4 The Simplex Method: Standard Maximization Problems The Simplex Method: Standard Minimization...

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The Simplex Method: The Simplex Method:

Standard Maximization ProblemsStandard Maximization Problems The Simplex Method: The Simplex Method:

Standard Minimization ProblemsStandard Minimization Problems The Simplex Method: The Simplex Method:

Nonstandard ProblemsNonstandard Problems

Linear Programming: An Algebraic Approach Linear Programming: An Algebraic Approach

4.14.1The Simplex Method: The Simplex Method: Standard Maximization ProblemsStandard Maximization Problems

xx yy uu vv PP ConstantConstant

11 00 3/53/5 ––1/51/5 00 4848

00 1 1 ––1/51/5 2/52/5 00 8484

0 0 00 9/259/25 7/257/25 11 148148 4/54/5

The Simplex MethodThe Simplex Method

The The simplex methodsimplex method is an is an iterative procedureiterative procedure.. Beginning at a Beginning at a vertexvertex of the of the feasible regionfeasible region SS, each , each

iterationiteration brings us to another brings us to another vertexvertex of of SS with an with an improvedimproved value of the value of the objective functionobjective function..

The The iterationiteration ends when the ends when the optimal solutionoptimal solution is reached. is reached.

A Standard Linear Programming ProblemA Standard Linear Programming Problem

A A standard maximization problemstandard maximization problem is one in which is one in which

1.1. The The objective functionobjective function is to be is to be maximizedmaximized..

2.2. All the All the variablesvariables involved in the problem are involved in the problem are nonnegativenonnegative..

3.3. All other All other linear constraintslinear constraints may be written so may be written so that the expression involving the variables is that the expression involving the variables is less less than or equal tothan or equal to a nonnegative constanta nonnegative constant..

Setting Up the Initial Simplex TableauSetting Up the Initial Simplex Tableau

1.1. Transform the Transform the system of linearsystem of linear inequalitiesinequalities into a into a system of linearsystem of linear equationsequations by by introducing introducing slack variablesslack variables..

2.2. Rewrite the Rewrite the objective functionobjective function

in the formin the form

where all the where all the variablesvariables are on the are on the leftleft and the and the coefficientcoefficient of of PP is is +1+1. Write this equation . Write this equation below the equations in below the equations in step 1step 1..

3.3. Write the Write the augmented matrixaugmented matrix associated with associated with this system of linear equations.this system of linear equations.

1 1 2 2 n nP c x c x c x 1 1 2 2 n nP c x c x c x

1 1 2 2 0n nc x c x c x P 1 1 2 2 0n nc x c x c x P

Applied Example 1:Applied Example 1: A Production Problem A Production Problem

Recall the production problem discussed in Recall the production problem discussed in Chapter 3Chapter 3, , which required us to which required us to maximizemaximize the the objective functionobjective function

subject tosubject to the the system of inequalitiessystem of inequalities

This is a This is a standard maximization problemstandard maximization problem and may be and may be solved by the solved by the simplex methodsimplex method..

Set upSet up the initial the initial simplex tableausimplex tableau for this linear for this linear programming problem.programming problem.

, 0x y , 0x y

61.2

5P x y P x y or equivalently,

61.2

5P x y P x y or equivalently,

3 300x y 3 300x y 2 180x y 2 180x y

Example 1, page 206

Applied Example 1:Applied Example 1: A Production Problem A Production ProblemSolutionSolution First, introduce the First, introduce the slack variablesslack variables uu and and vv into the into the

inequalities inequalities

and turn these into and turn these into equationsequations, getting, getting

Next, rewrite the Next, rewrite the objective functionobjective function in the form in the form

2 180

3 300

x y u

x y v

2 180

3 300

x y u

x y v

3 300x y 3 300x y

2 180x y 2 180x y

60

5x y P

60

5x y P

Example 1, page 206

Applied Example 1:Applied Example 1: A Production Problem A Production ProblemSolutionSolution Placing the restated Placing the restated objective functionobjective function below the system of below the system of

equations of the equations of the constraintsconstraints we get we get

Thus, the Thus, the initial tableauinitial tableau associated with this system is associated with this system is

