Game theory online calculator with steps

Number of player 1 strategies:

Number of player 2 strategies:

matrixA,0,0,matrixmatrixA,1,0,matrixmatrixA,2,0,matrixmatrixA,3,0,matrix

matrixA,0,1,matrixmatrixA,1,1,matrixmatrixA,2,1,matrixmatrixA,3,1,matrix

matrixA,0,2,matrixmatrixA,1,2,matrixmatrixA,2,2,matrixmatrixA,3,2,matrix

matrixA,0,3,matrixmatrixA,1,3,matrixmatrixA,2,3,matrixmatrixA,3,3,matrix

About Game theory calculator

Solve matrix games via game theory online — minimax, maximin, mixed strategies via linear programming, with detailed step-by-step solution. with complete, detailed, step-by-step description of solutions, that solves linear-programming problems up to 20×20 in size with coefficients of this type: decimal numbers and fractions.

To start the calculation, you need to first enter the problem dimensions in the input fields at the very top of the screen, and also choose the desired operation from the side menu.

A little below you will find an input window where you need to enter the coefficients, constraints, and right-hand-side values using the keyboard. The input control panel is also located here, which simplifies work with LP problems and contains the following control elements:

The first element allows you to expand the input window. This can be especially useful when the tableau does not fit completely on the screen. If the table is still not fully visible after expanding the window, you can change the scale using the + / - buttons;
The second element copies the current problem input to the memory buffer. This can be useful when you frequently solve the same LP problem or need to transfer data between operations;
And the last element pastes the previously copied input, which allows you to restore problem data in just a few clicks instead of re-entering it manually;

And further down you will find a toolbar that allows you to customize the calculator and make it easier to work with. It is visually divided into three parts, each of which is responsible for the following functionality:

The first part lets you select the number format used when displaying the solution result. You can also turn off step-by-step comments if you already understand the method and only need to check your own calculations, or hide the step-by-step solution entirely if you only need the final answer;
The second part contains buttons that let you modify the input table dimensions, clear individual coefficients or the entire input, and the main button with an equal sign that takes you to the solution screen. All these buttons are duplicated by keyboard shortcuts. Hover over a button to see the corresponding key in a tooltip. You can also use the arrow keys to move the cursor between input fields;
And the last part allows you to choose the number of digits after the decimal point for rounding non-integer results. A live preview shows how the rounded values will appear;

What is a matrix game (Game theory)?

A matrix game is a finite two-player zero-sum game where Player 1 chooses a row and Player 2 simultaneously chooses a column of a payoff matrix; Player 1 receives the corresponding entry and Player 2 pays it. The minimax theorem guarantees that every finite zero-sum game has a value — the amount Player 1 can guarantee regardless of Player 2's strategy — achieved either in pure strategies (a saddle point) or in mixed strategies (probability distributions over rows and columns).

How to find a solution in pure strategies (saddle point)?

For each row, find the row minimum (the worst payoff Player 1 can guarantee by choosing that row). For each column, find the column maximum (the worst Player 2 can guarantee by choosing that column). If the maximum of the row minima equals the minimum of the column maxima, a saddle point exists at that entry; the game value equals this common number and both players should use the corresponding pure strategy.

How to find a solution in mixed strategies?

When no saddle point exists, both players must randomize. The problem is converted to a linear program: shift the payoff matrix by a constant to ensure all entries are positive, then Player 1's optimal mixed strategy solves a standard LP whose variables are the (normalized) strategy probabilities. Our calculator solves the inner LP with the simplex method, running on the transposed payoff matrix (swapping player roles as in the Angular source), and recovers both Player 1 and Player 2 mixed strategies from the primal and dual solutions respectively.

Example of solving a matrix game

	B₁	B₂	B₃	min
A₁	2	4	3	2	t₁
A₂	3	1	4	1	t₂
A₃	4	3	2	2	t₃
max	4	4	4

Lower value (maximin)

α = 2

Upper value (minimax)

β = 4

α ≠ β — no saddle point. Solve the matrix game by reducing to a linear program.

F(x) = x₁+x₂+x₃ → min

F(x) = x₁+x₂+x₃+0x₄+0x₅+0x₆+Mx₇+Mx₈+Mx₉ → min

2x₁+3x₂+4x₃ ≥ 1

2x₁+3x₂+4x₃-x₄+x₇ = 1

4x₁+x₂+3x₃ ≥ 1

4x₁+x₂+3x₃-x₅+x₈ = 1

3x₁+4x₂+2x₃ ≥ 1

3x₁+4x₂+2x₃-x₆+x₉ = 1

The preliminary stage begins with the need to get rid of negative values in the right part of the restrictions. For what the corresponding restrictions are multiplied by -1. After this manipulation, the sign of inequality is reversed.

