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THE SIMPLEX METHOD

INDR 262-INTRODUCTION TO OPTIMIZATION METHODSMetin Türkay

Department of Industrial EngineeringKoç University, Istanbul

WYNDOR GLASS CO

x1

x2x1 ≤4

2x2≤12

3x1 +2x

2 ≤18

A

B C

D

E

INDR 262 – The Simplex Method Metin Türkay 2

max $ = 3'( + 5'+s.t.

'( ≤ 42'+ ≤ 12

3'( + 2'+ ≤ 18'(, '+ ≥ 0

ADJACENT CPF SOLUTIONS

ØDefinition: Adjacent CPF solutions

For any linear programming problem with ndecision variables, two CPF solutions are

adjacent to each other if they share n-1

constraint boundaries. The two adjacent CPF

solutions are connected by a line segment that

lies on these same shared constraint

boundaries. Such a line is referred to as an edge

on the feasible region.


EXAMPLE: WYNDOR GLASS CO.

Point (x1,x2) Adjacent CPF SolutionsA (0,0) B(0,6)

E(4,0)B (0,6) A(0,0)

C(2,6)C (2,6) B(0,6)

D(4,3)D (4,3) C(2,6)

E(4,0)E (4,0) A(0,0)

D(4,3)

x1

x2

x1 ≤4

2x2≤12

3x1 +2x

2 ≤18

A

B C

D

E


OPTIMALITY

Ø If a CPF solution has no adjacent CPF solutions that are better, then it must be an optimal solution.


OPTIMIZATION ALGORITHM

Ø0. Initialize: choose an initial CPF solution.

Ø1. Optimality Test: evaluate the performance measure at the current solution. If its value is larger than all of its adjacent CPF solutions; the current solution is optimal. Otherwise go to Step 2.

Ø2. Iteration: Move to a better adjacent CPF solution. Evaluate the objective function value. Go to Step 1.


WYNDOR GLASS CO

x1

x2

x1 ≤4

2x2≤12

3x1 +2x

2 ≤18

A

B C

D

E

0. Initialization:Select A as the initial CPF solution

1. Optimality Test:max z=3x1+5x2 can be rearranged asmax z-3x1−5x2=0Since an increase in x1 or x2 results in anincrease in z value, the current solution is not

optimal. 2. Iteration 1:

Both x1 or x2 directions improve the z value.We choose x2 direction since z increases fasterin x2 direction. That will take us to CPF B.z value at CPF solution B is 30.

1. Optimality Test:There are 2 extreme directions that we can follow at CPF solution B.B → A (does not improve the z value, prev. sol.)B → C (improves the z value, while x2 remains

the same, x1 increases)if we further increase x2, feasibility is violated


WYNDOR GLASS CO

x1

x2

x1 ≤4

2x2≤12

3x1 +2x

2 ≤18

A

B C

D

E

2. Iteration 2:The x1 direction improves the z value.That will take us to CPF C.z value at CPF solution C is 36.

1. Optimality Test:There are 2 extreme directions that we can follow at CPF solution C.C → B (does not improve the z value, prev. sol.)C → D (does not improve the z value,

the slope of the constraint is 3/2 while the slope of the objective function is 3/5.So, objective function value decreaseswhen we move in this in direction)

the other two directions violate feasibility.

THE OPTIMAL SOLUTION IS FOUND!z*=36

x1* = 2, x2*=6


CONSIDERATIONSØFocus solely on CPF solutions.ØConduct the search for the optimal solution

iteratively. ØWhenever possible, choose the origin as the initial

CPF solution.ØGiven a CPF solution, it is much quicker to gather

information about its adjacent CPF solutions than its non-adjacent CPF solutions.


CONSIDERATIONSØAfter the current CPF solution is identified, the simplex

method examines each of the vertices of the feasible region

that emerge from this CPF solution. The most promising

vertex is selected for the next iteration.

ØA positive rate of improvement in z implies that the adjacent

CPF solution is better; a negative rate of improvement in zimplies that the adjacent CPF solution is worse. The

optimality test consists of simply of checking whether any of

the vertices give a positive rate improvement in z. If none

do, then the current CPF solution is optimal.


