Capacity reallocation via sinking high-quality resource in a hierarchical healthcare system

Abstract

This paper studies the capacity reallocation in a hierarchical medical ecosystem via sinking high-quality resource from high-level hospitals to low-level hospitals. To facilitate the capacity sinking, we develop two payment schemes: fee-for-capacity (FFC) and performance payment (PP). Under the FFC scheme, the low-level hospital always pays a unit capacity sinking price to the high-level hospital, whereas under the PP scheme, the reallocation price is paid contingent on the increased patient visits at the low-level hospital due to capacity sinking. By considering the profit- and utility-maximizing behaviors of strategic parties, we build a four-stage Stackelberg sequential game model within a queuing framework to derive the equilibrium results in terms of the low-level hospital’s capacity, the high-level hospital’s capacity sinking rate, and the funder’s capacity sinking price. In the absence of funder’s coordination, it is shown that any increase in sinking price always reduces the capacity sinking rate. In the presence of funder’s coordination, we find that: (1) the payment schemes under study will not alter the efficiency or coordination of the overall healthcare systems; (2) for the setting with a high perceived value by patients, under each payment scheme, the capacity sinking program is efficient to increase the high-level hospital’s profit and the social welfare as well, but it lowers the patient visit rate at the low-level hospital; (3) for the setting with a higher difference between the patients’ perceived values at the two levels of hospitals, the capacity sinking program is efficient to increase the patient visit rate at the low-level hospital and the social welfare as well, but it sacrifices the high-level hospital’s profit. Finally, numerical studies provide more useful managerial insights.

Appendix

1.1 Proof of Proposition 1

According to $\left. {U_{1M} (x_{k} ) = U_{2} (x_{k} ,\;\mu_{k2} )} \right|_{{\lambda_{2} = \lambda^{\prime}_{2} }}$ and λ_k1M + φλ = min {Λ_k, λ} − λ′, if [α − β(x_k)]v − p ≠ 0, then we have

$$ \begin{aligned} \frac{{[\alpha - \beta (x_{k} )]v - p}}{d} & = \frac{1}{{(1 - x_{k} )\mu _{1} - (\Lambda _{k} \wedge \lambda - \lambda _{2} )}} - \frac{1}{{\mu _{2} + x\mu _{1} - \lambda _{2} }} \\ & \Leftrightarrow \frac{{[\alpha - \beta (x_{k} )]v - p}}{d}( = 1/\Delta ) \\ & = \frac{{[\mu _{{k2}} + (2x_{k} - 1)\mu _{1} + \Lambda _{k} ] - 2\lambda_{2} ^{\prime} }}{{\left\{ {[(1 - x_{k} )\mu _{1} - \Lambda _{k} \wedge \lambda ] + \lambda _{2}^{\prime} } \right\}[(\mu _{{k2}} + x_{k} \mu _{1} ) - \lambda _{2}^{\prime} ]}} \\ & \Leftrightarrow \left\{ {[(1 - x_{k} )\mu _{1} - \Lambda _{k} \wedge \lambda ] + \lambda _{2}^{\prime} } \right\}[(\mu _{{k2}} + x_{k} \mu _{1} ) - \lambda_{2} ^{\prime} ] \\ & = \;\Delta \left\{ {[\mu _{{k2}} + (2x_{k} - 1)\mu _{1} + \Lambda _{k} \wedge \lambda ] - 2\lambda _{2}^{\prime} } \right\} \\ & \Leftrightarrow \left\{ \begin{gathered} - \lambda _{2}^{\prime 2} + \left\{ {[\mu _{{k2}} + (2x_{k} - 1)\mu _{1} + \Lambda _{k} \wedge \lambda ] + 2\Delta } \right\}\lambda _{2}^{\prime} + \hfill \\ \left\{ {[(1 - x_{k} )\mu _{1} - \Lambda _{k} \wedge \lambda ](\mu _{{k2}} + x_{k} \mu _{1} ) - \Delta [\mu _{{k2}} + (2x_{k} - 1)\mu _{1} + \Lambda _{k} \wedge \lambda ]} \right\} \hfill \\ \end{gathered} \right\} = 0. \\ \end{aligned} $$

In light of Assumption 2, we have [α − β(x_k)]v − p > 0, thus, $\lambda^{\prime}_{2} = \tilde{\lambda }_{k 2} (x_{k} ,\;\mu_{k2} )$, where

$$ \begin{aligned} \tilde{\lambda }_{k 2} (x_{k} ,\;\mu_{k2} ) & = \Lambda_{k} \wedge \lambda + (x_{k} - 1)\mu_{1} + \left( {\frac{{\mu_{k2} + \mu_{1} - \Lambda_{k} \wedge \lambda }}{2} + \Delta } \right) \\ & \quad - \sqrt {\left\{ {\frac{{(\mu_{k2} + \mu_{1} ) - \Lambda_{k} \wedge \lambda }}{2}} \right\}^{2} + \Delta^{2} }. \\ \end{aligned} $$

Next, setting

$$ \begin{aligned} H_{1} (x_{k} ,\;\mu_{k2} ) & = \frac{{[\alpha - \beta (x_{k} )]v - p}}{d} - \left[ {\frac{1}{{(1 - x_{k} )\mu_{1} - \lambda }} - \frac{1}{{\mu_{k2} + x\mu_{1} }}} \right]; \\ H_{2} (x_{k} ,\;\mu_{k2} ) & = \frac{{[\alpha - \beta (x_{k} )]v - p}}{d} - \left[ {\frac{1}{{(1 - x_{k} )\mu_{1} - \Lambda_{k} }} - \frac{1}{{\mu_{k2} + x\mu_{1} }}} \right]. \\ H_{3} (x_{k} ,\;\mu_{k2} ) & = \frac{{[\alpha - \beta (x_{k} )]v - p}}{d} - \left[ {\frac{1}{{(1 - x_{k} )\mu_{1} - \varphi \lambda }} - \frac{1}{{\mu_{k2} + x\mu_{1} - (1 - \varphi )\lambda }}} \right]; \\ H_{4} (x_{k} ,\;\mu_{k2} ) & = \frac{{[\alpha - \beta (x_{k} )]v - p}}{d} - \left[ {\frac{1}{{(1 - x_{k} )\mu_{1} - \varphi \lambda }} - \frac{1}{{\mu_{k2} + x\mu_{1} - (\Lambda_{k} - \varphi \lambda )}}} \right]; \\ H_{5} (x_{k} ,\;\mu_{k2} ) & = \Lambda_{k} - \lambda ,\;\Lambda_{k} = \mu_{1} + \mu_{k2} - \frac{d}{\alpha v - p} - \frac{d}{{\beta (x_{k} ) \cdot v}}; \\ \end{aligned} $$

• If $(x_{k} ,\;\mu_{k2} ) \in \mathop { \arg }\limits_{{x_{k} \in [0,\;1],\mu_{k2} }} \left\{ {H_{1} (x_{k} ,\;\mu_{k2} ) \ge 0\;\text{and} \;H_{5} (x_{k} ,\;\mu_{k2} ) \ge 0} \right\}$, then we have λ_k1M(x_k, μ_k2) = (1 − φ)λ, λ_k2(x_k, μ_k2) = 0;

• if $(x_{k} ,\;\mu_{k2} ) \in \mathop { \arg }\limits_{{x_{k} \in [0,\;1],\;\mu_{k2} }} \left\{ {H_{2} (x_{k} ,\;\mu_{k2} ) \ge 0\;\text{and} \;H_{5} (x_{k} ,\;\mu_{k2} ) \le 0} \right\}$, then we have λ_k2M(x_k, μ_k2) = Λ_k(x_k, μ_k2) − φλ, λ_k2(x_k, μ_k2) = 0;

• if $(x_{k} ,\;\mu_{k2} ) \in \mathop { \arg }\limits_{{x_{k} \in [0,\;1],\;\mu_{k2} }} \left\{ {H_{1} (x_{k} ,\;\mu_{k2} ) \le 0,\;H_{3} (x_{k} ,\;\mu_{k2} ) \ge 0,\;\text{and} \;H_{5} (x_{k} ,\;\mu_{k2} ) \ge 0} \right\}$, then we have $\lambda_{k1M} (x_{k} ,\;\mu_{k2} ) = (1 - \varphi )\lambda - \tilde{\lambda }_{k2}^{1} (x_{k} ,\;\mu_{k2} ),\lambda_{k2} (x_{k} ,\;\mu_{k2} ) = \tilde{\lambda }_{k2}^{1} (x_{k} ,\;\mu_{k2} )$;

