Math 340
Assignment #2
Due Wednesday Feb 10, 2016 at the beginning of class.
1. An example of an LP and an associated dictionary where the basic solution is the optimal
solution and yet some coefficients in the z row are strictly positive must correspond to a case
where you are some number of degenerate pivots away from optimality.
An easy example is
max x1
x1 +x2 ≤ 0
x1 −2x2 ≤ 1
x1 , x2 ≥ 0
which has the initial dictionary
x3 = 0 −x1 −x2
x4 = 1 −x1 +2x2
z =
x1
One can easily argue that the current solution (all original variables 0) is optimal and is in fact
the only feasible solution. Many examples can be constructed which are only one degenerate
pivot away from being seen to be optimal.
2. Since I didn’t write a quiz 2, I will take the
to quiz 2.
Maximize −2x1
−x1 −x2
−x1
−2x1 −x2
with final dictionary
corresponding second problem from the practice
−x3
−x3
−x3
−x3
x4 = 6
x2 = 1 −x5
x3 = 2 +x5
z = −2 −x5
We read of the final basis as {x4 , x2 , x3 }.
x1 x2 x3 x4
x4 −1 −1 −1 1

A = x5  −1 0 −1 0
x6 −2 −1 −1 0
x5
0
1
0
x6
b



0
x4 3


0
 x5  −2 
1
x6 −3
x1 x2
−2 0
x5
0
x6 0

c=
x3 x4
−1 0
≤3
≤ −2
≤ −3
−x1
−x1
−x1
−x1
x1 , x2 , x3 ≥ 0
+x6
+x6
x4
x4 1

= x2  0
x3 0

basis: {x4 , x2 , x3 }, B −1
x5 x6

0 −1
1 −1 

−1 0
Note that B −1 is contained in the final dictionary as the columns of the slack variables
x4 , x5 , x6 but beware those pesky minus signs.
Solution:
x1 x5
T
T −1
cN − cB B AN = −2 0
x6 x4
0 − 0
x2
0
x4

x3 x4 1
−1 x2 
 0
x3 0
x5 x6
x1 x5


0 −1 x4 −1 0

1 −1 
 x5  −1 1
−1 0
x6 −2 0
x6

0
0

1
=
x1 x5 x6 −1 −1 0
Similarly
x4
x4 1
x4



x B =  x2  = x2  0
x3 0
x3



x5 x6
b
x4




0 −1 x4 3
x4 1



1 −1 
x
−2
−
 5
 x2  0
−1 0
x6 −3
x3 0
x5 x6
x1 x5


0 −1 x4 −1 0

1 −1 
 x5  −1 1
−1 0
x6 −2 0
x6
 

0
x1


0
  x5 
1
x6
yielding our final dictionary in a somewhat altered column order..
3. Theorem 5.5 is taken from page 65-66 of V. Chvátal’s book on Linear Programming. Consider
the LP:
n
cx
maximize
Pn j=1 j j
subject to
j=1 aij xj ≤ bi
xj ≥ 0
P
(i = 1, 2, . . . m) (5.24)
(j = 1, 2, . . . , n)
Theorem 5.5. If (5.24) has at least one non-degenerate basic feasible optimal solution, then
there is a positive with the property: If |ti | ≤ for all i = 1, 2, . . . , m, then the problem
n
maximize
j=1 cj xj
Pn
subject to
j=1 aij xj ≤ bi + ti
xj ≥ 0
P
(i = 1, 2, . . . m) (5.25)
(j = 1, 2, . . . , n)
has an optimal solution and its optimal value equals
z∗ +
Pm
i=1
∗
yi∗ ti
∗
with z standing for the optimal value of (5.24) and with y1∗ , y2∗ , . . . , ym
standing for the
optimal solution of its dual.
Now consider the following LP
max 12x1
4x1
x1
2x1
+20x2
+6x2
+(7/2)x2
+4x2
+18x3
+8x3
+2x3
+3x3
≤ 600
≤ 300
≤ 550
x1 , x2 , x3 ≥ 0
The final dictionary is:
x1
x2
x6
z
= 75/2 −2x3
= 75
= 175 +x3
= 1950 −6x3
−(7/16)x4
+(1/8)x4
+(3/8)x4
−(11/4)x4
+(3/4)x5
−(1/2)x5
+(1/2)x5
−x5
optimal basis
B −1
{x1 , x2 , x6 }
x4
x5
x6

x1 7/16 −3/4 0
1/2
0
= x2 
 −1/8

x6 −3/8 −1/2 1

Theorem 5.5 applies here because the current basic feasible solution is non degenerate. With
∆b = (t1 , t2 , . . . , t3 )T , the conclusions of Therem 5.5 are valid if B −1 (b + ∆b) ≥ 0. Thus we
need





