03-simplex-week2

The simplex algorithm — Discussions in class
Michel Bierlaire
1 Summary
The simplex algorithm solves a linear optimization problem by visiting the
vertices of the constraint polyhedron. It exploits the representation of ver-
tices by feasible basic solutions, and follows the basic directions to move from
one vertex to the next.
1.1 Main concepts
•Basic directions,
•reduced costs,
•solution on a vertex,
•from vertex to vertex,
•the simplex tableau,
B−1A B−1b
cT−cT
BB−1A−cT BB−1b
•the initial tableau: the simple case,
•the initial tableau: the general case.
For the multiple choice questions below, the correct responses are in bold
font.
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2 Reduced costs
Consider the condition “the reduced costs are all non negative”. Is it a
sufficient/necessary optimality condition?
•sufficient and necessary,
•sufficient and not necessary,
•not sufficient but necessary,
•neither sufficient, neither necessary.
3 Degeneracy: example
We propose an example where the optimal solution is degenerate, and the
reduced costs are non negative.
min x1+x2
subject to
x1≥x2,
x1≥0,
x2≥0.
Standard form:
min x1+x2
subject to
−x1+x2+x3=0,
x1≥0,
x2≥0,
x3≥0.
A= (−1 1 1), b =0, c =
1
1
0
.
Consider the feasible basic direction where x2is in the basis, and calculate
the reduced costs.
1. Is it an optimal vertex? Yes. If x2is in the basis, we have B=B−1=1,
and xB=B−1b=1·0=0. Therefore, all the variables are equal to
zero. The value of the objective function at (0, 0)is zero. It cannot
reach a lower value without violating the constraints.
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2. Why is there a negative reduced cost?
(a) Because the slack variable is out of the basis.
(b) Because the basis is degenerate. It is false, because a basis can
be degenerate, and have all the reduced costs non negative, as
illustrated by the last example below.
(c) Because it is a minimization problem.
(d) Because the corresponding basic direction is infeasible.
(e) Because there is only one basic variable.
If x2is in the basis, we have B=B−1=1. The basic entry of the basic
direction that tries to put x3in the basis is
(d3)B= −B−1A3= −(1)(1) = −1.
Therefore, the direction is
d3=
0
(d3)B
1
=
0
−1
1
and the reduced cost is
¯
c3=c3−cT
BB−1A3=0− (1)(1)(1) = −1.
Consequently, the objective function decreases along this direction.
The basic entry of the basic direction that tries to put x1in the basis
is
(d1)B= −B−1A1= −(1)(−1) = 1.
Therefore, the direction is
d1=
1
(d1)B
0
=
1
1
0
and the reduced cost is
¯
c1=c1−cT
BB−1A1=1− (1)(1)(−1) = 2.
Consequently, the objective function increases along this direction.
3

d3
d1
3. If, instead of minimizing the objective function, we maximize it on the
same set of constraints, what happens?
•As the polyhedron has only one vertex, it corresponds to the op-
timal solution of the maximization problem as well.
•There is no optimal solution.
•The optimal solution is not on a vertex.
4 Solution on a vertex
1. Consider a linear optimization problem, with at least one optimal so-
lution. Which one of the following statements is correct?
•There exists an optimal vertex of the constraints polyhe-
dron. Indeed, this is Theorem 16.2, p.365.
•An optimal solution is always at a vertex of the constraints poly-
hedron. No. It may happen that all solutions in a face of the
polyhedron are optimal. In that case, solutions that are not ver-
tices are also optimal.
•There exists a vertex of the constraints polyhedron which is op-
timal only if the problem is written in standard form. No. The
property is true irrespectively of how the polyhedron is written.
Indeed, any polyhedron can be written in standard form, and the
proof of Theorem 16.2 applies.
2. One of the following optimization problems has an optimal solution
which is not a vertex. Which one?
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(a) min xs.t. 0≤x≤1. The unique solution is x∗=0, which is on
a vertex.
(b) min x1+x2, s.t. x1, x2≥0. The unique solution is x1=x2=0,
which is on a vertex.
(c) min x1+2x2s.t. x1+2x2≤4,x1, x2≥0. The unique solution is
x1=x2=0, which is on a vertex.
(d) max x1+2x2s.t. x1+2x2≤4, x1, x2≥0. Note that it is
a maximization problem. All solutions such that x1+2x2=4
and x1, x2≥0are optimal. In particular, the solutions x1=0,
x2=2and x1=4,x2=0are optimal, and are on a vertex of the
polyhedron. But the solution x1=2,x2=1is also optimal, and
is not on a vertex.
(e) max xs.t. 0≤x≤1. The unique solution is x=1, which is on a
vertex.
(f) max x1+x2s.t. x1, x2≥0. This problem is not bounded, and has
no optimal solution.
5 Step along the direction
1. The length of the step αto perform along the basic direction dpis
determined using exactly one of the following conditions.
•The gradient at x+αdpis zero. No, the gradient is constant and
always equal to c. Therefore, except for a trivial problem where
c=0, the gradient is never zero.
•α=1. No, the value one is arbitrary.
•αis equal to the reduced cost associated with dp. No, the re-
duced cost is the slope, the directional derivative, of the objective
function in the direction dp.
•αcorresponds to the first constraint that is activated. Yes,
we follow the descent direction dpas far as we can, which is when
the first constraint is met. Beyond, the iterate would become
infeasible.
•αcorresponds to the last constraint that is activated. No, it must
be the first, as discussed above.
•There are m+1variables in the basis at x+αdp. No, there are
always mvariables in the basis.
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