SLACK VARIABLES

Augmented Form of the LP Model


x2

x10 2 4

x1 x3

!" ≤ 4introduce slack variable !%

!" + !% = 4and !% ≥ 0

2!+ ≤ 122!+ + !- = 12

!- ≥ 0

3!" + 2!+ ≤ 183!" + 2!+ + !0 = 18

!0 ≥ 0

max 4 = 3!" + 5!+s.t.

!" + !% = 42!+ + !- = 12

3!" + 2!+ + !0 = 18!", !+, !%, !-, !0 ≥ 0

DEFINITIONS

Ø Augmented Solution: a solution for the original variables that has been augmented by the corresponding values at the slack variables.§ Ex:

Ø Basic Solution: an augmented corner-point solution. § Ex: Points A, B, C, D, E are basic feasible solutions.

• Point A:


!"∗ = 2, !'∗ = 6!) = 2, !* = 2, !+ = 0

!" = 0, !' = 0, !) = 4, !* = 12, !+ = 18

BASIC SOLUTIONS

Ø Each variable is either a basic variable or non-basic variable.

Ø # of basic variables = # of functional constraints.Ø Non-basic variables are set equal to zero.Ø Values of basic variables are obtained by solving

system of equations.Ø If basic variables satisfy non-negativity

constraints, the basic solution is a BFS.


THE SIMPLEX METHOD

Ø The Simplex Method

Initialization

Optimality Test

Iteration

STOP


MODEL REPRESENTATION

Ø Model is rearranged as:


max $ = 3'( + 5'+s.t.

'( + '/ = 42'+ + '2 = 12

3'( + 2'+ + '4 = 18'(, '+, '/, '2, '4 ≥ 0

max $0 $ − 3'( − 5'+ =01 '( + '/ = 42 2'+ + '2 = 12(3) 3'( + 2'+ + '4 = 18

'(, '+, '/, '2, '4 ≥ 0

INITIALIZATION

Ø Since there are 3 equalities, we need to select 3 variables as basic and 2 as non-basic variables.

Ø select x3, x4 and x5 as basic variables


0 " − 3%& − 5%( =01 %& + %- = 42 2%( + %0 = 12(3) 3%& + 2%( + %3 = 18

%&, %(, %-, %0, %3 ≥ 0

%& = 0, %( = 0%- = 4, %0 = 12, %3 = 18

OPTIMALITY TEST

Ø Not Optimal!!!

Ø since ! − 3$% − 5$'=0

if $% or $' increase, ! will also increase.


ITERATION

Ø determine the direction of movementüConsider the objective function (row 0)ü if we choose to move in the direction of x1, the rate of

improvement in z will be 3 (δz/δx1=3)ü if we choose to move in the direction of x2, the rate of

improvement in z will be 5 (δz/δx2=5)üSo, we choose x2ü x2 is called the entering basic variable


ITERATION

Ø determine the limit on the movementüConsider the set of constraintsü x1=0, we will determine the maximum value that x2 will

get (as the entering basic variable) without violating any of the constraints.

1 "# + "% = 42 2") + "* = 12(3) 3"# + 2") + "/ = 18


191.38

THE MINIMUM RATIO

Ø Row1 "# + "% = 4 "% = 42 2") + "* = 12 if ") ≤ #)

) ≤ , the constraint is not violated(3) 3"# + 2") + "= = 18 if ") ≤ #?

) ≤ 9 the constraint is not violated

Ø Select the minimum value as the limit on the movement ⇒ ") = 6This would guarantee feasibility.