• if $(x_{k} ,\;\mu_{k2} ) \in \mathop { \arg }\limits_{{x_{k} \in [0,\;1],\;\mu_{k2} }} \left\{ {H_{2} (x_{k} ,\;\mu_{k2} ) \le 0,\;H_{4} (x_{k} ,\;\mu_{k2} ) \ge 0,\;\text{and} \;H_{5} (x_{k} ,\;\mu_{k2} ) \le 0} \right\}$, then we have $\lambda_{k2M} (x_{k} ,\;\mu_{k2} ) = \Lambda_{k} (x_{k} ,\;\mu_{k2} ) - \varphi \lambda - \tilde{\lambda }_{k2}^{2} (x_{k} ,\;\mu_{k2} ),\lambda_{k2} (x_{k} ,\;\mu_{k2} ) = \tilde{\lambda }_{k2}^{2} (x_{k} ,\;\mu_{k2} )$;

• if $(x_{k} ,\;\mu_{k2} ) \in \mathop { \arg }\limits_{{x_{k} \in [0,\;1],\;\mu_{k2} }} \left\{ {H_{4} (x_{k} ,\;\mu_{k2} ) \le 0\;\text{and} \;H_{5} (x_{k} ,\;\mu_{k2} ) \le 0} \right\}$, then we have λ_k1M(x_k, μ_k2) = 0, λ_k2(x_k, μ_k2) = Λ_k(x_k, μ_k2) − φλ;

• if $(x_{k} ,\;\mu_{k2} ) \in \mathop { \arg }\limits_{{x_{k} \in [0,\;1],\;\mu_{k2} }} \left\{ {H_{3} (x_{k} ,\;\mu_{k2} ) \le 0\;\text{and} \;H_{5} (x_{k} ,\;\mu_{k2} ) \ge 0} \right\}$, then we have λ_k1M(x_k, μ_k2) = 0, λ_k2(x_k, μ_k2) = (1 − φ)λ,

where

$$ \begin{aligned} \tilde{\lambda }_{k2}^{1} (x_{k} ,\;\mu_{k2} ) & = \lambda + (x_{k} - 1)\mu_{1} + \left( {\frac{{\mu_{k2} + \mu_{1} - \lambda }}{2} + \Delta } \right) - \sqrt {\left\{ {\frac{{(\mu_{k2} + \mu_{1} ) - \lambda }}{2}} \right\}^{2} + \Delta^{2} } ; \\ \tilde{\lambda }_{k2}^{2} (x_{k} ,\;\mu_{k2} ) & = \Lambda_{k} + (x_{k} - 1)\mu_{1} + \left( {\frac{{\mu_{k2} + \mu_{1} - \Lambda_{k} }}{2} + \Delta } \right) - \sqrt {\left\{ {\frac{{(\mu_{k2} + \mu_{1} ) - \Lambda_{k} }}{2}} \right\}^{2} + \Delta^{2} }. \\ \end{aligned} $$

Finally, this concludes the proof. □

1.2 Proof of Proposition 2

Under the fee-for-capacity scheme, if $\lambda_{f2} = \tilde{\lambda }_{f2}$, taking the first-and second-order derivative of $\tilde{\lambda }_{f2}$ with respect to μ_f2, then we have

$$ \frac{{\partial \tilde{\lambda }_{f2} }}{{\partial \mu_{f2} }} = \frac{1}{2}\left\{ { 1- {1 \mathord{\left/ {\vphantom {1 {\sqrt {1 + \left\{ {\frac{2d}{{\left\{ {[\alpha - \beta (x_{f} )]v - p} \right\}[\mu_{f2} + (2x_{f} - 1)\mu_{1} + \Lambda_{f} ]}}} \right\}^{2} } }}} \right. \kern-0pt} {\sqrt {1 + \left\{ {\frac{2d}{{\left\{ {[\alpha - \beta (x_{f} )]v - p} \right\}[\mu_{f2} + (2x_{f} - 1)\mu_{1} + \Lambda_{f} ]}}} \right\}^{2} } }}} \right\} \in \left( {0,\;\frac{1}{2}} \right), $$

$$ \frac{{\partial^{2} \tilde{\lambda }_{f2} }}{{\partial \mu_{f2}^{2} }} = - \frac{{\frac{{2d^{2} }}{{\left\{ {[\alpha - \beta (x_{f} )]v - p} \right\}^{2} [\mu_{2} + (2x_{f} - 1)\mu_{1} + \Lambda_{f} ]^{3} }}}}{{\left\{ {1 + \left\{ {\frac{2d}{{\left\{ {[\alpha - \beta (x_{f} )]v - p} \right\}[\mu_{f2} + (2x_{f} - 1)\mu_{1} + \Lambda_{f} ]}}} \right\}^{2} } \right\}^{{\frac{3}{2}}} }} < 0. $$

Hence, we further have $\arg\left\{ {\mathop {\hbox{max} }\nolimits_{{\mu_{f2} }}\left\{ \lambda_{f2} (\mu_{f2} )\right\}} \right\} = \hbox{max} \left\{ {\mathop {\arg}\nolimits_{{\mu_{f2} }} \left\{ {\Pi_{f2} (\mu_{f2} ) \ge 0} \right\}} \right\}$.

Since the expected profit of Hospital 2 is Π_f2(μ_f2) = B − c₂μ_f2 − p_fx_fμ₁, we obtain $\frac{{\partial \Pi_{f2} (\mu_{f2} )}}{{\partial \mu_{f2} }} = - c_{2} \le 0$. Denoting μ_f2*(x_f) as Hospital 2’s optimal capacity level for any given x_f, then we show that $\mu_{f2}^{ * } (x_{f} ) = \arg\mathop {\hbox{max} }\nolimits_{{\mu_{2} }} \{\lambda_{f2} (\mu_{f2} )\} = \mathop {\arg}\nolimits_{{\mu_{2} }} \left\{ {\Pi_{f2} (\mu_{f2} ) = 0} \right\} = {{(B - p_{f} x_{f} \mu_{1} )} \mathord{\left/ {\vphantom {{(B - p_{f} x_{f} \mu_{1} )} {c_{2} }}} \right. \kern-0pt} {c_{2} }}$.

Consequently, this completes the proof. □

1.3 Proof of Lemma 1

Since $\Lambda_{f1} (x_{f} ) = (1 - x_{f} )\mu_{1} - \frac{d}{\alpha v - p} \ge \varphi \lambda$, one has $x_{f} \le \bar{x}^{0} = 1 - {{\varphi \lambda } \mathord{\left/ {\vphantom {{\varphi \lambda } {\mu_{1} }}} \right. \kern-0pt} {\mu_{1} }} - {d \mathord{\left/ {\vphantom {d {[(\alpha v - p)\mu_{1} ]}}} \right. \kern-0pt} {[(\alpha v - p)\mu_{1} ]}}$.

Taking the first-and second-order derivatives of Λ_f(x_f) and Λ_f2(x_f) with respect to $x_{f}$, we obtain

$$ \begin{aligned} \frac{{\;\partial \Lambda_{f2} (x_{f} )}}{{\partial x_{f} }} & = - \left( {\frac{{p_{f} }}{{c_{2} }} - 1} \right)\mu_{1} + \frac{d(\alpha - 1)}{{ [ 1+ x_{f} (\alpha - 1)]^{2} \cdot v}}; \\ \frac{{\;\partial^{2} \Lambda_{f2} (x_{f} )}}{{\partial x_{f}^{2} }} & = - \frac{{2d(\alpha - 1)^{2} }}{{ [ 1+ x_{f} (\alpha - 1)]^{3} \cdot v}}; \\ \end{aligned} $$

$$ \begin{aligned} \frac{{\;\partial \Lambda_{f} (x_{f} )}}{{\partial x_{f} }} & = - \frac{{p_{f} \mu_{1} }}{{c_{2} }} + \frac{d(\alpha - 1)}{{ [ 1+ x_{f} (\alpha - 1)]^{2} \cdot v}}; \\ \frac{{\;\partial^{2} \Lambda_{f} (x_{f} )}}{{\partial x_{f}^{2} }} & = - \frac{{2d(\alpha - 1)^{2} }}{{ [ 1+ x_{f} (\alpha - 1)]^{3} \cdot v}}. \\ \end{aligned} $$

Therefore,

$$ \frac{{\;\partial \Lambda_{f2} (x_{f} )}}{{\partial x_{f} }} \ge 0 \Leftrightarrow x_{f} \le \bar{x}_{f}^{01} ;\,\,\,\,\frac{{\;\partial \Lambda_{f} (x_{f} )}}{{\partial x_{f} }} \ge 0 \Leftrightarrow x_{f} \le \bar{x}_{f}^{02} , $$

where $\bar{x}_{f}^{01} = \sqrt {\frac{d}{{v[\mu_{1} (p_{f} /c_{2} - 1)(\alpha - 1)]}}} - \frac{1}{(\alpha - 1)}$, and $\bar{x}_{f}^{02} = \sqrt {\frac{d}{{v[\mu_{1} (p_{f} /c_{2} )(\alpha - 1)]}}} - \frac{1}{(\alpha - 1)}.$

When x_f = 0, we have $\left. {\Lambda_{f2} (x_{f} )} \right|_{{x_{f} = 0}} = \frac{B}{{c_{2} }} - \frac{d}{v} \triangleq \Lambda_{f2}^{0}$.

