Kelsey McKee
1.
1’ 2 3
4 1’
6 9* 4’ 5 4’
5 2’ 6
5 5
0’ 0’ 1
1 4
1 1 4
3 2
9* 8 7*’ 8* 7*’
*Max in column ‘ min in row
2
6
7*
2
4
7*’
6
5
6
0’
0’
8*
*’ saddle point or equilibrium
Saddle point, value of game=7
Optimal strategy for the row player: (0,0,0,0,0,1)^T
Optimal strategy for the column player: (0,0,1,0,0,0,0)
2.
5 2 3 1 2 4 5
3 1 4 4 2 4 4
3 3 2 5 2 3 1
4 1 3 5 2 3 2
4 1 2 4 2 1 3
After domination C1, C3, C6, and R5 can be eliminated
Leaving this matrix:
2
1’
3*
1’
2’
1’
4
5*
5*
9/2
2*
2*
2*
2*
2*’
5*
4
1’
2
5/2
The mixed strategy for the row player is (R2+R3)/2
*max in column ‘min in row
Saddle point, value of game=2
Optimal strategy for the row player: (0,1/2 ,1/2, 0)^T
Optimal strategy for the column player: (0,0,1,0)
[ In the terms of the original problem,
Optimal Strategy for Row Player: (0,1/2,1/2,0.0)^T
Optimal Strategy for Column Player: (0,0,0,0,1,0,0)j-- ml5078@psu.edu]
3.
2 0
2 -2
0 -2 0 5
4 2
3 0
After domination C1, C3, and R1 can be eliminated
Leaving the matrix:
-2
5
2
0
Row player’s mixed strategy: (2R2+7R3)/9
Column player’s mixed strategy: (5C2+4C4)/9
Value of game=10/9
Optimal strategy for row player: (0, 2/9, 7/9)^T
Optimal strategy for the column player: (0, 5/9, 0, 4/9)
4. x+ty=3
4x+y=6t
x=3-ty
4(3-ty)+y=6t
y=(6t-12)/(1-4t)
substitute back into x=3-ty
x=3-t((6t-12)/(1-4t))
x=(3-6t^2)/(1-4t)
[This works when t is not ¼. The answer is wrong when t = ¼. If t = ¼, the system has no
solutions. –LV]
5. mean=29/12
Midrange=4
Central v
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From:
kmm5230@psu.edu
Subject: slight change to answers
Date: December 4, 2008 4:23:05 PM EST
I need to make a slight change to the answers I submitted
For #2 An optimal strategy for row player is: (0, 1/2 , 1/2 ,0,0)^T
An optimal strategy for column player is: (0,0,0,0,1,0,0)
In my previous answer I did not include enough zeros is the optimal strategies
alue=0