2 180

3 300

60

5

x y u

x y v

x y P

2 180

3 300

60

5

x y u

x y v

x y P


22 11 11 00 00 180180

11 3 3 00 11 00 300300

––1 1 –– 6/56/5 00 00 11 00

Example 1, page 206

The Simplex MethodThe Simplex Method

1.1. Set up the Set up the initial simplex tableauinitial simplex tableau..2.2. Determine whether the Determine whether the optimal solutionoptimal solution has has

been reached by been reached by examining all entriesexamining all entries in the in the last last rowrow to the to the leftleft of the of the vertical linevertical line..a.a. If all the entries are If all the entries are nonnegativenonnegative, the , the optimal optimal

solutionsolution hashas been reached been reached. Proceed to . Proceed to step 4step 4..b.b. If there are one or more If there are one or more negative entriesnegative entries, the , the

optimal solutionoptimal solution has nothas not been reached been reached. . Proceed to Proceed to step 3step 3..

3.3. Perform the Perform the pivot operationpivot operation. Return to . Return to step 2step 2..4.4. Determine the Determine the optimal solution(s)optimal solution(s)..

Applied Example 1:Applied Example 1: A Production Problem A Production Problem

Recall again the Recall again the production problemproduction problem discussed previously. discussed previously. We have already performed We have already performed step 1step 1 obtaining the obtaining the initial initial

simplex tableausimplex tableau::

Now, complete the Now, complete the solutionsolution to the problem. to the problem.


22 11 11 00 00 180180

11 3 3 00 11 00 300300

––1 1 –– 6/56/5 00 00 11 00

Example 2, page 208

Applied Example 1:Applied Example 1: A Production Problem A Production ProblemSolutionSolution

Step 2.Step 2. Determine whether the Determine whether the optimal solutionoptimal solution has been has been reached.reached.

✦ Since Since there there areare negative entries negative entries in the last row of the in the last row of the tableau, the tableau, the initial solutioninitial solution is is notnot optimal optimal..


22 11 11 00 00 180180

11 3 3 00 11 00 300300

––1 1 –– 6/56/5 00 00 11 00

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ Since the entry Since the entry –– 6/56/5 isis the the most negative entrymost negative entry to the left to the left

of the vertical line in the last row of the tableau, the of the vertical line in the last row of the tableau, the second columnsecond column in the tableau is the in the tableau is the pivot columnpivot column..


22 11 11 00 00 180180

11 3 3 00 11 00 300300

––1 1 –– 6/56/5 00 00 11 00

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ Divide each Divide each positive numberpositive number of the of the pivot columnpivot column into the into the

corresponding entrycorresponding entry in the in the column of constantscolumn of constants and and compare compare thethe ratiosratios thus obtained. thus obtained.

✦ We see that the We see that the ratioratio 300/3 = 100300/3 = 100 is is less thanless than the the ratioratio 180/1 = 180180/1 = 180, so , so row 2row 2 is the is the pivot rowpivot row..


22 11 11 00 00 180180

11 3 3 00 11 00 300300

––1 1 –– 6/56/5 00 00 11 00

1801

3003

180

100

1801

3003

180

100

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ The The entryentry 3 3 lying in the lying in the pivot columnpivot column and the and the pivot rowpivot row

is the is the pivot elementpivot element..


22 11 11 00 00 180180

11 3 3 00 11 00 300300

––1 1 –– 6/56/5 00 00 11 00

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ Convert the Convert the pivot elementpivot element into a into a 1 1..

123 R123 R


22 11 11 00 00 180180

11 3 3 00 11 00 300300

––1 1 –– 6/56/5 00 00 11 00

Example 2, page 208



123 R123 R


22 11 11 00 00 180180

1/31/3 1 1 00 1/31/3 00 100100

––1 1 –– 6/56/5 00 00 11 00

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ Use elementary Use elementary row operationsrow operations to convert the to convert the pivot pivot

columncolumn into a into a unit columnunit column..

1 2

63 25

R R

R R

1 2

63 25

R R

R R


22 11 11 00 00 180180

1/31/3 1 1 00 1/31/3 00 100100

––1 1 –– 6/56/5 00 00 11 00

Example 2, page 208




1 2

63 25

R R

R R

1 2

63 25

R R

R R


5/35/3 00 11 ––1/31/3 00 8080

1/31/3 1 1 00 1/31/3 00 100100

––3/5 3/5 00 00 2/52/5 11 120120

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ This This completes an iterationcompletes an iteration..

✦ The The last rowlast row of the tableau contains a of the tableau contains a negative numbernegative number, , so an so an optimal solutionoptimal solution hashas notnot been reached been reached..

