Math 484.1  Ocober 4, 2006   Name: SAMANVITHA VANGALA Corrected by L.V.

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

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

Solution: 2y = u, so y = u/2

x + z = v, so x = v - z

2. x + y ² -> max,

x² +  y ² = 1.

<>
Solution: x² +  y ² = 1, therefore y² =  1 - x² . Substituting this into the objective function, we have <>
x + 1 - x²       = 1.25 - (x-0.5)²    -> max.
So x = 0.5 , max = 1.25 , y = 3^1/2/2.

3.

x	y	1	3. Solve for x, y.
2 -1	3 2	-x y	>=  3 <= 2
1	2	0	- > min

Solution: 2x + 3y – x  >= 3

-x + 2y + y  < = 2,

x + 2y  ->  min

or

x + 3y >= 3  ,   x-3y >= - 2,   x + 2y  ->  min.
Now we can use ther graphical method.
Or we can multiply the first constrant by 5 and add this to the second constraint:
6x+12y >= 13, or    x+2y >= 13/6.
So min >= 13/6. Taking    x= 1/2, y = 5/6 we make   x+3y=3, x - 3y = -2, x+2y = 13/6 = min.

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

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

Solution: optimal tableau

a = b = c = d = 0 and min = -1

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

Solution : Bad  d-Row. Hence, Linear Program is infeasible

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

Solution: Bad a-column in a feasible tableau. Linear Program is unbounded.

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

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

Solution : (d) implies (a).

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

(a), (b), (d), (e) imply  (c).