Math 484.1  Ocober 4, 2007   Name:_________Vaserstein___________

 x y z 1. Solve for y, z 0 2 1 =u 1 0 1 =v

Solution. Pivot:

 x y z 0 2 1 =u 1 0 1* =v
-->
 x y v 1. Solve for y, z -1 2* 1 =u -1 0 1 =z
-->
 x u v 1. Solve for y, z 1/2 1/2 -1/2 =y -1 0 1 =z

2. x - y 3 -> min,

x2 2y 3 = 1.

Solution. Solve the equation for
y :
y 3 =(1-  x2 )/2.
Plug this  into the objective function:

x  - (1-  x2 )/2 =   ( (x+1)2 -2)/2   -> min.

Now it is clear that  min = -1 at  x = -1, y = 0 .

3.

 x y 1 3. Solve for x, y 1 -1 2 1 -x 2x >=  3 <= 2 -2 1 x - > min

Solution. It is given that  2y >= 3, x+y <= 2. The objective function, -x + y -> min.

We take the difference the given constraints and obtain that   -x+y >= 1.

Now it is clear that  min = 1 at  x =  1/2, y =3/2 .

4. Solve the linear programs given by the following standard tableaux:

 a b c 1 Problem 4a 1 0 -1 1 =d -1 0 1 -1 ->min

Solution. The tableau is feasible with a-column bad, so LP is unbounded.

 a b c 1 Problem 4b -1 -2 -1 -1 =d 1 0 1 0 ->min

Solution. The d-row is bad, so LP is infeasible.

 a b c 1 Problem 4c 1 0 -1 3 =d 2 0 1 -1 ->min

Solution. The tableau is optimal, so min = -1 at  a=b=c=0, d=3.

5. Find all logical implications between the following 5  constraints on  x, y:

(a)   x2 = y2, (b) x = y=   1, (c)  x >= y , (d) x + y = 2, (e) y > 0.

Solution. (b) implies (a),(c),(d),(e).