Home > Operation Research calculators > Game Theory >> Saddle Point example

6. Arithmetic method example ( Enter your problem )
  1. Method & Example-1
  2. Example-2
Other related methods
  1. Saddle point
  2. Dominance method
  3. Oddment method
  4. Algebraic method
  5. Calculus method
  6. Arithmetic method
  7. Matrix method
  8. 2Xn Games
  9. Graphical method
  10. Linear programming method
  11. Bimatrix method

5. Calculus method
(Previous method)
2. Example-2
(Next example)

1. Method & Example-1





Method
arithmetic method Steps (Rule)
Step-1: Find the difference between the two values of Row-1 and put this value against the Row-2, ignore the sign.
Step-2: Find the difference between the two values of Row-2 and put this value against the Row-1, ignore the sign.
Step-3: Find the difference between the two values of Column-1 and put this value against the Column-2, ignore the sign.
Step-4: Find the difference between the two values of Column-2 and put this value against the Column-1, ignore the sign.
Step-5: Find probabilities of each by dividing their sum
Step-6: Find value of the game by algebraic method.

Example-1
Find Solution of game theory problem using arithmetic method
Player A\Player BB1B2B3
A1105-2
A2131215
A3161410


Solution:
1. Saddle point testing
Players
Player `B`
`B_1``B_2``B_3`
Player `A``A_1` 10  5  -2 
`A_2` 13  12  15 
`A_3` 16  14  10 


We apply the maximin (minimax) principle to analyze the game.

Player `B`
`B_1``B_2``B_3`Row
Minimum
Player `A``A_1` 10  5  -2 `-2`
`A_2` 13  [12]  15 `[12]`
`A_3` 16  (14)  10 `10`
Column
Maximum
`16``(14)``15`


Select minimum from the maximum of columns
Column MiniMax = (14)

Select maximum from the minimum of rows
Row MaxiMin = [12]

Here, Column MiniMax `!=` Row MaxiMin

`:.` This game has no saddle point.



2. Dominance rule to reduce the size of the payoff matrix
Using dominance property
Player `B`
`B_1``B_2``B_3`
Player `A``A_1` 10  5  -2 
`A_2` 13  12  15 
`A_3` 16  14  10 


row-1 `<=` row-3, so remove row-1

Player `B`
`B_1``B_2``B_3`
Player `A``A_2` 13  12  15 
`A_3` 16  14  10 


column-1 `>=` column-2, so remove column-1

Player `B`
`B_2``B_3`
Player `A``A_2` 12  15 
`A_3` 14  10 




Player `B`
`B_2``B_3`
Player `A``A_2` 12  15 
`A_3` 14  10 


Using arithemetic method to get optimal mixed strategies for both the firms.
Player `B`
`B_2``B_3`
Player `A``A_2` 12  15 `|14-10|=4` `:. p_1=(4)/(3+4)=4/7`
`A_3` 14  10 `|12-15|=3` `:. p_2=(3)/(3+4)=3/7`
`|15-10|=5`
`:. q_1=(5)/(2+5)=5/7`
`|12-14|=2`
`:. q_2=(2)/(2+5)=2/7`


1. Find absolute difference between the two values in the first row and put it against second row of the matrix
`|12-15|=3`

2. Find absolute difference between the two values in the second row and put it against first row of the matrix
`|14-10|=4`

`:. p_1=(4)/(3+4)=4/7`

`:. p_2=(3)/(3+4)=3/7`


3. Find absolute difference between the two values in the first column and put it against second column of the matrix
`|12-14|=2`

4. Find absolute difference between the two values in the second column and put it against first column of the matrix
`|15-10|=5`

`:. q_1=(5)/(2+5)=5/7`

`:. q_2=(2)/(2+5)=2/7`


Hence, firm `A` should adopt strategy `A_2` and `A_3` with 57% of time and 43% of time respectively.

Similarly, firm `B` should adopt strategy `B_2` and `B_3` with 71% of time and 29% of time respectively.


Expected gain of Firm A
`(1)` `12 xx 4/7+14 xx 3/7 = 90/7 ,` Firm `B` adopt `B_2`

`(2)` `15 xx 4/7+10 xx 3/7 = 90/7 ,` Firm `B` adopt `B_3`


Expected loss of Firm B
`(1)` `12 xx 5/7+15 xx 2/7 = 90/7 ,` Firm `A` adopt `A_2`

`(2)` `14 xx 5/7+10 xx 2/7 = 90/7 ,` Firm `A` adopt `A_3`


This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here



5. Calculus method
(Previous method)
2. Example-2
(Next example)





Share this solution or page with your friends.


 
Copyright © 2023. All rights reserved. Terms, Privacy
 
 

.