Sustainable farming practices adoption in agriculture supply chain: the role of indirect support versus cost subsidy

Abstract

Due to the growing environmental concerns and associated health issues of using fertilisers and pesticides in agriculture, Zero Budget Natural Farming (ZNBF) is gaining traction to guide farmers toward sustainable farming practices. Therefore, to promote sustainability and also to safeguard the interests of small farmers who adopt ZNBF, the government in developing economies are adopting two prevalent intervention mechanisms: (i) cost subsidy and (ii) indirect support. However, little has been done to provide clear-cut directives on when to employ these intervention mechanisms, especially in the presence of heterogeneous farmers and yield uncertainty. Therefore, in this paper, we consider a supply chain comprising a government, small farmers, and consumers, and we perform a comparative analysis to understand the prevailing conditions under which a particular policy outperforms another in maximizing social welfare by promoting ZBNF practices. We develop a government-farmer game-theoretic model wherein the government first selects the intervening mechanism to maximise social welfare, followed by farmers’ sowing decisions. Our analysis reveals that both policies positively impact the adoption of ZBNF practices; however, in the scenario of higher environmental damage, cost subsidy outperforms the indirect support mechanism. On the contrary, for medium and lower environmental damage, the choice of intervention depends on the proportion of environmentally conscious farmers. Specifically, the government should extend indirect support only if there exists a critical mass of environmentally conscious farmers. Our findings from analytical and numerical analysis assist the government in identifying and designing the appropriate intervention mechanisms whereas to farmers in their showing decisions.

Appendix

Proof of Lemma 2 Using the incentive compatibility constraints of two types of farmers: environmentally conscious and neutral, we first prove that each farmer will cultivate either of the two crops, 1 or 2 in equilibrium. This proof is by contradiction. Let us assume a farmer of type $k\in \{E, N\}$ located at $x<\tau _k$ allocates $\beta$ proportion of her capacity to crop 2 and $1-\beta$ to crop 1. However, by the definition of $\tau _k$ given by Eq. (9), it is clear that all farmers at $x<\tau _k$ perceive higher utility by sowing crop 2 as compared to crop 1. Hence, by incentive compatibility, these farmers will allocate $\beta = 1$, implying that they will sow only one type of crop in equilibrium, which is crop 2. Similarly, we can prove that a farmer of type $k\in \{E, N\}$ located at $x>\tau _k$ will sow one and only one crop in equilibrium, which is crop 1. Additionally, without loss of generality, we can assume that a farmer of type $k\in \{E, N\}$ located at $x=\tau _k$, who is indifferent between crop 1 and 2, sows crop 2 in equilibrium.

Next, we obtain the fraction of farmers who sow crop 2 and crop 1 in equilibrium. Using the rational expectation equilibrium concept, each farmer believes ex-ante that $\phi$ fraction of farmers will grow crop 2 and $1-\phi$ fraction of farmers will grow crop 1 in equilibrium. Accordingly, we can get the fraction of farmers growing crop 2 in equilibrium by solving the fixed point equation given by Eq. (10). Substituting the obtained fraction of farmers who sows crop 2 in equilibrium in Eq. (9), we get the expressions for $\tau _k$.

Proof of Lemma 1 To ensure $\tau _k$ $\in$ $[-1, 1]$, the following two inequalities have to be satisfied: (a) $\tau _E \le 1$ and (b) $\tau _N \ge -1$. Simplifying these conditions, we get $-1+d\cdot \beta \le \tau _E \le 1$. Accordingly, for condition (b) to hold, i.e., $\tau _N \ge -1$ $\implies$ $\tau _E \ge -1+d\cdot \beta$. Using the equilibrium $\tau _E$ from Eq. (11), we can easily get $a>\frac{d\alpha (1+y^2+\frac{\theta ^2}{3})}{2(\gamma y-1)}-\frac{2}{(\gamma y-1)}\equiv a_1$. And to satisfy condition (a), i.e., $\tau _E \le 1$ $\implies$ $a>\frac{\alpha (1+y^2+\frac{\theta ^2}{3})}{(\gamma y-1)}-\frac{2}{(\gamma y-1)}\equiv a_2$. Therefore, if $a>min\{a_1,a_2\}$, then $\tau _k$ $\in$ $[-1, 1]$.

Proof of Lemma 3 Differentiating the aggregate threshold of the location of a farmer, $\tau$, with respect to the different parameter values, we get the desired result.

