SA��� – Linear Programming
Asst. Prof. Nelson Uhan
Spring ����
Lesson ��. �e Simplex Method – Example
Problem �. Consider the following LP
�x� + �x� + �x�
maximize
subject to �x� − x� + �x� ≤ ��
(�)
�x� + �x� + �x� ≤ ��
x� , x� , x� ≥ �
�e canonical form of this LP is
maximize
�x� + �x� + �x�
subject to �x� − x� + �x� + s� = ��
(�)
�x� + �x� + �x� + s� = ��
x� , x� , x� , s� , s� ≥ �
a. Use the simplex method to solve the canonical form LP (�). In particular:
● Use the initial BFS x� = (�, �, �, ��, ��) with basis B � = {s� , s� }.
● Choose your entering variable using Dantzig’s rule – that is, choose the improving simplex
direction with the most positive reduced cost. (If this was a minimization LP, you would choose
the improving simplex direction with the most negative reduced cost.)
b. What is the optimal value of the canonical form LP (�)? Give an optimal solution.
c. What is the optimal value of the original LP (�)? Give an optimal solution.
atb
I°=(
.
E
90,918,10
B°={
)
s,
,sz
}
€3
€2
IEx3=IaYoen0Ei×b
AE×2=o
⇒£×2=(
⇒E×3=(
Ea=3
:
Ex
AT
( 1,0 ,9ds ,dsD
'
-
'=O
1,90
,
-2
,
-4
MRI
⇒k
tmax
:
'
.
'
-
minstft
)
,
kottmaxd
Sz
,
"=(
't }=2
0,0
,
Ad
,=O
2+dsz=o
Ex ,=4
S,
,
.
,
:
4tdsz=0
⇒E×'=(
,ds ,dsD
:
991
0,91
5tdsz=0
4+9=0
E×3=(
E×2=(
0%0,1
ttds "=O
91,0
Ztds ,=0
:
,ds ,ds )
:
,
2,10 ,o
Sz
leaving
is
B
)
�
'={×3,s }
,
-2
)
:
,
-4, 5)
.
I
'
(0,931%0)
-
E :X '=(
'
1,0
B
'={
xns
,
}
,
dx3.dssDE2iX2-C0.1.dx3.dssDdI2ids2-Co.0.dx3.dssDAdxtoi2t4dx3tdg-oAoEiOi-1t4dx3tdg-0Aoeoi4dx3tdg-o4t5dxs-o2t5dxs-01t5dxs-o-Exha.oiEEs.D-ExECo.h-Itt.D-ExECo.atsts.DIy-OF@choosexzIsz.y
as
×}
MIT
1ma×=min{¥}=5
:
E=x" txmaxd
⇒
"
-
( 0,59230 )
E'=(
2(
leaving
is
×3
)
entering
B2={
xns
,
}
}
5+2×2=0
0,1×2,1
4t2d×z
B2={
1+2×2=0
⇒d→=(
4- 0,1×2,0
:d→=(
E2=(
0,5 "=O
at
3
,lI
0,230
,
E :X
'
xns
,
"
E
1,1×2,0 ,ds ,o )
,ds
,0 )
,ds
)
A££=O
⇒£h=(
,
AI×'=o
AI
dxztds ,=o
:
,
:
dxztds ,=o
:
:
-
,
dxztds ,=o
=O
⇒E×'
all
,
-2,0
,
-4,0 )
o
-551 ,
,
'
-
E ,o )
xtnisoptimalwithualuelbc
Ex ,=
No simplex
.
In
I×=
-2
directions
the
are
3
×
,=0
±
I
2
Sz
=
-
2
⇒
improving
original LP,
-
,o
,
-12=5
,
Xs=0
is
an
optimal
solution
with value
15
'
,
's
)