Linear Programming with Inexact Data is NP-Hard
Rohn, Ji
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1995
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INSTITUTE OF COMPUTER SCIENCE
ACADEMY OF SCIENCES OF THE CZECH REPUBLIC
Linear Programming with Inexact Data is
NP-Hard
Jir Rohn
Technical report No. 642
June 20, 1995
Institute of Computer Science, Academy of Sciences of the Czech Republic
Pod vo drenskou v 2, 182 07 Prague 8, Czech Republic
phone: (+422) 66414244 fax: (+422) 8585789
e-mail: uivt@uivt.cas.cz
INSTITUTE OF COMPUTER SCIENCE
ACADEMY OF SCIENCES OF THE CZECH REPUBLIC
Linear Programming with Inexact Data is
NP-Hard
1
Jir Rohn
2
Technical report No. 642
June 20, 1995
Abstract
We prove that the problem of checking whether all linear programming problems whose
data range in prescrib ed intervals have optimal solutions is NP-hard.
Keywords
Linear programming, inexact data, NP-hardness
1
This work was supp orted by the Czech Republic Grant Agency under grantGA
CR 201/95/1484
2
Faculty of Mathematics and Physics, Charles University, Prague (rohn@kam.ms.m.cuni.cz) and
Institute of Computer Science, Academy of Sciences, Prague, Czech Republic (rohn@uivt.cas.cz)
1 Intro duction
Consider a family of linear programming (LP) problems
min
f
c
T
x
Ax
=
b x
0
g
(1.1)
for all data satisfying
A
2
A
I
b
2
b
I
c
2
c
I
(1.2)
where
A
I
=
f
A
A
A
A
g
is an
m
n
interval matrix,
m
n
,and
b
I
=
f
b
b
b
b
g
c
I
=
f
c
c
c
c
g
are interval vectors of dimensions
m
and
n
, resp ectively (the inequalities are understood
componentwise). The family (1.1), (1.2) maybe interpreted as a linear programming
problem with inexact data, or as a fully parametrized parametric linear programming
problem.
The problem of existence of optimal solutions of all linear programming problems in
the family (1.1), (1.2) was addressed in 6]. There it was proved that each LP problem
(1.1) with data satisfying (1.2) has an optimal solution if and only if the LP problem
min
f
c
T
x
A
x
b Ax
bx
0
g
has an optimal solution and each of the 2
m
systems
Ax
=
b
whose eachrowiseither
of the form
(
A
x
)
i
=
b
i
or of the form
(
Ax
)
i
=
b
i
(
i
=1
:::m
) has a nonnegative solution. Hence, wehave a nitely veriable necessary
and sucient condition, but the numb er of systems to b e checked for nonnegative
solvability is exp onential in
m
.
In the main result of this pap er weshow that the problem in question is NP-hard
(cf. Garey and Johnson 1]). Hence, unless the famous conjecture "P
6
=NP" 1] is
false, there do es not exist a necessary and sucient condition for checking existence of
optimal solutions of all LP problems (1.1), (1.2) whichcouldbe veried in p olynomial
time. The pro of given in section 2 shows that even checking
feasibility
of all LP
problems in the family (1.1), (1.2) is NP-hard. Some concluding remarks are given in
section 3.
1
2 Main result
Theorem 1
The fol lowing problem is NP-hard:
Instance.
A
I
b
I
c
I
(with rational bounds).
Question.
Does each LP problem (1.1) with data (1.2) have an optimal solution?
Proof.
0) For the purpose of the proof, let us introduce
A
c
=
1
2
(
A
+
A
)
=
1
2
(
A
;
A
)
b
c
=
1
2
(
b
+
b
) and
=
1
2
(
b
;
b
), so that
A
I
=
A
c
;
A
c
+]
and
b
I
=
b
c
;
b
c
+
]
:
The proof goes through several steps.
1) First weprove that each system
Ax
=
b x
0 (2.1)
with data satisfying
A
2
A
I
b
2
b
I
(2.2)
has a solution if and only if
(
8
y
)(
A
T
c
y
+
T
j
y
j
0
)
b
T
c
y
;
T
j
y
j
0) (2.3)
holds. "Only if": Let each system (2.1) with data (2.2) have a solution, and let
A
T
c
y
+
T
j
y
j
0 for some
y
2
IR
m
. Dene a diagonal matrix
T
by
T
ii
=1 if
y
i
0,
T
ii
=
;
1if
y
i
<
0, and
T
ij
=0 if
i
6
=
j
(
i j
=1
:::m
), then
j
y
j
=
Ty
. Consider now
the system
(
A
c
+
T
)
x
=
b
c
;
Tx
0
:
(2.4)
Since
A
c
+
T
2
A
I
and
b
c
;
T
2
b
I
, the system (2.4) has a solution according to
the assumption, and (
A
c
+
T
)
T
y
=
A
T
c
y
+
T
j
y
j
0
hence Farkas lemma 3] applied
to (2.4) gives that
b
T
c
y
;
T
j
y
j
=(
b
c
;
T
)
T
y
0, which proves (2.3). "If": Assuming
that (2.3) holds, consider a system (2.1) with data satisfying (2.2). Let
A
T
y
0 for
some
y
then
A
T
c
y
+
T
j
y
j
(
A
c
+
A
;
A
c
)
T
y
=
A
T
y
0, hence (2.3) gives that
b
T
y
=(
b
c
+
b
;
b
c
)
T
y
b
T
c
y
;
T
j
y
j
0. Thus wehaveproved that for each
y
,
A
T
y
0
implies
b
T
y
0, and Fark
as lemma proves the existence of a solution to (2.1).
2) For a given
square
m
m
interval matrix
A
I
0
=
A
0
c
;
0
A
0
c
+
0
], construct an
m
2
m
interval matrix
A
I
=
A
c
;
A
c
+] (2.5)
with
A
c
=(
A
0
T
c
;
A
0
T
c
)
(2.6)
=(
0
T
0
T
)
(2.7)
and an interval
m
-vector
b
I
=
;
e e
]
(2.8)
2