begin{aligned} \mathrm{{Fee}}=\mathrm{{MinFee}}+\frac{1}{2}\left( {\mathrm{{r}}_\mathrm{{c}} \sigma ^{2}-\frac{\mathrm{{k}}^{2}}{\mathrm{{b}}}} \right) \mathrm{{n}}^{2} \end{aligned}$$

(D5)

Special cases

Contracts with a risk-neutral contractor

For the case where the contractor is risk-neutral, while the owner is either risk-neutral or risk-averse, the optimal sharing ratio and fixed fee are obtained by setting $\mathrm{{r}}_{\mathrm{{c}}}$ $=$ 0,

$$\begin{aligned} \mathrm{{n}}=1 \end{aligned}$$

(D6)

$$\begin{aligned} \mathrm{{Fee}}=\mathrm{{MinFee}}-\frac{\mathrm{{k}}^{2}}{2\mathrm{{b}}} \end{aligned}$$

(D7)

Contracts with a risk-neutral owner

For the case where the owner is risk-neutral while the contactor is risk-averse (ranging to risk-neutral) the optimal sharing ratio is obtained by setting $\mathrm{{r}}_{\mathrm{{o}}}$ $=$ 0,

$$\begin{aligned} \mathrm{{n}}=\frac{1}{1+\mathrm{{r}}_\mathrm{{c}} \sigma ^{2}\mathrm{{b}}/{\mathrm{{k}}^{2}}} \end{aligned}$$

(D8)

And the optimal fixed fee is obtained from Eq. (D5).

Appendix E. Contracts with a Consortium of Risk-Neutral Contractors

Derivation of the solution to the maximization problem presented in expressions (32), (33) and (34)

Differentiating expression (34) with respect to $\mathrm{{e}}_{\mathrm{{i}}}$ and setting to zero provides the optimal level of effort,

$$\begin{aligned} \mathrm{{e}}_\mathrm{{i}} =\frac{\mathrm{{k}}_\mathrm{{i}} }{\mathrm{{b}}}\mathrm{{n}}_\mathrm{{i}} \qquad \qquad \mathrm{{i}} = 1, 2 \end{aligned}$$

(E1)

The optimal value of $\mathrm{{F}}_{\mathrm{{i}}}$ would be such that expression (33) holds as an equality, that is,

$$\begin{aligned} \mathrm{{F}}_\mathrm{{i}} =\mathrm{{MinFee}}_\mathrm{{i}} -\mathrm{{n}}_\mathrm{{i}} \sum _{\mathrm{{j}}=1}^2 {\mathrm{{k}}_\mathrm{{j}} \mathrm{{e}}_\mathrm{{j}} } +\frac{\mathrm{{b}}}{2}\mathrm{{e}}_{_\mathrm{{i}} }^2 {{\qquad \mathrm {i}}} = 1, 2 \end{aligned}$$

(E2)

Substituting Eqs. (E1) and (E2) into (32), the owner’s problem can be restated as,

$$\begin{aligned} \mathop {\mathrm{{Max}}}\limits _{\mathrm{{n}}_1 ,\mathrm{{n}}_2 } \left\{ {\mathrm{{E}}\left[ {\mathrm{{U}}_{^{\mathrm{{O}}}} } \right] \mathop {=\mathrm{{Max}}}\limits _{\mathrm{{n}}_1 ,\mathrm{{n}}_2 } \left( {\sum _{\mathrm{{i}}=1}^2 {\frac{\mathrm{{k}}_{_\mathrm{{i}} }^2 }{\mathrm{{b}}}\mathrm{{n}}_\mathrm{{i}} } } \right) -\sum _{\mathrm{{i}}=1}^2 {\left( {\mathrm{{MinFee}}_\mathrm{{i}} +\frac{\mathrm{{k}}_{_\mathrm{{i}} }^2 }{2\mathrm{{b}}}\mathrm{{n}}_{_\mathrm{{i}} }^2 } \right) } } \right\} \end{aligned}$$