Next, you need to get rid of inequalities, for which we introduce compensating variables in the left-hand side of the inequalities. If an inequality of the form ≤, then the compensating variable has the sign +, if the inequality of the form ≥, then the compensating variable has the sign -. Compensating variables are included in the objective function of the problem with a zero coefficient.

Now in the constraint system it is necessary to find a sufficient number of basis variables. Each constraint must have one basis variable. The basic is a variable that has a coefficient of 1 with it and is found only in one constraint. If there are no basis variables in some restriction, then we add them artificially, and artificial variables enter the objective function with the coefficient -M if the objective function tends to max and M, if the objective function tends to min.

Description

Iteration 1

B	Cb	P	x₁↓	x₂	x₃	x₄	x₅	x₆	x₇	x₈	x₉	Q
B	Cb	P	1	1	1	0	0	0	M	M	M	Q
x₇	M	1	2	3	4	-1	0	0	1	0	0	12
x₈←	M	1	4	1	3	0	-1	0	0	1	0	14
x₉	M	1	3	4	2	0	0	-1	0	0	1	13
min		3M	9M-1	8M-1	9M-1	-M	-M	-M	0	0	0

Elements of the column basis (B)

Transfer to the table the basic elements that we identified in the preliminary stage:

B₁ = x₇;

B₂ = x₈;

B₃ = x₉;

Cb column items

Each cell of this column is equal to the coefficient, which corresponds to the base variable in the corresponding row.

Cb₁ = M;

Cb₂ = M;

Cb₃ = M;

Values of variable variables and column P

At this stage, no calculations are needed, just transfer the values from the preliminary stage to the corresponding table cells:

P₁ = 1;

P₂ = 1;

P₃ = 1;

x_1,1 = 2;

x_1,2 = 3;

x_1,3 = 4;

x_1,4 = -1;

x_1,5 = 0;

x_1,6 = 0;

x_1,7 = 1;

x_1,8 = 0;

x_1,9 = 0;

x_2,1 = 4;

x_2,2 = 1;

x_2,3 = 3;

x_2,4 = 0;

x_2,5 = -1;

x_2,6 = 0;

x_2,7 = 0;

x_2,8 = 1;

x_2,9 = 0;

x_3,1 = 3;

x_3,2 = 4;

x_3,3 = 2;

x_3,4 = 0;

x_3,5 = 0;

x_3,6 = -1;

x_3,7 = 0;

x_3,8 = 0;

x_3,9 = 1;

Objective function value

We calculate the value of the objective function by elementwise multiplying the column Cb by the column P, adding the results of the products.

Min_P = (Cb₁ * P₁) + (Cb₂ * P₂) + (Cb₃ * P₃) = (M * 1) + (M * 1) + (M * 1) = 3M;

Evaluated Control Variables

We calculate the estimates for each controlled variable, by element-wise multiplying the value from the variable column, by the value from the Cb column, summing up the results of the products, and subtracting the coefficient of the objective function from their sum, with this variable.

Min_x₁ = ((Cb₁ * x_1,1) + (Cb₂ * x_2,1) + (Cb₃ * x_3,1)) - k_x₁ = ((M * 2) + (M * 4) + (M * 3)) - 1 = 9M-1;

Min_x₂ = ((Cb₁ * x_1,2) + (Cb₂ * x_2,2) + (Cb₃ * x_3,2)) - k_x₂ = ((M * 3) + (M * 1) + (M * 4)) - 1 = 8M-1;

Min_x₃ = ((Cb₁ * x_1,3) + (Cb₂ * x_2,3) + (Cb₃ * x_3,3)) - k_x₃ = ((M * 4) + (M * 3) + (M * 2)) - 1 = 9M-1;

Min_x₄ = ((Cb₁ * x_1,4) + (Cb₂ * x_2,4) + (Cb₃ * x_3,4)) - k_x₄ = ((M * -1) + (M * 0) + (M * 0)) - 0 = -M;

Min_x₅ = ((Cb₁ * x_1,5) + (Cb₂ * x_2,5) + (Cb₃ * x_3,5)) - k_x₅ = ((M * 0) + (M * -1) + (M * 0)) - 0 = -M;

