Risk management of food health hazard by meat consumption reduction: a coopetitive game approach

Abstract

In this paper, we face the serious problem of food health hazard, also in connection with global food production scarcity and feeding sustainability, in view of important environmental issues and the severe incumbent climate change. Specifically, our innovative risk management approach considers cooperation among producers of vegan and non-vegan food, a strong commitment more and more observed, recently, in technologically advanced western countries. The novelty of our work consists in proposing possible quantitative agreements among complementary food producers, usually non-interacting, in order to develop a sustainable healthy food production for human population—also characterized by low impact on the planet. Another new feature of our approach lies in using coopetition and game theory together; we show, quantitatively, how to conjugate human health defense, environmental defense, economic interests and less government spending, needs which usually appear in contrast with each other. Another point of our coopetitive approach is the suggestion of an easier way to entry the global market for vegan food producers. Meanwhile, our model suggests to big producers/sellers of non-vegan food a way to smoothly and rapidly transit toward more sustainable production. Technically, we propose an innovative exemplary complex agreement among global food sellers and small (but strongly sustainable and innovative) vegan food producers. Moreover, our model implies a general saving for the countries, by mitigating the health expenditures. The result of our mathematical study suggests a novel win–win solution for global economy, world environment and governments, while improving human population sustainability and climate change effects.

Appendix: complete analysis of the parametric game

In this appendix, we study the parametric game

$$\begin{aligned} {\mathcal {G}} = (p, >) \end{aligned}$$

associated with the game G.

1.1 Preliminaries

We now consider the parametric game ${\mathcal {G}} = (p, >)$ associated with the coopetitive game $G= (f,>).$ Note that, fixed a cooperative strategy $z_1$ in the interval ${\mathcal {C}}_1$, the parametric game defined by

$$\begin{aligned} {\mathcal {G}}(z_1) = (p(z_1), >) \end{aligned}$$

with parametric payoff function

$$\begin{aligned} p(z_1):{\mathcal {E}}\times {\mathcal {F}} \times {\mathcal {C}}_2 \rightarrow {\mathbb {R}}^2, \end{aligned}$$

defined on the parallelepiped $T_2$ (cartesian product of ${\mathcal {E}}$, ${\mathcal {F}}$ and ${\mathcal {C}}_2$) by

$$\begin{aligned} p(z_1)(x, y,z_2) = f(x, y, z), \end{aligned}$$

is the translation of the game ${\mathcal {G}}(0)$ by the infinite family of vectors defined by

$$\begin{aligned} w_1(z_1) = z_1(0,1). \end{aligned}$$

So that we can study the game ${\mathcal {G}}(0)$ and then we can translate the various information of the game ${\mathcal {G}}(0)$ by the infinite family of vectors $w_1(z_1)$.

More specifically, we fix the intervals of the strategy space and the other constant of the game as follows.¹

$$\begin{aligned} {\mathcal {E}}= & {} [0,1], \qquad {\mathcal {F}}=[1,2],\\ {\mathcal {C}}= & {} {\mathcal {C}}_1\times {\mathcal {C}}_2 = [1,2] \times [0,2], \\ p _{10}= & {} 7, \qquad p_{31} = 1,\\ p_{23}= & {} 1, \qquad p_{20} = 5,\\ F=&0.5, \qquad k'= 0.4, \qquad a = 1.5. \end{aligned}$$

For our convenience, in the representation of payoff space, we consider a game translated by the vector $(-3.6,0)$, so that to the payoff of the first player we need to add, at the end of the day, exactly 3.6 equal 144 million dollars.

From (4), we have:

$$\begin{aligned} f(x,y,z)= & {} (6x - 4 - 1.5y,\, 0.5y - 0.5) + z_1(0,1) \nonumber \\&+ z_2\left( -\frac{7x}{2}, \, 5\right) . \end{aligned}$$

(5)

We define the new parametric game

$$\begin{aligned} H:{\mathcal {E}}\times {\mathcal {F}} \times {\mathcal {C}}_2 \rightarrow {\mathbb {R}}^2 \end{aligned}$$

by

$$\begin{aligned} H(x,y,z_2)=g(x,y)+z_2\left( -\frac{7x}{2}, \, 5\right) \end{aligned}$$

for every

$$\begin{aligned} (x,y,z_2) \in {\mathcal {E}}\times {\mathcal {F}} \times {\mathcal {C}}_2 , \end{aligned}$$

where

$$\begin{aligned} g(x, y) := (6x - 4 - 1.5y, \, 0.5y - 0.5). \end{aligned}$$

Observe that

$$\begin{aligned} H(x,y,0)=g(x,y), \end{aligned}$$

for every $(x,y)\in {\mathcal {E}}\times {\mathcal {F}}$.

