Math 484.001.   April 22, 2016.   Midterm 3.
10 problems, 5 pts each.   Write your name here __________Dr. V_____

(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

 3 3 2 1 2 2 1 2 3 3

is  (A) 2.5, (B) 28/9,  (C)   11/3,  (D) 2.

Solution. The first and last columns go by domination, without changing the value:

3   2   1

1   2  3

By graphical method, we find the equilibrium

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

 3 2 1 2* 1 2 3 2* 2 2 2 2*’

where the payoffs for  p and q are added as the last row and the last column.The column player has other optimal strategies.

(D).

37.  Consider the matrix game with payoff matrix

0  -2   1 -1

2   0  0   -1

-1  3   0  1

An optimal strategy for the column player is

(A)[4, 2, 3, 3]/12,  (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:

c1 c2 c3 c4       A         B      C        D

r1   0  -2   1 -1      -1/3     -1/4    0     -1/2

He     r2   2  0  0  -1      5/12*    -1/4  3/4*   1/4

r3  -1 3   0   1       5/12*     1* -1/2     3/4*

So A is better than  B. C, D.

If A is optimal, the value of game is  5/12  and his optimal strategy should be a mixture of r2 and r3.

But his payoff against c3 is 0 < 5/12.

(E)

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

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

1 -1  0  -1

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

4    5    2     4

5   15   4    9

2     4   2    4

(A) is a solution, but   (B), (C), and (D) are not

(A).

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 - 1| + | 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  7,  8, 11/3 , and 10/3. hence only D could be optimal.

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

= 2 +  2y  +  |x + 2y -  1|   + | x - y + 1|.

Since  |x + 2y -  1| ≥ - x - 2y +  1 and   | x - y + 1|  ≥  x -  y + 1, we have

|x + 2y -  1|   + | x - y + 1|  ≥  -3y +  2, hence  |x + 2y -  1|   + | x - y + 1|   ≥  2(|x + 2y -  1|   + | x - y + 1| )/3 ≥  -2y + 4/3

and  f ≥  2 +  2y   -2y + 4/3 = 10/3 near D. So D is optimal.

(D).

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) [-1/3,  2/3, 0].

Solution. Let f = max(|x+y +z - 3| ,  |x + 2y -  1| ,  |x+3y +z-1 | ,  | 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    4, 2.5, 7/3, and  8/3. hence only C could be optimal.

Near C, f = 3 - x - y - z.. So C is not optimal.

(E).

41.  An integer closest to the value of   matrix game  with payoff matrix

 -1 3 5 -4 6 -1 -1 -4 0 0 2 0 2 0 -3 0 0 -2 2 2 0 1 2 0 1 1 1 -2 -3 6 7 -4 0 1 2 -5

is  (A) 0, (B) 1,  (C)  2,  (D) 3.

Solution.    We have an saddle point in row 3 & column 6, hence the value of game is 0.

(A).

42.   Consider the matrix game with payoff matrix

0  -2   1    1    0

2   0   0  1   -2

-1  1   1 -1    1

Is   q =  [ 1, 1, 1, 1, 1]/5 an optimal strategy for the column player?

(A) Yes, (B) No.

Solution.    We add a column with the the payoff for q:

She

c1 c2 c3 c4  c5        q

r1    0  -2   1    1    0      0

He  r2    2   0   0    1   -2  0.2*

r3  -1  1    1 -1     1   0.2*

If q is optimal. his   optimal strategy  p  is a mixture of r2 and r3.

This p should give him ≥ 0.2 against any strategy of her.

Restricting him to mixture of r2 and r3 and deleting columns by domination, we obtain

c4   c5

r2  1 -2

r3 -1  1

Using slopes, the value of this game us -0.2. So he cannot get more than -0.2.

(B)

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

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

44. A median of the numbers 0,-1,-3, 1, 4, 3, 0, 0, 0, 1, -2    is

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