Min_x₆ = ((Cb₁ * x_1,6) + (Cb₂ * x_2,6) + (Cb₃ * x_3,6)) - k_x₆ = ((M * 0) + (M * 0) + (M * -1)) - 0 = -M;

Min_x₇ = ((Cb₁ * x_1,7) + (Cb₂ * x_2,7) + (Cb₃ * x_3,7)) - k_x₇ = ((M * 1) + (M * 0) + (M * 0)) - M = 0;

Min_x₈ = ((Cb₁ * x_1,8) + (Cb₂ * x_2,8) + (Cb₃ * x_3,8)) - k_x₈ = ((M * 0) + (M * 1) + (M * 0)) - M = 0;

Min_x₉ = ((Cb₁ * x_1,9) + (Cb₂ * x_2,9) + (Cb₃ * x_3,9)) - k_x₉ = ((M * 0) + (M * 0) + (M * 1)) - M = 0;

Q column items

Since there are positive values among the estimates of the controlled variables, the current table does not yet have an optimal solution. Therefore, in the basis we introduce the variable with the highest positive estimate.

The number of variables in the basis is always constant, so it is necessary to choose which variable to derive from the basis, for which we calculate Q.

The elements of the Q column are calculated by dividing the values from column P by the value from the column corresponding to the variable that is entered in the basis:

Q₁ = P₁x_1,1 = 12 = 12;

Q₂ = P₂x_2,1 = 14 = 14;

Q₃ = P₃x_3,1 = 13 = 13;

We deduce from the basis the variable with the least positive value of Q.

At the intersection of the line that corresponds to the variable that is derived from the basis, and the column that corresponds to the variable that is entered into the basis, is the resolving element.

This element will allow us to calculate the elements of the table of the next iteration.

Description

Iteration 2

B	Cb	P	x₁	x₂↓	x₃	x₄	x₅	x₆	x₇	x₈	x₉	Q
B	Cb	P	1	1	1	0	0	0	M	M	M	Q
x₇	M	12	0	212	212	-1	12	0	1	-12	0	15
x₁	1	14	1	14	34	0	-14	0	0	14	0	1
x₉←	M	14	0	314	-14	0	34	-1	0	-34	1	113
min		34M+14	0	534M-34	214M-14	-M	114M-14	-M	0	-214M+14	0

Elements of the column basis (B)

For the results of the calculations of the previous iteration, we remove the variable from the basis x₈ and put in her place x₁. All other cells remain unchanged.

Cb column items

Each cell of this column is equal to the coefficient, which corresponds to the base variable in the corresponding row.

Cb₁ = M;

Cb₂ = 1;

Cb₃ = M;

Values of variable variables and column P

(The data from the previous iteration is taken as the initial data)

Fill all cells with zeros corresponding to the variable that has just been entered into the basis:

(The resolution element remains unchanged)

x_1,1 = 0;

x_3,1 = 0;

We transfer the row with the resolving element from the previous table into the current table, elementwise dividing its values into the resolving element:

x_2,1 = x_2,1x_2,1 = 44 = 1;

x_2,2 = x_2,2x_2,1 = 14 = 14;

x_2,3 = x_2,3x_2,1 = 34 = 34;

x_2,4 = x_2,4x_2,1 = 04 = 0;

x_2,5 = x_2,5x_2,1 = -14 = -14;

x_2,6 = x_2,6x_2,1 = 04 = 0;

x_2,7 = x_2,7x_2,1 = 04 = 0;

x_2,8 = x_2,8x_2,1 = 14 = 14;

x_2,9 = x_2,9x_2,1 = 04 = 0;

P₂ = P₂x_2,1 = 14 = 14;

The remaining empty cells, except for the row of estimates and the column Q, are calculated using the rectangle method, relative to the resolving element:

x_1,2 = (x_1,2 * x_2,1) - (x_1,1 * x_2,2)x_2,1 = (3 * 4) - (2 * 1)4 = 212;

x_1,3 = (x_1,3 * x_2,1) - (x_1,1 * x_2,3)x_2,1 = (4 * 4) - (2 * 3)4 = 212;

x_1,4 = (x_1,4 * x_2,1) - (x_1,1 * x_2,4)x_2,1 = (-1 * 4) - (2 * 0)4 = -1;

x_1,5 = (x_1,5 * x_2,1) - (x_1,1 * x_2,5)x_2,1 = (0 * 4) - (2 * -1)4 = 12;

x_1,6 = (x_1,6 * x_2,1) - (x_1,1 * x_2,6)x_2,1 = (0 * 4) - (2 * 0)4 = 0;

x_1,7 = (x_1,7 * x_2,1) - (x_1,1 * x_2,7)x_2,1 = (1 * 4) - (2 * 0)4 = 1;

x_1,8 = (x_1,8 * x_2,1) - (x_1,1 * x_2,8)x_2,1 = (0 * 4) - (2 * 1)4 = -12;

x_1,9 = (x_1,9 * x_2,1) - (x_1,1 * x_2,9)x_2,1 = (0 * 4) - (2 * 0)4 = 0;

