chapter three: the simplex method & sensitivity...

Chapter Three: The Simplex Method & sensitivity analysis

Hamdy A. Taha, Operations Research: An introduction,

8th Edition

Introduction

• Most real-life LP problems have more than two variables and cannot be solved using the graphical procedure

• The simplex method

• Examines the corner points in a systematic fashion

– An iterative process

– Each iteration improves the value of the objective function

– Yields optimal solution and other valuable economic information

Introduction

It is a general algebraic method to solve a set of linear

equations.

We use simplex method to get extreme (or corner) point

solution.

We must first convert the model into the standard LP form by using slack or surplus variables to convert the inequality constraints into equations.

Our interest in the standard LP form lies in the basic

solutions of the simultaneous linear equations.

Standard LP Form and its Basic Solution Example

Express the following LP model in standard form:

Maximize Z= 5X1 + 6 X2

S.T 2X1 + 3X2 >= 18

2X1 +X2 <= 12

3X1 + 3X2 = 24

X1, X2 >=0

• Standard LP Form

How to Set Up the

Initial Simplex Solution

• Flair Furniture Company problem

T = number of tables produced

C = number of chairs produced

Maximize profit = $70T + $50C (objective function)

subject to 2T + 1C ≤ 100 (painting hours constraint)

4T + 3C ≤ 240 (carpentry hours constraint)

T, C ≥ 0 (nonnegativity constraints)

Converting the Constraints to

Equations

• Convert inequality constraints into an equation

• Add a slack variable to less-than-or-equal-to constraints

S1 = slack variable representing unused hours in the painting department

S2 = slack variable representing unused hours in the carpentry department

2T + 1C + S1 = 100

4T + 3C + S2 = 240

Equations

• If Flair produces T = 40 and C = 10

2T + 1C + S1 = 100

2(40) + 1(10) + S1 = 100

S1 = 10

• Adding all variables into all equations

2T + 1C + 1S1 + 0S2 = 100

4T + 3C + 0S1 + 1S2 = 240

T, C, S1, S2 ≥ 0

• If Flair produces T = 40 and C = 10

2T + 1C + S1 = 100

2(40) + 1(10) + S1 = 100

S1 = 10

Equations

• Adding all variables into all equations

2T + 1C + 1S1 + 0S2 = 100

4T + 3C + 0S1 + 1S2 = 240

T, C, S1, S2 ≥ 0

Objective function becomes

Maximize profit = $70T + $50C + $0S1 + $0S2

Finding an Initial Solution

Algebraically Corner Points of the Flair Furniture Company Problem

100 –

80 –

60 –

40 –

20 –

| | | | | | | | | | | |

0 20 40 60 80 100 T

Number of Tables

2T + 1C ≤ 100

4T + 3C ≤ 240

D = (50,0)

B = (0,80)

(0,0) A

C = (30,40)

Finding an Initial Solution

Algebraically Corner Points of the Flair Furniture Company Problem

100 –

80 –

60 –

40 –

20 –

| | | | | | | | | | | |

0 20 40 60 80 100 T

Number of Tables

2T + 1C ≤ 100

4T + 3C ≤ 240

D = (50,0)

B = (0,80)

(0,0) A

C = (30,40)

Find a basic feasible solution by setting all real variables to 0

The First Simplex Tableau

• Constraint equations

– Place all of the coefficients into tabular form, the first simplex tableau

SOLUTION

QUANTITY

(RIGHT-HAND-SIDE)

S1 2 1 1 0 100

S2 4 3 0 1 240

Cj SOLUTION MIX

$0 S2 QUANTITY

Zj -$70 -$50 $0 $0 $0

$0 S1 2 1 1 0 100

$0 S2 4 3 0 1 240

Profit per unit row

Constraint equation

Gross profit row

Flair Furniture’s Initial Simplex Tableau

• Begin the initial solution procedure at the origin

• Slack variables are nonzero and are the initial solution mix

• Values in the quantity column

• Initial solution is the basic feasible solution

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• Variables in the solution mix, the basis, are called basic variables

• Those not in the basis are called nonbasic variables

• Optimal solution in vector form

• T and C are the final basic variables

• S1 and S2 are nonbasic variables

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• Substitution rates

Under T 2

ø÷ Under C

ø÷ Under S1

ø÷ Under S2

– For every unit of T introduced into the current solution, 2 units of S1 and 4 units of S2 must be removed