7/16 −3/4 0
600 + t1
75/2 + (7/16)t1 − (3/4)t2





−1
B (b + ∆b) =  −1/8 1/2 0   300 + t2  =  75 − (1/8)t1 + (1/2)t2  ≥ 0.
−3/8 −1/2 1
550 + t3
175 − (3/8)t1 − (1/2)t2 + t3
We must choose so that for all choices for each ti satisfying − ≤ ti ≤ , the inequalities
are true. From an inequality of the form a + a1 t1 + a2 t2 + a3 t3 ≥ 0 with a ≥ 0, we have form
a1 t1 +a2 t2 +a3 t3 ≥ −a. Now using − ≤ ti ≤ , we have a1 t1 +a2 t2 +a3 t3 ≥ −(|a1 |+|a2 |+|a3 |)
which can be achieved by appropriate choices of t1 , t2 , t3 (e.g. if a1 < 0 take t1 = and if a1 ≥ 0
take t1 = −). Thus a1 t1 + a2 t2 + a3 t3 ≥ −(|a1 | + |a2 | + |a3 |) ≥ −a and so ≤ |a1 |+|aa2 |+|a3 | .
From the first inequality 75/2 + (7/16)t1 − (3/4)t2 ≥ 0 we deduce that the worst case would
be to have t1 = −, t2 = and then deduce that −((7/16) + (3/4)) ≥ −75/2 and hence
≤ 600/19. From the second inequality 75 − (1/8)t1 + (1/2)t2 ≥ 0 we deduce that ≤ 600
.
5
280
From the third inequality 175 − (3/8)t1 − (1/2)t2 + t3 ≥ 0 we deduce that ≤ 3 . Thus the
largest possible for which if |ti | ≤ for all i = 1, 2, 3, then the conclusions of Theorem 5.5
.
hold, is to take = 600
19
4. Consider our two phase method in the case that the LP is infeasible. We begin with the
primal LP
max
z
Ax ≤ b .
x≥0
We introduce an artificial variable x0 and give the new LP (in Phase 1)
max
−x
0 x
[−1 | A] 0 ≤ b ,
x
x ≥ 0, x0 ≥ 0
min
b·y
(−1)T y ≥ −1
AT y ≥ 0
y≥0
a) Its dual is
,
We have assumed the maximum value of the objective function in the new LP is strictly
negative and so by Strong Duality, the objective function in the dual is strictly negative.
Hence for an optimal dual solution y∗ we have b · y∗ < 0 and y∗ feasible and hence AT y∗ ≥ 0
and y∗ ≥ 0. These conditions yield that if we take the sum of yi∗ times the ith inequality
of the primal for i = 1, 2, . . . , m, then we obtain a new inequality (the yi∗ ≥ 0 ensure no
inequalities flip) whose coefficients for each xj are all positive (because AT y∗ ≥ 0) and whose
right hand side is strictly negative (b · y∗ < 0) which shows that the inequalities have no
feasible solution.
b) An optimal primal solution for our new LP has x0 > 0 and hence by Complementary Slackness we deduce that for any optimal dual solution y∗ we have that the inequality (−1)T y ≥ −1
∗ T
∗
is an equality (−1)T y∗ = −1 which for y∗ = (y1∗ , y2∗ , . . . , ym
) becomes y1∗ + y2∗ + · · · + ym
= 1.
5.
a) Let A be an m × n matrix and let u be an n × 1 vector (of upper bounds). We wish to show
that either:
i) there exists an x with Ax ≤ b, x ≤ u
or
ii) there exists vectors y, z with AT y + z = 0,
y, z ≥ 0,
b·y+u·z<0
but not both.
Proof: Consider the following primal dual pair of Linear Programs:
max
primal:
0·x
Ax
≤
b,
x
≤
u
min
b·y+u·z
A y + Iz
=
0.
y≥0
z≥0
T
dual:
We note that y = 0 and z = 0 yields a feasible solution to the Dual (these vectors have m
and n coordinates respectively). By the Fundamental Theorem of Linear Programming, we
deduce that the dual either has an optimal solution or is unbounded.
CASE 1: Dual has an optimal solution y∗ , z∗ .
Thus, by Strong Duality, the primal has an optimal solution x∗ which is a feasible solution
and hence i) holds. Also 0 = 0 · x∗ = b · y∗ + u · z∗ which means by Weak Duality that every
feasible solution y, z to the dual has 0 ≤ b · y + u · z and hence ii) does not hold
CASE 2: Dual is unbounded.
Thus there is a feasible solution y, z to the dual has b · y + u · z ≤ −1 < 0 and hence ii)
holds. The dual being unbounded implies, using Weak Duality, that the primal has no feasible
solution and hence i) does not hold.
Thus we have established the theorem in both cases.
b) Use a) to show that the system of equalities with bounded variables given below has no
solution:
−x1 −x2 ≤ −6
−x1 +x2 ≤ −8
x1 −2x2 ≤ 3
x1
≤4
x2
≤2
We can quess and take y1 = 3, y2 = 1, y3 = 0, z1 = 4, and z2 = 2. We have
3
 
−1 −1 1 1
−1 1 −1 0
1

0 
 
0

1 
4
0
0
=
2
and −6y1 − 8y2 + 3y3 + 4z1 + 2z2 = −6 < 0 and so by c), no solution to the 5 inequalities in
x1 , x2 exist. The choices for y, z are not unique.