Linear Programming Lecture Notes - Penn State Personal Web Server

Expression 5.20 is called the minimum ratio test. We are interested in which index i is the
minimum ratio.
Suppose that in executing the minimum ratio test, we find that xj = bk/ajk . The variable
xj (which was non-basic) becomes basic and the variable xBk becomes non-basic. All other
basic variables remain basic (and positive). In executing this procedure (of exchanging one
basic variable and one non-basic variable) we have moved from one extreme point of X to
another.
Theorem 5.6. If zj − cj ≥ 0 for all j ∈ J , then the current basic feasible solution is
optimal.
Proof. We have already shown in Theorem 5.1 that if a linear programming problem
has an optimal solution, then it occurs at an extreme point and we’ve shown in Theorem
5.5 that there is a one-to-one correspondence between extreme points and basic feasible
solutions. If zj − cj ≥ 0 for all j ∈ J , then ∂z/∂xj ≤ 0 for all non-basic variables xj.
That is, we cannot increase the value of the objective function by increasing the value of any
non-basic variable. Thus, since moving to another basic feasible solution (extreme point)
will not improve the objective function, it follows we must be at the optimal solution.
Theorem 5.7. In a maximization problem, if aji ≤ 0 for all i = 1, . . . , m, and zj −cj < 0,
then the linear programming problem is unbounded.
Proof. The fact that zj − cj < 0 implies that increasing xj will improve the value of the
objective function. Since aji < 0 for all i = 1, . . . , m, we can increase xj indefinitely without
violating feasibility (no basic variable will ever go to zero). Thus the objective function can
be made as large as we like.
Remark 5.8. We should note that in executing the exchange of one basic variable and
one non-basic variable, we must be very careful to ensure that the resulting basis consist
of m linearly independent columns of the original matrix A. The conditions for this are
provided in Lemma 3.38. Specifically, we must be able to write the column corresponding
to xj, the entering variable, as a linear combination of the columns of B so that:
(5.21) α1b1 + . . . αmbm = A·j
and further if we are exchanging xj for xBi (i = 1, . . . , m), then αi = 0.
We can see this from the fact that aj = B −1 A·j and therefore:
Baj = A·j
and therefore we have:
A·j = B·1aj1 + · · · + B·majm
which shows how to write the column A·j as a linear combination of the columns of B.
Exercise 49. Consider the linear programming problem given in Exercise 48. Under
what conditions should a non-basic variable enter the basis? State and prove an analogous
theorem to Theorem 5.6 using your observation. [Hint: Use the definition of reduced cost.
Remember that it is −∂z/∂xj.]
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