Hence, $\Lambda_{f2} (x_{f} ) - \Lambda_{f2}^{0} = \frac{d}{v} - \left[ {\left( {\frac{{p_{f} }}{{c_{2} }} - 1} \right)\mu_{1} x_{f} + \frac{d}{{\beta (x_{f} ) \cdot v}}} \right]$. Next, setting Λ_f2(x_f) − Λ ⁰_f2  = 0, we have

$$ \begin{aligned} & \frac{d}{v} = \left[ {\left( {\frac{{p_{f} }}{{c_{2} }} - 1} \right)\mu_{1} x_{f} + \frac{d}{{\beta (x_{f} ) \cdot v}}} \right] \\ & \quad \Leftrightarrow x_{f} (\alpha - 1) = \frac{{(p_{f} - c_{2} )\mu_{1} v}}{{dc_{2} }}x_{f} \left[ {1 + x_{f} (\alpha - 1)} \right] \\ & \quad \Leftrightarrow \bar{x}_{f}^{ 1} = 0,\;\text{or} \;\bar{x}_{f}^{ 2} = \frac{{dc_{2} }}{{(p_{f} - c_{2} )\mu_{1} v}} - \frac{1}{(\alpha - 1)}. \\ \end{aligned} $$

When x_f = 0, we have $\left. {\Lambda_{f} (x_{f} )} \right|_{{x_{f} = 0}} = \mu_{1} + \frac{B}{{c_{2} }} - \frac{d}{\alpha v - p} - \frac{d}{v} \triangleq \Lambda_{f}^{0}$.

Hence, $\Lambda_{f} (x_{f} ) - \Lambda_{f}^{0} = \frac{d}{v} - \left[ {\frac{{p_{f} x_{f} \mu_{1} }}{{c_{2} }} + \frac{d}{{\beta (x_{f} ) \cdot v}}} \right]$. Next, setting Λ_f(x_f) − Λ ⁰_f  = 0, we have

$$ \begin{aligned} \frac{d}{v} & = \left[ {\frac{{p_{f} x_{f} \mu_{1} }}{{c_{2} }} + \frac{d}{{\beta (x_{f} ) \cdot v}}} \right] \Leftrightarrow [\beta (x_{f} ) - 1] \\ & = \frac{{p_{f} \mu_{1} v}}{{dc_{2} }}x_{f} \beta (x_{f} ) \\ & \Leftrightarrow x_{f} (\alpha - 1) = \frac{{p_{f} \mu_{1} v}}{{dc_{2} }}x_{f} \left[ {1 + x_{f} (\alpha - 1)} \right] \\ & \Leftrightarrow \bar{x}_{f}^{ 3} = 0,\;\text{or} \;\bar{x}_{f}^{ 4} = \frac{{dc_{2} }}{{p_{f} \mu_{1} v}} - \frac{1}{(\alpha - 1)}. \\ \end{aligned} $$

• (i) If $\bar{x}_{f}^{ 4} = \frac{{dc_{2} }}{{p_{f} \mu_{1} v}} - \frac{1}{(\alpha - 1)} \le 0\Leftrightarrow p_{f} \ge \frac{{dc_{2} (\alpha - 1)}}{{\mu_{1} v}}$, we have Λ_f2(x_f) ≤ Λ ⁰_f2 and Λ_f(x_f) ≤ Λ ⁰_f for $x_{f} \in [ 0 ,\;\bar{x}_{f}^{ 0} ]$.

• (ii) If $\bar{x}_{f}^{ 4} = \frac{{dc_{2} }}{{p_{f} \mu_{1} v}} - \frac{1}{(\alpha - 1)} \ge 0\Leftrightarrow p_{f} \le \frac{{dc_{2} (\alpha - 1)}}{{\mu_{1} v}}$,

  • (a) when $\bar{x}_{f}^{2} = \frac{{dc_{2} }}{{(p_{f} - c_{2} )\mu_{1} v}} - \frac{1}{(\alpha - 1)} \le \bar{x}^{0}$ $\Leftrightarrow p_{f} \ge \frac{{dc_{2} }}{{\mu_{1} v\left\{ {1 - {{\varphi \lambda } \mathord{\left/ {\vphantom {{\varphi \lambda } {\mu_{1} }}} \right. \kern-0pt} {\mu_{1} }} - d / [(\alpha v - p)\mu_{1} ]+ 1/(\alpha - 1)} \right\}}} + c_{2}$,

    • if $x_{f} \in [ 0 ,\;\bar{x}_{f}^{4} ]$, then we have Λ_f(x_f) ≥ Λ ⁰_f and Λ_f2(x_f) ≥ Λ ⁰_f2 ;

    • if $x_{f} \in [\bar{x}_{f}^{4} ,\;\bar{x}_{f}^{2} ]$, then we have Λ_f(x_f) ≤ Λ ⁰_f and Λ_f2(x_f) ≥ Λ ⁰_f2 ; and

    • if $x_{f} \in [\bar{x}_{f}^{2} ,\;\bar{x}^{0} ]$, then we have Λ_f(x_f) ≤ Λ ⁰_f and Λ_f2(x_f) ≤ Λ ⁰_f2 .

  • (b) when

    $$ \begin{aligned} \bar{x}_{f}^{2} & = \frac{{dc_{2} }}{{(p_{f} - c_{2} )\mu_{1} v}} - \frac{1}{(\alpha - 1)} \ge \bar{x}^{0} \Leftrightarrow p_{f} \\ & \le \frac{{dc_{2} }}{{\mu_{1} v\{ 1 - {{\varphi \lambda } \mathord{\left/ {\vphantom {{\varphi \lambda } {\mu_{1} }}} \right. \kern-0pt} {\mu_{1} }} - d / [(\alpha v - p)\mu_{1} ]+ 1/(\alpha - 1)\} }} + c_{2} \\ \end{aligned} $$

    and $\bar{x}_{f}^{4} = \frac{{dc_{2} }}{{p_{f} \mu_{1} v}} - \frac{1}{(\alpha - 1)} \le \bar{x}^{0} \Leftrightarrow p_{f} \ge \frac{{dc_{2} }}{{\mu_{1} v\{ 1 - {{\varphi \lambda } \mathord{\left/ {\vphantom {{\varphi \lambda } {\mu_{1} }}} \right. \kern-0pt} {\mu_{1} }} - d / [(\alpha v - p)\mu_{1} ]+ 1/(\alpha - 1)\} }}$,

    • if $x_{f} \in [ 0 ,\;\bar{x}_{f}^{4} ]$, then we have Λ_f(x_f) ≥ Λ ⁰_f and $\Lambda_{f2} (x_{f} ) \ge \Lambda_{f2}^{0}$;

    • if $x_{f} \in [\bar{x}_{f}^{4} ,\;\bar{x}^{0} ]$, then we have Λ_f(x_f) ≤ Λ ⁰_f and Λ_f2(x_f) ≥ Λ ⁰_f2

  • (c) when $\bar{x}_{f}^{4} = \frac{{dc_{2} }}{{p_{f} \mu_{1} v}} - \frac{1}{(\alpha - 1)} \ge \bar{x}^{0} \Leftrightarrow p_{f} \le \frac{{dc_{2} }}{{\mu_{1} v\{ 1 - {{\varphi \lambda } \mathord{\left/ {\vphantom {{\varphi \lambda } {\mu_{1} }}} \right. \kern-0pt} {\mu_{1} }} - d / [(\alpha v - p)\mu_{1} ]+ 1/(\alpha - 1)\} }}$, for any $x_{f} \in [0 ,\;\bar{x}^{0} ]$, we have Λ_f(x_f) ≥ Λ ⁰_f and Λ_f2(x_f) ≥ Λ ⁰_f2 .

This completes the proof. □

1.4 Proof of Lemma 2

When x_f = 0, we have $\left. {\Lambda_{f} (x_{f} )} \right|_{{x_{f} = 0}} = \hbox{min} \left\{ {\mu_{1} + B/c_{2} - d /(\alpha v - p) - {d \mathord{\left/ {\vphantom {d v}} \right. \kern-0pt} v},\;\lambda } \right\} = \lambda$.

Hence, Λ_f(x_f) − λ = μ₁ + B/c₂ − (p_f/c₂)x_fμ₁ − d/(αv − p) − d/[β(x_f) · v] − λ.

