Question

In the simplex algorithm can you have a degenerate pivot without cycling occuring? I know that Bland's rule prevents cycling but what about degenrate pivots? Or are degenerate pivots the same as cycling?

Also, when a variables leaves the tableau im assuming it cant come back the tableau for bland's?

Answer 1

Degenerate Pivots and Cycling

A rotate in the Simplex Method is supposed to be degenerate when it

doesn't change the essential arrangement. This happens when we get a proportion of 0 in picking the leaving variable.

Degenerate turns are very normal, and typically innocuous. Be that as it may, it's workable for cycling to happen in an arrangement of savage turns. This implies that a similar scene happens more than once. At the point when that occurs, the Simplex Method would continue rehashing a succession of turns until the end of time.

The accompanying model shows cycling, utilizing our standard turning rules.

maximize    z = 10x₁ − 57x₂ − 9x₃ − 24x₄

subject to            1/2x₁ − 11/2x₂ − 5/2x₃ + 9x₄ ≤ 0

1/2x₁ − 3/2x₂ − 1/2x₃ + x₄ ≤ 0

x₁ + x₂ + x₃ + x₄   ≤ 1

x₁, x₂, x₃, x₄ ≥ 0

Initial tableau:

z x₁                x₂             x₃x₄s₁s₂s₃            rhs

1	−10	57	9	24	0	0	0	0	=	z
0 0 0	1/2 1/2 1	−11/2 −3/2 1	−5/2 −1/2 1	9 1 1	1 0 0	0 1 0	0 0 1	0 0 1	= = =	s1 s2 s3

xx₁ enters, s₁ leaves

z x₁           x₂           x₃x₄   s₁   s₂   s₃            rhs

1	0	−53	−41	204	20	0	0	0	=	z
0 0 0	1 0 0	−11 4 12	−5 2 6	18 −8 −17	2 −1 −2	0 1 0	0 0 1	0 0 1	= = =	x1 s2 s3

xx₂ eters (since −53 < −41), s₂ leaves

z x₁x₂           x₃   x₄              s₁   s₂   s₃ rhs

1	0	0	−29/2	98	27/4	53/4	0	0	=	z
0 0 0	1 0 0	0 1 0	1/2 1/2 0	−4 −2 7	−3/4 −1/4 1	11/4 1/4 −3	0 0 1	0 0 1	= = =	x1 x2 s3

x₃ enters, x₁ leaves

z x₁   x₂   x₃           x₄              s₁              s₂   s₃ rhs

1	29	0	0	−18	−15	93	0	0	=	z
0 0 0	2 −1 0	0 1 0	1 0 0	−8 2 7	−3/2 1/2 1	11/2 −5/2 −3	0 0 1	0 0 1	= = =	x3 x2 s3

xx₄ enters (−18 < −15), x₂ leaves

z          x₁             x₂x₃x₄                s₁               s₂s₃ rhs

1	20	9	0	0	−21/2	141/2	0	0	=	z
0 0 0	−2 −1/2 7/2	4 1/2 −7/2	1 0 0	0 1 0	1/2 1/4 −3/4	−9/2 −5/4 23/4	0 0 1	0 0 1	= = =	x3 x4 s3

s1 s1 enters, x3 leaves (another tie for littlest proportion)

z        x₁             x₂   x₃x₄       s₁           s₂s₃ rhs

1	−22	93	21	0	0	−24	0	0	=	z
0 0 0	1−4 /2 1/2	8 −3/2 5/2	2 −1/2 3/2	0 1 0	1 0 0	−9 1 −1	0 0 1	0 0 1	= = =	s1 x4 s3

At that point x3 would have entered, and s3 left. This would be, finally, a nondegenerate turn, creating an ideal scene:

z        x₁                x₂             x₃x₄   s₁       s₂s₃ rhs   

1	−10	57	9	24	0	0	0	0	=	z
0 0 0	1/2 1/2 1	−11/2 −3/2 1	−5/2 −1/2 1	9 1 1	1 0 0	0 1 0	0 0 1	0 0 1	= = =	s1 s2 s3

One strategy for forestalling cycling is to utilize Bland's Rule rather than the Most Negative Coefficient Rule. Given a fixed requesting of the factors, for example x1, x2, . . . , xn, s1, . . . , sm, Bland's Rule says:

(a)    When picking the entering variable, take the first in the requesting that has a negative section in the goal line.

(b)   When picking the leaving variable, if there is a tie for least proportion, take the competitor that is first in the requesting.

On the off chance that we were utilizing Bland's Rule, everything would have been the equivalent up to that last rotate, where x1 would have entered rather than s2, and x4 would have left.

z x₁         x₂         x₃x₄   s₁         s₂s₃ rhs

1	0	27	−1	44	0	20	0	0	=	z
0 0 0	0 1 0	−4 −3 4	−2 −1 2	8 2 −1	1 0 0	−1 2 −2	0 0 1	0 0 1	= = =	s1 x1 s3

At that point x3 would have entered, and s3 left. This would be, finally, a nondegenerate turn, creating an ideal scene:

At that point x3 would have entered, and s3 left. This would be, finally, a nondegenerate turn, creating an ideal scene:

z x₁         x₂   x₃           x₄   s₁         s₂          s₃ rhs

1	0	29	0	87/2	0	19	1/2	1/2	=	z
0 0 0	0 1 0	0 −1 2	0 0 1	7 3/2 −1/2	1 0 0	−3 1 −1	1 1/2 1/2	1 1/2 1/2	= = =	s1 x1 x3