The image of the function

$$\begin{aligned} H:{\mathcal {E}}\times {\mathcal {F}} \times {\mathcal {C}}_2 \rightarrow {\mathbb {R}}^2 \end{aligned}$$

is the union

$$\begin{aligned} \text {Im}(H)=\bigcup _{z_2 \in {\mathcal {C}}_2} \text {Im} (H(.,.,z_2)). \end{aligned}$$

The function

$$\begin{aligned} p:{\mathcal {C}}_1 \mapsto p({\mathcal {C}}_1) \end{aligned}$$

is defined by

$$\begin{aligned} z_1 \mapsto H+z_1(0,1). \end{aligned}$$

Now, we immediately see that

$$\begin{aligned} \text {Im}(f)=\bigcup _{z_1 \in {\mathcal {C}}_1} \left( \text {Im} (H) + z_1(0,1) \right) . \end{aligned}$$

Since

$$\begin{aligned} {[}(z_1-1)+1] (0,1) = (z_1-1)(0,1) + (0,1) \end{aligned}$$

for every $z_1 \in {\mathcal {C}}_1$, then

$$\begin{aligned} \text {Im}(f)=\bigcup _{z_1 \in {\mathcal {C}}_1} \left( \text {Im} (H) + (z_1-1)(0,1) \right) +(0,1). \end{aligned}$$

So, let us consider the game ${\mathcal {G}}(0)=(H,>)$ and let

$$\begin{aligned} g : {\mathcal {E}} \times {\mathcal {F}} \rightarrow {\mathbb {R}}^2 , \end{aligned}$$

defined by

$$\begin{aligned} g(x, y) = (6x - 4 - 1.5y, \, 0.5y - 0.5), \end{aligned}$$

the function such that

$$\begin{aligned} H(x,y,0)=g(x,y). \end{aligned}$$

In order to graph the payoff space

$$\begin{aligned} H(T_2)=\text {Im}(H), \end{aligned}$$

we transform all the vertices of the bi-strategy square Q by the function $H(.,z_2)$.

In particular, we have

$$\begin{aligned} \text {Im}(H)= & {} \bigcup _{z_2 \in {\mathcal {C}}_2} \text {Im} (H(.,.,z_2)) \\= & {} \bigcup _{z_2 \in {\mathcal {C}}_2} \text {conv} (A'(0,z_2),(B'(0,z_2),\\&\quad (C'(0,z_2),(D'(0,z_2)) \end{aligned}$$

where

$$\begin{aligned} A'(0,z_2)= & {} H(A,z_2)=f(A,0,z_2), \\ B'(0,z_2)= & {} H(B,z_2)=f(B,0,z_2),\\ C'(0,z_2)= & {} H(C,z_2)=f(C,0,z_2), \\ D'(0,z_2)= & {} H(D,z_2)=f(D,0,z_2). \end{aligned}$$

1.2 Payoff space of game $(g,>)$

The transformation of segment [A, B] by means of the function g is the segment

$$\begin{aligned}{}[A'(0,0), B'(0,0)], \end{aligned}$$

where

$$\begin{aligned} A'(0,0) := g(A) \end{aligned}$$

is the payoff (− 5.5, 0) and

$$\begin{aligned} B'(0,0) := g(B) \end{aligned}$$

is the payoff (0.5, 0).

The transformation of segment [B, C] by means of the function g is the segment

$$\begin{aligned} {[}B'(0,0), C'(0,0)], \end{aligned}$$

where

$$\begin{aligned} C'(0,0) := g(C) = (-1, 0.5). \end{aligned}$$

The transformation of segment [C, D] by means of the function g is the segment

$$\begin{aligned} {[}C'(0,0), D'(0,0)], \end{aligned}$$

where

$$\begin{aligned} D'(0,0) := g(D) \end{aligned}$$

is the payoff (− 7, 0.5).