We have

$$ \begin{aligned} \Lambda_{f} \left( {x_{f} } \right) - \lambda & \ge 0 \Leftrightarrow [\mu_{1} + B/c_{2} - \lambda - d /(\alpha v - p)] \\ & \ge {{(p_{f} } \mathord{\left/ {\vphantom {{(p_{f} } {c_{2} }}} \right. \kern-0pt} {c_{2} }})x_{f} \mu_{1} + {d \mathord{\left/ {\vphantom {d {\{ [1 + x_{f} (\alpha - 1)] \cdot v\} }}} \right. \kern-0pt} {\{ [1 + x_{f} (\alpha - 1)] \cdot v\} }} \\ & \Leftrightarrow b_{f} [{1 \mathord{\left/ {\vphantom {1 {(\alpha - 1)}}} \right. \kern-0pt} {(\alpha - 1)}} + x_{f} ] \ge \mu_{1} (\alpha - 1)[{1 \mathord{\left/ {\vphantom {1 {(\alpha - 1)}}} \right. \kern-0pt} {(\alpha - 1)}} + x_{f} ]x_{f} \\ & \quad + (d/v)(c_{2} /p_{f} ) \\ & \Leftrightarrow \mu_{1} (\alpha - 1)x_{f}^{2} + (\mu_{1} - b_{f} )x_{f} + (d/v)(c_{2} /p_{f} ) \\ & \quad - {{b_{f} } \mathord{\left/ {\vphantom {{b_{f} } {(\alpha - 1)}}} \right. \kern-0pt} {(\alpha - 1)}} \le 0 \Leftrightarrow 0 \le x_{f} \le \bar{x}_{f}^{1} , \\ \end{aligned} $$

where

$$ \begin{aligned} b_{f} & = (\alpha - 1)(c_{2} /p_{f} )[\mu_{1} + B/c_{2} - \lambda - {d \mathord{\left/ {\vphantom {d {(\alpha v - p)]}}} \right. \kern-0pt} {(\alpha v - p)]}}; \\ \bar{x}_{f}^{1} & = \frac{{(b_{f} - \mu_{1} ) + \sqrt {(\mu_{1} + b_{f} )^{2} - 4\mu_{1} (d/v)(\alpha - 1)(c_{2} /p_{f} )} }}{{2(\alpha - 1)\mu_{1} }}. \\ \end{aligned} $$

We further have

$$ \bar{x}_{f}^{1} \le \bar{x}^{0} \Leftrightarrow p_{f} \ge \hat{p}_{f}^{0} \triangleq \frac{{c_{2} }}{{\mu_{1} \bar{x}^{0} }}\left\{ {\mu_{1} + \frac{B}{{c_{2} }} - \lambda - \frac{d}{\alpha v - p} - \frac{d}{{v[(\alpha - 1)\bar{x}^{0} + 1]}}} \right\}, $$

where

$$ \hat{p}_{f}^{0} = \frac{{c_{2} }}{{\mu_{1} \bar{x}^{0} }}\left\{ {\mu_{1} + \frac{B}{{c_{2} }} - \lambda - \frac{d}{\alpha v - p} - \frac{d}{{v[(\alpha - 1)\bar{x}^{0} + 1]}}} \right\}. $$

• If $p_{f} \ge \hat{p}_{f}^{0}$, then we have Λ_f(x_f) ≤ λ for any $\bar{x}_{f}^{1} \le x_{f} \le \bar{x}^{0}$, and Λ_f(x_f) ≥ λ for any $0 \le x_{f} \le \bar{x}_{f}^{1}$.

• If $0 < p_{f} < \hat{p}_{f}^{0}$, then we have Λ_f(x_f) > λ for any $0 \le x_{f} \le \bar{x}^{0}$.

This completes the proof. □

1.5 Proof of Proposition 3

When x_f = 0, we have Π ⁰₁  = p(λ − λ ⁰₂ ) − c₁μ₁.

When $x_{f} = \bar{x}_{f}^{1}$, we have $\tilde{\Pi }_{f1}^{ * } = \tilde{\Pi }_{f1}^{1 * }$,

$$ \tilde{\Pi }_{f1}^{1 * } = p(1 - \bar{x}_{f}^{1} )\mu_{1} - \frac{{dc_{2} }}{{[1 + (\alpha - 1)\bar{x}_{f}^{1} ]v}} - \frac{{d(p + c_{ 2} )}}{\alpha v - p} - c_{1} \mu_{1} + B + c_{2} (\mu_{1} - \lambda ). $$

$\tilde{\Pi }_{f1}^{ * } \ge \Pi_{1}^{0} \Leftrightarrow$$p_{f}^{ * } \ge \hat{p}_{f}^{1} \triangleq \frac{{c_{2} }}{{\mu_{1} \hat{x}^{1} }}\left\{ {(\mu_{1} - \lambda ) + \frac{B}{{c_{2} }} - \frac{d}{\alpha v - p} - \frac{d}{{v[(\alpha - 1)\hat{x}^{1} + 1]}}} \right\}$, where

$$ \begin{aligned} \hat{x}^{1} & = \frac{{(Y - \mu_{1} ) + \sqrt {(Y + \mu_{1} )^{2} - 4\mu_{1} (d/v)(\alpha - 1)(c_{2} /p)} }}{{2(\alpha - 1)\mu_{1} }}; \\ Y & = (\alpha - 1)\left\{ {\mu_{1} - (\lambda - \lambda_{2}^{0} ) - \frac{d}{\alpha v - p} + \frac{{c_{2} }}{p}\left[ {(\mu_{1} - \lambda ) + \frac{B}{{c_{2} }} - \frac{d}{\alpha v - p}} \right]} \right\}. \\ \end{aligned} $$

When $x_{f} = \bar{x}^{0}$, we have $\Pi_{f1}^{1} = p\varphi \lambda - c_{1} \mu_{1} + p_{f} \mu_{1} \bar{x}^{0}$, and then $\Pi_{f1}^{1} \ge \Pi_{1}^{0} \Leftrightarrow$ $p_{f} \in [\hat{p}_{f}^{2} ,\;\infty )$, where $\hat{p}_{f}^{2} = \frac{{p[(1 - \varphi )\lambda - \lambda_{2}^{0} ]}}{{\mu_{1} \bar{x}^{0} }}$.

Considering the impact of B, when $\tilde{B}^{0} < B \le \tilde{B}^{1}$, we have $\hat{p}_{f}^{2} \ge \hat{p}_{f}^{0}$, where

$$ \begin{aligned} X_{1} & = \frac{{c_{2} }}{p}\left\{ {\mu_{1} - \lambda - \frac{d}{\alpha v - p} - \frac{d}{{v[(\alpha - 1)\bar{x}^{0} + 1]}}} \right\} \\ & \quad - (\mu_{1} - \varphi \lambda ) + \frac{d}{(\alpha - 1)v - p}; \\ \tilde{B}^{1} & = \frac{{\sqrt {\left[ {\frac{1}{{c_{2} }}X_{1} - \frac{1}{p}\left( {\mu_{1} - \lambda } \right)} \right]^{2} + \frac{1}{p}\left( {\frac{1}{{c_{2} }} + \frac{1}{p}} \right)\frac{4d}{{[(\alpha - 1)v - p]^{2} }}} - \left[ {X_{1} \left( {\frac{1}{{c_{2} }} + \frac{2}{p}} \right) + \frac{1}{p}\left( {\mu_{1} - \lambda } \right)} \right]}}{{\frac{2}{p}\left( {\frac{1}{{c_{2} }} + \frac{1}{p}} \right)}}. \\ \end{aligned} $$

Consequently, this completes the proof for Proposition 3. □

1.6 Proof of Proposition 4

Based on Proposition 2, we have $\arg\mathop {\hbox{max} }\nolimits_{{\mu_{2} }} \{\lambda_{q2} (\mu_{2} )\} = \hbox{max} \mathop {\arg}\nolimits_{{\mu_{2} }} \left\{ {\Pi_{q2} (\mu_{2} ) \ge 0} \right\}$.

Since the expected profit of Hospital 2 is $\Pi_{q2} (\mu_{2} ) = B - c_{2} \mu_{2} - p_{q} (\lambda_{q 2} - \lambda_{2}^{0} )$, thus, if $\lambda_{q2} = \tilde{\lambda }_{q2}$, we get $\frac{{\partial \Pi_{q2} (\mu_{2} )}}{{\partial \mu_{2} }} = - c_{2} - p_{q} \frac{{\partial \lambda_{q 2} }}{{\partial \mu_{2} }}$.