For any variable ever to appear in the solution mix column, it must have the number 1 someplace in its column and 0s in every other place in that column

• Substitution rates

Under T 2

ø÷ Under C

ø÷ Under S1

ø÷ Under S2

– For every unit of T introduced into the current solution, 2 units of S1 and 4 units of S2 must be removed

For any variable ever to appear in the solution mix column, it must have the number 1 someplace in its column and 0s in every other place

For three less-than-or-equal-to constraints

SOLUTION MIX S1 S2 S3

S1 1 0 0

S2 0 1 0

S3 0 0 1

Simplex Solution Procedures

1. Determine which variable to enter into the solution mix next. One way of doing this is by identifying the column, and hence the variable, with the largest negative number in the Zj row of the preceding tableau. The column identified in this step is called the pivot column.

2. Determine which variable to replace. Decide which basic variable currently in the solution will have to leave to make room for the new variable. Divide each amount in the quantity column by the corresponding number in the column selected in step 1. The row with the smallest nonnegative number calculated in this fashion will be replaced in the next tableau. This row is often referred to as the pivot row. The number at the intersection of the pivot row and pivot column is referred to as the pivot number.

Simplex Solution Procedures

3. Compute new values for the pivot row by dividing every number in the row by the pivot number.

4. Compute the new values for each remaining row. All remaining row(s) are calculated as follows:

(New row numbers) = (Numbers in old row)

Number aboveor belowpivot number

÷÷´

Corresponding number inthe new row, that is, therow replaced in step 3

5. Compute the Zj rows, as demonstrated in the initial tableau. If all numbers in the Zj row are 0 or positive, an optimal solution has been reached. If this is not the case, return to step 1.

The Second Simplex Tableau

• Step 1 – Select variable to enter, T

Pivot Column Identified in the Initial Simplex Tableau

Zj 0 -15 35 0 3500

T 1 0.5 0.5 0 50 100

S2 0 1 -2 1 40 40Iter

• Step 2 – Select the variable to be replaced, S1

For S1

100 hours of painting time available( )2 hours required per table( )

= 50 tables

For S2

240 hours of painting time available( )4 hours required per table( )

= 60 tables

• Step 2 – Select the variable to be replaced, S1

Cj SOLUTION

QUANTITY

$0 S1 2 1 1 0 100

$0 S2 4 3 0 1 240

Zj - $70 -50 $0 $0 $0

Cj – Zj $70 $50 $0 $0

Pivot row

Pivot column Pivot number

TABLE– Pivot Row and Pivot Number Identified in the Initial Simplex Tableau

The Second Simplex Tableau – Graph of the Flair Furniture Company Problem

100 –

80 –

60 –

40 –

20 –

| | | | | | | | | | | |

0 20 40 60 80 100 T

Number of Tables

D = (50,0)

B = (0,80)

(0,0) A

C = (30,40)

F = (0,100)

E = (60,0)

• Step 3 – Compute the replacement for the pivot row

2 = 0.5

2 = 50

Cj SOLUTION MIX T C S1 S2 QUANTITY

$70 T 1 0.5 0.5 0 50

• The entire pivot row

• Step 4 – Compute new values for the S2 row

Number in

New S2 Row =

Number in

Old S2 Row –

Number Below

Pivot Number x

Corresponding Number

in the New T Row

0 = 4 – (4) x (1)

1 = 3 – (4) x (0.5)

-2 = 0 – (4) x (0.5)

1 = 1 – (4) x (0)

40 = 240 – (4) x (50)

• Step 4 – Compute new values for the S2 row

Number in

New S2 Row =

Number in

Old S2 Row –

Number Below

Pivot Number x

Corresponding Number

in the New T Row

0 = 4 – (4) x (1)

1 = 3 – (4) x (0.5)

-2 = 0 – (4) x (0.5)

1 = 1 – (4) x (0)

40 = 240 – (4) x (50)

$70 T 1 0.5 0.5 0 50

$0 S2 0 1 –2 1 40

• The second tableau

Interpreting the Second Tableau

• Current solution and resources – Production of 50 tables (T) and 0 chairs (C)