Ø These calculations are called the minimum ratio test. Also identify the basic variable associated with the row that is selected in the minimum ratio test as the leaving basic variable ⇒ "* = 0


ITERATION

Ø Solve the new basic feasible solution

ØWe want the column that corresponds to the entering basic variable as:New Old

Row x2 x2(0) 0 -5(1) 0 0(2) 1 2(3) 0 2

1 2

1 3

2 4

1 2 5

1

(0) 3 5 0 (1) 4 (2) 2 12 (3) 3 2 18 ,

z x xx x

x xx x xx x

- - =+ =

+ =+ + =

2 3 4 5, , , 0x x x ³

Pivot element


PIVOTING

Ø Multiply row 2 by (2)-1:

Ø Multiply row 2 by 5 and add to row 0:

1 2

1 3

2 4

1 2

Row Equation (0) 3 5 0 (1) 4

1 (2) 62

(3) 3 2

z x xx x

x x

x x

- - =+ =

+ =

+ 5 18x+ =

1 4

1 3

2 4

1

Row Equation 5 (0) 3 302

(1) 41 (2) 62

(3) 3 2

z x x

x x

x x

x

- + =

+ =

+ =

+ 2 5 18x x+ =INDR 262 – The Simplex Method Metin Türkay 22

PIVOTING

Ø Multiply row 2 by -2 and add to row 3:

Ø BF Solution:1

2

3

4

5

06406

xxxxx

é ù é ùê ú ê úê ú ê úê ú ê ú=ê ú ê úê ú ê úê ú ê úë ûë û

1 4

1 3

2 4

1

Row Equation 5 (0) 3 + 302

(1) 41 (2) 62

(3) 3

z x x

x x

x x

x

- =

+ =

+ =

4 5 6x x- + =


OPTIMALITY TEST

Ø z = 30+3x1− 5/2x4Ø a move in the direction of x1 would increase

the objective function value because it has a positive cost coefficient.


ITERATION #2

Ø x1 is the entering basic variable.Ø Minimum ratio test:

Ø x1=2, x5 is the leaving variable.

Row Equation Ratio

(1) x1 + x3 = 4 x3 = 4− x1 ≥ 0⇒ x1 ≤41= 4

(2) x2 + 12x4 = 6 x2 = 6 ≥ 0 no upper limit on x1

(3) 3x1 − x4 + x5 = 6 x5 = 6−3x1 ≥ 0⇒ x1 ≤63= 2


NEW BF SOLUTION

Ø ero’s:ü Multiply row 3 by (3)-1

ü Multiply row 3 by 3 and add to row 0ü Multiply row 3 by -1 and add to row 1

1 4

1 3

2 4

1

Row Equation 5 (0) 3 302

(1) 41 (2) 62

(3) 3

z x x

x x

x x

x

- + =

+ =

+ =

4 5 6x x- + =


BF SOLUTION

Ø New BF solution:

Ø The current BF solution is OPTIMAL!Ø z*=36; x1=2, x2=6, x3=2, x4=x5=0

4 5

3 4 5

2 4

1 4 5

3 (0) 3621 1 (1) 23 31 (2) 621 1 (3) 23 3

z x x

x x x

x x

x x x

+ + =

+ =

+ =

- + =

-


TABLEAU FORMAT

Ø All of the information on the iterations of the simplex method can be represented in a systematic form

Ø The most natural representation uses the matrix form (in numerical values) of the objective function and the constraints


TABLEAU FORMAT

xB z x1 ... xn+1 ... RHS Ratio

z 1 −c1 ... 0 ... 0 ---

xn+1 0 a11 ... 1 ... b1

xn+2 0 a21 ... 0 ... b2

... 0 ... ... 0 ... ...

INITIAL TABLEAUX

Basic Variables

Objective FunctionVariable

Original Variables

Slack Variables

Right Hand Side

Table Header

Row 0

Row 1

Row 2

Row ...