• If $\hbox{min} \{ \Lambda_{q} ,\;\lambda \} = \lambda \Leftrightarrow \mu_{2} \ge \lambda - \mu_{1} + \frac{d}{\alpha v - p} + \frac{d}{{[1 + x_{q} (\alpha - 1)] \cdot v}}$, we have

  $$ \frac{{\partial \tilde{\lambda }_{q2} }}{{\partial \mu_{2} }} = \frac{ 1}{ 2}\left[ { 1- \frac{{\left( {\frac{{\mu_{1} + \mu_{2} - \lambda }}{2}} \right)}}{{\sqrt {\left( {\frac{{\mu_{1} + \mu_{2} - \lambda }}{2}} \right)^{2} + \left\{ {\frac{d}{[\alpha - \beta (x)]v - p}} \right\}^{2} } }}} \right] \ge 0; $$

• if $\hbox{min} \{ \Lambda_{q} ,\;\lambda \} = \Lambda_{q} \Leftrightarrow \mu_{2} \le \lambda - \mu_{1} + \frac{d}{\alpha v - p} + \frac{d}{{[1 + x_{q} (\alpha - 1)] \cdot v}}$, we have $\frac{{\partial \tilde{\lambda }_{q2} }}{{\partial \mu_{2} }} = 1\ge 0.$ Hence, $\frac{{\partial \Pi_{q2} (\mu_{2} )}}{{\partial \mu_{2} }} \le 0$. Denoting μ_q2*(x_q) as Hospital 2’s optimal capacity level under the PP scheme for any given x_q, then we show that $\mu_{q2}^{ * } (x_{q} ) = \arg\mathop {\hbox{max} }\nolimits_{{\mu_{2} }} \{\lambda_{q2} (\mu_{2} )\} = \mathop {\arg}\nolimits_{{\mu_{2} }} \left\{ {\Pi_{q2} (\mu_{2} ) = 0} \right\}$.

• (1) If $\hbox{min} \{ \Lambda_{q} ,\;\lambda \} = \Lambda_{q} \Leftrightarrow \mu_{2} \le \lambda - \mu_{1} + \frac{d}{\alpha v - p} + \frac{d}{{[1 + x_{q} (\alpha - 1)] \cdot v}}$, under the condition of $x_{q} \le \bar{x}^{ 0}$, we have $B - c_{2} \mu_{2} - p_{q} (\lambda_{q 2} - \lambda_{2}^{0} ) = 0 \Leftrightarrow \lambda_{q 2} = \lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} }}} \right. \kern-0pt} {p_{q} }} - ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }})\mu_{2} \Leftrightarrow$

  $$\begin{aligned} & \mu_{q2}^{ * } (x_{q} ) = \frac{{\lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} }}} \right. \kern-0pt} {p_{q} }} - x_{q} \mu_{1} + \sqrt {h_{1}^{2} + h_{2}^{2} } + h_{1} - h_{2} }}{{ 1+ {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }}}}\;{\text{and}}\;x_{q} \in X_{q}^{ 1} ,\\ & \quad {\text{where}}\;h_{1} = \frac{d}{2(\alpha v - p)} + \frac{d}{{2[1 + x_{q} (\alpha - 1)] \cdot v}};\quad h_{2} = \frac{d}{{[(1 - x_{q} )(\alpha - 1)]v - p}}; \hfill \\ & \quad X_{q}^{1} = \left\{ {\left. {x_{q} } \right|0 < \sqrt {h_{1}^{2} + h_{2}^{2} } + h_{1} - h_{2} - x_{q} \mu_{1} \le ( 1+ {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}(\lambda - \mu_{1} + 2h_{1} ) - \lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} \;{\text{and}}\;x_{q} < \tilde{x}_{q}^{0} }}} \right. \kern-0pt} {p_{q} \;{\text{and}}\;x_{q} < \tilde{x}_{q}^{0} }}} \right\}. \hfill \\ \end{aligned} $$

• (2) If $\hbox{min} \{ \Lambda_{q} ,\;\lambda \} = \lambda \Leftrightarrow \mu_{2} \ge \lambda - \mu_{1} + \frac{d}{\alpha v - p} + \frac{d}{{[1 + x_{q} (\alpha - 1)]v}}$, with $x_{q} \le \bar{x}^{ 0}$, we have

  $$ \begin{aligned} B - c_{2} \mu _{2} - p_{q} (\lambda _{{q{\text{2}}}} - \lambda _{{q2}}^{0} ) & = 0 \Leftrightarrow \lambda _{{q{\text{2}}}} = \lambda _{{q2}}^{0} + \frac{{B - c_{2} \mu _{2} }}{{p_{q} }} \Leftrightarrow \frac{{\mu _{2} + \mu _{1} - \lambda }}{2} \\ & \quad + \frac{d}{{[\alpha - \beta (x_{q} )]v - p}} + (x_{q} - 1)\mu _{1} + \lambda - \left( {\lambda _{{q2}}^{0} + \frac{{B - c_{2} \mu _{2} }}{{p_{q} }}} \right)\left( { > 0} \right) \\ & = \sqrt {\left( {\frac{{\mu _{1} + \mu _{2} - \lambda }}{2}} \right)^{2} + \left\{ {\frac{d}{{[\alpha - \beta (x_{q} )]v - p}}} \right\}^{2} } \Leftrightarrow \\ \end{aligned} $$
  $$ \begin{aligned} & \left\{ {\left[ {(x_{q} - 1)\mu _{1} + \lambda } \right] - \left( {\lambda _{2}^{0} + \frac{{B - c_{2} \mu _{2} }}{{p_{q} }}} \right)} \right\}^{2} + \left\{ {\left[ {(x_{q} - 1)\mu _{1} + \lambda } \right] - \left( {\lambda _{2}^{0} + \frac{{B - c_{2} \mu _{2} }}{{p_{q} }}} \right)} \right\} \\ & \quad \cdot \left\{ { - \frac{{p_{q} }}{{c_{2} }}\left( {\lambda _{2}^{0} + \frac{{B - c_{2} \mu _{2} }}{{p_{q} }}} \right) + \left\{ {\frac{{p_{q} }}{{c_{2} }}\left( {\lambda _{2}^{0} + \frac{B}{{p_{q} }}} \right) + \mu _{1} - \lambda + \frac{{{\text{2}}d}}{{[\alpha - \beta (x_{q} )]v - p}}} \right\}} \right\} \\ & \quad + \frac{{d\left\{ { - \frac{{p_{q} }}{{c_{2} }}\left( {\lambda _{2}^{0} + \frac{{B - c_{2} \mu _{2} }}{{p_{q} }}} \right) + \left[ {\frac{{p_{q} }}{{c_{2} }}\left( {\lambda _{2}^{0} + \frac{B}{{p_{q} }}} \right) + \mu _{1} - \lambda } \right]} \right\}}}{{[\alpha - \beta (x_{q} )]v - p}} = 0 \Leftrightarrow \\ \end{aligned} $$
  $$\begin{aligned} & a\left( {\lambda _{2}^{0} + \frac{{B - c_{2} \mu _{2} }}{{p_{q} }}} \right)^{2} - b\left( {\lambda _{2}^{0} + \frac{{B - c_{2} \mu _{2} }}{{p_{q} }}} \right) + c = 0\; \Leftrightarrow \\ & \mu _{{q2}}^{ * } (x_{q} ) = \frac{{p_{q} }}{{c_{2} }} \cdot \left( {\lambda _{2}^{0} + \frac{B}{{p_{q} }} - \frac{{b - \sqrt {b^{2} - 4ac} }}{{2a}}} \right){\text{ and }}\;x_{q} \in X_{q}^{2} , \\ \end{aligned} $$
  $$\begin{aligned} & \text{where} \; \\ & \quad a = 1 + \frac{{p_{q} }}{{c_{2} }};b = \left\{ {\left( {2 + \frac{{p_{q} }}{{c_{2} }}} \right)\left\{ {\left[ {(x_{q} - 1)\mu _{1} + \lambda } \right] + \frac{d}{{\left[ {(1 - x_{q} )(\alpha - 1)} \right]v - p}}} \right\} + \frac{{p_{q} }}{{c_{2} }}\left( {\lambda _{2}^{0} + \frac{B}{{p_{q} }}} \right) + \mu _{1} - \lambda } \right\}; \\ & \quad c = \left[ {(x_{q} - 1)\mu _{1} + \lambda } \right]\left\{ {\frac{{p_{q} }}{{c_{2} }}\left( {\lambda _{2}^{0} + \frac{B}{{p_{q} }}} \right) + x_{q} \mu _{1} + \frac{{{\text{2}}d}}{{\left[ {(1 - x_{q} )(\alpha - 1)} \right]v - p}}} \right\} + \frac{{d\left[ {\frac{{p_{q} }}{{c_{2} }}\left( {\lambda _{2}^{0} + \frac{B}{{p_{q} }}} \right) + \mu _{1} - \lambda } \right]}}{{\left[ {(1 - x_{q} )(\alpha - 1)} \right]v - p}}; \\ & \quad X_{q}^{2} = \left\{ {\left. {x_{q} } \right|\frac{{p_{q} }}{{c_{2} }}\left( {\lambda _{2}^{0} + \frac{B}{{p_{q} }} - \frac{{b - \sqrt {b^{2} - 4ac} }}{{2a}}} \right) > \max \left\{ {\lambda - \mu _{1} + 2h_{1} ,\;\frac{{\lambda _{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} }}} \right. \kern-\nulldelimiterspace} {p_{q} }}}}{{1 + {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-\nulldelimiterspace} {p_{q} }}}}} \right\}{\text{and}}\;x_{q} < \tilde{x}_{q}^{0} } \right\}. \\ \end{aligned} $$

