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Chapter 3. Pitfalls

Initialization Ambiguity in an iteration Finite termination?

Resolving ambiguity in each iterationSelection of the entering variable:

set of nonbasic variables

Any nonbasic variable with can enter basis.

(However, performance of the algorithm depends on the choice of the en-tering variable.

examples of the rules : largest coefficient rule, maximum increase rule, steepest ascent rule, …)

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Selection of leaving variable:

a) No restriction in minimum ratio test : can increase the value of the en-tering variable indefinitely while satisfying the constraints (including nonnegativity), hence problem is unbounded

Ex)

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Current point is (0, 5, 0, 0, 7). We want to increase from 0 to , while keeping and

at 0, then direction of movement should be (0, 2, 1, 0, 0) 0 so that the new point satisfies the equations. ( let ( 0, 2, 1, 0, 0) )

Also, new point satisfies nonnegativity of variables for any since (0, 2, 1, 0, 0) . Objective value increases by ( 1 is coefficient of in the zeroth equation). Hence objective value becomes as , so the problem is unbounded. (Unboundedness of LP can be shown by identifying a direction such that is feasible for any , and ob-jective value becomes as .)

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b) In case of ties in the minimum ratio test :

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Ties (=1/2)

Some basic variables have value 0 after pivot.

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( entering, leaving )

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Though cannot be increased, we perform the pivot as usual, making as basic and as nonbasic. No change in the solution, but basic and nonbasic status changes for two variables. Later, we will examine what this means in geometry.

In the next iteration,

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Terminology:

degenerate solution ( 퇴화해 ) : basic feasible solution with one or more basic variables having 0 values.

degenerate iteration : simplex iteration that does not change the cur-rent basic solution (only basis changes).

Observations:If we have a nondegenerate b.f.s., the simplex iteration is nondegener-

ate. We move to a different point and the objective value strictly in-creases.

Given a nondegenerate b.f.s., we must have ties in the minimum ratios so that we have a degenerate solution after the pivot.

A degenerate iteration occurs only if we have a degenerate solution, but the converse is not true (i.e. we may have a nondegenerate iteration al-though we have a degenerate solution).

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Geometric meaning of a degenerate iteration

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x1=0

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( used) ( used)

A A

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(ex- continued)

In the first dictionary, the point is identified by using the three equa-tions in and setting at equalities. The nonnegativity constraints for nonbasic variables (here are used to identify point

( provides 5 equalities together with ).

After pivot, the solution point is not changed. But it is identified now by using at equalities together with .

Hence a degenerate iteration changes the defining equations one at a time, but the solution point is not changed.

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Getting out of degenerate iterations:

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x1=0

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( x3=0, x5=0 used)

A

Here, we changed the data in the previous dictionary a little bit to explain the example of getting out of degenerate iterations. Now, enters (takes value 1) and leaves the basis, and it is a nondegenerate pivot.

Geometrically, we still satisfy the equation , but do not need to satisfy . As we increase the value of up to 1, becomes 0, hence the equation is now

used (together with and ) to define a new point.

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Degenerate iteration is the process of identifying the same point (solu-tion) using different defining equations (different nonnegativity con-straints). If we are lucky enough to obtain defining equations that cor-rectly guides the moving direction, we move to a different point with a nondegenerate pivot.

If we have a degenerate solution, pivot may continue indefinitely (Ex-ample in text p.31, pivoting rule is largest coefficient for entering variable and smallest subscript for leaving variable in case of ties. Then we have the initial dictionary again after 6 pivots.)

Terminology:Cycling : appearance of the same dictionary (tableau) again in the sim-

plex iterations.

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In practice, cycling hardly occurs. However, during the degenerate it-erations, the algorithm stalls and it may hamper the performance of the algorithm. Such phenomenon is of practical concern and affects the performance of the algorithm (especially, for problems with some special structures and large problems).

We also need mechanisms to avoid cycling for any problem in-stances. Otherwise, the simplex method may not terminate finitely.

Cycling is the only reason that simplex method may fail to terminate (i.e. simplex method terminates in a finite number of iterations as long as cycling is avoided).

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Thm: If simplex method fails to terminate, it must cycle.

Pf) Number of ways to choose basic variables are finite. Hence, if sim-plex fails to terminate, same basis must appear again.

Show same basis same dictionary. Hence same dictionary appears, i.e. cycles.

Suppose the same basis appears again and consider the dictionaries

(1) (2),

where is the set of basic variables.

Dictionaries (1) and (2) have the same set of solutions (ignore nonnega-tivity constraints here)

Let be a nonbasic variable and be a number. Consider a solution to (1) given as the following:

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This solution satisfies (1). So it also satisfies (2), hence

( )

( )

These equalities must hold for any real number .

Therefore, we have

We can use any nonbasic variable in the above proof, so dictionaries are identical.

The theorem can be proved easily if we use matrices, but the above proof uses only algebraic arguments.

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The proof of the theorem shows that if we have the same basis, then the dictionaries (tableaus) are the same. Since there are only a finite number of ways to choose the basis, the simplex method terminates fi-nitely if the same basis (the same dictionary) does not appear again, i.e. cycling is avoided.

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Avoid cycling

Perturbation method, Lexicographic method Smallest-subscript rule (Bland’s rule)

Smallest-subscript rule (Bland’s rule) : Choose the variable having the smallest index among possible candidates (any nonbasic variable with ) as the entering variable.

Also, if ties occur while choosing the leaving variable, select the smallest indexed variable among the candidates (tied basic variables in the mini-mum ratio test) as the leaving variable.

See the proof in the text.

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Examples of smallest subscript rule:

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enters, leaves

are candidates for leaving.)

Ties (=1/2)

enters, leaves.

are candidates for entering and

are candidates for leaving.)

Ties (=0)