Using price–volume agreements to manage pharmaceutical leakage and off-label promotion

Abstract

Unapproved or “off-label” uses of prescription drugs are quite common. The extent of this use may be influenced by the promotional efforts of manufacturers. This paper investigates how a manufacturer makes promotional decisions in the presence of a price–volume agreement. We developed an optimization model in which the manufacturer maximizes its expected profit by choosing the level of marketing effort to promote uses for different indications. We considered several ways a volume threshold is determined. We also compared models in which off-label uses are reimbursed and those in which they are forbidden to illustrate the impact of off-label promotion on the optimal decisions and on the decision maker’s performance. We found that the payer chooses a threshold which may be the same as the manufacturer’s optimal decision. We also found that the manufacturer not only considers the promotional cost in promoting off-label uses but also considers the health benefit of off-label uses. In some situations, using a price–volume agreement to control leakage may be a better idea than simply preventing leakage without using the agreement, from a social welfare perspective.

Appendix

Proof for proposition 2

By taking the derivative of E[NMB(T)] = 0 for both sides with respect to h _i , we have $\frac{{\partial E[{\text{NMB}}]}}{\partial T}\frac{{{\text{d}}T^{M,N} }}{{{\text{d}}h_{i} }} + \frac{\partial E[NMB]}{{\partial h_{i} }} = 0.$ That is, $\frac{{{\text{d}}T^{M,N} }}{{{\text{d}}h_{i} }} = - {{\frac{\partial E[NMB]}{{\partial h_{i} }}} \mathord{\left/ {\vphantom {{\frac{\partial E[NMB]}{{\partial h_{i} }}} {\frac{\partial E[NMB]}{\partial T}}}} \right. \kern-0pt} {\frac{\partial E[NMB]}{\partial T}}} > 0$ because $\frac{{\partial E[{\text{NMB}}]}}{\partial T} < 0$ at $T^{M,N}$. By following the same logic, we obtain the properties of $T^{M,N}$ with respect to other parameters. $\frac{{{\text{d}}m^{M,N} }}{{{\text{d}}h_{i} }} = \frac{{\partial m^{E,*} }}{{\partial T^{M,N} }}\frac{{\partial T^{M,N} }}{{\partial h_{i} }} + \frac{{\partial m^{E,*} }}{{\partial h_{i} }} \ge 0$, and similarly we have $\frac{{\partial m^{M,N} }}{\partial c} \le 0$.

Proof to problem (P)

Given $m_{i}^{E,*}$ (which is a function of T), the payer’s objective becomes a function of T:

$$D = Q_{1} + \sum\limits_{i = 1}^{3} {l_{i1} } m_{i}^{E,*} - \kappa \sum\limits_{j = 2}^{3} {\left( {Q_{j} + \sum\limits_{i = 1}^{3} {l_{ij} } m_{i}^{E,*} } \right)} .$$

(9)

The first-order and second-order derivatives of D with respect to T are

$$\frac{{{\text{d}}D}}{{{\text{d}}T}} = (1 - \kappa l_{12} - \kappa l_{13} )\frac{{{\text{d}}m_{1}^{E,*} }}{{{\text{d}}T}} + (l_{21} - \kappa - \kappa l_{23} )\frac{{{\text{d}}m_{2}^{E,*} }}{{{\text{d}}T}} + (l_{31} - \kappa l_{32} - \kappa )\frac{{{\text{d}}m_{3}^{E,*} }}{{{\text{d}}T}},$$

(10)

and

$$\frac{{{\text{d}}^{2} D}}{{{\text{d}}T^{2} }} = (1 - \kappa l_{12} - \kappa l_{13} )\frac{{{\text{d}}^{2} m_{1}^{E,*} }}{{{\text{d}}T^{2} }} + (l_{21} - \kappa - \kappa l_{23} )\frac{{{\text{d}}^{2} m_{2}^{E,*} }}{{{\text{d}}T^{2} }} + (l_{31} - \kappa l_{32} - \kappa )\frac{{{\text{d}}^{2} m_{3}^{E,*} }}{{{\text{d}}T^{2} }}.$$