Consequently, this completes the proof. □

1.7 Proof of Proposition 5

When min {Λ_q(x_q), λ} = λ, i.e., Λ_q(x_q) ≥ λ, we have Π_q1(x_q) = p_qλ ⁰₁  − (p_q − p)λ_q1(x_q) − c₁μ₁. Since Hospital 1’s expected profit under no performance payment scheme is $\Pi_{ 1}^{ 0} = p\lambda_{1}^{0} - c_{1} \mu_{1} ,$ we further have

$$ \Pi_{q1} (x_{q} ) > \Pi_{ 1}^{ 0} \Leftrightarrow (p_{q} - p)[\lambda_{1}^{0} - \lambda_{q1} (x_{q} )] > 0 \Leftrightarrow p_{q} > p; $$

$\Pi_{q1} (x_{q} ) = \Pi_{ 1}^{ 0} \Leftrightarrow p_{q} = p;$ $\Pi_{q1} (x_{q} ) < \Pi_{ 1}^{ 0} \Leftrightarrow p_{q} < p.$When min {Λ_q(x_q), λ} = Λ_q(x_q), we have $\Pi_{q1} (x_{q} ) = p\Lambda_{q} (x_{q} ) - c_{1} \mu_{1} + (p_{q} - p)\lambda_{q 2} (x_{q} ) - p_{s} \lambda_{2}^{0}.$ Taking the first-order derivative of Π_q1(x_q) with respect to x_q, we get

$$ \begin{aligned} {{\partial \Pi_{q1} (x_{q} )} \mathord{\left/ {\vphantom {{\partial \Pi_{q1} (x_{q} )} {\partial x_{q} }}} \right. \kern-0pt} {\partial x_{q} }} & = p{{\partial \lambda_{q 1} (x_{q} )} \mathord{\left/ {\vphantom {{\partial \lambda_{q 1} (x_{q} )} {\partial x_{q} }}} \right. \kern-0pt} {\partial x_{q} }} + p_{q} {{\partial \lambda_{q 2} (x_{q} )} \mathord{\left/ {\vphantom {{\partial \lambda_{q 2} (x_{q} )} {\partial x_{q} }}} \right. \kern-0pt} {\partial x_{q} }} \\ & \le p{{\partial \mu_{q 2}^{ * } (x_{q} )} \mathord{\left/ {\vphantom {{\partial \mu_{q 2}^{ * } (x_{q} )} {\partial x_{q} }}} \right. \kern-0pt} {\partial x_{q} }} - p_{q} \cdot ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }}) \cdot {{\partial \mu_{q 2}^{ * } (x_{q} )} \mathord{\left/ {\vphantom {{\partial \mu_{q 2}^{ * } (x_{q} )} {\partial x_{q} }}} \right. \kern-0pt} {\partial x_{q} }} \le 0. \\ \end{aligned} $$

When $\begin{aligned} \lambda_{q 2} (x_{q} ) & = \lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{s} }}} \right. \kern-0pt} {p_{s} }} - ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }})\mu_{q 2}^{ * } (x_{q} ) \\ & = \mu_{q 2}^{ * } (x_{q} ) + x_{q} \mu_{1} - \frac{d}{{[1 + x_{q} (\alpha - 1)] \cdot v}} \\ \end{aligned}$, we have $\mu_{q 2}^{ * } (x_{q} ) = \frac{{\lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{s} }}} \right. \kern-0pt} {p_{s} }} - x_{q} \mu_{1} }}{{[1 + ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }})]}} + \frac{{{d \mathord{\left/ {\vphantom {d v}} \right. \kern-0pt} v}}}{{[1 + ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }})][1 + x_{q} (\alpha - 1)]}}$, when $\Lambda_{q} (x_{q} ) - \lambda = \mu_{1} + \mu_{q 2}^{ * } (x_{q} ) - \frac{d}{\alpha v - p} - \frac{d}{{\beta (x_{q} ) \cdot v}} - \lambda \ge 0$, we have

$$ \lambda \le \mu_{1} - \frac{d}{\alpha v - p} + \frac{{\lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} }}} \right. \kern-0pt} {p_{q} }}}}{{ 1+ {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }}}} - \frac{{x_{q} \mu_{1} }}{{ 1+ {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }}}} - \frac{{{{(c_{2} } \mathord{\left/ {\vphantom {{(c_{2} } {p_{q} )}}} \right. \kern-0pt} {p_{q} )}} \cdot {{(d} \mathord{\left/ {\vphantom {{(d} {v)}}} \right. \kern-0pt} {v)}}}}{{( 1+ {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }})[1 + x_{q} (\alpha - 1)]}} \Leftrightarrow $$

$$ {{(c_{2} } \mathord{\left/ {\vphantom {{(c_{2} } {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}{{(d} \mathord{\left/ {\vphantom {{(d} {v)}}} \right. \kern-0pt} {v)}} + x_{q} \mu_{1} [1 + x_{q} (\alpha - 1)] \le ( 1+ {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }})\left[ {\left( {\mu_{1} - \frac{d}{\alpha v - p} + \frac{{\lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} }}} \right. \kern-0pt} {p_{q} }}}}{{ 1+ {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }}}}} \right) - \lambda } \right][1 + x_{q} (\alpha - 1)] $$

$\Leftrightarrow \mu_{1} (\alpha - 1)x_{q}^{2} + (\mu_{1} - b_{q} )x_{q} - {{b_{q} } \mathord{\left/ {\vphantom {{b_{q} } {(\alpha - 1)}}} \right. \kern-0pt} {(\alpha - 1)}} + {{(c_{2} } \mathord{\left/ {\vphantom {{(c_{2} } {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}{{(d} \mathord{\left/ {\vphantom {{(d} {v)}}} \right. \kern-0pt} {v)}} \le 0 \Leftrightarrow 0 < x_{q}^{{}} \le \bar{x}_{q}^{1} ,$ where

$$ b_{q} = (\alpha - 1)\left\{ {\frac{1}{{p_{q} }}\left[ {c_{2} \left( {\mu_{1} - \lambda - \frac{d}{\alpha v - p}} \right) + B} \right] + \left( {\mu_{1} - \lambda - \frac{d}{\alpha v - p} + \lambda_{2}^{0} } \right)} \right\}; $$

$$ \bar{x}_{q}^{1} = \frac{{\sqrt {(\mu_{1} + b_{q} )^{2} - {{4(\mu_{1} } \mathord{\left/ {\vphantom {{4(\mu_{1} } {v)}}} \right. \kern-0pt} {v)}}(\alpha - 1)(c_{2} /p_{q} )} - (\mu_{1} - b_{q} )}}{{2(\alpha - 1)\mu_{1} }}. $$

$\bar{x}_{q}^{1} < \bar{x}^{0} \Leftrightarrow p_{q} > \hat{p}_{q}^{0}$, where

$$ \begin{aligned} \hat{p}_{q}^{0} & = c_{2} \cdot \frac{{\left\{ {{B \mathord{\left/ {\vphantom {B {c_{2} }}} \right. \kern-0pt} {c_{2} }} + \mu_{1} - \lambda - {d \mathord{\left/ {\vphantom {d {(\alpha v - p)}}} \right. \kern-0pt} {(\alpha v - p)}} - {d \mathord{\left/ {\vphantom {d {\{ v[1 + (\alpha - 1)\bar{x}^{0} ]\} }}} \right. \kern-0pt} {\{ v[1 + (\alpha - 1)\bar{x}^{0} ]\} }}} \right\}}}{{\lambda - \{ [(1 - \bar{x}^{0} )\mu_{1} - {d \mathord{\left/ {\vphantom {d {(\alpha v - p)}}} \right. \kern-0pt} {(\alpha v - p)}}] + \lambda_{2}^{0} \} }}, \\ \lambda_{2}^{0} & = \frac{{{B \mathord{\left/ {\vphantom {B {c_{2} }}} \right. \kern-0pt} {c_{2} }} - \mu_{1} + \lambda }}{2} + \frac{d}{(\alpha - 1)v - p} - \sqrt {\left\{ {\frac{{{B \mathord{\left/ {\vphantom {B {c_{2} }}} \right. \kern-0pt} {c_{2} }} + \mu_{1} - \lambda }}{2}} \right\}^{2} + \frac{{d^{2} }}{{[(\alpha - 1)v - p]^{2} }}}. \\ \end{aligned} $$

Thus, when p_q ∊ [0, p), we have x_q* = 0; when $p_{f} \in [p,\;\hat{p}_{q}^{0} ]$, we have $x_{q}^{ * } = \bar{x}^{0}$; when $p_{q} \in [\hat{p}_{q}^{0} \wedge p,\;\infty )$, we have $x_{q}^{ * } = \bar{x}_{q}^{1}$. This concludes the proof. □

