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THE SIMPLEX METHODSTARTING SOLUTIONS

INDR 262-INTRODUCTION TO OPTIMIZATION METHODS

Metin TürkayDepartment of Industrial Engineering

Koç University, Istanbul

ORIGIN IS NOT BASIC FEASIBLE

Ø If the origin IS NOT basic feasible, how should we address the initialization?

If b1, b2, b3 ≥ 0, then we have an initialization problem. Since xe ≥ 0, selecting them would be violation of non-negativity constraints, and equality constraints do not have easily assigned basic variables.

INDR 262 – The Simplex Method: Starting Solutions Metin Türkay 2

max $ = &'(s.t.

,-( ≤ /-,0( ≥ /0,2( = /2( ≥ 0

max $$ − &'( =0

,-( + 6(7 = /-,0( − 6(8 = /0,2( = /2(, (7, (8 ≥ 0

ARTIFICIAL VARIABLES

Ø Define artificial variables to address the initialization of complicating constraints (≥ and =)

Ø When areselected as initial basic variables, the initialization problem is solved.

Ø However, the articial variables are not natural to the problem and should be forced to become non-basic.

2 3, , and sx x x

There are 2 different approaches to forcing the articial variables to become non-basic.1. big-M formulation2. 2-Phase method

max $$ − &'( =0

+,( + .(/ = 0,+1( − .(2 + .(̅1 = 01+4( + .(̅4 = 04(, (/, (2, (̅1, (̅4 ≥ 0


BIG-M FORMULATION

Ø The big-M formulation introduces very large penalty (M) for the artificial variables in the modified objective function.

Then, the problem is solved with the simplex method.

max $$ − &'( +M(̅,+M(̅-=0

01( + 2(3 = 410,( − 2(5 + 2(̅, = 4,0-( + 2(̅- = 4-(, (3, (5, (̅,, (̅- ≥ 0


2-PHASE METHOD

Ø The problem is solved in 2 phases:1. Feasibility Problem where the objective function is

replaced by minimization of the sum of artificial variables.

When the optimal solution is found, then the feasibility is checked by considering the optimal z value (it should be 0 for a feasible solution to the original problem) and the artificial variables cannot be basic.

min $$ − '̅( − '̅)=0

,-' + /'0 = 1-,(' − /'2 + /'̅( = 1(,)' + /'̅) = 1)', '0, '2, '̅(, '̅) ≥ 0


2-PHASE METHOD

2. Optimality Problem where the original objective function is restored, the artifial variables are eliminated and the optimal solution of the Phase 1 problem is used (where [Aʹ Sʹ Eʹ] and bʹ indicate the constraint matrix and right-hand-side at the optimal solution of the Phase 1 problem).

The optimal solution is found starting from this basic feasible solution in Phase 1 to the original problem.

min $$ − &'( =0

+,- ( + /,-(0 + 1,-(2 = 3,-+4- ( + /4-(0 + 14-(2 = 34-+5- ( + /5-(0 + 15-(2 = 35-(, (0, (2 ≥ 0


EXAMPLE

1 2

1 2

1 2

1 2

max =2 3s.t. + 2 4 + 3 , 0

z x x

x xx xx x

+

£=

³

1 2

1 2 3

1 2

1 2 3

max s.t. 2 3 0 +2 + 4 + 3 , , 0

z

z x xx x xx xx x x

- - ===

³


BIG-M

xB z x1 x2 x3 x4 RHS Ratio

z 1 −2 −3 0 M 0 ---

x3 0 1 2 1 0 4

x4 0 1 1 0 1 3

1 2

1 2 3

1 2

1 2 3

max s.t. 2 3 0 +2 + 4 + 3 , , 0

z

z x xx x xx xx x x

- - ===³

1 2 4

1 2 3

1 2 4

1 2 3 4

max s.t. 2 3 0 +2 + 4 + + 3 , , , 0

z

z x x M xx x xx x xx x x x

- - + ===

³

z 1 −M−2 −M−3 0 0 −3M ---

x3 0 1 2 1 0 4 4/2=2

x4 0 1 1 0 1 3 3/1=3

Ø M in row 0 for a basic variable is inconsistent with the definition of basic variable!