P₁ = (P₁ * x_2,1) - (x_1,1 * P₂)x_2,1 = (1 * 4) - (2 * 1)4 = 12;

x_3,2 = (x_3,2 * x_2,1) - (x_3,1 * x_2,2)x_2,1 = (4 * 4) - (3 * 1)4 = 314;

x_3,3 = (x_3,3 * x_2,1) - (x_3,1 * x_2,3)x_2,1 = (2 * 4) - (3 * 3)4 = -14;

x_3,4 = (x_3,4 * x_2,1) - (x_3,1 * x_2,4)x_2,1 = (0 * 4) - (3 * 0)4 = 0;

x_3,5 = (x_3,5 * x_2,1) - (x_3,1 * x_2,5)x_2,1 = (0 * 4) - (3 * -1)4 = 34;

x_3,6 = (x_3,6 * x_2,1) - (x_3,1 * x_2,6)x_2,1 = (-1 * 4) - (3 * 0)4 = -1;

x_3,7 = (x_3,7 * x_2,1) - (x_3,1 * x_2,7)x_2,1 = (0 * 4) - (3 * 0)4 = 0;

x_3,8 = (x_3,8 * x_2,1) - (x_3,1 * x_2,8)x_2,1 = (0 * 4) - (3 * 1)4 = -34;

x_3,9 = (x_3,9 * x_2,1) - (x_3,1 * x_2,9)x_2,1 = (1 * 4) - (3 * 0)4 = 1;

P₃ = (P₃ * x_2,1) - (x_3,1 * P₂)x_2,1 = (1 * 4) - (3 * 1)4 = 14;

Objective function value

We calculate the value of the objective function by elementwise multiplying the column Cb by the column P, adding the results of the products.

Min_P = (Cb₁ * P₁) + (Cb₂ * P₂) + (Cb₃ * P₃) = (M * 12) + (1 * 14) + (M * 14) = 34M+14;

Evaluated Control Variables

Min_x₁ = ((Cb₁ * x_1,1) + (Cb₂ * x_2,1) + (Cb₃ * x_3,1)) - k_x₁ = ((M * 0) + (1 * 1) + (M * 0)) - 1 = 0;

Min_x₂ = ((Cb₁ * x_1,2) + (Cb₂ * x_2,2) + (Cb₃ * x_3,2)) - k_x₂ = ((M * 212) + (1 * 14) + (M * 314)) - 1 = 534M-34;

Min_x₃ = ((Cb₁ * x_1,3) + (Cb₂ * x_2,3) + (Cb₃ * x_3,3)) - k_x₃ = ((M * 212) + (1 * 34) + (M * -14)) - 1 = 214M-14;

Min_x₄ = ((Cb₁ * x_1,4) + (Cb₂ * x_2,4) + (Cb₃ * x_3,4)) - k_x₄ = ((M * -1) + (1 * 0) + (M * 0)) - 0 = -M;

Min_x₅ = ((Cb₁ * x_1,5) + (Cb₂ * x_2,5) + (Cb₃ * x_3,5)) - k_x₅ = ((M * 12) + (1 * -14) + (M * 34)) - 0 = 114M-14;

Min_x₆ = ((Cb₁ * x_1,6) + (Cb₂ * x_2,6) + (Cb₃ * x_3,6)) - k_x₆ = ((M * 0) + (1 * 0) + (M * -1)) - 0 = -M;

Min_x₇ = ((Cb₁ * x_1,7) + (Cb₂ * x_2,7) + (Cb₃ * x_3,7)) - k_x₇ = ((M * 1) + (1 * 0) + (M * 0)) - M = 0;

Min_x₈ = ((Cb₁ * x_1,8) + (Cb₂ * x_2,8) + (Cb₃ * x_3,8)) - k_x₈ = ((M * -12) + (1 * 14) + (M * -34)) - M = -214M+14;

Min_x₉ = ((Cb₁ * x_1,9) + (Cb₂ * x_2,9) + (Cb₃ * x_3,9)) - k_x₉ = ((M * 0) + (1 * 0) + (M * 1)) - M = 0;

Q column items

The number of variables in the basis is always constant, so it is necessary to choose which variable to derive from the basis, for which we calculate Q.

