Math 484 Ocober 4, 2007 Mayur Aras
corrected by L.V.
x 
y 
z 
1.
Solve for y, z 
0 
2 
1 
=u 
1 
0 
1 
=v 
Solution:
The
second row reads x + z = v hence
z= v  x.
The first row reads 2y + z = u hence 2y= uz = u (vx) =
uv+x and
y = u/2 v/2 +x/2.
So the answer is
u 
v 
x 

1/2 
1/2 
1/2 
=y 
0 
1 
1 
=z 
This tableau can be also obtained from the initial tableau by two pivot
steps.
2. x  2y^{
3} > min,
x^{2 }+ y^{ 3 }= 1.
Solution:
Solve
the equation for y^{3}:
y^{ 3 }=(1 x^{2 }).
Plug into the objective function:
x  2(1 x^{2 }) = x 2 +2x^{2}
= 2(x+1/4)^{2} 1/82 > min.
Now it
is clear that min = 17/8 at x
= 1/4, y^{3} = 15/16.
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, f = x  y > min.
The
constraints say that 2 <= f <= 1/2.
Now it
is clear that min = 2 at x
= 1, y =1 (there are other optimal solutions).
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 drow is bad so
the LP is infeasible.
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) x^{3} = y^{3}, (b) x = y= 0,
(c) x >= y , (d) x + y = 0, (e) y > 1.
Solution:
a implies c
b implies a, c, d, e