Ø M can be enforced to 0 with a Type 2 ero.


BIG-M


z 1 −1/2(M+1) 0 1/2(M+3) 0 −M+6 ---

x2 0 1/2 1 1/2 0 2 2/(1/2)=4

x4 0 1/2 0 −1/2 1 1 1/(1/2)=2

z 1 0 0 1 M+1 7 ---

x2 0 0 1 1 −1 1

x1 0 1 0 −1 2 2

OPTIMAL SOLUTION: z*= 7, x1*=2, x2*=1


2-PHASE


z -1 0 0 0 1 0 ---

x3 0 1 2 1 0 4

x4 0 1 1 0 1 3

1 2

1 2 3

1 2

1 2 3

max s.t. 2 3 0 +2 + 4 + 3 , , 0

z

z x xx x xx xx x x

- - ===

³

4

1 2 3

1 2 4

1 2 3 4

max s.t. 0 +2 + 4 +

(

+ 3 , , , 0

min )z

z xx x xx x xx x

z

x x

-

- + ==

º

=³

z -1 −1 −1 0 0 −3 ---

x3 0 1 2 1 0 4 4/1=4

x4 0 1 1 0 1 3 3/1=3

Ø 1 in row 0 for a basic variable is inconsistent with the definition of basic variable!

Ø 1 can be enforced to 0 with a Type 2 ero.

Phase 1:


2-PHASE


z −1 0 0 0 1 0 ---

x3 0 0 1 1 −1 1

x1 0 1 1 0 1 3

Ø OPTIMAL SOLUTION for the Phase 1 problem.Ø This solution is basic feasible to the original problem.Ø We can construct the Phase 2 problem from the optimal solution to the

Phase 1 problem.

Ø Objective function value for Phase 1 problem is 0.

Ø No artificial variables in the basis.

1 2

2 3

1 2

1 2 3

max s.t. 2 3 0 + 1 + 3 , , 0

z

z x xx x

x xx x x

- - ===

³

Phase 2:


2-PHASExB z x1 x2 x3 RHS Ratio

z 1 −2 −3 0 0 ---

x3 0 0 1 1 1

x1 0 1 1 0 3

z 1 0 −1 0 6 ---

x3 0 0 1 1 1 1/1=1

x1 0 1 1 0 3 3/1=3

OPTIMAL SOLUTION: z*= 7, x1*=2, x2*=1

Ø −2 in row 0 for a basic variable is inconsistent with the definition of basic variable!


z 1 0 0 1 7 ---

x2 0 0 1 1 1

x1 0 1 0 −1 2


ANOTHER EXAMPLE

1 2

1 2

1 2

1 2

1 2

min =4s.t. 3 + 3 4 +3 6 +2 4 , 0

z x x

x xx xx xx x

+

=³£³

1 2

1 2

1 2 3

1 2 4

1 2 3 4

min =4s.t. 3 + 3 4 +3 6 +2 4 , , , 0

z x x

x xx x xx x xx x x x

+

=- =

+ =³


BIG-M

xB z x1 x2 x3 x4 x5 x6 RHS Ratio

z −1 4 1 0 0 M M 0 ---x5 0 3 1 0 0 1 0 3x6 0 4 3 −1 0 0 1 6x4 0 1 2 0 1 0 0 4

1 2 5 6

1 2 5

1 2 3 6

1 2 4

max s.t. 4 0 3 + 3 4 +3 6

( min

+2 4

)

z

z x x Mx Mxx x xx x

x

z

x xx x

º-

- + + + + =+ =

- + =+ =

1 2 3 4 5 6 , , , , , 0x x x x x x ³

Ø M in row 0 for a basic variable is inconsistent with the definition of basic variable!

Ø M can be enforced to 0 with a Type 2 ero.