Ratio Calculation

Update the Tableaux for each iteration of the Simplex method.INDR 262 – The Simplex Method Metin Türkay 29

WYNDOR GLASS CO.

xB z x1 x2 x3 x4 x5 RHS Ratio

z 1 −3 −5 0 0 0 0 ---

x3 0 1 0 1 0 0 4 4/0=∞

x4 0 0 2 0 1 0 12 12/2=6

x5 0 3 2 0 0 1 18 18/2=9

Entering Basic Variable, also the Pivot Column(the variabe with the most negative Row 0 coefficient)

Leaving Basic Variable, also the Pivot Row(the row with the minimum ratio) Pivot Element

(all ero’s will be defined with this element)

INITIAL TABLEAUX


WYNDOR GLASS CO.


z 1 −3 0 0 5/2 0 30 ---

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

x2 0 0 1 0 1/2 0 6 6/0=∞

x5 0 3 0 0 −1 1 6 6/3=2

ITERATION (TABLEAUX) 1:Update the initial tableaux with the following ero’s:New Row 2 = (1/2) Old Row 2New Row 0 = (+5) New Row 2 + Old Row 0New Row 1 = (0) New Row 2 + Old Row 1New Row 3 = (-2) New Row 2 + Old Row 3


WYNDOR GLASS CO.


z 1 0 0 0 3/2 1 36 ---

x3 0 0 0 1 1/3 −1/3 2

x2 0 0 1 0 1/2 0 6

x1 0 1 0 0 −1/3 1/3 2

ITERATION (TABLEAUX) 2:Update the initial tableaux with the following ero’s:New Row 3 = (1/3) Old Row 3New Row 0 = (+3) New Row 3 + Old Row 0New Row 1 = (-1) New Row 3 + Old Row 1New Row 2 = (0) New Row 3 + Old Row 2

OPTIMAL SOLUTION: z*=36, x1*=2, x2*=6INDR 262 – The Simplex Method Metin Türkay 32

ANOTHER EXAMPLE

1 2 3

1 2 3

1 3

1 2 3

1 2 3

1 2 3

max =5 4 3s.t. +3 3 3 2 2 2 4 2 +3 2 , , 0

z x x x

x x xx xx x xx x xx x x

+ +

+ £- + £

+ + £+ £³

1 2 3

1 2 3 4

1 3 5

1 2 3 6

max s.t. 5 4 3 =0 +3 3 3 2 2 2 4

z

z x x xx x x xx x xx x x x

- - -

+ + =- + + =

+ + + =

1 2 3 7

1 2 3 4 5 6 7

2 +3 2 , , , , , , 0

x x x xx x x x x x x

+ + =

³


ANOTHER EXAMPLExB z x1 x2 x3 x4 x5 x6 x7 RHS Ratio

z 1 −5 −4 −3 0 0 0 0 0 ---

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

x5 0 −1 0 3 0 1 0 0 2 ---

x6 0 2 1 2 0 0 1 0 4 4/2=2

x7 0 2 3 1 0 0 0 1 2 2/2=1

z 1 0 7/2 −1/2 0 0 0 5/2 5 ---

x4 0 0 3/2 1/2 1 0 0 −1/2 2 2/(1/2)=4

x5 0 0 3/2 7/2 0 1 0 1/2 3 3/(7/2)=6/7

x6 0 0 −2 1 0 0 1 -1 2 2/1=2

x1 0 1 3/2 1/2 0 0 0 1/2 1 1/(1/2) =2


ANOTHER EXAMPLE

xB z x1 x2 x3 x4 x5 x6 x7 RHS

z 1 0 26/7 0 0 1/7 0 18/7 38/7

x4 0 0 9/7 0 1 −1/7 0 −4/7 11/7

x3 0 0 3/7 1 0 2/7 0 1/7 6/7

x6 0 0 −17/7 0 0 −2/7 1 −8/7 8/7

x1 0 1 9/7 0 0 −1/7 0 3/7 4/7

OPTIMAL SOLUTION: z*=38/7, x1*=4/7, x2*=0, x3*=6/7


COMPLICATIONS

Ø Bounds• Positive Upper Bounds• Negative Values

Ø Ties:• Entering Basic Variable• Leaving Basic Variable – Degeneracy

Ø No Leaving Basic Variable – UnboundednessØ Multiple Optimal SolutionsØ Origin IS NOT Basic Feasible


BOUNDSØ Bounds on Decision Variables:

üLower Bounds (! ≥ !#):use variable substutition,

Solve the problem and perform backsubstutitionExample:

üUpper Bounds (! ≤ !%):treat them as functional constraints (for now!!!)