• 1. $\frac{d\tau }{d\theta } = -\frac{12 \theta (2 + d \alpha \beta +a (y \gamma -1))}{(9 + 3 y^2 +\theta ^2)^2)}<0$.

• 2. $\frac{d\tau }{dd} = \frac{6\alpha \beta }{9+3y^2+\theta ^2}>0$ and $\frac{d\tau }{d\gamma } = \frac{6ay}{9+3y^2+\theta ^2}>0$.

• 3. $\frac{d\tau }{dy} = \frac{6 (6 y (-2 + a - d\alpha \beta ) - 3 a y^2 \gamma + a \gamma (9 + \theta ^2))}{(9+3y^2+\theta ^2)^2}$, which is positive if

  $y<\frac{-12 + 6 a - 6 d \alpha \beta + \sqrt{36 (2 - a + d \alpha \beta )^2 + 12 a^2 \gamma ^2 (9 + \theta ^2)}}{6a\gamma }\equiv \tilde{y}$ and negative otherwise.

• 4. $\frac{d\tau }{d\beta } = \frac{6d\alpha }{9+3y^2+\theta ^2} >0$.

• 5. $\frac{d\tau _E}{d\alpha } = -d\beta \left( 1-\frac{6}{9+3y^2+\theta ^2}\right) <0$ and $\frac{d\tau _N}{d\alpha } = -d\beta \left( 1-\frac{6}{9+3y^2+\theta ^2}\right) <0$ .

• 6. $\frac{d\tau }{d\alpha } = \frac{6d\beta }{9+3y^2+\theta ^2} >0$.

Proof of Proposition 1: The total social welfare under the benchmark case is:

$$\begin{aligned} TSW_{Benchmark} = \frac{\begin{matrix}-2 + 4 a - 4 c - 4 d - 4 d \alpha \beta + d^2 \alpha \beta ^2 -d^2 \alpha ^2 \beta ^2 &{}\\ + \frac{6 (2 + d \alpha \beta + a (-1 + y\gamma )) (2 + d (2 + \alpha \beta ) + a (-1 + y \gamma ))}{{9+3y^2+\theta ^2}}\end{matrix}}{4} \end{aligned}$$

Differentiating with respect to $\theta$, we get

$$\begin{aligned} \frac{dTSW_{Benchmark}}{d\theta } = - \frac{3 (2 + d \alpha \beta + a (-1 + y \gamma )) (2 + d (2 + \alpha \beta ) + a (-1 + y \gamma )) \theta }{(9+3y^2+\theta ^2)^2}<0. \end{aligned}$$

Proof of Theorem 1 It is easy to verify that the government objective is concave in the decision variable s, i.e., $\frac{d^2TSW_{SD}}{ds^2} =-\frac{3}{9+3y^2+\theta ^2}<0$. Hence, the optimal subsidy can be obtained by equating $\frac{dTSW_{SD}}{ds}=0$, when solved for s, we get the optimal subsidy as $min\{d,\bar{s}\}$, where $\bar{s}$ is obtained from the constraint of the government problem and is given in Theorem 1.

Proof of Proposition 2 Differentiating the optimal subsidy, $s^*$, with respect to the different parameter values, we get the desired result.

• 1. $\frac{ds^*}{dd}=1>0$. Hence, as the environmental damage increases, the government increases the subsidy and should stop at $\tilde{d}$ to ensure both crops are grown in equilibrium, where $\tilde{d}=\frac{6(1+y^2+\frac{\theta ^2}{3}-a(\gamma y-1))}{6+3\beta (3+y^2(1-\alpha )-\alpha )+\beta \theta ^2(1-\alpha )}$.

• 2. $\frac{ds^*}{d\theta }=\frac{(2-d(1-\alpha )\beta ) \theta }{2}>0$.

• 3. $\frac{ds^*}{d\alpha }= \frac{d\beta (1+y^2+\frac{\theta ^2}{3})}{2}>0$.

• 4. $\frac{ds^*}{d\beta }= -\frac{3 d (3 + y^2 (1 - \alpha ) - \alpha ) + d (1 - \alpha ) \theta ^2}{6}<0$.

• 5. $\frac{ds^*}{d\gamma }= -ay<0$.

Proof of Corollary 1 It is easy to verify that $\tau _E$ and $\tau _N$ are interior points, i.e., $\in$ $[-1, 1]$. And also, the difference between $\tau _E$ and $\tau _N$ is $d\beta$, which is positive. Hence, the aggregate threshold of the location of a farmer, $\tau$ $\in$ $[-1, 1]$ as it is a convex combination of $\tau _E$ and $\tau _N$. Therefore, there exists a few farmers located beyond the threshold who grows crop 1 in equilibrium despite the presence of subsidies in crop 2.