– Profit = $3,500

– T is a basic variable, C nonbasic

– Slack variable S2 = 40, S1 nonbasic

• Substitution rates – Marginal rates of substitution

– Negative substitution rate • If 1 unit of a column variable is added to the solution, the

value of the corresponding solution (or row) variable will increase

– Positive substitution rate • If 1 unit of the column variable is added to the solution, the

row variable will decrease

Developing the Third Tableau

• Not all numbers in the Zj row are 0 or Positive

– Previous solution is not optimal

– Repeat the five simplex steps

Developing the Third Tableau • Step 2 – Identifying the pivot row

Pivot column

Pivot row

TABLE M7.5 – Pivot Row, Pivot Column, and Pivot Number Identified in the

Second Simplex Tableau

Cj $70 $50 $0 $0

SOLUTION

MIX T C S1 S2 QUANTITY

Z 0 -15 35 0 3500

$70 T 1 0.5 0.5 0 50

$0 S2 0 1 –2 1 40

Pivot number

For the T row: 50

0.5=100 chairs For the S2 row:

1= 40 chairs

• Step 3 – The pivot row is replaced

1 = –2

1 = 40

$50 C 0 1 –2 1 40

• The new C row

• Step 4 – New values for the T row

Number in

new T row =

Number in

old T row –

Number above

pivot number x

Corresponding number

in new C row

1 = 1 – (0.5) x (0)

0 = 0.5 – (0.5) x (1)

1.5 = 0.5 – (0.5) x (–2)

–0.5 = 0 – (0.5) x (1)

30 = 50 – (0.5) x (40)

$70 T 1 0 1.5 –0.5 30

$50 C 0 1 –2 1 40

• Step 5 – Calculate the Zj rows

TABLE–Final Simplex Tableau for the Flair Furniture Problem

Zj 0 0 5 15 4100

T 1 0 1.5 -0.5 30

C 0 1 -2 1 40Iter

• Step 5 – Calculate the Zj rows

TABLE– Final Simplex Tableau for the Flair Furniture Problem

Since every number in the tableau’s Zj row is 0 or

positive, an optimal solution has been reached

Zj 0 0 5 15 4100

T 1 0 1.5 -0.5 30

C 0 1 -2 1 40Iter

• Verifying the solution

First constraint: 2T + 1C ≤ 100 painting dept hours

2(30) + 1(40) ≤ 100

100 ≤ 100 ✓

Second constraint: 4T + 3C ≤ 240 carpentry dept hours

4(30) + 3(40) ≤ 240

240 ≤ 240 ✓

Objective function profit = $70T + $50C

= $70(30) + $50(40)

= $4,100

Review of Procedures

I. Formulate the LP problem’s objective function and constraints.

II. Add slack variables to each less-than-or-equal-to constraint and to the problem’s objective function.

III. Develop an initial simplex tableau with slack variables in the basis and the decision variables set equal to 0. Compute the Zj values for this tableau.

IV. Follow these five steps until an optimal solution has been reached:

1. Choose the variable with the greatest negative Zj to enter the solution. This is the pivot column.

2. Determine the solution mix variable to be replaced and the pivot row by selecting the row with the smallest (nonnegative) ratio of the quantity-to-pivot column substitution rate. This row is the pivot row.

3. Calculate the new values for the pivot row.

4. Calculate the new values for the other row(s).

5. Calculate the Zj values for this tableau. If there are any Zj numbers less than 0, return to step 1. If there are no Zj numbers that are less than 0, an optimal solution has been reached.

Surplus and Artificial Variables

• Conversions are made for ≥ and = constraints

• For Surplus Variables

Constraint 1: 5X1 + 10X2 + 8X3 ≥ 210

rewitten: 5X1 + 10X2 + 8X3 – S1 = 210

For X1 = 20, X2 = 8, and X3 = 5

5(20) + 10(8) + 8(5) – S1 = 210

– S1 = 210 – 220

S1 = 10 surplus units

• For Surplus Variables

Constraint 1: 5X1 + 10X2 + 8X3 ≥ 210

rewitten: 5X1 + 10X2 + 8X3 – S1 = 210

For X1 = 20, X2 = 8, and X3 = 5

5(20) + 10(8) + 8(5) – S1 = 210

– S1 = 210 – 220

S1 = 10 surplus units

• If X1 and X2 = 0, then the S1 variable is negative which violates the nonnegative condition