The elements of the Q column are calculated by dividing the values from column P by the value from the column corresponding to the variable that is entered in the basis:

Q₁ = P₁x_1,2 = 12212 = 15;

Q₂ = P₂x_2,2 = 1414 = 1;

Q₃ = P₃x_3,2 = 14314 = 113;

We deduce from the basis the variable with the least positive value of Q.

This element will allow us to calculate the elements of the table of the next iteration.

Description

Iteration 3

B	Cb	P	x₁	x₂	x₃↓	x₄	x₅	x₆	x₇	x₈	x₉	Q
B	Cb	P	1	1	1	0	0	0	M	M	M	Q
x₇←	M	413	0	0	2913	-1	-113	1013	1	113	-1013	435
x₁	1	313	1	0	1013	0	-413	113	0	413	-113	310
x₂	1	113	0	1	-113	0	313	-413	0	-313	413	∞
min		413M+413	0	0	2913M-413	-M	-113M-113	1013M-313	0	-1213M+113	-11013M+313

Elements of the column basis (B)

For the results of the calculations of the previous iteration, we remove the variable from the basis x₉ and put in her place x₂. All other cells remain unchanged.

Cb column items

Each cell of this column is equal to the coefficient, which corresponds to the base variable in the corresponding row.

Cb₁ = M;

Cb₂ = 1;

Cb₃ = 1;

Values of variable variables and column P

(The data from the previous iteration is taken as the initial data)

Fill all cells with zeros corresponding to the variable that has just been entered into the basis:

(The resolution element remains unchanged)

x_1,2 = 0;

x_2,2 = 0;

We transfer the row with the resolving element from the previous table into the current table, elementwise dividing its values into the resolving element:

x_3,1 = x_3,1x_3,2 = 0314 = 0;

x_3,2 = x_3,2x_3,2 = 314314 = 1;

x_3,3 = x_3,3x_3,2 = -14314 = -113;

x_3,4 = x_3,4x_3,2 = 0314 = 0;

x_3,5 = x_3,5x_3,2 = 34314 = 313;

x_3,6 = x_3,6x_3,2 = -1314 = -413;

x_3,7 = x_3,7x_3,2 = 0314 = 0;

x_3,8 = x_3,8x_3,2 = -34314 = -313;

x_3,9 = x_3,9x_3,2 = 1314 = 413;

P₃ = P₃x_3,2 = 14314 = 113;

The remaining empty cells, except for the row of estimates and the column Q, are calculated using the rectangle method, relative to the resolving element:

x_1,1 = (x_1,1 * x_3,2) - (x_1,2 * x_3,1)x_3,2 = (0 * 314) - (212 * 0)314 = 0;

x_1,3 = (x_1,3 * x_3,2) - (x_1,2 * x_3,3)x_3,2 = (212 * 314) - (212 * -14)314 = 2913;

x_1,4 = (x_1,4 * x_3,2) - (x_1,2 * x_3,4)x_3,2 = (-1 * 314) - (212 * 0)314 = -1;

x_1,5 = (x_1,5 * x_3,2) - (x_1,2 * x_3,5)x_3,2 = (12 * 314) - (212 * 34)314 = -113;

x_1,6 = (x_1,6 * x_3,2) - (x_1,2 * x_3,6)x_3,2 = (0 * 314) - (212 * -1)314 = 1013;

x_1,7 = (x_1,7 * x_3,2) - (x_1,2 * x_3,7)x_3,2 = (1 * 314) - (212 * 0)314 = 1;

x_1,8 = (x_1,8 * x_3,2) - (x_1,2 * x_3,8)x_3,2 = (-12 * 314) - (212 * -34)314 = 113;

x_1,9 = (x_1,9 * x_3,2) - (x_1,2 * x_3,9)x_3,2 = (0 * 314) - (212 * 1)314 = -1013;

