Math 484.3.  Dec. 5, 2014.   Midterm 3.  
10 problems,5 pts each.   Write your name here ____________________________  

Return the scantron 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 like the given answers, explain why here and choose (E) at scantron.



36.  The value of the matrix game  with payoff matrix


1

5

5

4

1

6

2

0

2

5

is  (A) 2.5, (B) 28/9,  (C)   11/3,  (D) 25/9.


Solution. The first two columns go by domination:


5

4

1

0

2

5


By graphical method, we find the equilibrium

(p, q) = ([5, 4]T/9, [4, 0, 5]/9) with the value of game 25/9:


5

4

1

25/9*

0

2

5

25/9*

25/9'

28/9

25/9'

25/9*'


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



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]/4,    (D) [1, 1, 1, 1]/4.


Solution. We compute payoff for the proposed strategies:


                       A    B    C         D

0  -2   1 -1      0*  -1/4   0   -1/2

2  0  -3  1       0* -5/4  1/2*     0

-1 3   0  1       0*    1* -1/4   3/4*

So the value of game is ≤ 0 hence B, C  and D are not optimal.

To prove that (A) is optimal we match it with its transpose and get an equilibrium:


                            A  

      0  -2   1 -1      0  

      2   0  -3  1       0  

     -1   3   0  1       0   

AT   0  0   0  0     0*’


(A).



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

x+y = 3,  x + 2y = 5,  x+3y =  7, x - 6y = -11      is

(A) (3,1), (B) (6.5, 0.5),  (C)  (19/3, 1/2), (D) (1,2).


Solution. x = 1, y = 2 is an exact solution for the system. (D)

Here is the standard way to solve the problem (it is longer).

The augmented matrix  for the system is

1   1   3

1   2   5

1   3   7

1 -6 -11

The system for the least-squares fit has the augmented matrix

4    0      4

0   50 100


The solution is x = 1, y = 2.



39.    The optimal solution for

 |2x| +  |x - 1| +   |x/2 - 5|   ->  min 

 is  (A) 0, (B)  3 (C)   1, (D) -1.


Solution. The optimal solutions x  is  the is the same as for

4 |x| + 2 |x - 1| +    |x - 10|  ->  min which are the medians of 7 numbers

0, 0, 0, 0, 1, 1, 10, i.e., x = 0.   (A).



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

The optimal solution is (A) 1, (B) 1.5,  (C)   1.8, (D) 3.


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

So the last term can be dropped because it is ≤ 0. The second term is never maximal.

Thus,  that max( |2x|,   |x-1| ,   |x - 3| ,  3(x- 5)) = max( |2x|,  |x - 3 |) (assuming that  x ≤ 5).

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

So  2x = 3 - x for optimal  x, hence  x =1. (A)



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

1


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


Solution.   The columns  1, 2 and 5 are dominated by the last column. Column 3 is dominated by column 6.


     c2 c4 c7

 r1 -4   5 -1

 r2  2  -3  0

 r3   1   0  1


Computing max min  and min max in pure strategies, we obtain

0 ≤  the value of game ≤ 1 (with r3 and c7 being the best pure strategies), hence  only (C) could be a correct answer. But the direction from r3 to r1 is improving  (e.g., the worst case payoff for (10r3+r1)/11 is > 0),

so (C)    is not correct.  (E)



42.   Consider the matrix game with payoff matrix

0  -2   1 -1   0

2   0  -3  1  -2

-1  3    0 1    1

An optimal strategy for the column player is

(A) [3, 1, 2, 0, 0]/6,  (B)   [0, 1, 2, 1, 0]/4 ,    (C)  [2, 0, 1, 1, 1]/5,    (D) [1,1,1,1]/4.


Solution.  D is not a mixed strategy (has only 4 entries).

Using A or C,  the column player pays  0 (in all cases). So the value of game  is ≤ 0. Can it be improved?

No, it cannot because using  [3, 1, 2]T/6, the row player gets   0 (in all cases).  So the value is 0.

(A), (C).

If you found that both A and C are optimal and did not make any mistakes in your solution of this problem,  you get  6  bonus points.



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

(A) 0.5, (B) 0, (C) 3/7, (D) 3/8.


Answer (D).


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

(A) 0.5, (B) 0, (C) 3/7, (D) 1.


Answer (B).


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

(A) 0.5, (B) 0, (C) 3/7, (D) 1.


Answer (B).