1.8 Proof of Proposition 6

By Eqs. (16) and (23), taking the first-order derivative of $\bar{x}_{f}^{1}$ and $\bar{x}_{q}^{1}$ with respect to p_f and p_q, respectively, we have

$$ \frac{{\partial \bar{x}_{f}^{1} }}{{\partial p_{f} }} = \frac{{\left\{ \begin{aligned} \frac{{\left\{ \begin{aligned} \mu_{1} + (\alpha - 1)\left( {{{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{f} }}} \right. \kern-0pt} {p_{f} }}} \right)\left[ {\mu_{1} + B/c_{2} - \lambda - d/\left( {\alpha v - p} \right)} \right] \hfill \\ \cdot \left( {\alpha - 1} \right)\left[ {\mu_{1} + B/c_{2} - \lambda - d/\left( {\alpha v - p} \right)} \right] - 4\mu_{1} \left( {d/v} \right)\left( {\alpha - 1} \right) \hfill \\ \end{aligned} \right\}}}{{\sqrt {\left\{ {\mu_{1} + (\alpha - 1)\left[ {( 1+ \frac{{c_{2} }}{{p_{q} }})\left( {\mu_{1} - \lambda - \frac{d}{\alpha v - p}} \right) + \lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} }}} \right. \kern-0pt} {p_{q} }}} \right]} \right\}^{2} - \frac{{4\mu_{1} (\alpha - 1)}}{v} \cdot \frac{{c_{2} }}{{p_{q} }}} }} \hfill \\ + (\alpha - 1)\left[ {\left( {\mu_{1} - \lambda - \frac{d}{\alpha v - p}} \right)\frac{{\partial ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}}}{{\partial p_{q} }} + \frac{{\partial ({B \mathord{\left/ {\vphantom {B {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}}}{{\partial p_{q} }}} \right] \hfill \\ \end{aligned} \right\}}}{{2(\alpha - 1)\mu_{1} }} \le 0. $$

$$ \frac{{\partial \bar{x}_{q}^{1} }}{{\partial p_{q} }} = \frac{{\left\{ \begin{aligned} \frac{{\left\{ \begin{aligned} \left\{ {\mu_{1} + (\alpha - 1)\left[ {( 1+ {{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} }}} \right. \kern-0pt} {p_{q} }})\left( {\mu_{1} - \lambda - \frac{d}{\alpha v - p}} \right) + \left( {\lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} }}} \right. \kern-0pt} {p_{q} }}} \right)} \right]} \right\}(\alpha - 1) \hfill \\ \cdot \left[ {\left( {\mu_{1} - \lambda - \frac{d}{\alpha v - p}} \right)\frac{{\partial ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}}}{{\partial p_{q} }} + \frac{{\partial ({B \mathord{\left/ {\vphantom {B {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}}}{{\partial p_{q} }}} \right] - \frac{{4\mu_{1} }}{v}(\alpha - 1)\frac{{\partial ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}}}{{\partial p_{q} }} \hfill \\ \end{aligned} \right\}}}{{\sqrt {\left\{ {\mu_{1} + (\alpha - 1)\left[ {( 1+ \frac{{c_{2} }}{{p_{q} }})\left( {\mu_{1} - \lambda - \frac{d}{\alpha v - p}} \right) + \lambda_{2}^{0} + {B \mathord{\left/ {\vphantom {B {p_{q} }}} \right. \kern-0pt} {p_{q} }}} \right]} \right\}^{2} - \frac{{4\mu_{1} (\alpha - 1)}}{v} \cdot \frac{{c_{2} }}{{p_{q} }}} }} \hfill \\ + (\alpha - 1)\left[ {\left( {\mu_{1} - \lambda - \frac{d}{\alpha v - p}} \right)\frac{{\partial ({{c_{2} } \mathord{\left/ {\vphantom {{c_{2} } {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}}}{{\partial p_{q} }} + \frac{{\partial ({B \mathord{\left/ {\vphantom {B {p_{q} )}}} \right. \kern-0pt} {p_{q} )}}}}{{\partial p_{q} }}} \right] \hfill \\ \end{aligned} \right\}}}{{2(\alpha - 1)\mu_{1} }} \le 0. $$

When $x_{k}^{ * } (p_{k} ) = \bar{x}_{k}^{1} > \hbox{max} \left\{ {0,\;\sqrt {\frac{d}{v} \cdot \frac{{c_{2} }}{p} \cdot \frac{ 1}{{\mu_{1} (\alpha - 1)}}} - \frac{1}{(\alpha - 1)}} \right\}$, we have $\lambda_{k1}^{ * } (p_{k} ) = [1 - x_{k}^{ * } (p_{k} )]\mu_{1} - {d \mathord{\left/ {\vphantom {d {(\alpha v - p)}}} \right. \kern-0pt} {(\alpha v - p)}},$ λ_k2*(p_k) = λ − λ_k1*(p_k), 

$$ \mu_{k2}^{ * } (p_{k} ) = \lambda - \mu_{1} + \frac{d}{\alpha v - p} + \frac{d}{{[1 + (\alpha - 1) \cdot x_{k}^{ * } (p_{k} )]v}}, $$

$$ \begin{aligned} \Pi_{k1}^{ * } (p_{k} ) & = p\left\{ {[1 - x_{k}^{ * } (p_{k} )]\mu_{1} - \frac{d}{\alpha v - p}} \right\} - c_{1} \mu_{1} \\ & \quad + B - c_{2} \left\{ {\lambda - \mu_{1} + \frac{d}{\alpha v - p} + \frac{d}{{[1 + (\alpha - 1)x_{k}^{ * } (p_{f} )]v}}} \right\}, \\ \end{aligned} $$

where k = f, q. Hence, we get $\frac{{\partial \lambda_{k1}^{ * } (p_{k} )}}{{\partial p_{k} }} \le 0,\;\frac{{\partial \mu_{k2}^{ * } (p_{k} )}}{{\partial p_{k} }} \ge 0,\;\frac{{\partial \Pi_{k1}^{ * } (p_{k} )}}{{\partial p_{k} }} \ge 0$.

Consequently, this completes the proof. □

1.9 Proof of Proposition 7

First, we solve and obtain the optimal capacity sinking rate for the healthcare funder x^o. By Eq. (28), we set $I(x) = [p - ({1 \mathord{\left/ {\vphantom {1 \theta }} \right. \kern-0pt} \theta } - 1)p_{s} ](1 - x)\mu_{1} - \frac{{dc_{2} }}{[1 + (\alpha - 1)x]v}$ and

$$ S_{f} (x) = \theta \left\{ {I(x) - \left\{ {\frac{{d[p + c_{2} - ({1 \mathord{\left/ {\vphantom {1 \theta }} \right. \kern-0pt} \theta } - 1)p_{s} ]}}{\alpha v - p} + c_{1} \mu_{1} - B - c_{2} (\mu_{1} - \lambda ) - ({1 \mathord{\left/ {\vphantom {1 \theta }} \right. \kern-0pt} \theta } - 1)p_{s} \lambda } \right\}} \right\},$$

with $0 < x < \bar{x}^{0}$, and thus, we get max_x{I(x)} = max_x{S_f(x)}.

Taking the first-and second-order derivatives of I(x). with respect to x, we have

$$ \begin{aligned} \frac{\partial I(x)}{\partial x} & = \theta \left\{ {\frac{{{{(d} \mathord{\left/ {\vphantom {{(d} v}} \right. \kern-0pt} v})c_{2} (\alpha - 1)}}{{[1 + (\alpha - 1)x]^{2} }} - [p - ({1 \mathord{\left/ {\vphantom {1 \theta }} \right. \kern-0pt} \theta } - 1)p_{s} ]\mu_{1} } \right\}, \\ \frac{{\partial^{2} I(x)}}{{\partial x^{2} }} & = - \frac{{{{2(d} \mathord{\left/ {\vphantom {{2(d} v}} \right. \kern-0pt} v})\theta c_{2} (\alpha - 1)^{2} }}{{[1 + (\alpha - 1)x]^{3} }} \le 0. \\ \end{aligned} $$

Based on

$$ \begin{aligned} \mathop {\hbox{max} }\limits_{{0 \le x \le \bar{x}^{0} }} \left\{ \frac{\partial I(x)}{\partial x} \right\}& = {{(d} \mathord{\left/ {\vphantom {{(d} v}} \right. \kern-0pt} v})c_{2} (\alpha - 1) - \left( {p + p_{s} - \frac{{p_{s} }}{\theta }} \right) \\ \mu_{1} & = 0 \Leftrightarrow {\hat{\text{U}}}\,\theta = \tilde{\theta }^{1} = \frac{{p_{s} \mu_{1} }}{{(p + p_{s} )\mu_{1} - {{(d} \mathord{\left/ {\vphantom {{(d} v}} \right. \kern-0pt} v})c_{2} (\alpha - 1)}}; \\ \end{aligned} $$

$$ \begin{aligned} \mathop {\hbox{min} }\limits_{{0 \le x \le \bar{x}^{0} }} \left\{ \frac{\partial I(x)}{\partial x}\right\} & = \frac{{{{(d} \mathord{\left/ {\vphantom {{(d} v}} \right. \kern-0pt} v})c_{2} (\alpha - 1)}}{{[1 + (\alpha - 1)\bar{x}^{0} ]^{2} }} - \left( {p + p_{s} - \frac{{p_{s} }}{\theta }} \right)\mu_{1} = 0 \\ \Leftrightarrow \theta = \tilde{\theta }^{2} & = \frac{{p_{s} \mu_{1} }}{{(p + p_{s} )\mu_{1} - {{(d} \mathord{\left/ {\vphantom {{(d} v}} \right. \kern-0pt} v})c_{2} (\alpha - 1)[1 + (\alpha - 1)\bar{x}^{0} ]^{ - 2} }}; \\ \end{aligned} $$

thus, when $1 > \theta \ge \tilde{\theta }^{1}$, we have $\frac{\partial I(x)}{\partial x} \le 0 \Rightarrow x^{o} = 0$.