1 2

1 2

1 2

max =3 2s.t. 2 6 0, 1

z x x

x xx x

+

+ £³ ³

'1 2

'1 2

'1 2

max =3 2( 1)s.t. 2( 1) 6 , 0

z x x

x xx x

+ +

+ + £³

'1 2

'1 2

'1 2

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

z x x

x xx x

+ +

+ £³


max ) = +,-s.t.

1- ≤ 2- ≥ -3

max ) = +,(-5 + -3)s.t.

1(-5 + -3) ≤ 2-5 ≥ 0

- = -5 + -3

BOUNDSüNegative Values are Allowed (−∞ ≤ $ ≤ ∞):

use variable substutition,

Solve the problem and perform backsubstutitionExample:

1 2

1 2

1 2

1 2

max =2 3s.t. 2 2 2 8 0, -

z x x

x xx x

x x

+

- £+ £

³ ¥ £ £ ¥


1 2 2

1 2 2

1 2 2

1 2 2

max =2 3 3s.t. 2 2 2 2 2 8 , , 0

z x x x

x x xx x xx x x

+ -

+ -

+ -

+ -

+ -

- + £+ - £

³

% = %' − %(

max , = -.%s.t.

2% ≤ 3−∞ ≤ % ≤ ∞

max , = -.(%' − %()s.t.

2(%' − %() ≤ 3%', %( ≥ 0

TIESØ Two Types of Ties:

üTie for Entering Basic VariableüTie for Leaving Basic Variable

Ø Tie for Entering Basic Variable: This situation is observed when two or more variables have the same most negative (for max problems) row 0 coefficient.

Ø Options:1. Select one of the variables in tie randomly as

the entering basic variable2. Select one of the variables in tie with the

smallest variable index as the entering basic variable

3. More advanced rules that requires probing or application of the lexicographic rule


DEGENERACY

Ø Tie for Leaving Basic Variable: This situation is observed when two or more rows have the same minimum ratio. In this case, we have to conduct at least one DEGENERATE iteration where the objective function value does not change after conducting iteration. There are 2 possible outcomes when DEGENERACY is encountered.

1. The objective function improves after 1 or more iterations,

2. The objective function does not improve during a number of iterations where the basic feasible solutions circle back to the iteration where the degeneracy is encountered for the first time resulting in an infinite loop of iterations.


DEGENERACY EXAMPLE1 2

1 2

1 2

1 2

1 2

max =3 2s.t. 4 8 4 3 12 4 8 , 0

z x x

x xx xx xx x

+

- £+ £+ £

³

1 2

1 2 3

1 2 4

1 2 5

1 2 3 4 5

max s.t. 3 2 0 4 8 4 3 12 4 8 , , , , 0

z

z x xx x xx x xx x xx x x x x

- - =- + =

+ + =+ + =

³


z 1 −3 −2 0 0 0 0 ---

x3 0 4 −1 1 0 0 8 8/4=2

x4 0 4 3 0 1 0 12 12/4=3

x5 0 4 1 0 0 1 8 8/4=2

TIE


DEGENERACY EXAMPLE


z 1 0 −11/4 3/4 0 0 6 ---

x1 0 1 −1/4 1/4 0 0 2 ---

x4 0 0 4 −1 1 0 4 4/4=1

x5 0 0 2 −1 0 1 0 0/2=0

TIE fromprev. iter.

z 1 0 0 −5/8 0 11/8 6 ---

x1 0 1 0 1/8 0 1/8 2 2/(1/8)=16

x4 0 0 0 1 1 −2 4 4/1=4

x2 0 0 1 −1/2 0 1/2 0 ---


DEGENERACY EXAMPLE


z 1 0 0 0 5/8 1/8 68/8 ---

x1 0 1 0 0 −1/8 3/8 3/2

x3 0 0 0 1 1 −2 4

x2 0 0 1 0 1/2 −1/2 2

OPTIMAL SOLUTION: z*=68/8, x1*=3/2, x2*=2

We were lucky to solve this problem with 1 degenerate iteration only. Most problems require more degenerate iterations.


