We consider a standard problem min c·x subject to Ax = b, x � 0 where x ∈ Rn, b ∈ Rm etc.
How to start?
1
Finding the initial BFS
Assume b � 0 and consider min(y1 + . . . + ym) Ax + y = b
x, y � 0.
Big M method: Assume b � 0 and consider min M(y1 + . . . + ym)
Ax + y = b x, y � 0.
Is the simplex algorithm (SA) correct?
If the problem has only non-degenerated BFS then SA ter- minates after a finite number of steps.
There are pathological examples showing that the might be a loop.
There are pivoting rules avoiding loops also in degenerate cases (will be discussed later).
Complexity of SA
Each tableau requires O(mn) operations But how many iterations are needed?
There are two answers:
• In practice, SA is fast, typically m iterations suffice.
• Theoretically, SA is awful: it may lead you through all the vertices and the numer of them may grow expo- nentially!
Hirsh problem: Find the upper bound for the distance between two vertices of the polyhendron Ax = b.x � 0.
General concept of duality
Consider f : A → R and min f(x) for x ∈ A.
Suppose that we have g : B → R and we know that g(y) � f(x) for every y ∈ B and x ∈ A. Then we say that the problems
(i) min f(x) for x ∈ A;
(ii) max g(y) for y ∈ B;
are dual in the weak sense.
Definition. There is a strong duality between them if minx∈Af(x) = maxy∈B g(y).
Note that, given two dual problems, if we find x0 ∈ A and y0 ∈ B such that f(x0) = g(y0) then
f(x) � g(y0) = f(x0) � g(x),
for all x ∈ A and y ∈ B so x0 and y0 are optimal!
Dual linear problems
Example
Definition. Given a standard problem (P) min c·x subject to Ax = b, x � 0 we define the dual problem (DP) as max b · y subject to ATy � c.
Weak and strong duality
Lemma. (Ax) · y = x · (ATy).
Lemma. If x ∈ Rn and y ∈ Rm are feasible for (P) and (DP), respectively then b · y � c · x.
Theorem. If x∗ and y∗ are optimal solutions to (P) and (DP) then b · y∗ = c · x∗.