P₁ = (P₁ * x_3,2) - (x_1,2 * P₃)x_3,2 = (12 * 314) - (212 * 14)314 = 413;

x_2,1 = (x_2,1 * x_3,2) - (x_2,2 * x_3,1)x_3,2 = (1 * 314) - (14 * 0)314 = 1;

x_2,3 = (x_2,3 * x_3,2) - (x_2,2 * x_3,3)x_3,2 = (34 * 314) - (14 * -14)314 = 1013;

x_2,4 = (x_2,4 * x_3,2) - (x_2,2 * x_3,4)x_3,2 = (0 * 314) - (14 * 0)314 = 0;

x_2,5 = (x_2,5 * x_3,2) - (x_2,2 * x_3,5)x_3,2 = (-14 * 314) - (14 * 34)314 = -413;

x_2,6 = (x_2,6 * x_3,2) - (x_2,2 * x_3,6)x_3,2 = (0 * 314) - (14 * -1)314 = 113;

x_2,7 = (x_2,7 * x_3,2) - (x_2,2 * x_3,7)x_3,2 = (0 * 314) - (14 * 0)314 = 0;

x_2,8 = (x_2,8 * x_3,2) - (x_2,2 * x_3,8)x_3,2 = (14 * 314) - (14 * -34)314 = 413;

x_2,9 = (x_2,9 * x_3,2) - (x_2,2 * x_3,9)x_3,2 = (0 * 314) - (14 * 1)314 = -113;

P₂ = (P₂ * x_3,2) - (x_2,2 * P₃)x_3,2 = (14 * 314) - (14 * 14)314 = 313;

Objective function value

We calculate the value of the objective function by elementwise multiplying the column Cb by the column P, adding the results of the products.

Min_P = (Cb₁ * P₁) + (Cb₂ * P₂) + (Cb₃ * P₃) = (M * 413) + (1 * 313) + (1 * 113) = 413M+413;

Evaluated Control Variables

Min_x₁ = ((Cb₁ * x_1,1) + (Cb₂ * x_2,1) + (Cb₃ * x_3,1)) - k_x₁ = ((M * 0) + (1 * 1) + (1 * 0)) - 1 = 0;

Min_x₂ = ((Cb₁ * x_1,2) + (Cb₂ * x_2,2) + (Cb₃ * x_3,2)) - k_x₂ = ((M * 0) + (1 * 0) + (1 * 1)) - 1 = 0;

Min_x₃ = ((Cb₁ * x_1,3) + (Cb₂ * x_2,3) + (Cb₃ * x_3,3)) - k_x₃ = ((M * 2913) + (1 * 1013) + (1 * -113)) - 1 = 2913M-413;

Min_x₄ = ((Cb₁ * x_1,4) + (Cb₂ * x_2,4) + (Cb₃ * x_3,4)) - k_x₄ = ((M * -1) + (1 * 0) + (1 * 0)) - 0 = -M;

Min_x₅ = ((Cb₁ * x_1,5) + (Cb₂ * x_2,5) + (Cb₃ * x_3,5)) - k_x₅ = ((M * -113) + (1 * -413) + (1 * 313)) - 0 = -113M-113;

Min_x₆ = ((Cb₁ * x_1,6) + (Cb₂ * x_2,6) + (Cb₃ * x_3,6)) - k_x₆ = ((M * 1013) + (1 * 113) + (1 * -413)) - 0 = 1013M-313;

Min_x₇ = ((Cb₁ * x_1,7) + (Cb₂ * x_2,7) + (Cb₃ * x_3,7)) - k_x₇ = ((M * 1) + (1 * 0) + (1 * 0)) - M = 0;

Min_x₈ = ((Cb₁ * x_1,8) + (Cb₂ * x_2,8) + (Cb₃ * x_3,8)) - k_x₈ = ((M * 113) + (1 * 413) + (1 * -313)) - M = -1213M+113;

Min_x₉ = ((Cb₁ * x_1,9) + (Cb₂ * x_2,9) + (Cb₃ * x_3,9)) - k_x₉ = ((M * -1013) + (1 * -113) + (1 * 413)) - M = -11013M+313;

Q column items

The number of variables in the basis is always constant, so it is necessary to choose which variable to derive from the basis, for which we calculate Q.

The elements of the Q column are calculated by dividing the values from column P by the value from the column corresponding to the variable that is entered in the basis:

Q₁ = P₁x_1,3 = 4132913 = 435;

Q₂ = P₂x_2,3 = 3131013 = 310;

Q₃ = P₃x_3,3 = 113-113 = ∞;

We deduce from the basis the variable with the least positive value of Q.

This element will allow us to calculate the elements of the table of the next iteration.

