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OR 3310/5310/5311 Spring 2012
Supplemental Handout
II.15 Theorem:
With least index rules, the simplex algorithm will not cycle.
Proof:
Assume least index rules are used, but a cycle occurs; we show this is impossible by deriving
a contradiction. During the cycle we pass through a sequence of degenerate pivots, eventually
repeating a basis; in these iterations certain variables enter and leave the basis at level zero. Among
these variables, let
x
‘
be the one with the largest
index.
Denote by
A,b,c
the data for the representation when
x
‘
leaves
the basis. Suppose
x
‘
is basic
in row
r
of this representation (hence
x
‘
=
b
r
= 0) and that
x
k
enters the basis to replace
x
‘
on
this iteration. We then have:
c
k
>
0 (
x
k
enters)
(1)
a
rk
>
0 (pivot on
a
rk
)
.
(2)
Denote by
¯
A,
¯
b,
¯
c
the data for the representation when
x
‘
enters
the basis. At this iteration we
have:
¯
c
‘
>
0 (
x
‘
enters)
.
(3)
And since
x
k
also enters/leaves during the cycle, we have
k < ‘
, so the least index choice of
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 Spring '08
 BLAND

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