When $0 < \theta \le \tilde{\theta }^{2}$, we have $\frac{\partial I(x)}{\partial x} \ge 0 \Rightarrow x^{o} = \bar{x}^{0}.$

When $\tilde{\theta }^{1} < \theta < \tilde{\theta }^{1}$, if $x \le \tilde{x}^{o}$, then $\frac{\partial I(x)}{\partial x} \ge 0$; if $x \ge \tilde{x}^{o}$, then $\frac{\partial I(x)}{\partial x} \le 0$; hence we get $x^{o} = \tilde{x}^{o}$, where $x^{o} = \tilde{x}^{o} \triangleq \sqrt {\frac{{{{(d} \mathord{\left/ {\vphantom {{(d} v}} \right. \kern-0pt} v})c_{2} }}{{(\alpha - 1)[p - ({1 \mathord{\left/ {\vphantom {1 \theta }} \right. \kern-0pt} \theta } - 1)p_{s} ]\mu_{1} }}} - \frac{1}{\alpha - 1}$.

By solving the equation in which $\tilde{x}^{o} = \bar{x}_{f}^{1}$ with respect to p_f, we have $p_{f} = \tilde{p}_{f}^{ * }$, where

$$ \tilde{p}_{f}^{ * } = \frac{{c_{2} }}{{\mu_{1} \tilde{x}^{o} }}\left\{ {(\mu_{1} - \lambda ) + \frac{B}{{c_{2} }} - \frac{d}{\alpha v - p} - \frac{d}{{v[(\alpha - 1)\tilde{x}^{o} + 1]}}} \right\}. $$

According to Proposition 3, considering the impacts of B, we show:

• (1) If $B \ge \tilde{B}^{\text{o}}$, we have $\tilde{\Pi }_{f1}^{o} \ge \Pi_{1}^{ 0}$, where

  $$ \begin{aligned} \tilde{\Pi }_{f1}^{o} & = p(1 - \tilde{x}^{o} )\mu_{1} - \frac{{dc_{2} }}{{[1 + (\alpha - 1)\tilde{x}^{o} ]v}} - \frac{{d(p + c_{ 2} )}}{\alpha v - p} - c_{1} \mu_{1} + B + c_{2} (\mu_{1} - \lambda ); \\ X_{o} & = \frac{{c_{2} }}{p}\left\{ {\mu_{1} - \lambda - \frac{d}{\alpha v - p} - \frac{d}{{v[(\alpha - 1)\tilde{x}^{o} + 1]}}} \right\} - \mu_{1} \tilde{x}^{o} - \frac{d}{\alpha v - p} + \frac{d}{(\alpha - 1)v - p}; \\ \tilde{B}^{o} & = \frac{{\sqrt {\left[ {\frac{1}{{c_{2} }}X_{o} - \frac{1}{p}\left( {\mu_{1} - \lambda } \right)} \right]^{2} + \frac{1}{p}\left( {\frac{1}{{c_{2} }} + \frac{1}{p}} \right)\frac{4d}{{[(\alpha - 1)v - p]^{2} }}} - \left[ {X_{o} \left( {\frac{1}{{c_{2} }} + \frac{2}{p}} \right) + \frac{1}{p}\left( {\mu_{1} - \lambda } \right)} \right]}}{{\frac{2}{p}\left( {\frac{1}{{c_{2} }} + \frac{1}{p}} \right)}}. \\ \end{aligned} $$

• (2) If $\tilde{B}^{ 0} \le B \le \tilde{B}^{o}$, from the results of Proposition 3, we have Π_f1*(p_f) ≥ Π ⁰₁ only when $x_{f}^{ * } (p_{f} ) \le \hat{x}^{1} \le \tilde{x}^{o}$. By solving the equation in which $x_{f}^{ * } (p_{f} ) \le \hat{x}^{ 1}$ and then get $p_{f}^{ * } \ge \hat{p}_{f}^{1}.$

In sum, when $\tilde{\theta }^{ 2} < \theta < \tilde{\theta }^{ 1}$, we show that the equilibrium capacity sinking price satisfies as follows:

• (a) if $\tilde{B}^{0} \le B < \tilde{B}^{o}$, we have $p_{f}^{ * } = \hat{p}_{f}^{1}$;

• (b) if $B \ge \tilde{B}_{f}^{o}$, we have $p_{f}^{ * } = \tilde{p}_{f}^{ * }$.

When $0 < \theta \le \tilde{\theta }^{2}$, from Eq. (28) and Proposition 5, we have:

• if $\tilde{B}^{0} \le B < \tilde{B}^{1}$, then $p_{f}^{ * } = \hat{p}_{f}^{1}$; if $B \ge \tilde{B}_{f}^{1}$, then $p_{f}^{ * } = \hat{p}_{f}^{0}$.

When $1 > \theta \ge \tilde{\theta }^{1}$, we have $p_{f}^{ * } \in [0,\;\hat{p}_{f}^{1} )$:

• if $\tilde{B}^{0} \le B < \tilde{B}^{1}$, then $p_{f}^{ * } \in [0,\;\hat{p}_{f}^{1} )$; if $B \ge \tilde{B}_{f}^{1}$, then $p_{f}^{ * } \in [0,\;\hat{p}_{f}^{2} )$.

Consequently, this completes the proof. □

1.10 Proof of Proposition 8

In view of Proposition 7, we first have: when $\tilde{\theta }^{ 2} < \theta < \tilde{\theta }^{ 1}$, the optimal capacity sinking rate for the funder satisfies $x^{o} = \tilde{x}^{o}$. In this case, by solving the equation in which $\tilde{x}^{o} = \bar{x}_{q}^{1} (p_{q} )$ with respect to p_q, we have $p_{q} = \tilde{p}_{q}^{ * }$, where

$$ \tilde{p}_{q}^{ * } = \frac{{B + c_{2} \left\{ {(\mu_{1} - \lambda ) - {d \mathord{\left/ {\vphantom {d {(\alpha v - p)}}} \right. \kern-0pt} {(\alpha v - p)}} - {d \mathord{\left/ {\vphantom {d {\{ v[1 + (\alpha - 1)\tilde{x}^{o} ]\} }}} \right. \kern-0pt} {\{ v[1 + (\alpha - 1)\tilde{x}^{o} ]\} }}} \right\}}}{{\mu_{1} \tilde{x}^{o} - [(\mu_{1} - \lambda ) - {d \mathord{\left/ {\vphantom {d {(\alpha v - p)}}} \right. \kern-0pt} {(\alpha v - p)}} + \lambda_{2}^{0} ]}}. $$

According to Proposition 5, considering the impacts of B, we show:

• if $p \le \tilde{p}_{q}^{ * }$, we have $B \ge \tilde{B}^{o}.$

Hence, when $\tilde{\theta }^{ 2} < \theta < \tilde{\theta }^{ 1}$, we show that the equilibrium capacity sinking price satisfies as follows:

• (1) if $\tilde{B}^{0} \le B < \tilde{B}^{o}$, we have p_q* = p;

• (2) if $\tilde{B}^{o} \le B$, we have $p_{q}^{ * } = \tilde{p}_{q}^{ * }$.

When $0 < \theta \le \tilde{\theta }^{2}$,

• (1) if $\tilde{B}^{0} \le B < \tilde{B}^{1}$, we have p_q* = p;

• (2) if $\tilde{B}^{1} \le B$, we have $p_{q}^{ * } = \hat{p}_{q}^{ 0}$.

When $1 > \theta \ge \tilde{\theta }^{1}$, we have $p_{q}^{ * } \in [ 0,\;p)$.

This completes the proof. □