# DOMINANCE PROPERTY - Quantitative Techniques for management

## Dominance property in Operation Research

The principle of dominance states that if one strategy of a player dominates over the other strategy in all conditions then the later strategy can be ignored. A strategy dominates over the other only if it is preferable over other in all conditions. The concept of dominance is especially useful for the evaluation of two-person zero-sum games where a saddle point does not exist.

In case of pay-off matrices larger than 2 × 2 size, the dominance property can be used to reduce the size of the pay-off matrix by eliminating the strategies that would never be selected.

## Dominance Principle in Game Theory problems

The dominance principle in game theory problems are explained below

Example : Solve the game given below in Table after reducing it to 2 × 2 game:

Game Problem Solution: Reduce the matrix by using the dominance property. In the given matrix for player A, all the elements in Row 3 are less than the adjacent elements of Row 2. Strategy 3 will not be selected by player A, because it gives less profit for player A. Row 3 is dominated by Row 2. Hence delete Row 3, as shown in table.

Reduced the Matrix by Using Dominance Property For Player B, Column 3 is dominated by column 1 (Here the dominance is opposite because Player B selects the minimum loss). Hence delete Column 3. We get the reduced 2 × 2 matrix as shown below in table.

Reduced 2 × 2 Matrix Now, solve the 2 × 2 matrix, using the maximin criteria as shown below in table.

Maximin Procedure Therefore, there is no saddle point and the game has a mixed strategy. Applying the probability formula,

p1 =
2-6
(1+2)-(7+6)
=
-43-13
=
4
10
=
2
5
q1 =
2-7
(1+2)-(7+6)
=
-53-13
=
5
10
=
1
2
q2 =1- q1 =1-
1
2
=
1
2
Value of the game, v =
(1*2)-(7*6)
(1+2)-(7+6)
=
2-42
3-13
=
40
10
= 4

The optimum strategies are shown in table

Optimum Strategies Example : Is the following two-person zero-sum game stable? Solve the game given below in table.

Two-person Zero-sum Game Problem Solution: Solve the given matrix using the maximin criteria as shown in table.

Table 14.25: Maximin Procedure Therefore, there is no saddle point and hence it has a mixed strategy.

The pay-off matrix is reduced to 2×2 size using dominance property. Compare the rows to find the row which dominates other row. Here for Player A, Row 1 is dominated by Row 3 (or row 1 gives the minimum profit for Player A), hence delete Row 1. The matrix is reduced as shown in table.

Use Dominance Property to Reduce Matrix (Deleted Row 1) When comparing column wise, column 2 is dominated by column 4. For Player B, the minimum profit column is column 2, hence delete column 2. The matrix is further reduced as shown in table.

Matrix Further Reduced to 3×3 (2 Deleted Column) Now, Row 2 is dominated by Row 3, hence delete Row 2, as shown in table.

Reduced Matrix (Row 2 Deleted) Now, as when comparing rows and columns, no column or row dominates the other. Since there is a tie while comparing the rows or columns, take the average of any two rows and compare. We have the following three combinations of matrices as shown in table.

Matrix Combinations When comparing column 1 and the average of column 3 and column 4, column 1 is dominated by the average of column 3 and 4. Hence delete column 1. Finally, we get the 2 × 2 matrix as shown in table.

2×2 Matrix After Deleting Column 1 The strategy for the arrived matrix is a mixed strategy; using probability formula, we find p1, p2 and q1, q2.

p1 =
4-(-1)
(15+4)-(1+(-1))
=
5
19
p2= 1- =
5
9
=
14
9
q1
4-1
19
=
3
19
q2 = 1-
3
19
=
16
19
Value of the game, v =
(15*4)-(1*(-1))
(15+4)-(1+(-1))
=
60+1
19
=
61
19

The optimum mixed strategies are given below in table.

Optimum Mixed Strategies Quantitative Techniques for management Topics