✦ Therefore, we Therefore, we repeatrepeat the the iteration stepiteration step..


5/35/3 00 11 ––1/31/3 00 8080

1/31/3 1 1 00 1/31/3 00 100100

––3/5 3/5 00 00 2/52/5 11 120120

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation again. again.✦ Since the entry Since the entry –– 3/53/5 isis the the most negative entrymost negative entry to the left to the left

of the vertical line in the last row of the tableau, the of the vertical line in the last row of the tableau, the first first columncolumn in the tableau is now the in the tableau is now the pivot columnpivot column..


5/35/3 00 11 ––1/31/3 00 8080

1/31/3 1 1 00 1/31/3 00 100100

––3/5 3/5 00 00 2/52/5 11 120120

Example 2, page 208



corresponding entrycorresponding entry in the in the column of constantscolumn of constants and and compare the ratioscompare the ratios thus obtained. thus obtained.

✦ We see that the We see that the ratioratio 80/(5/3) = 4880/(5/3) = 48 is is less thanless than the the ratioratio 100/(1/3) = 300100/(1/3) = 300, so , so row 1row 1 is the is the pivot rowpivot row now. now.


5/35/3 00 11 ––1/31/3 00 8080

1/31/3 1 1 00 1/31/3 00 100100

––3/5 3/5 00 00 2/52/5 11 120120

805/3

1001/3

48

300

805/3

1001/3

48

300

Ratio

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ The The entryentry 55//3 3 lying in the lying in the pivot columnpivot column and the and the pivot pivot

rowrow is the is the pivot elementpivot element..


5/35/3 00 11 ––1/31/3 00 8080

1/31/3 1 1 00 1/31/3 00 100100

––3/5 3/5 00 00 2/52/5 11 120120

Example 2, page 208




5/35/3 00 11 ––1/31/3 00 8080

1/31/3 1 1 00 1/31/3 00 100100

––3/5 3/5 00 00 2/52/5 11 120120

315 R315 R

Example 2, page 208




11 00 3/53/5 ––1/51/5 00 4848

1/31/3 1 1 00 1/31/3 00 100100

––3/5 3/5 00 00 2/52/5 11 120120

315 R315 R

Example 2, page 208




12 13

33 15

R R

R R

12 13

33 15

R R

R R


11 00 3/53/5 ––1/51/5 00 4848

1/31/3 1 1 00 1/31/3 00 100100

––3/5 3/5 00 00 2/52/5 11 120120

Example 2, page 208




12 13

33 15

R R

R R

12 13

33 15

R R

R R


11 00 3/53/5 ––1/51/5 00 4848

00 1 1 ––1/51/5 2/52/5 00 8484

0 0 00 9/259/25 7/257/25 11 148148 4/54/5

Example 2, page 208


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ The The last rowlast row of the tableau contains of the tableau contains nono negative negative

numbersnumbers, so an , so an optimal solutionoptimal solution hashas been reached been reached..


11 00 3/53/5 ––1/51/5 00 4848

00 1 1 ––1/51/5 2/52/5 00 8484

0 0 00 9/259/25 7/257/25 11 148148 4/54/5

Example 2, page 208


Step 4.Step 4. Determine the Determine the optimal solutionoptimal solution..✦ Locate the Locate the basic variablesbasic variables in the final tableau. in the final tableau.

In this case, the In this case, the basic variablesbasic variables are are xx, , yy, and , and PP.. The The optimal valueoptimal value for for xx is is 4848.. The The optimal valueoptimal value for for yy is is 8484.. The The optimal valueoptimal value for for PP is is 148.8148.8..

✦ Thus, the firm will Thus, the firm will maximize profitsmaximize profits at at $148.80$148.80 by by producing producing 4848 type-A type-A souvenirs and souvenirs and 84 84 type-Btype-B souvenirs. souvenirs.This This agreesagrees with the results obtained in with the results obtained in chapter 3chapter 3..


