1. (40 points) Consider the following linear program.
max 2x1 + 3x2 − x3
s.t.
x1 + 2x2 + x3
−x1 + x2 + x3
x1 + x2 + 2x3
x1
x2
=5
≥1
≤8
≥0
≥0
(a)
(b)
(c)
(d)
(e)
(a) (5 points) Which constraints are active at the point (2, 0, 3)? Use the letters next to each
constraint to indicate your answer.
Active constraints are constraints satisfied at equality. Here, (a), (b), (c), and (e) are all active.
(b) (5 points) Is the vector d> = (−1, 1, −1) a feasible direction at (2, 0, 3)? Justify your answer.
For d to be a feasible direction at (2, 0, 3) there must be a small > 0 so that (2, 0, 3) + λd is
feasible for all λ ∈ (0, ). To check this, it suffices to make sure that the active equality constraint
(a),
(−1) + 2(1) + (−1) = 0,
at the active inequality constraints (b) and (e):
−(−1) + (2) + (−1) = 2 > 0,
(1) > 0,
and at the active inequality constraint (c)
(−1) + (2) + 2(−1) = −1 < 0.
Thus, d is a feasible direction.
(c) (5 points) Is the vector d> = (−1, 1, −1) an improving direction at (2, 0, 3)?
The gradient of the objective is the vector (2, 3, −1) and the gradient test indicates that
(2, 3, −1) · d = −2 + 3 + (−1)(−1) = 2 > 0
so d is improving.
(d) (5 points) Could the point (2, 0, 3) be an optimal solution to the linear program? Justify your
answer.
No it could not be optimal because there is an improving and feasible direction at (2, 0, 3).
1
(e) (4 points) Is the point (2, 0, 3) a basic solution? Justify your answer.
Yes. Here, n = 3 and there are four active constraints. From these four active constraints, (2, 0, 3)
is a unique solution to (b), (c), and (e) as the corresponding constraint matrix is


−1 1 1
 1 1 2
0 1 0
which has a determinant of -3.
(f) (4 points) Is the point (2, 0, 3) an extreme point? Justify your answer.
Yes, as it as a basic feasible solution, and so, is an extreme point.
(g) (4 points) Is the point (2, 0, 3) degenerate?
Yes, as noted before, there are four active constraints and n = 3.
(h) (6 points) Convert the linear program to canonical form.
−
max 2x1 + 3x2 − x+
3 + x3
s.t.
−
x1 + 2x2 + x+
3 − x3
−
−x1 + x2 + x+
3 − x3 + s 2
−
x1 + x2 + 2x+
3 − 2x3 − s3
−
x1 , x2 , x+
3 , x3 , s1 , s2
=5
=1
=8
≥0
(a)
(b)
(c)
(e)
(i) (2 points) Suppose the linear program was made nonlinear by changing the objective to
max x21 + 3x22 − x3
Is d> = (−1, 1, −1) an improving direction at (2, 0, 3)? Justify your answer.
Here, the gradient of the objective is the vector (2x1 , 6x2 , −1) and so, at (2, 0, 3) is (4, 0, −1). The
gradient test then indicates
(4, 0, −1) · d = −4 + 0(1) + 1 = −3,
so d is not improving.
2
2. (30 points) Consider the following linear program:
max 2x1 + x2 + 3x3 − 20x4
s.t.
x1 − x2 + 3x3 + 5x4 = 2
−x1 + 2x2 + 3x3 + 4x4 = 3
x1 , x2 , x3 , x4 ≥ 0
(a) (7 points) At the basic solution (7, 5, 0, 0) what are the basic and nonbasic variables?
The basic variables are {x1 , x2 } and the nonbasic variables are {x3 , x4 }.
(b) (7 points) At the basic solution (7, 5, 0, 0), one of the simplex directions is (−9, −6, 1, 0). What
is the other simplex direction? Are any of the directions improving?
The given simplex direction corresponds to adding in the x3 variable. The other simplex direction
corresponds to adding in the x4 variable. Thus, it is (d(x4 ) )> = (d(x4 ) , d(x4 ) , 0, 1) and we must solve
Ad(x4 ) = 0 to determine the two unknowns. Solving this system reveals that


−14
 −9 

d(x4 ) = 
 0 
1
and gradient tests on both reveal that
>

>
2
2
 1  (x )



 d 3 = −18 − 6 + 3 = −21, and  1  d(x4 ) = −28 − 9 − 20 = −57,
 3 
 3 
−20
−20

so neither direction is improving.
3
(c) (7 points) At the basic solution (0, 13 , 79 , 0), calculate the step size if the direction d> = (1, 32 , − 91 , 0)
is used. Also, what are the entering and basic variables?
The stepsize is calculated to be the maximum positive number λ so that
 
   
0
1
0
1
 2  0
 37  + λ  3 1  ≥  
 
−  0
9
9
0
0
0
which means that λ = 7. The leaving variable is, thus, x3 and the entering variable is x1 .
(d) (7 points) Write the Phase I linear program.
min a1 + a2
s.t.
x1 − x2 + 3x3 + 5x4 + a1 = 2
−x1 + 2x2 + 3x3 + 4x4 + a2 = 3
x1 , x2 , x3 , x4 ≥ 0
(e) (2 points) Professor May B. Wright says, “At the basic solution, (0, 0, 37 , −1), we could compute Simplex directions even though it’s an infeasible direction! Then, we just choose an improving
direction and see if we become feasible!” The simplex directions at this point are (1, 0, 3, 2) and
, 3). What goes wrong if you try this?
(0, 1, − 14
3
Professor Wright is incorrect. The problem is that both directions are not improving as the
gradient test reveals that
>  

> 

1
2
0
2
 1  0

 


   = 2 + 9 − 40 = −29 and  1   114  = 1 − 14 − 60 = −63.
 3  3
 3  − 
3
−20
2
−20
3

Thus, doing what Prof. Wright suggests would result in the infeasible basic solution (0, 0, 73 , −1) being
reported as the optimal solution.
4
y
2
1
x
0
-2
-1
0
1
2
-1
-2
Figure 1: The set S = {(x, y) : −1 ≤ x − y ≤ 1}.
3. (10 points) True or False. No justification needed.
True To prove that the function f (x) = −|x| is not convex, we can simply note that
f ((1/3)1 + (2/3)(−1)) = −| − 1/3| = −1/3 > −1 = (1/3)f (1) + (2/3)f (−1).
False To prove that the set {(x, y) : −1 ≤ x − y ≤ 1} is convex we can simply note that
f ((1/3)1 + (2/3)(−1)) = −| − 1/3| = −1/3 > (1/3)f (1) + (2/3)f (−1) = −1.
True The set S = {(x, y) : −1 ≤ x − y ≤ 1}, drawn in Figure 1, is the feasible region of a linear
program but has no extreme points.
False Converting the constraints that define S = {(x, y) : −1 ≤ x − y ≤ 1} (depicted in Figure
1) into canonical form results in the constraints
x+ − x− − y + + y − − s1 = −1
x + − x− − y + + y − + s 2 = 1
x+ , x− , y + , y − , s1 , s2 ≥ 0.
A linear program using these constraints has no extreme points.
False Let S = {(x, y) : −1 ≤ x − y ≤ 1} (depicted in Figure 1). The following optimization
problem is unbounded:
max y − x s.t. (x, y) ∈ S.
5