(E3)

The sum of the contractors’ sharing ratios, namely $\mathrm{{n}}_{1} + \mathrm{{n}}_{2}$, is equal to the outcome sharing ratio to the consortium, that is the proportion the consortium receives in the consortium-owner relationship. The consortium proportion of outcome share takes values in the range 0 to 1. Thus the solution of the above maximization needs to satisfy,

$$\begin{aligned} \sum _{\mathrm{{i}}=1}^2 {\mathrm{{n}}_\mathrm{{i}} } \le 1 \end{aligned}$$

(E4)

Differentiating expression (E3) with respect to $\mathrm{{n}}_{\mathrm{{i}}}$, i $=$ 1, 2, and setting to zero,

$$\begin{aligned} \mathrm{{n}}_\mathrm{{i}} =1 {\qquad \mathrm{{i}}} = 1, 2 \end{aligned}$$

(E5)

This result does not satisfy Eq. (E4). Accordingly, the maximum of expression (E3) lies on the line $\mathrm{{n}}_{1} + \mathrm{{n}}_{2}$ = 1 which is the boundary of the admissible region of the maximization problem.

Introducing a Lagrange multiplier $\lambda$, the maximization becomes,

$$\begin{aligned} \mathop {\mathrm{{Max}}}\limits _{\mathrm{{n}}_1 ,\mathrm{{n}}_2 } \left\{ {\left( {\sum _{\mathrm{{i}}=1}^2 {\frac{\mathrm{{k}}_{_\mathrm{{i}} }^2 }{\mathrm{{b}}}\mathrm{{n}}_\mathrm{{i}} } } \right) -\sum _{\mathrm{{i}}=1}^2 {\left( {\mathrm{{MinFee}}_\mathrm{{i}} +\frac{\mathrm{{k}}_{_\mathrm{{i}} }^2 }{2\mathrm{{b}}}\mathrm{{n}}_{_\mathrm{{i}} }^2 } \right) } +\lambda \left( {\sum _{\mathrm{{i}}=1}^2 {\mathrm{{n}}_\mathrm{{i}} } -1} \right) } \right\} \end{aligned}$$

(E6)

Differentiating expression (E6) with respect to $\mathrm{{n}}_{\mathrm{{i}}}$, i $=$ 1, 2, and $\lambda$, setting to zero, and simplifying leads to the optimal sharing ratios of Eqs. (59) and (60).

Substituting Eq. (E1) into (E2), leads to the optimal fixed fees of Eqs. (61) and (62).

Appendix F. Contracts with a Consortium of Risk-Averse Contractors

Derivation of the solution to the maximization problem presented in expressions (32), (35) and (36)

Differentiating Eq. (36) with respect to $\mathrm{{e}}_{\mathrm{{i}}}$ and setting to zero provides the optimal level of effort, and this leads to Eq. (E1).

The optimal value of $\mathrm{{F}}_{\mathrm{{i}}}$ would be such that expression (35) holds as an equality, that is,

$$\begin{aligned} \mathrm{{F}}_{_\mathrm{{i}} } = \mathrm{{MinFee}}_\mathrm{{i}} -\mathrm{{n}}_\mathrm{{i}} \sum _{\mathrm{{j}}=1}^2 {\mathrm{{k}}_\mathrm{{j}} \mathrm{{e}}_\mathrm{{j}} } +\frac{\mathrm{{b}}}{2}\mathrm{{e}}_{_\mathrm{{i}} }^2 +\frac{1}{2}\mathrm{{n}}_{_\mathrm{{i}} }^2 \mathrm{{r}}_\mathrm{{i}} \sigma ^{2} {\quad \mathrm{{i}}} = 1, 2 \end{aligned}$$

(F1)

