Math 484.1 Ocober 4, 2007 Name:_________Vaserstein___________

Midterm 1, 5 problems, 15 points each. Return this page.

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 y³  :
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).