11 00 3/53/5 ––1/51/5 00 4848

00 1 1 ––1/51/5 2/52/5 00 8484

0 0 00 9/259/25 7/257/25 11 148148 4/54/5

Example 2, page 208

4.24.2The Simplex Method: The Simplex Method: Standard Minimization ProblemsStandard Minimization Problems

3030

––1/501/50

3/1003/100

xx

00

11

00

vv

450450

11/1011/10

––3/203/20

ww

0 0

00

11

uu

1140114011120120

13/2513/25002/252/25

1/501/5000––1/501/50

ConstantConstantPPyy

3030

––1/501/50

3/1003/100

xx

00

11

00

vv

450450

11/1011/10

––3/203/20

ww

0 0

00

11

uu

1140114011120120

13/2513/25002/252/25

1/501/5000––1/501/50

ConstantConstantPPyy

SolutionSolution for thefor theprimal problemprimal problem

Minimization with Minimization with Constraints Constraints

In the last section we developed the In the last section we developed the simplex methodsimplex method to to solve linear programming problems that satisfy solve linear programming problems that satisfy three three conditionsconditions::

1.1. The The objective functionobjective function is to be is to be maximizedmaximized..

2.2. All the All the variables involvedvariables involved are are nonnegativenonnegative..

3.3. Each Each linear constraintlinear constraint may be written so that the may be written so that the expression involving the variables is expression involving the variables is less than or equal to less than or equal to a nonnegative constanta nonnegative constant..

We will now see how the simplex method can be used to We will now see how the simplex method can be used to solve solve minimization problemsminimization problems that that meet the second and meet the second and third conditionsthird conditions listed above. listed above.

ExampleExample

Solve the following Solve the following linear programming problemlinear programming problem::

This problem involves the This problem involves the minimizationminimization of the objective of the objective function and so is function and so is notnot a a standard maximization problemstandard maximization problem..

Note, however, that Note, however, that all the other conditionsall the other conditions for a for a standard standard maximizationmaximization hold truehold true..

2 3Minimize C x y 2 3Minimize C x y

5 4 32

2 10

, 0

subject to x y

x y

x y

5 4 32

2 10

, 0

subject to x y

x y

x y

Example 1, page 226

ExampleExample

We can use the We can use the simplex methodsimplex method to solve this problem by to solve this problem by convertingconverting the the objective functionobjective function from from minimizingminimizing CC to its to its equivalentequivalent of of maximizingmaximizing PP = –= – CC. .

Thus, the Thus, the restatedrestated linear programming problemlinear programming problem is is

This problem can now be solved using the This problem can now be solved using the simplex methodsimplex method as discussed in as discussed in section 4.1section 4.1..

2 3Maximize P x y 2 3Maximize P x y

5 4 32

2 10

, 0

subject to x y

x y

x y

5 4 32

2 10

, 0

subject to x y

x y

x y

Example 1, page 226

ExampleExampleSolutionSolutionStep 1.Step 1. Set up the Set up the initial simplex tableauinitial simplex tableau..

✦ Turn the Turn the constraintsconstraints into into equationsequations adding to them the adding to them the slack variablesslack variables uu and and vv. Also . Also rearrangerearrange the the objective objective functionfunction and place it below the constraints: and place it below the constraints:

✦ Write the Write the coefficientscoefficients of the system in a of the system in a tableautableau::

5 4 32

2 10

2 3 0

x y u

x y v

x y P

5 4 32

2 10

2 3 0

x y u

x y v

x y P

2 3Maximize P x y 2 3Maximize P x y xx yy uu vv PP ConstantConstant

55 44 11 00 00 3232

11 2 2 00 11 00 1010

––2 2 ––33 00 00 11 00

Example 1, page 226

ExampleExampleSolutionSolution

Step 2.Step 2. Determine whether the Determine whether the optimal solutionoptimal solution has been has been reached.reached.

✦ Since Since there there areare negative entries negative entries in the last row of the in the last row of the tableau, the tableau, the initial solutioninitial solution is is notnot optimal optimal..


55 44 11 00 00 3232

11 2 2 00 11 00 1010

––2 2 ––33 00 00 11 00

Example 1, page 226


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ Since the entry Since the entry –– 33 isis the the most negative entrymost negative entry to the left to the left

of the vertical line in the last row of the tableau, the of the vertical line in the last row of the tableau, the second columnsecond column in the tableau is the in the tableau is the pivot columnpivot column..


55 44 11 00 00 3232

11 2 2 00 11 00 1010

––2 2 ––33 00 00 11 00

Example 1, page 226




✦ We see that the We see that the ratioratio 10/2 = 510/2 = 5 is is less thanless than the the ratio ratio 32/4 = 832/4 = 8, so , so row 2row 2 is the is the pivot rowpivot row..