The transformation of segment [A, D] by means of the function g is the segment

$$\begin{aligned} {[}A'(0,0), D'(0,0)]. \end{aligned}$$

In Fig. 5, we show the payoff space g(Q).

Fig. 5

Payoff space of game H(., 0)

1.3 Payoff space of game ${(H(.,z_2),>)}$

We now consider the payoff section

$$\begin{aligned} {H(.,z_2)}: {\mathcal {E}}\times {\mathcal {F}} \rightarrow {\mathbb {R}}^2, \end{aligned}$$

defined by

$$\begin{aligned} H(x,y,z_2)=g(x,y)+z_2\left( -\frac{7x}{2}, \, 5\right) , \end{aligned}$$

for every $(x,y) \in {\mathcal {E}}\times {\mathcal {F}}$, with $z_2 \in {\mathcal {C}}_2$.

The transformation of segment [A, B] by means of the function ${H(.,z_2)}$ is the segment

$$\begin{aligned} {[}A'(0,z_2), B'(0,z_2)], \end{aligned}$$

where

$$\begin{aligned} A'(0,z_2) := H(A,z_2)= g(A) + z_2\left( 0, \, 5\right) \end{aligned}$$

is the payoff

$$\begin{aligned} (-5.5, 0)+z_2\left( 0, \, 5\right) = (-5.5, 5 z_2) \end{aligned}$$

and

$$\begin{aligned} B'(0,z_2) := H(B,z_2) \end{aligned}$$

is the payoff

$$\begin{aligned} (0.5, 0)+z_2\left( -\frac{7}{2}, \, 5\right) =\left( 0.5-\frac{7}{2}z_2, 5 z_2\right) . \end{aligned}$$

The transformation of segment [B, C] by means of the function ${H(.,z_2)}$ is the segment

$$\begin{aligned} {[}B'(0,z_2), C'(0,z_2)], \end{aligned}$$

where

$$\begin{aligned} C'(0,z_2) := H(C,z_2) \end{aligned}$$

is the payoff

$$\begin{aligned} (-1, 0.5)+z_2\left( -\frac{7}{2}, \, 5\right) = \left( -1 -\frac{7}{2} z_2, 0.5 + 5 z_2\right) . \end{aligned}$$

The transformation of segment [C, D] by means of the function ${H(.,z_2)}$ is the segment

$$\begin{aligned} {[}C'(0,z_2), D'(0,z_2)], \end{aligned}$$

where

$$\begin{aligned} D'(0,z_2) := H(D,z_2) \end{aligned}$$

is the payoff

$$\begin{aligned} (- 7, 0.5)+z_2\left( 0, \, 5\right) = (-7, 0.5 + 5 z_2). \end{aligned}$$

The transformation of segment [A, D] by means of the function ${H(.,z_2)}$ is the segment

$$\begin{aligned}{}[A'(0,z_2), D'(0,z_2)]. \end{aligned}$$

In Fig. 6, we show the payoff space $H(T_2)$.

Fig. 6

Payoff space $\mathrm {Im}(H)$

We see in Fig. 6 the moving segment

$$\begin{aligned} {[}B',C']&: {\mathcal {C}}_1 \times {\mathcal {C}}_2 \rightarrow {\mathcal {P}}({\mathbb {R}}^2): (z_1,z_2)\mapsto \\&[B'(z_1,z_2),C'(z_1,z_2)]. \end{aligned}$$

This segment $[B',C']$ comes from the transformation of the segment [B, C], which shows the first coordinate of all its points equal to one. The segment $[B',C']$ is moving progressively toward the left, as the strategy $z_2$ increases. We observe that the moving segment $[A',D']$, coming from [A, D], lies always at the left of the moving segment $[B',C']$, until the value of the strategy $z_2$ reaches the level 12/7.

Remark

Observe that, we have defined four parametric curves in the Cartesian plane, namely the four curves $A'$, $B'$, $C'$ and $D'$. These four curves are defined on the compact interval [0, 4] even if, for our coopetitive analysis, we consider only the parts of curves with endpoints corresponding to the values 1 and 4 of the coopetitive parameter z.