UNBOUNDEDNESS

1 2

1 2

1

1 2

max =2s.t. 10 2 40 , 0

z x x

x xxx x

+

- ££³

1 2

1 2 3

1 4

1 2 3 4

max s.t. 2 0 10 2 40 , , , 0

z

z x xx x xx xx x x x

- - =- + =

+ =³


UNBOUNDEDNESS

xB z x1 x2 x3 x4 RHS Ratio

z 1 −2 −1 0 0 0 ---

x3 0 1 −1 1 0 10 10/1=10

x4 0 2 0 0 1 40 40/2=20

z 1 0 −3 2 0 20 ---

x1 0 1 −1 1 0 10 ---

x4 0 0 2 −2 1 20 20/2=10

z 1 0 0 −1 3/2 50 ---

x1 0 1 0 0 1/2 20

x2 0 0 1 −1 1/2 10

NO LEAVING VARIABLE, UNBOUNDED OBJ. FN. VALUE

Question: If the feasible region is unbounded, is the objective function value ALWAYS unbounded?


MULTIPLE OPTIMA

1 2

1 2

1 2

1 2

max = 2s.t. 2 4 2 4 6 , 0

z x x

x xx xx x

+

+ £+ £

³

1 2

1 2 3

1 2 4

1 2 3 4

max s.t. 2 0 2 4 2 4 6 , , , 0

z

z x xx x xx x xx x x x

- - =+ + =+ + =

³


MULTIPLE OPTIMAxB z x1 x2 x3 x

4RHS Ratio

z 1 −1 −2 0 0 0 ---

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

x4 0 2 4 0 1 6 6/4=3/2

z 1 0 0 0 1/2 3 ---

x3 0 3/2 0 1 −1/4 5/2 (5/2)/(3/2)=5/3

x2 0 1/2 1 0 1/4 3/2 (3/2)/(1/2)=3

z 1 0 0 0 1/2 3

x1 0 1 0 2/3 −1/6 5/3

x2 0 0 1 −1/3 1/3 2/3

Ø A row 0 coefficient of a non-basic variable with a value of 0 at the optimal solution indicates multiple optimal solutions.

Ø The objective function is parallel to a binding constraints at the optimal solution and two adjacent CPF solutions on the binding constraint are optimal.


OPTIMAL SOLUTIONS: z*=3, 1:x1*=0, x2*=3/22:x1*=5/3, x2*=2/3

INTERESTING PROBLEM

1 2

1 2

1 2 3

1 2 3

1 2 3

max =3s.t. 2 5 2 7 3 5 20 , , 0

z x x

x xx x xx x xx x x

+

+ £+ - £

+ - £

³

1 2

1 2 4

1 2 3 5

1 2 3 6

max s.t. 3 0 2 5 2 7 3 5 20

z

z x xx x xx x x xx x x x

- - =+ + =+ - + =+ - + =

1 2 3 4 5 6 , , , , , 0x x x x x x ³

xB z x1 x2 x3 x4 x5 x6 RHS Ratio

z 1 −3 −1 0 0 0 0 0 ---

x4 0 1 2 0 1 0 0 5 5/1=5

x5 0 1 1 −1 0 1 0 2 2/1=2

x6 0 7 3 −5 0 0 1 20 20/7


INTERESTING PROBLEMxB z x1 x2 x3 x4 x5 x6 RHS Ratio

z 1 0 2 −3 0 3 0 6 ---

x4 0 0 1 1 1 −1 0 3 3/1=3

x1 0 1 1 −1 0 1 0 2 ---

x6 0 0 −4 2 0 −7 1 6 6/2=3

z 1 0 5 0 3 0 0 15 ---

x3 0 0 1 1 1 −1 0 3

x1 0 1 2 0 1 0 0 5

x6 0 0 −6 0 −2 −5 1 0

WHAT IS THE VERDICT?INDR 262 – The Simplex Method Metin Türkay 49

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