2 3Maximize P x y 2 3Maximize P x y 324

102

8

5

324

102

8

5


55 44 11 00 00 3232

11 2 2 00 11 00 1010

––2 2 ––33 00 00 11 00

Example 1, page 226

Ratio





55 44 11 00 00 3232

11 2 2 00 11 00 1010

––2 2 ––33 00 00 11 00

Example 1, page 226




55 44 11 00 00 3232

11 2 2 00 11 00 1010

––2 2 ––33 00 00 11 00

122 R122 R

Example 1, page 226




55 44 11 00 00 3232

1/21/2 1 1 00 1/21/2 00 55

––2 2 ––33 00 00 11 00

122 R122 R

Example 1, page 226




2 3Maximize P x y 2 3Maximize P x y 1 2

3 2

4

3

R R

R R

1 2

3 2

4

3

R R

R R


55 44 11 00 00 3232

1/21/2 1 1 00 1/21/2 00 55

––2 2 ––33 00 00 11 00

Example 1, page 226




2 3Maximize P x y 2 3Maximize P x y 1 2

3 2

4

3

R R

R R

1 2

3 2

4

3

R R

R R


33 00 11 ––22 00 1212

1/21/2 1 1 00 1/21/2 00 55

––1/2 1/2 00 00 3/23/2 11 1515

Example 1, page 226


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ This This completes an iterationcompletes an iteration..

✦ The The last rowlast row of the tableau contains a of the tableau contains a negative numbernegative number, , so an so an optimal solutionoptimal solution hashas notnot been reached been reached..

✦ Therefore, we Therefore, we repeatrepeat the the iteration stepiteration step..


33 00 11 ––22 00 1212

1/21/2 1 1 00 1/21/2 00 55

––1/2 1/2 00 00 3/23/2 11 1515

Example 1, page 226


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ Since the entry Since the entry –1/2–1/2 isis the the most negative entrymost negative entry to the left to the left

of the vertical line in the last row of the tableau, the of the vertical line in the last row of the tableau, the first first columncolumn in the tableau is now the in the tableau is now the pivot columnpivot column..


33 00 11 ––22 00 1212

1/21/2 1 1 00 1/21/2 00 55

––1/2 1/2 00 00 3/23/2 11 1515

Example 1, page 226




✦ We see that the We see that the ratioratio 12/3 = 412/3 = 4 is is less thanless than the the ratioratio 5/(1/2) = 105/(1/2) = 10, so , so row 1row 1 is now the is now the pivot rowpivot row..


33 00 11 ––22 00 1212

1/21/2 1 1 00 1/21/2 00 55

––1/2 1/2 00 00 3/23/2 11 1515

123

51/2

4

10

123

51/2

4

10

Example 1, page 226





33 00 11 ––22 00 1212

1/21/2 1 1 00 1/21/2 00 55

––1/2 1/2 00 00 3/23/2 11 1515

Example 1, page 226




33 00 11 ––22 00 1212

1/21/2 1 1 00 1/21/2 00 55

––1/2 1/2 00 00 3/23/2 11 1515

113 R113 R

Example 1, page 226




11 00 1/31/3 ––2/32/3 00 44

1/21/2 1 1 00 1/21/2 00 55

––1/2 1/2 00 00 3/23/2 11 1515

113 R113 R

Example 1, page 226




12 12

13 12

R R

R R

12 12

13 12

R R

R R


11 00 1/31/3 ––2/32/3 00 44

1/21/2 1 1 00 1/21/2 00 55

––1/2 1/2 00 00 3/23/2 11 1515

Example 1, page 226




12 12

13 12

R R

R R

12 12

13 12

R R

R R


11 00 1/31/3 ––2/32/3 00 44

00 1 1 ––1/61/6 5/65/6 00 33

00 00 1/61/6 7/67/6 11 1717

Example 1, page 226


Step 3.Step 3. Perform the Perform the pivot operationpivot operation..✦ The The last rowlast row of the tableau contains of the tableau contains nono negative negative

numbersnumbers, so an , so an optimal solutionoptimal solution has beenhas been reached reached..


11 00 1/31/3 ––2/32/3 00 44

00 1 1 ––1/61/6 5/65/6 00 33

00 00 1/61/6 7/67/6 11 1717

Example 1, page 226


Step 4.Step 4. Determine the Determine the optimal solutionoptimal solution..✦ Locate the Locate the basic variablesbasic variables in the final tableau. in the final tableau.

In this case, the In this case, the basic variablesbasic variables are are xx, , yy, and , and PP.. The The optimal valueoptimal value for for xx is is 44.. The The optimal valueoptimal value for for yy is is 33.. The The optimal valueoptimal value for for PP is is 1717, which means that , which means that

the the minimized valueminimized value for for CC is is –17–17..


