Handout for Math 164: Simplex Method Formulae, Study notes of Optimization Techniques in Engineering

The general formulae for the simplex method in linear programming. It includes the reordering of variables and the expression of basic variables and the objective function in terms of non-basic variables. The document also shows how to solve for xb and use it to find the value of z.

Typology: Study notes

Pre 2010

Uploaded on 08/30/2009

koofers-user-ltg
koofers-user-ltg 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 164 - Handout Used for 5.2.1 - General Formulae
for the Simplex Method
Andrea Brose, May 19, 2005
Consider the linear programming problem in standard form
minimize z=cTx
subject to Ax =b
x O
()
with b O.
1. Reordering
Write1
x=xB
xNand A=B N
then
z=cTx
=cT
BcT
NxB
xN
=cT
BxB+cT
NxN(1)
We then can rewrite problem () as follows
minimize z=cT
BxB+cT
NxN(2)
subject to BxB+N xN=b(3)
x O
Bis basis matrix, hence non-singular, i.e. B1exists.
1see section 4.3 for notation
1
pf2

Partial preview of the text

Download Handout for Math 164: Simplex Method Formulae and more Study notes Optimization Techniques in Engineering in PDF only on Docsity!

Math 164 - Handout Used for 5.2.1 - General Formulae

for the Simplex Method

Andrea Brose, May 19, 2005

Consider the linear programming problem in standard form

minimize z = cT^ x

subject to Ax = b

x ≥ O

with b ≥ O.

  1. Reordering

Write^1

x =

[

xB

xN

]

and A =

[

B N

]

then

z = c

T x

[

c

T B c

T N

]

[

xB

xN

]

= c

T B xB^ +^ c

T N xN^ (1)

We then can rewrite problem (∗) as follows

minimize z = c

T B xB^ +^ c

T N xN^ (2)

subject to BxB + N xN = b (3)

x ≥ O

B is basis matrix, hence non-singular, i.e. B

− 1 exists.

(^1) see section 4.3 for notation

  1. Express Basic Variables and Objective in terms of the Non-Basic Vari-

ables

Solving (3) for xB yields

xB = B

− 1 b − B

− 1 N xN. (4)

Using (4) in (1) yields

z = c

T B xB^ +^ c

T N xN

= c

T B (B

− 1 b − B

− 1 xN ) + c

T N xN^ by (4)

= c

T B B

− 1 ︸ ︷︷ ︸ def = yT

b + (c

T N −^ c

T B B

− 1 ︸ ︷︷ ︸ def = yT

N )xN