• An artificial variable, R1 is added to resolve this problem

Constraint 1 completed: 5X1 + 10X2 + 8X3 – S1 + R1 = 210

• For equalities, add only an artificial variable

Constraint 2: 25X1 + 30X2 = 900

Constraint 2 completed: 25X1 + 30X2 + R2 = 900

Artificial variables have no physical meaning

and drop out of the solution mix before the

final tableau if a feasible solution exists

in the Objective Function

• Assign a very high cost $M to artificial variables in the objective function to force them out before the final solution is reached

Minimize cost = $5X1 + $9X2 + $7X3 + $0S1 + $MR1 + $MR2

subject to 5X1 + 10X2 + 8X3 – 1S1 + 1R1 + 0R2 = 210

25X1 + 30X2 + 0X3 + 0S1 + 0R1 + 1R2 = 900

Minimize cost = $5X1 + $9X2 + $7X3

Solving Minimization Problems

• The Muddy River Chemical Company

Minimize cost = $5X1 + $6X2

subject to X1 + X2 = 1,000 lb

X1 ≤ 300 lb

X2 ≥ 150 lb

X1, X2 ≥ 0

X1 = number of pounds of phosphate

X2 = number of pounds of potassium

Graphical Analysis

1,000 –

800 –

600 –

400 –

200 –

100 –

– | | | | |

0 200 400 600 800 1,000

F G H X2 ≥ 150

X1 ≤ 300

X1 + X2 = 1,000

FIGURE– Muddy River Chemical Corporation’s Feasible Region Graph

Converting the Constraints and

Objective Function

Minimize cost = $5X1 + $6X2 + $0S1 + $0S2 + $MR1 + $MR2

subject to 1X1 + 1X2 + 0S1 + 0S2 + 1R1 + 0R2 = 1,000

1X1 + 0X2 + 1S1 + 0S2 + 0R1 + 0R2 = 300

0X1 + 1X2 + 0S1 – 1S2 + 0R1 + 1R2 = 150

X1, X2, S1, S2, R1, R2 ≥ 0

Rules of the Simplex Method for

Minimization Problems 1. Choose the variable with a positive Zj that indicates

the largest decrease in cost to enter the solution. The corresponding column is the pivot column.

2. Determine the row to be replaced by selecting the one with the smallest (nonnegative) quantity-to-pivot column substitution rate ratio. This is the pivot row.

3. Calculate new values for the pivot row.

4. Calculate new values for the other rows.

5. Calculate the Zj values for this tableau. If there are any Zj numbers more than 0, return to step 1.

Fourth and Optimal Solution to

the Muddy River Chemical

Corporation Problem

$5 $6 $0 $0 $M $M

SOLUTION MIX X1 X2 S1 S2 A1 A2 QUANTITY

Zj $5,700

S2 550

X1 300

X2 700

Review of Procedures I. Formulate the LP problem’s objective

function and constraints.

II. Include slack variables in each less-than-or-equal-to constraint, artificial variables in each equality constraint, and both surplus and artificial variables in each greater-than-or-equal-to constraint. Then add all of these variables to the problem’s objective function.

III. Develop an initial simplex tableau with artificial and slack variables in the basis and the other variables set equal to 0. Compute the Zj values for this tableau.

IV. Follow these five steps until an optimal solution has been reached:

1. Choose the variable with the positive Zj indicating the greatest improvement to enter the solution. This is the pivot column.

2. Determine the row to be replaced by selecting the one with the smallest (nonnegative) quantity-to-pivot column substitution rate ratio. This is the pivot row.

3. Calculate the new values for the pivot row.

4. Calculate the new values for the other row(s).

5. Calculate the Zj values for this tableau. If there are any Zj numbers more than 0, return to step 1. If there are no Zj numbers that are more than 0, an optimal solution has been reached.

In the Simplex method we have used the slack variables as the starting solution. These

were coming from the standardized form of constraints that are type of “”

However, if the original constraint is a “≥” or “=” type of constraint, we no longer have an

easy starting solution.

Therefore, Artificial Variables are used in such cases. An artificial variable is a variable

introduced into each equation that has a surplus variable.

To ensure that we consider only basic feasible solutions, an artificial variable is required to

satisfy the nonnegative constraint.