Proof of Proposition 3 The total social welfare under the benchmark case is:

$$\begin{aligned} TSW_{Benchmark} = \frac{\begin{matrix}-2+4a-4c-4d-4d\alpha \beta + d^2\alpha \beta ^2-d^2\alpha ^2 \beta ^2&{}\\ +\frac{6(2+d\alpha \beta +a(-1+y\gamma ))\cdot (2+d(2+\alpha \beta )+a(-1+y \gamma ))}{{9+3y^2+\theta ^2}}\end{matrix}}{4} \end{aligned}$$

The total social welfare under the SD policy is:

$$\begin{aligned} TSW_{SD}=\frac{\begin{matrix}-2+4a-4c-4d-4d\alpha \beta + d^2\alpha \beta ^2-d^2\alpha ^2\beta ^2 &{}\\ +\frac{6 (2+d +d\alpha \beta +a(-1+y\gamma ))^2}{{9+3y^2+\theta ^2}}\end{matrix}}{4} \end{aligned}$$

Subtracting these two, we get

$$\begin{aligned} TSW_{SD}-TSW_{Benchmark}=\frac{3d^2}{2(9+3y^2+\theta ^2)}>0 \end{aligned}$$

Proof of Proposition 4: We first calculate the total social under each intervention, followed by subtracting each other to compare the performance of the two interventions.

$$\begin{aligned} TSW_{SD}&= \frac{-2 + 4 a - 4 c - 4 d - 4 d \alpha \beta + d^2 \alpha \beta ^2 -d^2 \alpha ^2 \beta ^2+\frac{6 (2+d + d \alpha \beta + a (-1 + y\gamma ))^2}{{9+3y^2+\theta ^2}}}{4}\\ TSW_{IS}&= \frac{\begin{matrix}-2+4a-4c-4d-4d\alpha \beta + d^2\alpha \beta ^2-d^2\alpha ^2 \beta ^2&{}\\ +\frac{6(2+d\alpha \beta +a(-1+y\gamma ))\cdot (2-2g+d(2+\alpha \beta )+a(-1+y \gamma ))}{{9+3y^2+\delta ^2\theta ^2}}\end{matrix}}{4} \end{aligned}$$

Let $\Delta TSW = TSW_{IS}-TSW_{SD}$, we get

$\Delta TSW = \frac{\frac{6(2+d\alpha \beta +a(-1+y\gamma ))\cdot (2-2g+d(2+\alpha \beta )+a(-1+y \gamma ))}{9+3y^2+\delta ^2\theta ^2}-\frac{6 (2+d + d \alpha \beta + a (-1 + y\gamma ))^2}{9+3y^2+\theta ^2}}{4}$

In case $\Delta TSW$ is positive, indirect support would yield a higher total social welfare as compared to the subsidy. Hence to find the conditions under which IS policy performs better, we solve $\Delta TSW(g) = 0$. After solving this, we observe that $\Delta TSW$ is positive if $g<F(\alpha )$, negative otherwise. This implies that if $g<F(\alpha )$ holds, then total social welfare in case of indirect support exceeds that of subsidy otherwise, total social welfare in case of subsidy is more than that of indirect support, where $F(\alpha )$ is given below.

$F(\alpha )\!=\!\frac{2d(1+\alpha \beta )(2+a(-1+y\gamma ))(1 -\delta ^2)\theta ^2+(2+a(-1 + y \gamma ))^2(1-\delta ^2) \theta ^2 - d^2(9+3y^2+(-\alpha \beta (2+\alpha \beta )+(\delta + \alpha \beta \delta )^2) \theta ^2)}{2(2+d\alpha \beta +a(-1+y\gamma ))(9+3y^2+\theta ^2)}.$

Next, we identify the necessary conditions under which $g<F(\alpha )$ holds. It is easy to verify that $F(\alpha )$ is a continuous monotonously increasing function in $\alpha$, which can be proved by the construction technique. We plot $\frac{dF(\alpha )}{d\alpha }$ over the entire range of $\alpha$ and we find the derivate to be positive for all $\alpha \in [0,1]$. This suggests $F(\alpha )$ is a monotonously increasing function in $\alpha$. Therefore, the maximum and minimum value is obtained at $F(\alpha =1)$ and $F(\alpha =0)$, respectively. For further analysis, we take $F(\alpha =1)=\bar{g}$ and $F(\alpha =0)=\underline{g}$ and consider three cases.

