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 y21 - 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 = 31/2/2.

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).