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
Row D is bad row
min = + infinity

4C:

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

Tableau (The LP) is unbounded
Col C is bad column
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.)