Math 484 Ocober 4, 2007      Mayur Aras

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

2y + z = u à z = u – 2y

x + z = v

x + (u-2y) = v à x = v – u +2y

Solution:

z = u – 2y

x = v – u +2y

2. x - 2y ³ -> min,

x² +  y ³ = 1.

Solution:

Solve the equation for y³: y ³ =(1-  x² ).
Plug into the objective function:

x - 2(1-  x² ) =   x – 2 +x²-> 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	>=  1 <= 2
-2	-1	x	- > min

Solution: It is given that  2y + 2x >= 1, x+y <= 2. The objective function, -x - y -> min.

We take the difference the given constraints (multiply the second constraint by 3 and then subtract it from the first) and obtain that   -x-y >=5.

Now it is clear that  min = 5 at  x =  1/2, y =3/2 .

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

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

Solution: The tableau is feasible with column b bad, so LP is unbounded.

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

Solution: Tableaux is already optimal: min=0 at a=b=c=0, d=1

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

Solution: Tableaux is feasible with column a bad, so LP is unbounded.

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

(a)   x³ = y³, (b) x = y= 0, (c)  x >= y , (d) x + y = 0, (e) y > -1.

Solution:

a implies c

b implies a, c, d, e