Description

Iteration 4

B	Cb	P	x₁	x₂	x₃	x₄	x₅	x₆	x₇	x₈	x₉	Q
B	Cb	P	1	1	1	0	0	0	M	M	M	Q
x₃	1	435	0	0	1	-1335	-135	27	1335	135	-27
x₁	1	17	1	0	0	27	-27	-17	-27	27	17
x₂	1	335	0	1	0	-135	835	-27	135	-835	27
min		1235	0	0	0	-435	-335	-17	-M+435	-M+335	-M+17

Elements of the column basis (B)

For the results of the calculations of the previous iteration, we remove the variable from the basis x₇ and put in her place x₃. All other cells remain unchanged.

Cb column items

Each cell of this column is equal to the coefficient, which corresponds to the base variable in the corresponding row.

Cb₁ = 1;

Cb₂ = 1;

Cb₃ = 1;

Values of variable variables and column P

(The data from the previous iteration is taken as the initial data)

Fill all cells with zeros corresponding to the variable that has just been entered into the basis:

(The resolution element remains unchanged)

x_2,3 = 0;

x_3,3 = 0;

We transfer the row with the resolving element from the previous table into the current table, elementwise dividing its values into the resolving element:

x_1,1 = x_1,1x_1,3 = 02913 = 0;

x_1,2 = x_1,2x_1,3 = 02913 = 0;

x_1,3 = x_1,3x_1,3 = 29132913 = 1;

x_1,4 = x_1,4x_1,3 = -12913 = -1335;

x_1,5 = x_1,5x_1,3 = -1132913 = -135;

x_1,6 = x_1,6x_1,3 = 10132913 = 27;

x_1,7 = x_1,7x_1,3 = 12913 = 1335;

x_1,8 = x_1,8x_1,3 = 1132913 = 135;

x_1,9 = x_1,9x_1,3 = -10132913 = -27;

P₁ = P₁x_1,3 = 4132913 = 435;

The remaining empty cells, except for the row of estimates and the column Q, are calculated using the rectangle method, relative to the resolving element:

x_2,1 = (x_2,1 * x_1,3) - (x_2,3 * x_1,1)x_1,3 = (1 * 2913) - (1013 * 0)2913 = 1;

x_2,2 = (x_2,2 * x_1,3) - (x_2,3 * x_1,2)x_1,3 = (0 * 2913) - (1013 * 0)2913 = 0;

x_2,4 = (x_2,4 * x_1,3) - (x_2,3 * x_1,4)x_1,3 = (0 * 2913) - (1013 * -1)2913 = 27;

x_2,5 = (x_2,5 * x_1,3) - (x_2,3 * x_1,5)x_1,3 = (-413 * 2913) - (1013 * -113)2913 = -27;

x_2,6 = (x_2,6 * x_1,3) - (x_2,3 * x_1,6)x_1,3 = (113 * 2913) - (1013 * 1013)2913 = -17;

x_2,7 = (x_2,7 * x_1,3) - (x_2,3 * x_1,7)x_1,3 = (0 * 2913) - (1013 * 1)2913 = -27;

x_2,8 = (x_2,8 * x_1,3) - (x_2,3 * x_1,8)x_1,3 = (413 * 2913) - (1013 * 113)2913 = 27;

x_2,9 = (x_2,9 * x_1,3) - (x_2,3 * x_1,9)x_1,3 = (-113 * 2913) - (1013 * -1013)2913 = 17;

P₂ = (P₂ * x_1,3) - (x_2,3 * P₁)x_1,3 = (313 * 2913) - (1013 * 413)2913 = 17;

x_3,1 = (x_3,1 * x_1,3) - (x_3,3 * x_1,1)x_1,3 = (0 * 2913) - (-113 * 0)2913 = 0;

x_3,2 = (x_3,2 * x_1,3) - (x_3,3 * x_1,2)x_1,3 = (1 * 2913) - (-113 * 0)2913 = 1;

x_3,4 = (x_3,4 * x_1,3) - (x_3,3 * x_1,4)x_1,3 = (0 * 2913) - (-113 * -1)2913 = -135;

x_3,5 = (x_3,5 * x_1,3) - (x_3,3 * x_1,5)x_1,3 = (313 * 2913) - (-113 * -113)2913 = 835;

x_3,6 = (x_3,6 * x_1,3) - (x_3,3 * x_1,6)x_1,3 = (-413 * 2913) - (-113 * 1013)2913 = -27;

x_3,7 = (x_3,7 * x_1,3) - (x_3,3 * x_1,7)x_1,3 = (0 * 2913) - (-113 * 1)2913 = 135;

x_3,8 = (x_3,8 * x_1,3) - (x_3,3 * x_1,8)x_1,3 = (-313 * 2913) - (-113 * 113)2913 = -835;

x_3,9 = (x_3,9 * x_1,3) - (x_3,3 * x_1,9)x_1,3 = (413 * 2913) - (-113 * -1013)2913 = 27;

P₃ = (P₃ * x_1,3) - (x_3,3 * P₁)x_1,3 = (113 * 2913) - (-113 * 413)2913 = 335;

Objective function value

We calculate the value of the objective function by elementwise multiplying the column Cb by the column P, adding the results of the products.

