Math 484.003   December 4, 2015.   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 5 2 0 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, without changing the value of game:

 5 2 0 0 2 5

By graphical method, we find the equilibrium

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

 5 2 0 2* 0 2 5 2* 2.5 2’ 2.5 2*’

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

(E).

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) [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  0  -1       1*    -1/4  3/4*   1/4

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

Using C or D, the column player pays ≤  3/4 So the value of game is ≤ 3/4 hence A and B are not optimal.

If C is optimal, the value of game is 3/4 and the row player’s optimal strategy should be  the second row,  but 3/4 is not minimal in the row.

If D is optimal, the value of game is 3/4 and the row player’s optimal strategy should be  the third row,  but 3/4 is not minimal in the row.

(E)

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

x+y +z = 3,  x + 2y = 1,  x+3y +z =  0, x - y =  -1      is

(A) [1, 1, 1], (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  0

1 -1  0  -1

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

4    5    2     3

5   15   4     6

2     4   2     3

The solution is x = 0, y = 0, z = 3/2 which is very easy to check. But   A, B, and C are not solutions.

(D).

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) [0, 0, 1.5].

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

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

=   |x+y +z - 3|  + 3 x + 4 y + z

It is clear that the value 8 at A is not minimal. Just keep  x = z = 1 and  decrease y a little bit,

say, y = 0.99. Then f = 0.01  +3  +4 - 0.04  + 1 = 8 - 0.03 < 8.

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

so B is not optimal.

At C,    f = 4 + 2/3.    Near C,  f =     4 + x + y   + |x + 2y -  1|          Replacing C =  [1/3, 1/3, 0]  by

[1/3 - 0.02, 1/3+0.01, 0], we improve f hence C is not optimal.

At D, f =  1.5 + 1 + 1.5 + 1 =5. This value is not optimal because f(C) < 5.

(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 | ,  | 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.

Near A, f = x+3y +z.

It is clear that the value  5 at A is not minimal. Just keep  x = y = 1 and  decrease  y a little bit,  say, y = 0.99. Than f =  5 - 0.01 < 5.

At B, f = 3.5 (another reason for dropping A). Near B , f = x+3y +z. So B is not optimal.

At  C ,  f =  max(7/3,  0, 4/3, 1) = 7/3. Near C, f =  3 - x - y - z.  So C is not optimal.

At D, f = max(1.5, 1,  1.5,  1) = 1.5. Near D, f =  max(3-x -y - z,  x+3y +z).

Now we take  x = 0, y close to 0 and set z = 1.5 - 2y. Then f = max(1.5+y, 1.5 + y) = 1.5 + y.

We can make f < 1.5 taking y to be a negative  number close to 0, say, y = -0.1.

Thus,  D is not optimal.

If we started with D, we would not need to consider A, B, C because f takes bigger values there.

(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 row 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   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    is

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