A strategy, say s, can also be dominated if it is inferior to a convex linear combination of several other pure strategies.
Method
of convex linear combination
A strategy, say s, can also be
dominated if it is inferior to a convex linear combination of several other
pure strategies. In this case if the domination is strict, then the strategy s
can be deleted. If strategy s dominates the convex linear combination of some
other pure strategies, then one of the pure strategies involved in the
combination may be deleted. The domination will be decided as per the above
rules. Let us consider an example to illustrate this case.
Problem 2
Solve
the game with the following pay-off matrix for firm A
Solution First
consider the minimum of each row
Maximum
of {-2, 0, -6, -3, -1} = 0 Next consider the maximum of each
column.
Minimum
of { 4, 8, 6, 8, 6}= 4 Hence, Maximum of {row minima} ≠
minimum of {column maxima} So we see that there is no saddle
point. Compare the second row with the fifth row. Each element in the second
row exceeds the corresponding element in the fifth row. Therefore, A_{2} dominates A_{5} . The choice is for firm A. It
will retain strategy A_{2} and give up strategy A_{5} . Therefore the game reduces to the following.
Compare the second and fourth rows. We see that A_{2} dominates A_{4} . So, firm A will retain the
strategy A_{2} and give up the strategy A_{4} . Thus the game reduces to the following:
Compare the first and fifth columns. It is observed
that B1 dominates B5. The choice is for firm B. It will retain the strategy B_{1} and give up the strategy B_{5} . Thus the game reduces to the
following
Compare the first and fourth columns. We notice
that B1 dominates B4. So firm B will discard the strategy B_{4} and retain the strategy B_{1} . Thus the game reduces to the
following:
For this
reduced game, we check that there is no saddle point. Now none of the pure strategies of firms A and B is
inferior to any of their other strategies. But, we observe that convex linear
combination of the strategies B_{2} and B_{3} dominates B_{1} , i.e. the averages of payoffs
due to strategies B_{2} and B_{3} ,
dominate B_{1} . Thus B_{1} may be omitted from consideration. So we have the reduced matrix
The value
of the game = 3. Thus the game is favourable to firm A. Problem 3 For the game with the following pay-off matrix, determine the saddle
point
The
choice is with the player B. He has to choose between strategies II and III. He will lose more in strategy III than in strategy II,
irrespective of what strategy is followed by A. So he will drop strategy III
and retain strategy II. Now the given game
reduces to the following game.
Maximum of row minimum = Minimum
of column maximum = 0. So, a saddle point exists for the
given game and the value of the game is 0. Interpretation No player gains and no player loses. i.e., The game is not favourable to
any player. i.e. It is a fair game.
Since Maximum of { Row Minima} and Minimum of { Column Maxima } are
different, it follows that the given game has no saddle point.
Denote the strategies of player A by A_{1} , A_{2} , A_{3} . Denote the strategies of
player B by B_{1} , B_{2} , B_{3} . Compare the first and third
columns of the given matrix.
The pay-offs in B_{3} are greater than or equal to the corresponding pay-offs in B_{1} . The player B has to make a choice between his strategies 1 and 3. He
will lose more if he follows strategy 3 rather than strategy 1. Therefore he
will give up strategy 3 and retain strategy 1. Consequently, the given game is
transformed into the following game
The pay-offs in A_{1} are greater than or equal to the corresponding pay-offs in A_{3} . The player A has to make a choice between his strategies 1 and 3. He
will gain more if he follows strategy 1 rather than strategy 3. Therefore he
will retain strategy 1 and give up strategy 3. Now the given game is
transformed into the following game.
Maximum {row minima} and Minimum {column maxima }
are not equal Therefore, the reduced game has no saddle point. So, it is a
mixed game
The probability that player A will use his first strategy is p. This is
calculated as
The probability that player B will use his first strategy is r. This is
calculated asOut of 3 trials, player A will
use strategy 1 once and strategy 2 once. Out of 4 trials, player B will use
strategy 1 thrice and strategy 2 once. The game is favourable to player A. Problem 5 Dividing a game into sub-games Solve the game with the following
pay-off matrix.
We see that Maximum of { row
minima} ≠ Minimum of { column maxima}. So the game has no saddle point. Hence it
is a mixed game. Compare the first and third columns
We assert that Player B will retain the first strategy and give up the
third strategy. We get the following reduced matrix.
We check that it is a game with
no saddle point. Sub games Let us consider the 2x2 sub
games. They are:
The value of 0 will be preferred
by the player A. For this value, the first and third strategies of A correspond
while the first and second strategies of the player B correspond to the value 0
of the game. So it is a fair game.
Tags : Operations Management - Game Theory, Goal Programming & Queuing Theory
Last 30 days 229 views