Math 484.2.  April 25, 2014.   Midterm 3.
10 problems,5 pts each.   Write your name here ____________________________

Return the scantron and this  page with both answers and details

36.  The value of the matrix game  with payoff matrix

2   4  5   4

7   3  0   4

6   3  0   3

is  (A) 3.5, (B) 8/3,  (C)   2.6,  (D) none of the above.

Solution. The last row and column go by domination:

2   4  5

7   3  0

Now we can solve the game by graphical method:

c1   c2   c3            (c1+c3)/2

r1                     2      4     5              3.5

r2                     7      3     0              3.5

0.7r1+0.3r2    3.5   3.7   3.5            3.5*'       (A)

37.  Consider the matrix game with payoff matrix

0  -1    2     2

1   0   -1     3

-2  1    0     4

An optimal strategy for the column player is

(A) [1, 1, 1, 1]/4, (B)   [1,2, 1, 0]/4 ,    (C)  [2, 1, 1, 0]    (D) none of the above.

Solution. The last column goes by domination:

0  -1    2

1   0   -1

-2  1    0

Now the game is symmetric, so its value is 0. (C) is not a mixed strategy.  In the worst case, the column player pays

3/4 and 0  using (A) and  (B)  resp.     (B)

38.   The least-squares solution (x, y)  for the system

x+y = 7,  x + 2y = 8,  x+3y =  8, x - 6y = 0    is

(A) (5.75, 9.4), (B) (6.5, 0.5),  (C)  (19/3, 1/2), (D) none of the above.

Solution. In matrix form AX = b, we have  the augmented matrix [A, b] =

1   1   7

1   2   8

1   3  8

1 -6   0

hence  ATA = diag(4, 50) and     ATb = [23, 47] T,  so the system   for  the  least-squares solution (x, y)   is  4x = 23, 50y = 47. (D)

39.     |0.2x| +  |x-1| +  |x - 3| +  |0.5x- 0.5| ->  min.

The optimal solution is  (A) 0, (B)  0 ≤ x ≤ 1,  (C)   1, (D) 2

Solution. We multiply the objective function by 10. Now it is clear that we are looking for the LAD fit  for the following

27 numbers: 0,0, 1 (fifteen times), 3 (ten times). The median is 1. (C)

40.   max( |2x|,   |x-1| ,   |x - 3| ,   |5x- 5| ) ->  min.

The optimal solution is (A)  0.9 , (B) 1.5,  (C)  1,  (D) 5/3.

Solution.  For optimal x, we have  0 ≤ x ≤ 3 (otherwise, we can decrease all 4 terms).

Since   |5x- 5|   |x- 1|  we have   max( |2x|,   |x-1| ,   |x - 3| ,   |5x- 5| ) -= max( |2x|,    |x - 3| ,   |5x- 5| ) .

Near  x = 1, the  objective function  f  = max( |2x|,   |x - 3| ). Moreover   f = 3 -x  on the left of 1 and f = 2x

on the right side. (C)

41.  The value of the matrix game  with payoff matrix

 1 3 5 4 6 5 4 6 2 0 2 0 -3 0 1 2 0 1 2 0 0

is  (A) 2.5, (B) 8/3,  (C)  0,  (D) none of the above

Solution.  Columns 3, 4, and    5 are dominated by the last column. Row 3 is dominated by Row 1.

c1 c2 c6 c7

r1  1   3   5  4

r2  6   2  -3  0

Now we can solve the game by graphical method:

c1  c2     c6   c7     (8c1 + 5c6)/13

r1               1     3       5    4            33/13*

r2               6     2      -3    0            33/13*

(9r1+4r2)/13   33/13 35/13  33/13 36/13             33/13*’

(D).

42.   Consider the matrix game with payoff matrix

0  -2   1 -1   9

2   0  -3  1  -2

0   1   3   0  1

An optimal strategy for the column player is

(A) [3, 1, 2, 0]/6,  (B)    [0, 1, 1, 1, 0]/4 ,    (C)  [2, 0. 1, 1, 1],    (D) none of the above.

Solution.

(A) and (C) are not mixed strategies.  The payoffs for (B) are  [-2, -2,4]T /4. If (B) is optimal, the value of game is 1

and the last row is the only optimal

strategy for the row player. But 1 is not the minimum in this row.

(D).

43. The central value of the numbers 0,-1,-2, 1, 2, 3, 0, 0   is

(A)0.5, (B) 0, (C) 3/2, (D) none of the above.

44. The midrange of the numbers 0,-1,-2, 1, 2, 3, 0, 0    is

(A) 0.5, (B) 0, (C) 3/7, (D) none of the above.