Estimating the long-term impact of market power on the welfare gains from groundwater markets

Abstract

Water markets are considered an effective groundwater management instrument. However, the emergence of market power during their operation, i.e., price manipulation, cannot be excluded. In this paper, a simple water market between two groups of farmers is modeled and an attempt is made firstly to quantify the loss of aggregate total benefits during a given planning period from the occurrence of market power in this market using a "structural model" to describe the market conditions and solving an optimal control problem and secondly to determine the initial allocation of water rights that limits the loss of benefits due to market power. The results of simulations based on hydro-economic data of a region in Northern Greece lead to two conclusions. The first conclusion is that the loss of aggregate total benefit is likely to reach even $10\%$ compared to perfect competition when there is a full monopoly or monopsony in the water market. The second conclusion is that an initial allocation close to the quantities consumed by each group under perfect competition leads to a limitation of the impact of market power on the aggregate total benefit gained by the two groups of farmers during the planning period.

Appendix

The current value Hamiltonian is:

$$\mathcal{H}={NB}_{tot,t}+\mu \frac{1}{AS}\left[R-\left(1-a\right){Q}_{tot,t}\right]$$

(A1)

where $\mu$ is a costate variable that expresses the shadow value of groundwater, i.e., the marginal value of the groundwater that stays into the aquifer.

For an interior solution to exist, the Hamiltonian has to be concave at ${Q}_{tot,t}$. The necessary conditions for the solution of the optimal control problem according to Pontryagin’s Maximum Principle are the following (Hoy et al. 2001):

$$\frac{\partial \mathcal{H}}{\partial {Q}_{tot,t}}=0 \,and\, \dot{\mu }=r\mu -\frac{\partial \mathcal{H}}{\partial H(t)}$$

(A.2)

Applying the first necessary condition, it follows that

$$\mu =\frac{AS}{(1-a)}\left\{\left(\frac{{{s}_{1}}^{2}}{{k}_{1}}+\frac{{s}^{2}}{{k}_{2}}\right)\frac{R}{\left(1-a\right)}+\frac{{s}_{1}}{{k}_{1}}\left({\theta }_{1}-{g}_{1}\right)+\frac{{s}_{2}}{{k}_{2}}\left({\theta }_{2}-{g}_{2}\right)-{c}_{0}{s}_{L}+\dots \right\}\left\{\dots +{c}_{0}H\left(t\right)-\frac{AS}{(1-a)}\left(\frac{{{s}_{1}}^{2}}{{k}_{1}}+\frac{{{s}_{2}}^{2}}{{k}_{2}}\right)\dot{H}\right\}$$

(A.3)

Differentiating equation (A.3) with respect to time $t$, the following equation can be obtained:

$$\dot{\mu }=\frac{AS}{(1-a)}\left[{c}_{0}\dot{H}-\frac{AS}{(1-a)}\left(\frac{{{s}_{1}}^{2}}{{k}_{1}}+\frac{{{s}_{2}}^{2}}{{k}_{2}}\right)\ddot{H}\right]$$

(A.4)

To apply the second necessary condition, the partial derivative $\frac{\partial \mathcal{H}}{\partial H(t)}$ has to be calculated, that is:

$$\frac{\partial \mathcal{H}}{\partial {H}_{t}}={c}_{0}{Q}_{tot,t}$$

(A.5)

From Eq. (9) it follows that:

$${Q}_{tot,t}=\frac{R}{(1-a)}-\frac{AS}{(1-a)}\dot{H}$$

(A.6)

Applying the second necessary condition the following differential equation results:

$$\ddot{H}+\dot{\frac{{h}_{2}}{{h}_{1}}H}+\frac{{h}_{3}}{{h}_{1}}H=\frac{{h}_{4}}{{h}_{1}}$$

(A.7)

where ${h}_{1}=-\frac{AS}{(1-a)}\left(\frac{{{s}_{1}}^{2}}{{k}_{1}}+\frac{{{s}_{2}}^{2}}{{k}_{2}}\right)$, ${h}_{2}=\frac{rAS}{(1-a)}\left(\frac{{{s}_{1}}^{2}}{{k}_{1}}+\frac{{{s}_{2}}^{2}}{{k}_{2}}\right)$,${h}_{3}=-r{c}_{0}$ and ${h}_{4}=-\frac{R{c}_{0}}{AS}+r\left[\left(\frac{{{s}_{1}}^{2}}{{k}_{1}}+\frac{{{s}_{2}}^{2}}{{k}_{2}}\right)\frac{R}{(1-a)}+\frac{{s}_{1}}{{k}_{1}}\left({\theta }_{1}-{g}_{1}\right)+\frac{{s}_{2}}{{k}_{2}}\left({\theta }_{2}-{g}_{2}\right)-{c}_{0}{s}_{L}\right]$.

Equation (A.7) is a second order differential equation with constant coefficients. The general solution to this equation is the sum of the homogeneous solution ${H}_{h}(t)$ and particular solution $\overline{H }(t)$, i.e., it is:

$$H\left(t\right)={H}_{h}\left(t\right)+\overline{H }\left(t\right)$$

(A.8)

The homogenous solution is the following:

$${H}_{h}\left(t\right)={X}_{1}{e}^{{\lambda }_{1}t}+{X}_{2}{e}^{{\lambda }_{2}t}$$

(A.9)

where ${\mathrm{X}}_{1}$, ${\mathrm{X}}_{2}$ are coefficients that depend on the problem’s boundary conditions and ${\lambda }_{1},{\lambda }_{2}$ are the roots of the homogenous differential equation’ characteristic equation, which are ${\lambda }_{\mathrm{1,2}}=-\frac{{h}_{2}}{2{h}_{1}}\pm \frac{\sqrt{{(\frac{{h}_{2}}{{h}_{1}})}^{2}-4\frac{{h}_{3}}{{h}_{1}}}}{2}$.

The partial solution is the following:

$$\overline{H }\left(t\right)=\frac{{h}_{4}}{{h}_{3}}$$

(A.10)

Thus, the general solution will be:

$$H\left(t\right)={X}_{1}{e}^{{\lambda }_{1}t}+{X}_{2}{e}^{{\lambda }_{2}t}+\frac{{h}_{4}}{{h}_{3}}$$

(A.11)

Applying boundary conditions $H\left(0\right)={H}_{0}$ and $H\left(T\right)={H}_{min}$ it is obtained that.

${X}_{1}=\frac{{e}^{{\lambda }_{2}T}\left({H}_{o}-\frac{{h}_{4}}{{h}_{3}} \right)-{H}_{min}+\frac{{h}_{4}}{{h}_{3}} }{{e}^{{\lambda }_{2}T}-{e}^{{\lambda }_{1}T}}$ and ${X}_{2}=\frac{{H}_{min}-\frac{{h}_{4}}{{h}_{3}} -{e}^{{\lambda }_{1}T}({H}_{o}-\frac{{h}_{4}}{{h}_{3}} )}{{e}^{{\lambda }_{2}T}-{e}^{{\lambda }_{1}T}}$.

Based on equations (A.6) and (A.11), the time path ${Q}_{tot,t}$ for the amount of water pumped by the two groups of farmers during year $t$ can be easily derived. Taking Eq. (11) into account, the time path for the total net benefit ${NB}_{tot,t}$ from water consumption during year $t$ can also be derived.