Math 484.1.  December 1  2011.   Midterm 3.  
5 problems, 15 pts each. Name____________________________

On the scantron, choose one of 5 answers.  Use #2 pencil. 
For me, write  down details.

6-8.   Matrix game is given by its payoff matrix.
6.
0
 2
 3
 4
 5
 3
 3
0
 1
 4
 5
 4
 6
 3
1
 2
 6
 5
 5
 7
 0
0
 0
 1
 1
 4
 2
 0
1
 1
 4
 3
 2
 4
 0
3
 2
 2
 3
 3
 6
 3

The value of game is (A) 0 , (B) 1, (C) 2, (D) 3, (E) 4.

Solution. There is a saddle points at  the last row & second column.  The value of game is 2.  (C)

7.
1
 2
 3
 1
 2
 4
 3
3
 1
 4
 4
 2
 4
 0
3
 3
 2
 0
 4
 3
 1
4
 1
 4
 0
 2
 3
 2
3
 1
 2
 4
 2
 1
 3
An optimal strategy p   for  the row player is the transpose  of
(A) [0,0,0.5,0,0.5], (B) [1/2, 1/2, 0, 0, 0],   (C) [0,0, 0,0.5,0.5],  (D) [1/2, 0, 0, 0, 1/2], (E) [1/2, 1/2, 0].

Solutrion . We add as the last row r6 the payoof corresponding to  p = [0,0,0.5,0,0.5]T:
1
 2
 3
 1
 2
 4
 3
3
 1
 4
 4
 2
 4
 0
3
 3
 2
 0
 4
 3
 1
4
 1
 4
 0
 2
 3
 2
3
 1
 2
 4
 2
 1
 3
3
 2
 2
 2
 3
 2
 2

So the worst-case payoff for (A) is 2. For (B) or (D) it is 1.5 (the second entry) so (B) and (D) are wrong answers. For (C), it in 1, so (C) is a wrong answer.

The answer (D) is wrong since it has only 3 entries. The only remaining answer is (A). Since the test was graded by computer, you got 15 pts for (A).

To prove that (A) is correct, we have to find a strategy for her where she pays him at most 2. Such a strategy cannot use the columns 1, or 5.

(0,2/3,0,1/3,0,0,0) is such a strategy. (A)


8.

2
 0
 2
 2
2
 1
 1
 2
4
 2
 0
 1

The value of game is (A) 0 , (B) 1, (C) 2, (D) 3, (E) 4.

Solution. By domination, the first and the last column can be eliminated.
Then we can solve the game by graphical method. (B).

9.   Consider the system of two equations for two unknowns x, y, where t is a given number:

x+ty = 3
2x+t2y=6

Then  (A) the system is not linear because the second equation is not linear, (B) there is t such that the system  has no solutions,
(C) there is  t such that the system has infinitely many solutions.
(D) the system cannot be solved, (E) 0 = 1.

Solution. It t = 0, then x = 3, y is arbitrary. If t = 2, then x = 3 - 2y, y is arbitrary. If t ≠ 0, 2, then the system has exactly one solution. (C)



10.  An optimal solution x for
2x2 + (x-5)2 + (x-3)2 -> min
is (A) 0 , (B) 1, (C) 2, (D) 3, (E) 2.5.

Solution. Completing the square in the objective function, we obtain
4x2 - 16x +25+9 = 4(x-2)2 + constant
hence the optimal x is 2.  Another approach: x  is the least-squres fit to  4 numbers 0, 0, 5, 3 which is their mean 2. (C)