Math 484.002   April 24, 2015.   Midterm 3.
10 problems, 5 pts each.   Write your name here ____________________________

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 4 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:

 5 4 0 0 2 5

By graphical method, we find the equilibrium

(p, q) = ([1,1]T/2, [1, 0, 1]/2) with the value of game 2.5:

 5 4 0 2.5* 0 2 5 2.5* 2.5' 3 2.5' 2.5*'

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

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]/9,  (B)    [0, 1, 2, 1]/4 ,    (C)  [2, 0, 1, 1]/4,    (D) ]1, 1, 1, 1]/4.

Solution. (A) is not a mixed strategy. We compute payoff for the other proposed strategies:

B    C         D

0  -2   1 -1         -1/4   0*   -1/2

2  0  -3  -1          -7/4  0*    -1/2

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

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

If C is optimal, the value of game is 0 and the row player’s optimal strategy should be a mixture of the first two rows, and the payoff for the mixture  should be ≥ 0.

But the payoff against the last column is -1. So C is not  optimal. (E)

P.S. It is easy to improve C by moving to the last column.

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

x+y = 3,  x + 2y = 5,  x+3y =  7, x - 6y = -9      is

(A) (3,1), (B) (1.5,1.76),  (C)  (19/3, 1/3), (D) (1,2).

Solution

The augmented matrix  for the system is

1   1   3

1   2   5

1   3   7

1 -6   -9

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

4    0      6

0   50   88

The solution is x = 1.5, y =1.76. (B).

39.    An optimal solution x   for

|x| + 2 |x + 1| +   3|x - 5|   ->  min

is  (A) -2, (B)  3 (C)  6, (D) -1.

Solution. The optimal solutions x  are  the medians of 6 numbers

-1, -1, 0,  5, 5, 5, i.e.,  the numbers x such that  0 ≤ x ≤ 5.   (B).

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

An optimal solution is (A) 1, (B) 2,  (C)  -1, (D) -2.

Solution.   The objective function f is convex and piece-wise linear.

The slope of   f at  x =1 is 2. So any optimal x  < 1.   For  x close to 1, f = 2x.

At x = -1, the slope is -1  so optimal  x > - 1. When x    -1 and close to -1, f = 1-x.

(E)

To find an optimal  x, let us try the point x where  2x = 1- x, that is, x =  1/3.

At this point, the left slope is -1 and the right slope is 2, so x = 1/3 is the only optimal solution.

41.  The value of the matrix game  with payoff matrix

 1 3 5 -4 6 5 -1 -4 0 6 2 0 2 0 -3 0 0 -2 1 2 0 1 2 0 1 1 0 -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 9.  (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, 2, 1, 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.  If the column player does not use the last two columns, she has a symmetric game so  using her optimal strategy she pays him 0 in the worst case.

Using the proposed strategies, she pays a positive amount in worst case. So none of them is optimal.  (E).

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

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

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

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