# Task 10. Game Theory

Problem text. An oil and gas company has set the task of revising its sales strategy for natural and liquefied gas during the winter period. In a calm winter, there is a possibility of supplying 2200 cubic meters of natural gas and 3500 cubic meters of liquefied gas, while in the event of severe frosts, 3800 cubic meters of natural gas and 2450 cubic meters of liquefied gas. The cost of extraction and transportation of one cubic meter of natural gas is 19 monetary units, and for liquefied gas - 25. The final price for the consumer will be 35 and 58 monetary units, respectively. Determine the optimal strategy for the company to maximize its profit regardless of weather conditions.

Note. The answer is the planned amount of cubic meters of liquefied and natural gas to be extracted and supplied.
Solution to the problem. Let's model the possible scenarios. Due to unpredictable weather fluctuations, the company can apply two "pure" strategies:

$\mathrm{F}_{1}=(2200,3500) -$ procurement of $2200 \mathrm{m}^{3}$ of natural gas and $3500 \mathrm{m}^{3}$ of liquefied gas

$\mathrm{F}_{2}=(3800,2450) -$ procurement of $3800 \mathrm{m}^{3}$ of natural gas and $2450 \mathrm{m}^{3}$ of liquefied gas

Nature (the market) also has two strategies:

$\mathrm{D}_{1}=$ warm weather

$\mathrm{D}_{2}=$ cool weather

If the company applies the pure strategy $\mathrm{F}_{1}$ and the weather is warm $\left(D_{1}\right)$, the extracted gas will be fully sold, and the revenue will be:

$$
\mathrm{W}_{11}=2200 *(35-19)+3500 *(58-25)=150700
$$

On the other hand, if the company chooses strategy $\mathrm{F}_{1}$, but the winter is cold ( $\mathrm{D}_{2}$ ), the gas will be sold only partially, and the revenue will be:

$\mathrm{W}_{12}=2200 *(35-19)+2450 *(58-25)-(3500-2450) * 25=89900$.

Similarly, if the company applies strategy $\mathrm{F}_{2}$, and nature strategy $\mathrm{D}_{1}$, the revenue will be:

$\mathrm{W}_{21}=2200 *(35-19)+2450 *(58-25)-(3800-2200) * 19=85860$,

and if nature chooses strategy $\mathrm{D}_{2}$, then

$\mathrm{W}_{22}=3800 *(35-19)+2450 *(58-25)=141650$.

Considering the firm and nature as two players, we get the payoff matrix of the game, which will serve as the game model of the problem:

$$
W=\left(\begin{array}{cc}
150700 & 89900 \\
85860 & 141660
\end{array}\right)
$$

Since the maximin strategy of the game: $a=\max (85860,89900)=89900$, and the minimax strategy $b=\min (150700,141650)=141650$, the value of the game lies in the range:

$$
89900 \text { units }<v<141650 \text { units }
$$

There is no solution to the game in pure strategies, so we will look for a solution in mixed strategies. We will solve this game analytically. The average payoff of the first player, if he uses the optimal mixed strategy $x^{*}=\left(x_{1}^{*}, x_{2}^{*}\right)$, and the second player uses a pure strategy corresponding to the first column of the payoff matrix, is equal to the value of the game $v$:

$$
150700 x_{1}^{*}+85860 x_{2}^{*}=v
$$

The same average payoff is obtained by the first player if the second player applies a strategy corresponding to the second column of the payoff matrix, that is,

$$
89900 x_{1}^{*}+141650 x_{2}^{*}=v
$$

Considering that $x_{1}^{*}+x_{2}^{*}=1$, we get a system of equations to determine the optimal strategy of the first player and the value of the game:

$$
\left\{\begin{array}{c}
150700 x_{1}^{*}+85860 x_{2}^{*}=v \\
89900 x_{1}^{*}+141650 x_{2}^{*}=v \\
x_{1}^{*}+x_{2}^{*}=1
\end{array}\right.
$$

Solving this system, we find:

$$
\left\{\begin{array}{c}
x_{1}^{*}=0.48 \\
x_{2}^{*}=0.52 \\
v=116983
\end{array}\right.
$$

The optimal strategy for the company:

$$
P^{*}=x_{1}^{*} * F_{1}+x_{2}^{*} * F_{2}=0.48(2200 ; 3500)+0.52(3800 ; 2450)=(3032 ; 2954)
$$

Thus, it is optimal for the company to supply $3032 \mathrm{~m}^{3}$ of natural gas and 2954 m $^{3}$ of liquefied gas.