### Math 484.2     October 2, 2008     Name: ______Patrick Johnson_____corr by L.V and _E.L.__________ Midterm 1, 5 problems, 15 points each. Return this page.

1. Solve for x,y:

 x y z 1 2 0 = u 1 0 2 = v

Solution — Pivoting

 x y z 1 2 0 = u 1* 0 2 = v

-->

 v y z 1 2* -2 = u 1 0 -2 = x

-->

 v u z -1/2 1/2 1 = y 1 0 -2 = x

x = v – 2z
y = -v/2 + u/2 + z

2. 4x + y^2 –> min
x^2 + y^2 = 1

Solution
solve eq. 2 for y^2:
y^2= 1 – x^2
substitute into eq. 1:
4x + (1 – x^2) –> min
-x^2 + 4x + 1 –> min
Objective funcion is now defined only by x
Eq. 2 is the formula for a circle of radius 1, so our feasible range of x is -1 <= x <= 1.
The objective function is minimal when x = -1, with an optimal solution
value of -4. Then y = 0.

3. Solve for x,y

 x y 1 1 1 -x >= 3 -1 1 2 <= 22 1 2 0 –> min

Solution: write equation from row 1:
x + y – x >= 3
y >= 3
From row 2:
-x + y + 2 <= 2
-x +y <= 0
y <= x
x >= y
Combining the equations, x >= y >= 3
Objective function is 2x + y, so graphical method can be used or
it should be obvious that minimal values of x and y will result in
minimal objective function, so:
min = 9 when x = y = 3

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

4A:

 a b c 1 1 0 -1 1 = d 1 0 1 -2 –> min

Tableau is Optimal
min = -1(in fact, -2)
when a,b,c = 0, and d = 1

4B:

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

Tableaux is infeasible
min = + infinity

4C:

 a b c 1 -1 0 1 0 = d -2 1 -1 -1 –> min

Tableau (The LP)   is unbounded
min = – infinity

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

(a) x_ = y_
(b) 0 > 1
(c) 2y > x
(d) 1/3 > 1/2
(e) y >= x

Solution:
(b) is unconditionally true, so (b) is implied by anything
(d) is unconditionally false, so (d) implies everything
a => b
c => b
d => a
d => b
d => c
d => d
d => e
e => b

a => e  (added by L.V.)