Note that $\frac{{{\text{d}}m_{1}^{E,u} }}{{{\text{d}}T}} = \frac{{\varphi_{2}^{''} \varphi_{3}^{''} (1 + l_{12} + l_{13} )}}{{\left( {\varphi_{2}^{''} \varphi_{3}^{''} \sum\nolimits_{j = 1}^{3} {l_{1j} + \varphi_{1}^{''} \varphi_{3}^{''} } \sum\nolimits_{j = 1}^{3} {l_{2j} + \varphi_{1}^{''} \varphi_{2}^{''} } \sum\nolimits_{j = 1}^{3} {l_{3j} } } \right) + {{\varphi_{1}^{''} \varphi_{2}^{''} \varphi_{3}^{''} } \mathord{\left/ {\vphantom {{\varphi_{1}^{''} \varphi_{2}^{''} \varphi_{3}^{''} } {\alpha pf}}} \right. \kern-0pt} {\alpha pf}}}}.$ Similar formulas are obtained for ${\text{d}}m_{2}^{E,u} /{\text{d}}T$ and ${\text{d}}m_{3}^{E,u} /{\text{d}}T,$ which are substituted into (10). Thus, when $T \ge \overline{T} = \hbox{max} (T^{1} ,T^{2} ,T^{3} ),$ we have

$$\frac{{{\text{d}}D}}{{{\text{d}}T}} = \frac{{(1 - \kappa l_{12} - \kappa l_{13} )\sum\nolimits_{j = 1}^{3} {l_{1j} } \varphi_{2}^{''} \varphi_{3}^{''} + (l_{21} - \kappa - \kappa l_{23} )\sum\nolimits_{j = 1}^{3} {l_{2j} } \varphi_{1}^{''} \varphi_{3}^{''} + (l_{31} - \kappa l_{32} - \kappa )\sum\nolimits_{j = 1}^{3} {l_{3j} } \varphi_{1}^{''} \varphi_{2}^{''} }}{{\left( {\varphi_{2}^{''} \varphi_{3}^{''} \sum\nolimits_{j = 1}^{3} {l_{1j} } + \varphi_{1}^{''} \varphi_{3}^{''} \sum\nolimits_{j = 1}^{3} {l_{2j} + \varphi_{1}^{''} \varphi_{2}^{''} } \sum\nolimits_{j = 1}^{3} {l_{3j} } } \right) + {{\varphi_{1}^{''} \varphi_{2}^{''} \varphi_{3}^{''} } \mathord{\left/ {\vphantom {{\varphi_{1}^{''} \varphi_{2}^{''} \varphi_{3}^{''} } {\alpha pf}}} \right. \kern-0pt} {\alpha pf}}}}.$$

(11)

The denominator of (11) is always positive because $\varphi_{i}^{''} > 0.$ Thus, the sign of dD/dT depends on the sign of the numerator. Let $\varGamma$ represent the numerator of (11), i.e., $\varGamma = (1 - \kappa l_{12} - \kappa l_{13} )\sum\nolimits_{j = 1}^{3} {l_{1j} } \varphi_{2}^{''} \varphi_{3}^{''} + (l_{21} - \kappa - \kappa l_{23} )\sum\nolimits_{j = 1}^{3} {l_{2j} } \varphi_{1}^{''} \varphi_{3}^{''} + (l_{31} - \kappa l_{32} - \kappa )\sum\nolimits_{j = 1}^{3} {l_{3j} } \varphi_{1}^{''} \varphi_{2}^{''} .$ $\varGamma$ is a constant over T and is decreasing in $\kappa$. Let $\kappa_{0}$ be the value of $\kappa$ where $\varGamma = 0.$ Thus,

$$\kappa_{0} = \frac{{\varphi_{2}^{''} \varphi_{3}^{''} \sum\nolimits_{j = 1}^{3} {l_{1j} } + \varphi_{1}^{''} \varphi_{3}^{''} l_{21} \sum\nolimits_{j = 1}^{3} {l_{2j} } + \varphi_{1}^{''} \varphi_{2}^{''} l_{31} \sum\nolimits_{j = 1}^{3} {l_{3j} } }}{{(l_{12} + l_{13} )\varphi_{2}^{''} \varphi_{3}^{''} \sum\nolimits_{j = 1}^{3} {l_{1j} } + (1 + l_{23} )\varphi_{1}^{''} \varphi_{3}^{''} \sum\nolimits_{j = 1}^{3} {l_{2j} } + (1 + l_{32} )\varphi_{1}^{''} \varphi_{2}^{''} \sum\nolimits_{j = 1}^{3} {l_{3j} } }}.$$

dD/dT > 0 if $\kappa < \kappa_{0} ,$ dD/dT < 0 if $\kappa > \kappa_{0}$ and dD/dT = 0 if $\kappa = \kappa_{0} .$ That is, the payer would be indifferent about T if $\kappa = \kappa_{0} .$ Without the NMB constraint, the payer will set $T = \infty$ when $\kappa < \kappa_{0}$ and T = 0 otherwise. Since $1 + l_{23} \ge l_{21}$ and $1 + l_{32} \ge l_{31} ,$ the second and third components of the denominator of $\kappa_{0}$ are always larger than those of the numerator. Thus, $\kappa_{0} \le 1$ if $l_{12} + l_{13} \ge 1.$