Math 484.2. 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 |
2 |
3 |
3 |
|
0 |
1 |
4 |
2 |
4 |
6 |
3 |
|
1 |
2 |
2 |
5 |
2 |
7 |
0 |
|
0 |
0 |
1 |
1 |
4 |
2 |
0 |
|
1 |
1 |
4 |
3 |
2 |
4 |
0 |
|
3 |
3 |
3 |
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 & column. The value of game is 3. (D)
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 q for the column player is
(A) [0,0,1/2,0,1/3,0,0], (B) [0, 2/3, 0, 1/3,0,0,0], (C) [0,0, 0,1/2,1/2, 0,0], (D) [1/2, 0, 0, 0, 0, 0, 1/2], (E) [1/2, 1/2, 0,0,0,0,0].
Solution . In (A), the sum of entries is not 1. We add as the last column c8 the payooff corresponding to q = [0, 2/3, 0, 1/3,0,0,0]:
|
1 |
2 |
3 |
1 |
2 |
4 |
3 |
5/3 |
|
3 |
1 |
4 |
4 |
2 |
4 |
0 |
2 |
|
3 |
3 |
2 |
0 |
4 |
3 |
1 |
2 |
|
4 |
1 |
4 |
0 |
2 |
3 |
2 |
2/3 |
|
3 |
1 |
2 |
4 |
2 |
1 |
3 |
2 |
If q is optimal, the value of game is 2 and every optimal strategy for the row player is a mixture of the rows 2, 3, and 5.
This mixture should have every entry at least 2. The halfsum of the rows 3 and 5 is such a strategy. (B)
8.
|
2 |
2 |
1.5 |
1.5 |
|
2 |
1 |
1 |
2 |
|
4 |
2 |
2 |
1 |
The value of game is (A) 0 , (B) 1, (C) 2, (D) 3, (E) 1.5.
Solution. By domination, the first and second column can be eliminated.
Then we can solve the game by graphical method. (E).
9. Consider the system of two equations for two unknowns x, y, where t is a given number:
x+y = 3
x+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 = 1 or -1, then there are no solutions. . If t ≠1, -1, then the system has exactly one solution. (B)
10. An optimal solution x for
2x2 + (x-1)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 - 8x +10 = 4(x-1)2 + constant
hence the optimal x is 1. Another approach: x is the least-squres fit to 4 numbers 0, 0, 1, 3 which is their mean 1. (B)