Math 484.002.   April 22, 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 like the given answers, explain why here and choose (E) at scantron.

36.  The value of the matrix game  with payoff matrix

 1 5 2 3 2 5 1 2 2 3

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.  Consider the matrix game with payoff matrix

0  0  0  1  1

2  1  3  1  2

0  1  2  1  3

3  1  1  0  1

An optimal strategy for the column player is

(A) [4, 2, 3, 3]/12,  (B)    [0, 1, 2, 1, 0]/4,    (C)  [0, 2, 0,  1, 0]/3,    (D) ]1, 1, 1, 1, 1]/5.

Solution.    There are saddle points in pure strategies, so

the value of game is 1. Computing the worst-case payoffs, we see that (C) is optimal but   (B), D are not. (A) has only 4 entries so it is not clear what mixed strategy for the column paler it represents.

(C)

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

x+y +z = 3,  x + 2y = 1,  x+3y +z =  2, x - y =  -1      is

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

Solution

The augmented matrix  for the system is

1   1   1  3

1   2   0  1

1   3   1  2

1 -1  0  -1

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

4    5    2     5

5   15   4     12

2     4   2     5

None of A, B, C, D is a solution. The solution is  [x,y,z] = [-4, 8, 53]/26.

(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. Let f = |x+y +z - 3| + |x + 2y -  1| +   |x+3y +z - 2| + | x - y + 1|.

The function  f  is convex and piece-wise linear. We want to check whether it reaches its minimum at A, B, C,  and D.

The values of  f at A, B, C, and D are  6,  7, 14/3 , and 3. hence only D could be optimal.

Near D,  f =     3 - x - y - z  + |x + 2y -  1|   + 2 - x - 3y -z     + | x - y + 1|

= 5  -  2x  - 4y  - 2z +  |x + 2y -  1| +  | x - y + 1|.

It is easy to decrease this: keep    x=-1/3 and  y=2/3   the same and and  set z =  0.1.

(E).

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.

Let f = max(|x+y +z - 3| ,  |x + 2y -  1| ,  |x+3y +z-2 | ,  | x - y + 1|).

The function  f  is convex and piece-wise linear. We want to check whether it reaches its minimum at A, B, C, D.

The values of  f at A, B, C, and D are 3, 2.5, 7/3.and  1.5, hence only D could be optimal.

Near D, f = 3 - x - y - z so it is easy to decrease.

(E).

41.  The value of the matrix game  with payoff matrix

 -1 3 5 -4 6 5 -1 -4 0 0 2 0 2 0 -3 0 0 -2 0 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 3, column 1.  (C)

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

Solution.  Using A, she pays him 0 so the value of game is ≤ 0.

If he (the column player) uses [3,1,2]T/6.  his worst-case payoff is  0. So   ( [3,1,2]T/6. ,  [3, 1, 2, 0,  0]/6) is an equilibrium, hence A is optimal.

(A).

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

(A) 0.5, (B) 0, (C) 3/7, (D) 2/7.

44. A median of the numbers  -1,-3, 1, 4, 3, 0, 0, 0, 1    is

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

45. The midrange of the numbers  -3,4, 1, 2, 4, 0, -3    is

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