Math 484.002.   Dec 2, 2016..   Midterm 3.  
10 problems, 5 pts each.   Write your name here __________Dr. V_____  

Return the scantron with answers and this  page with both answers and details (how you got your answers). Do not talk with other students in class even after finishing your midterm.

If you do not agree with the given answers, explain why here and choose (E) at scantron.


36.  The value of the matrix game  with payoff matrix


1

5

2

1

5

5

1

2

6

1

is  (A) 2.5, (B) 28/9,  (C)   2.  (D) 25/9.


Solution. The last two columns go by domination  without changing the value:


1

5

2

5

1

2


By graphical method, we find an equilibrium

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


1

5

2

2*

5

1

2

2*

3

3

2’

2*’


where the payoffs for  p and q are added as the last row and the last column. 

(C)



37. The value of  the matrix game with payoff matrix

0  0  0  1  1  1

2  1  3  1  2  0

0  1  2  1  3  0

3  1  1  0  1  0

 is  (A) -1,  (B)   0,    (C)  1/2,    (D) 1


Solution.     Let v be the value. Computing max min and min max in pure strategies, we see that  0 ≤ v ≤ 1.

If the row player uses the half sum of the first two rows, he gets at least 1/2, so v ≥ 1/2.

If the column player uses the half sum of the columns  2 and 6, she pays him 1/2, so v ≤ 1/2.

Thus, v = 1/2.

(C)


38.   The least-squares solution [x, y, z]  for the system

x+y +z = 3,  x + 2y = 3,  x+3y +z =  5, x - y = 0      is

(A) [-1, 2, 23]/13, (B) [1.5, 1, -1],  (C)  [1/3, 1/3, 0], (D) [0, 0, 1.5].


Solution

The system has the unique solution (x, y, z] = [1, 1, 1]. So it is the best fit for every criteria we know.

(E).


39.    The LAD solution [x,y,z]   for the system in Problem 38  is 

(A) [1, 1, 1], (B) [1.5, 1, -1],  (C)  [1/3, 1/3, 0], (D) [-1/3,  2/3, 0].


Solution. The system has the unique solution (x, y, z] = [1, 1, 1]. So it is the best fit for every criteria we know.

(A).   



40.   The  uniform  solution [x,y,z]   for the system in Problem 38  is 

(A) [1, 1, 1], (B) [1.5, 1, -1],  (C)  [1/3, 1/3, 0], (D) [0, 0, 1.5].


Solution. 

The system has the unique solution (x, y, z] = [1, 1, 1]. So it is the best fit for every criteria we know.

(A).




41.  The value of the matrix game  with payoff matrix 


2

3

0

4

2

1

3

1

0

0

2

0

2

0

-3

0

0

-2

3

2

0

1

2

0

1

1

-1

-2

-3

-6

7

-4

0

1

2

 5


is  (A) 2.5, (B) 8/3,  (C)  0,  (D) -1.


Solution.  There is a saddle point at row 1, column 3.  

(C)


42.   An  optimal solution x  for  max(2|x+1|, |x|, 3|x-1|) -> min  is

(A) -1,  (B) 0, (C) 1/6, (D) 1.


Solution.  It is clear that    max(2|x+1|, |x|, 3|x-1|) =  max(2|x+1|,   3|x-1|) for all x and that

-1 ≤ x ≤ 1 for optimal x.  For optimal x,  we have  2|x+1| =   3|x-1|  hence 2(x+1) = -3(x-1), so x = 1/5 = 0.2.

(E).


43. An  optimal solution x  for  2(x+1)2   + x2  + 3(x-1)2   -> min is 

(A) -1,  (B) -1/3, (C) 1/6, (D) 4/3.


Solution. The optimal solution is the mean of -1,-1,0,1,1,1, hence x = 1/6.

 (C).


44. An  optimal solution x  for  2|x+1|   + |x|  + 3|x-1|   -> min is 

(A) -1,  (B) -1/3, (C) 1/6, (D) 4/3.


Solution. The optimal solutions are  the medians  of -1,-1,0,1,1,1, i.e., 0 ≤ x ≤ 1.

(C).


45.  An  optimal solution x  for  2(x+1)0   + x0  + 3(x-1)0   -> min  (where  00= 0)   is 

(A) -1,  (B) 0, (C) 1/6, (D) 1.


Solution. The optimal solutions are  the modes  of -1,-1,0,1,1,1, i.e., x = 1.

(D).