1 2

1 2

1 2 3

1 2 4

1 2 3 4

min =4s.t. 3 + 3 4 +3 6 +2 4 , , , 0

z x x


+

=- =

+ =³

z −1 −7M+4 −4M+1 M 0 0 0 -9M ---x5 0 3 1 0 0 1 0 3 3/3=1x6 0 4 3 −1 0 0 1 6 6/4=3/2

x4 0 1 2 0 1 0 0 4 4/1=4INDR 262 – The Simplex Method: Starting Solutions Metin Türkay 14

BIG-M


z −1 0 −1/3(5M+1) M 0 1/3(7M-4) 0 −2M−4 ---x1 0 1 1/3 0 0 1/3 0 1 1/(1/3)=3

x6 0 0 5/3 −1 0 −4/3 1 2 2/(5/3)=6/5

x4 0 0 5/3 0 1 −1/3 0 3 3/(5/3)=9/5

z −1 0 0 −1/5 0 M-8/5 M+1/5 -18/5 ---x1 0 1 0 1/5 0 3/5 −1/5 3/5 3x2 0 0 1 −3/5 0 −4/5 3/5 6/5 ---x4 0 0 0 1 1 4/3 1 1 1

z −1 0 0 0 1/5 M-4/3 M+2/5 -17/5 ---x1 0 1 0 0 −1/5 1/3 −2/5 2/5x2 0 0 1 0 3/5 0 6/5 9/5x3 0 0 0 1 1 4/3 1 1

OPTIMAL SOLUTION: z*= 17/5x1*=2/5, x2*=9/5

2-PHASE


z −1 0 0 0 0 1 1 0 ---x5 0 3 1 0 0 1 0 3x6 0 4 3 −1 0 0 1 6x4 0 1 2 0 1 0 0 4

5 6

1 2 5

1 2 3 6

1 2 4

max s.t. 0 3 + 3 4 +3 6 +2 4

( min

)z

z x xx x xx x x xx x

z

x

-

- + + =+ =

- +

º

=+ =

1 2 3 4 5 6 , , , , , 0x x x x x x ³

Ø 1 in row 0 for a basic variable is inconsistent with the definition of basic variable!


INDR 262 – The Simplex Method Metin Türkay 16

1 2

1 2

1 2 3

1 2 4

1 2 3 4

min =4s.t. 3 + 3 4 +3 6 +2 4 , , , 0

z x x


+

=- =

+ =³

z −1 −7 −4 1 0 0 0 −9 ---x5 0 3 1 0 0 1 0 3 3/3=1x6 0 4 3 −1 0 0 1 6 6/4=3/2

x4 0 1 2 0 1 0 0 4 4/1=4

Phase 1:

2-PHASE


z −1 0 −5/3 1 0 7/3 0 −2 ---x1 0 1 1/3 0 0 1/3 0 1 1/(1/3)=3

x6 0 0 5/3 −1 0 −4/3 1 2 2/(5/3)=6/5

x4 0 0 5/3 0 1 −1/3 0 3 3/(5/3)=9/5


z −1 0 0 0 0 1 1 0 ---x1 0 1 0 1/5 0 3/5 −1/5 3/5x2 0 0 1 −3/5 0 −4/5 3/5 6/5x4 0 0 0 1 1 1 −1 1

Ø Objective function value for Phase 1 problem is 0.

Ø No artificial variables in the basis.

Ø OPTIMAL SOLUTION for the Phase 1 problem.Ø This solution is basic feasible to the original problem.Ø We can construct the Phase 2 problem from the optimal solution to the

Phase 1 problem.

2-PHASE


z −1 4 1 0 0 0 ---x1 0 1 0 1/5 0 3/5x2 0 0 1 −3/5 0 6/5x4 0 0 0 1 1 1


1 2

1 3

2 3

3 4

max s.t. 4 0 +1/5 3/5 3/5 6 / 5

( min )z

z x xx x

x x

z

x x

-

- + + ===

+

º

-=

1 2 3 4

1 , , , 0x x x x ³

Phase 2:

Ø Positive values in row 0 for

basic variables is inconsistent

with the definition of a basic

variable!

Ø They can be enforced to 0with a Type 2 ero. z −1 0 0 −1/5 0 −18/5 ---

x1 0 1 0 1/5 0 3/5 3x2 0 0 1 −3/5 0 6/5 ---x4 0 0 0 1 1 1 1

2-PHASE


z −1 0 0 0 1/5 −17/5 ---x1 0 1 0 0 −1/5 2/5x2 0 0 1 0 3/5 9/5x3 0 0 0 1 1 1

OPTIMAL SOLUTION: z*= 17/5x1*=2/5, x2*=9/5


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