The two method used are:

• The M- method

• The Two-phase method

The M-Method

The M-method starts with the LP in the standard form

For any equation (i) that does not have slack, we augment an

artificial variable Ri

Given M is sufficiently large positive value, The variable Ri is

penalized in objective function using (-M Ri ) in case of maximization

and (+ M Ri ) in case of minimization. ( Penalty Role)

The M-Method ( Example)

Minimized 214 xxz

Subject to:

The M-Method ( Solution)

Minimized 214 xxz

Subject to:

By subtracting surplus x3 in second constraint and adding slack x4 in third

constraint, thus we get:

By using artificial variables in equations that haven’t slack variables and

penalized them in objective function, we got:

Minimized

21214 MRMRxxz

Subject to:

0,,,,,

214321

RRxxxx

Then, we can use R1 , R2 and x4 as the starting basic feasible

solution.

Solution x4 R2 R1 x3 x2 x1 Basic

0 0 -M -M 0 -1 -4 z

3 0 0 1 0 1 3 R1

6 0 1 0 -1 3 4 R2

4 1 0 0 0 2 1 x4

New z-row = Old z-row + M* R1-row + M*R2 -row

9M 0 0 0 -M -1+4M -4+7M z

3 0 0 1 0 1 3 R1

6 0 1 0 -1 3 4 R2

4 1 0 0 0 2 1 x4

Artificial variables become zero

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

1 0 0 1/3 0 1/3 1 x1

2 0 1 -4/3 -1 5/3 0 R2

3 1 0 -1/3 0 5/3 0 x4

Thus, the entering value is (-4+7M) because it the most positive coefficient in

the z-row.

The leaving variable will be R1 by using the ratios of the feasibility condition

After determining the entering and leaving variable, the new tableau can be

computed by Gauss-Jordan operations as follow:

The last tableau shows that x2 is the entering variable and R2 is the

leaving variable. The simplex computation must thus continued for two

more iteration to satisfy the optimally condition.

The results for optimality are:

2321 zandxxx

Observation regarding the Use of M-method:

1. The use of penalty M may not force the artificial variable to zero level in

the final simplex iteration. Then the final simplex iteration include at least

one artificial variable at positive level. This indication that the problem has

no feasible condition.

2.( M )should be large enough to act as penalty, but it should not be too large

to impair the accuracy of the simplex computations.

Example:

Maximize 21 5.02.0 xxz

Subject to:

Using Computer solution, apply the simplex method M=10, and repeat it using

M=999.999. the first M yields the correct solution x1 =1 and x2 =1.5, whereas the second

gives the incorrect solution x1 =4 and x2 =0

Multiplying the objective function by 1000 to get z= 200x1 + 500x2 and solve the

problem using M=10 and M=999.999 and observe the second value is the one that

yields the correct solution in this case

The conclusion from two experiments is that the correct choice of the value

of M is data dependent.

When a basic feasible solution is not readily available, the two-phase simplex method may be used as an alternative to the big M method.

In the two-phase simplex method, we add artificial variables to the same constraints as we did in big M method. Then we find a basic feasible solution to the original LP by solving the Phase I LP.

In the Phase I LP, the objective function is to minimize the sum of all artificial variables.

At the completion of Phase I, we use Phase II and reintroduce the original LP’s objective function and determine the optimal solution to the original LP.

Two-phase method

In Phase I, If the optimal value of sum of the artificial variables are

greater than zero, the original LP has no feasible solution which ends

the solution process. Other wise, We move to Phase II

Example

Minimized 214 xxz

Subject to:

Solution:

I: Phase

Minimize: 21 RRr

Subject to:

0,,,,,

214321

RRxxxx

0 0 -1 -1 0 0 0 r

3 0 0 1 0 1 3 R1

6 0 1 0 -1 3 4 R2

4 1 0 0 0 2 1 x4

New r-row = Old r-row + 1* R1-row + R2 -row

9 0 0 0 -1 4 7 r

3 0 0 1 0 1 3 R1

6 0 1 0 -1 3 4 R2

4 1 0 0 0 2 1 x4

0 0 -1 -1 0 0 0 r

3/5 0 -1/5 3/5 1/5 0 1 x1

6/5 0 3/5 -4/5 -3/5 1 0 x2

1 1 -1 1 1 0 0 x4

By using new r-row, we solve Phase I of the problem which yields the

following optimum tableau

Because minimum r=0, Phase I produces the basic feasible solution:

6, 425

31 xandxx

Phase II

After eliminating artificial variables column, the original problem can be

written as:

Minimize:

214 xxz

Subject to:

Solution x4 x3 x2 x1 Basic

0 0 0 -1 -4 z

3/5 0 1/5 0 1 x1

6/5 0 -3/5 1 0 x2

1 1 1 0 0 x4

Again, because basic variables x1 and x2 have nonzero coefficient in

he z row, they must be substituted out, using the following

computation:

New z-row = Old z-row + 4* x1-row + 1*x2 -row

Solution x4 x3 x2 x1 Basic

18/5 0 1/5 0 0 z

3/5 0 1/5 0 1 x1

6/5 0 -3/5 1 0 x2

1 1 1 0 0 x4

The initial tableau of Phase II is as the following:

The removal of artificial variables and their column at the end of Phase I

can take place only when they are all nonbasic. If one or more artificial

variables are basic ( at zero level) at the end of Phase I, then the

following additional steps must be under taken to remove them prior to

start Phase II

and solution basic the leave to artificial variable zero a Select . 1Step

designate its row as pivot row. The entering variable can be any nonbasic

(nonartificial) variable with nonzero (positive or negative) coefficient in the

pivot row. Perform the associated simplex iteration.

the from artificial ) leaving-Just(Remove the column of the . 2Step

tableau. If all the zero artificial variables have been removed , go to

Phase II. Otherwise, go back to Step I.

Remarks:

Special Cases

• Infeasibility

– Infeasibility occurs when there is no solution that satisfies all of the problem’s constraints

SOLUTION

MIX X1 X2 S1 S2 A1 A2 QUANTITY

Zj $0 $0 $2 $M – 31 $2M + 21 $0 $1,800 + 20M

X1 1 0 –2 3 –1 –0 200

X2 0 1 1 2 –2 0 100

A2 0 0 0 –1 –1 1 20

– Illustration of Infeasibility

Special Cases

• Unbounded Solutions

– Unboundedness describes linear programs that do not have finite solutions

SOLUTION MIX X1 X2 S1 S2 QUANTITY

Zj -$15 $0 $18 $0 $270

X1 –1 1 2 0 30

S2 –2 0 –1 1 10

- Problem with an Unbounded Solution

Pivot column

SOLUTION MIX X1 X2 S1 S2 QUANTITY

Zj -$15 $0 $18 $0 $270

X1 –1 1 2 0 30

S2 –2 0 –1 1 10

• Unbounded Solutions

– Unboundedness describes linear programs that do not have finite solutions

Special Cases

TABLE Problem with an Unbounded Solution

Pivot column

Ratio for the X2 row : 30

Ratio for the S2 row : 10

Negative ratios

unacceptable

Special Cases

• Degeneracy

– Degeneracy develops when three constraints pass through a single point

SOLUTION

MIX X1 X2 X3 S1 S2 S3 QUANTITY

Zj -$3 $0 $6 $16 $0 $0 $80

X2 0.25 1 1 –2 0 0 10

S2 4 0 0.33 –1 1 0 20

S1 2 0 2 0.4 0 1 10

- Problem Illustrating Degeneracy

Pivot column

• Degeneracy

– Degeneracy develops when three constraints pass through a single point

Special Cases

Problem Illustrating Degeneracy

For the X2 row : 10

0.25= 40

For the S2 row : 20

For the S3 row : 10

Tie for the smallest

ratio indicates

degeneracy

SOLUTION

MIX X1 X2 X3 S1 S2 S3 QUANTITY

Zj -$3 $0 $6 $16 $0 $0 $80

X2 0.25 1 1 –2 0 0 10

S2 4 0 0.33 –1 1 0 20

S1 2 0 2 0.4 0 1 10

Pivot column

Special Cases • More Than One Optimal Solution

– If the Zj value is equal to 0 for a variable that is not in the solution mix, more than one optimal solution exists

SOLUTION

MIX X1 X2 S1 S2 QUANTITY

Zj $0 $0 –$2 $0 $12

X2 1.5 1 1 0 6

S2 1 0 0.5 1 3

– Problem with Alternate Optimal Solutions

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