Min_P = (Cb₁ * P₁) + (Cb₂ * P₂) + (Cb₃ * P₃) = (1 * 435) + (1 * 17) + (1 * 335) = 1235;

Evaluated Control Variables

Min_x₁ = ((Cb₁ * x_1,1) + (Cb₂ * x_2,1) + (Cb₃ * x_3,1)) - k_x₁ = ((1 * 0) + (1 * 1) + (1 * 0)) - 1 = 0;

Min_x₂ = ((Cb₁ * x_1,2) + (Cb₂ * x_2,2) + (Cb₃ * x_3,2)) - k_x₂ = ((1 * 0) + (1 * 0) + (1 * 1)) - 1 = 0;

Min_x₃ = ((Cb₁ * x_1,3) + (Cb₂ * x_2,3) + (Cb₃ * x_3,3)) - k_x₃ = ((1 * 1) + (1 * 0) + (1 * 0)) - 1 = 0;

Min_x₄ = ((Cb₁ * x_1,4) + (Cb₂ * x_2,4) + (Cb₃ * x_3,4)) - k_x₄ = ((1 * -1335) + (1 * 27) + (1 * -135)) - 0 = -435;

Min_x₅ = ((Cb₁ * x_1,5) + (Cb₂ * x_2,5) + (Cb₃ * x_3,5)) - k_x₅ = ((1 * -135) + (1 * -27) + (1 * 835)) - 0 = -335;

Min_x₆ = ((Cb₁ * x_1,6) + (Cb₂ * x_2,6) + (Cb₃ * x_3,6)) - k_x₆ = ((1 * 27) + (1 * -17) + (1 * -27)) - 0 = -17;

Min_x₇ = ((Cb₁ * x_1,7) + (Cb₂ * x_2,7) + (Cb₃ * x_3,7)) - k_x₇ = ((1 * 1335) + (1 * -27) + (1 * 135)) - M = -M+435;

Min_x₈ = ((Cb₁ * x_1,8) + (Cb₂ * x_2,8) + (Cb₃ * x_3,8)) - k_x₈ = ((1 * 135) + (1 * 27) + (1 * -835)) - M = -M+335;

Min_x₉ = ((Cb₁ * x_1,9) + (Cb₂ * x_2,9) + (Cb₃ * x_3,9)) - k_x₉ = ((1 * -27) + (1 * 17) + (1 * 27)) - M = -M+17;

Answer

Since there are no positive values among the estimates of the controlled variables, the current table has an optimal solution.

The value of the objective function:

F* = 1235;

The variables that are present in the basis are equal to the corresponding cells of the column P, all other variables are equal to zero:

x₁ = 17;

x₂ = 335;

x₃ = 435;

Description

Answer

v → max min

F* = 1235

X* = (17; 335; 435)

Player I strategies3Player II strategies3TypeMixed strategy

Frequently asked questions

What does this game theory calculator solve?

It solves two-player zero-sum games from a payoff matrix, finding the value of the game and each player's optimal strategy — a pure strategy at a saddle point, or a mixed strategy when no saddle point exists.

What is a saddle point?

A saddle point is an entry that is the minimum of its row and the maximum of its column. When one exists, the game has a solution in pure strategies and the saddle value is the game's value.

How are mixed strategies computed?

When there is no saddle point, the game is converted into a pair of linear programs and solved with the simplex method. The resulting probabilities tell each player how often to play each option to guarantee the game's value on average.

What does the "value of the game" represent?

It is the average payoff the first player can secure — and the second player can hold them to — under optimal play. A positive value favors the row player, a negative value favors the column player.

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x_i ↓	- we enter a variable into the basis;
x_i←	- print the variable from the base;
x_i	- permissive element;
x_i	- basic element;
B	- basis;
Cb	- coefficient at the base variable;
P	- plan;

About Game theory calculator

What is a matrix game (Game theory)?

How to find a solution in pure strategies (saddle point)?

How to find a solution in mixed strategies?

Example of solving a matrix game

Frequently asked questions

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