• 1. When $0<d<\frac{(2+a(-1+y\gamma ))(1-\delta ^2)\theta ^2 - \sqrt{(2+a(-1+y\gamma ))^2(1-\delta ^2)\theta ^2(9+3y^2+\theta ^2)}}{9+3y^2+\delta ^2\theta ^2}\equiv d_1$.

  Under the above condition, $F(\alpha =0)>0$. Hence the for all $\alpha \in [0, 1]$ $F(\alpha =0)>0$. However, the LHS of $g>F(\alpha )$ is a constant term and if we closely look at the administration costs g, it can take three cases:

  • (a) If $g>\bar{g}$, then g is always greater than $F(\alpha )$, Hence $\Delta TSW < 0$, implying $TSW_{SD}>TSW_{IS}$.

  • (b) If $\underline{g}\le g \le \bar{g}$, then according to intermediate value theorem there exists a $\tilde{\alpha } \in [0, 1]$ such that if $\alpha >\tilde{\alpha }$ then $g<F(\alpha )$ implying $\Delta TSW > 0$, hence $TSW_{SD}<TSW_{IS}$. On the contrary, if $\alpha <\tilde{\alpha }$ then $g>F(\alpha )$ implying $\Delta TSW < 0$, hence $TSW_{SD}>TSW_{IS}$.

  • (c) If $0<g<\underline{g}$, then g will always be lower than $F(\alpha )$, implying $\Delta TSW > 0$, hence $TSW_{SD}<TSW_{IS}$ for all $\alpha \in [0,1]$.

• (2.) When $d_1\le d\le \frac{(-2+a-ay\gamma )\sqrt{1-\delta ^2}\theta }{(1+\beta )\sqrt{1-\delta ^2}\theta -\sqrt{9+3y^2+\theta ^2}} \equiv d_2$. Under the above condition, $F(\alpha =0)<o$, however, $F(\alpha =1)>o$. Hence there exist two cases:

  • (a) If $g>\bar{g}$, then g is always greater than $F(\alpha )$, implying $TSW_{SD}>TSW_{IS}$.

  • (b) If $0<g<\bar{g}$, then according to intermediate value theorem there exists a $\tilde{\alpha } \in [0, 1]$ such that if $\alpha >\tilde{\alpha }$ then $g<F(\alpha )$ implying $TSW_{SD}<TSW_{IS}$. On the contrary, if $\alpha <\tilde{\alpha }$ then $g>F(\alpha )$ implying $TSW_{SD}>TSW_{IS}$.

• 3. When $d>d_2$. Under the above condition, $F(\alpha =1)<o$. Hence the entire range is negative, implying $g>F(\alpha ) \implies TSW_{SD}>TSW_{IS}$.

Proof of Proposition 5: Similar to Proposition 4, we compare the performance of two intervention mechanisms using $\delta$ as a parameter as compared to $\alpha$ in Proposition 4. We find that SD generates higher social welfare only when $g<F(\delta )$. It is easy to verify that $F(\delta )$ is a monotonous decreasing function in $\delta$. We used the proof by construction to show $F(\delta )$ is a monotonous decreasing function. For further analysis, we take $F(\delta =0)= \tilde{g}$ and consider two cases:

• 1. When $d<\frac{(1+\alpha \beta )(2+a(-1+y\gamma ))\theta ^2+\sqrt{(2+a(-1+y\gamma ))^2 \theta ^2(9+3y^2+\theta ^2)}}{9+3y^2-\alpha \beta (2+\alpha \beta )\theta ^2}\equiv d_0$. Under this condition, we find that $\tilde{g}>0$ and $F(\delta =1)<0$. Hence, by using the intermediate value theorem, we can say that if $0<g<\tilde{g}$, then there exists a $\tilde{\delta } \in [o, 1)$ at which $g=F(\delta )$. Accordingly, if $\delta <\tilde{\delta }$ then $g<F(\delta )$ will hold and otherwise for $\delta >\tilde{\delta }$, $g>F(\delta )$ will hold.

• 2. When $d\ge d_0$. Under this condition, we find that $\tilde{g}<0$, which suggests that $F(\delta )<0$ for every $\delta \in [0,1]$. Hence $g>F(\delta )$ will always hold.