1.4 Payoff space of the coopetitive game G

Now, we consider the translation of the game $H(., .,z_2)$ by the vector

$$\begin{aligned} w_1(z_1)= z_1(0,1) \end{aligned}$$

with $z_1 \in [1,2]$, in order to obtain the total payoff space of the coopetitive game (see Fig. 7).

The final payoff space of our coopetitive game is the union

$$\begin{aligned} \mathrm {Im}(f) = \bigcup _{z_1 \in [1,2]} \Big (\mathrm {Im}(H) + z_1 (0,1)\Big ). \end{aligned}$$

Fig. 7

Payoff space of game G

We show in Fig. 8 the filled payoff space of the game G.

Fig. 8

Payoff space of the coopetitive game G

1.5 The Pareto maximal boundary of the payoff space f(S)

The Pareto maximal boundary of the payoff space f(S) of the coopetitive game G, where S denotes the strategy space of the game Cartesian product of the strategy square Q and of the intervals ${\mathcal {C}}_1$ and ${\mathcal {C}}_2$, is the union of the segments

$$\begin{aligned} {[}B'(2,0), B'(2,12/7)) \cup [A'(2,2), D'(2,2)], \end{aligned}$$

where the point

$$\begin{aligned} B'(2,12/7) = A'(2,12/7). \end{aligned}$$

Fig. 9

Pareto boundary of the payoff space of game $G(z_1,z_2)$

In Fig. 9, we see the Pareto maximal boundary of the payoff space.

1.6 Study of the Nash zone

Recalling the payoff function of the game defined by (5) for every $(x,y,z) \in S$. The Nash equilibrium strategies of the first player are determined solving the optimum problem

$$\begin{aligned} \max _{x \in {\mathcal {E}}} f_1(x,{y},z), \end{aligned}$$

for every $z \in {\mathcal {C}}$ and ${y} \in {\mathcal {F}}$. The Nash strategy set of the first player $N_1(z)$ is the set of the solutions of the optimization problem.

The Nash equilibrium strategies of the second player are determined solving the optimum problem

$$\begin{aligned} \max _{y \in {\mathcal {F}}} f_2({x},y,z), \end{aligned}$$

for every $z \in {\mathcal {C}}$ and ${x} \in {\mathcal {E}}$. The Nash strategy set of the first player $N_1(z)$ is the set of the solutions of the first optimization problem.

Fig. 10

Nash zone of the payoff space of game $G(z_1,z_2)$

Let us consider the positiveness of the first argument derivative of $f_1$ (with $x \in [0,1]$). From that we obtain

$$\begin{aligned} \partial _1 f_1 (x,y,z) = 6-\frac{7}{2}z_2>0 \end{aligned}$$

that gives

$$\begin{aligned} N_1(z)=\left\{ \begin{array}{lll} 1 &{} \quad \text { if } &{} \quad z_2<\frac{12}{7}\\ \left[ 0,1\right] &{} \quad \text { if } &{} \quad z_2=\frac{12}{7}\\ 0 &{} \quad \text { if } &{} \quad z_2>\frac{12}{7} \end{array} \right. \end{aligned}$$

The second argument derivative of $f_2$,

$$\begin{aligned} \partial _2 f_2 (x,y,z) = \frac{1}{2}>0, \end{aligned}$$

results positive for every $(x,y,z) \in S$, so the Nash strategy for the second player (with $y \in [1,2]$) is 2. So, the multivalued coopetitive Nash path is

$$\begin{aligned} N(z)=\left\{ \begin{array}{lll} (1,2)=C &{} \quad \text { if } &{} \quad z_2<\frac{12}{7}\\ \left[ 0,1\right] \times 2 = [C,D]&{} \quad \text { if } &{} \quad z_2=\frac{12}{7}\\ (0,2)=D &{} \quad \text { if } &{} \quad z_2>\frac{12}{7} \end{array} \right. , \end{aligned}$$

for every $z \in {\mathcal {C}}$. From that we calculate the Nash zone in the payoff space (see Fig. 10).