i)
(B)
5 10 20 Row Min.
5 -5 10 20 -5
(A) 10 5 -10 -10 -10
20 5 -20 -20 -20
Column max 5 10 20
(A) 2 4 Column max
8 9 (minimax)
ii)Here maximin of -5 is not equal to minimax of 5. Hence, no saddle point.
iii) The last column is dominated by second columns - we delete it
and the last row is dominated by second row - we delete it
Now consider a matrix
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c}\n & 5 & 10 & oddmeters \\\\ \\hline\n 5 & -5 & -10 & 15 \\\\\n \\hdashline\n 10 & 5 & -10 & 15\\\\\n\\hdashline\n oddmeters & 10 & 20\n\\end{array}"
p1 = p2 = 15/(15+15) = 1/2
q2 = 10/(10+20) = 1/3
q1 = 20/(10+20) = 2/3
Value of game = (a11 * a22 - a21 * a12)/[(a11+a22)-(a21+a12)] = ((-5)*(-10) - 5*10)/[(-5+(-10)) - (5+10)] = 0
Comments
Leave a comment