Math 484.1.  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

 1 3 5 4 6 6 2 0 2 0

is  (A) 2.5, (B) 8/3,  (C)   2.6,  (D) none of the above

Solution. The last two columns go by domination:

 1 3 5 6 2 0

By graphical method, we find the equilibrium

(p, q) = ([2, 1]T/3, [1, 5, 0]/6) with the value of game 8/3:

 1 3 5 8/3 6 2 0 8/3 8/3 8/3 10/3 8/3*'

where p and q are added as the last row and the last column. (B)

37.  Consider the matrix game with payoff matrix

0  -2   1 1

2   0  -3 1

-1  3   0 1

An optimal strategy for the column player is

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

Solution. The last column goes by domination:

0  -2   1

2  0  -3

-1 3   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

0 and 1  using (A) and  (B)  resp.  So (B) is not optimal.  (A)

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

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

(A) (3,1), (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   7

1  3  8

1 -6   0

hence  ATA = diag(4, 50) and     ATb = [22, 45] T,  so the system   for  the  least-squares solution (x, y)

is  4x = 22, 50y = 45. (D)

39.     |2x| +  |x-1| +  |x - 3| +  |3x- 5| ->  min.

The optimal solution is  (A) 0, (B)  3 (C)   5/3, (D) none of the above.

Solution. The optimal solution is  the LAD fit  x  for  the numbers 0, 0, 1,3, 5/3, 5/3, 5/3, which is the median 5/3. (C)

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

The optimal solution is (A) 1, (B) 1.5,  (C)   1.8, (D) none of the above.

Solution.  For optimal x, we have  0 ≤ x ≤ 5 (otherwise, we can decrease all 4 terms).

It is clear that max( |2x|,   |x-1| ,   |x - 3| ,  3|x- 5| ) = max( |2x|,   3|x- 5| ).

If these two terms are not equal, we can decrease the bigger one.

So  2x = 3(5-x) for optimal  x, hence  x = 3. (D)

41.  The value of the matrix game  with payoff matrix

 1 3 5 -4 6 5 -1 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.

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

(C).

42.   Consider the matrix game with payoff matrix

0  -2   1 -1   9

2   0  -3  1  -2

-1  3    0 1    0

An optimal strategy for the column player is

(A) [3, 1, 2, 0]/6,  (B)    [0, 1, 2, 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  [-1, -5,4]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 is this row. (D).

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

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

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

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