Substituting Eqs. (E1) and (F1) into expression (32), the owner’s problem can be restated as,

$$\begin{aligned} \mathop {\mathrm{{Max}}}\limits _{\mathrm{{n}}_1 ,\mathrm{{n}}_2 } \left\{ {\left( {\sum _{\mathrm{{i}}=1}^2 {\frac{\mathrm{{k}}_\mathrm{{i}}^2 }{\mathrm{{b}}}\mathrm{{n}}_\mathrm{{i}} } } \right) -\sum _{\mathrm{{i}}=1}^2 {\left( {\mathrm{{MinFee}}_\mathrm{{i}} +\frac{\mathrm{{k}}_\mathrm{{i}}^2 }{2\mathrm{{b}}}\mathrm{{n}}_\mathrm{{i}}^2 +\frac{1}{2}\mathrm{{n}}_\mathrm{{i}}^2 \mathrm{{r}}_\mathrm{{i}} \sigma ^{2}} \right) } } \right\} \end{aligned}$$

(F2)

Differentiating expression (F2) with respect to $\mathrm{{n}}_{\mathrm{{i}}}$, i $=$ 1, 2, and setting to zero, leads to the optimal sharing ratio of Eq. (63).

Substituting Eq. (E1) into (F1), leads to the optimal fixed components of Eqs. (64) and (65).

Where the contractors’ levels of risk aversion approach zero and the contractors become risk-neutral, the solutions of expression (F2) lie on the line $\mathrm{{n}}_{1} + \mathrm{{n}}_{2}$ = 1, and the optimal sharing ratios are obtained by Eqs. (59) and (60).

Appendix G. Contracts with Multiple Outcomes

Derivation of the solution to the maximization problem presented in expressions (41), (42) and (43)

Maximizing expression (43) with respect to e yields the optimal effort levels,

$$\begin{aligned} \mathbf{{e}}=\mathbf{{B}}^{-1}\mathbf{{K}}^{\mathrm{{T}}}\mathbf{{n}} \end{aligned}$$

(G1)

where the superscript $-$1 denotes the inverse of the matrix.

The optimal value of F is such that expression (42) holds as an equality, that is,

$$\begin{aligned} \mathrm{{F}}=\mathrm{{MinFee}}-\mathbf{{n}}^{\mathrm{{T}}}\mathbf{{Ke}}+\frac{1}{2}\mathbf{{e}}^{\mathrm{{T}}}\mathbf{{Be}}+\frac{1}{2}\mathrm{{r}}\mathbf{{n}}^{\mathrm{{T}}}\mathbf{{Pn}} \end{aligned}$$

(G2)

Substituting Eqs. (G1) and (G2) into expression (41), the owner’s problem can be restated as,

$$\begin{aligned} \mathop {\mathrm{{Max}}}\limits _\mathbf{{n}} \mathbf{{q}}^{\mathrm{{T}}}\mathbf{{KB}}^{-1}\mathbf{{K}}^{\mathrm{{T}}}\mathbf{{n}}-\mathrm{{MinFee}}-\frac{1}{2}\mathbf{{\mathbf{{n}}}}^{\mathrm{{T}}}\mathbf{{KB}}^{-1}\mathbf{{K}}^{\mathrm{{T}}}\mathbf{{n}}-\frac{1}{2}\mathrm{{r}}\mathbf{{n}}^{\mathrm{{T}}}\mathbf{{Pn}} \end{aligned}$$

(G3)

Differentiating expression (G3) with respect to n and setting to zero leads,

$$\begin{aligned} \mathbf{{KB}}^{-1}\mathbf{{K}}^{\mathrm{{T}}}\mathbf{{q}}-\mathbf{{KB}}^{-1}\mathbf{{K}}^{\mathrm{{T}}}\mathbf{{n}}-\mathrm{{r}} \mathbf{{Pn}}=0 \end{aligned}$$

(G4)

Multiplying by (KB $^{-1}\mathbf{{K}}^{\mathrm{{T}}})^{-1}$ leads to the optimal n of Eq. (66). Substituting Eq. (G1) into (G2) leads to the optimal fixed fee of Eq. (67).