11 00 1/31/3 ––2/32/3 00 44

00 1 1 ––1/61/6 5/65/6 00 33

00 00 1/61/6 7/67/6 11 1717

Example 1, page 226

The Dual ProblemThe Dual Problem

Another Another special classspecial class of of linear programming problemslinear programming problems we we encounter in practical applications is characterized by the encounter in practical applications is characterized by the following following conditionsconditions::

1.1. The The objective functionobjective function is to be is to be minimizedminimized..

2.2. All the All the variables involvedvariables involved are are nonnegativenonnegative..

3.3. All other All other linear constraintslinear constraints may be written so that the may be written so that the expression involving the variables is expression involving the variables is greatergreater than or than or equal to a nonnegative constantequal to a nonnegative constant..

Such problems are called Such problems are called standard minimization standard minimization problemsproblems..

The Dual ProblemThe Dual Problem

In solving this kind of In solving this kind of linear programming problemslinear programming problems, it , it helps to note that each helps to note that each maximizationmaximization problemproblem is associated is associated with a with a minimizationminimization problemproblem, and vice versa., and vice versa.

The The given problemgiven problem is called the is called the primal problemprimal problem, and the , and the related problemrelated problem is called the is called the dual problemdual problem..

ExampleExample Write the Write the dual problemdual problem associated with this problem: associated with this problem:

We first write down a We first write down a tableautableau for the for the primal problemprimal problem::

6 8Minimize C x y 6 8Minimize C x y 40 10 2400

10 15 2100

5 15 1500

, 0

subject to x y

x y

x y

x y

40 10 2400

10 15 2100

5 15 1500

, 0

subject to x y

x y

x y

x y

xx yy ConstantConstant

4040 1010 24002400

1010 1515 21002100

55 1515 15001500

6 6 88

Primal Primal ProblemProblem

Example 2, page 228

ExampleExample

Next, we Next, we interchangeinterchange the the columnscolumns and and rowsrows of the tableau of the tableau and and headhead the three columns of the resulting array with the the three columns of the resulting array with the three variablesthree variables uu, , vv, and , and ww, obtaining, obtaining

xx yy ConstantConstant

4040 1010 24002400

1010 1515 21002100

55 1515 15001500

6 6 88

uu vv ww ConstantConstant

4040 1010 55 66

1010 1515 1515 88

2400 2400 2100 2100 15001500Example 2, page 228

ExampleExample

Consider the resulting tableau as if it were the Consider the resulting tableau as if it were the initial initial simplex tableausimplex tableau for a for a standard maximization problemstandard maximization problem. .

From it we can reconstruct the required From it we can reconstruct the required dual problemdual problem::

uu vv ww ConstantConstant

4040 1010 55 66

1010 1515 1515 88

2400 2400 2100 2100 15001500

2400 2100 1500Maximize P u v w 2400 2100 1500Maximize P u v w

40 10 5 6

10 15 15 8

, , 0

subject to u v w

u v w

u v w

40 10 5 6

10 15 15 8

, , 0

subject to u v w

u v w

u v w

Dual Dual ProblemProblem

Example 2, page 228

Theorem 1Theorem 1

The Fundamental Theorem of DualityThe Fundamental Theorem of Duality A A primal problemprimal problem has a has a solutionsolution if and only if the if and only if the

corresponding corresponding dual problemdual problem has a has a solutionsolution.. Furthermore, if a solution exists, then:Furthermore, if a solution exists, then:

a.a. The The objective functionsobjective functions of both the of both the primalprimal and and the the dual problemdual problem attain the attain the same optimal valuesame optimal value..

b.b. The The optimal solutionoptimal solution to the to the primal problemprimal problem appears under the appears under the slack variablesslack variables in the last row in the last row of the final simplex tableau associated with the of the final simplex tableau associated with the dual problemdual problem..

ExampleExample Complete the solutionComplete the solution of the problem from our of the problem from our last examplelast example::

2400 2100 1500Maximize P u v w 2400 2100 1500Maximize P u v w

40 10 5 6

10 15 15 8

, , 0

subject to u v w

u v w

u v w

40 10 5 6

10 15 15 8

, , 0

subject to u v w

u v w

u v w

Dual Dual ProblemProblem

Example 3, page 229

ExampleExampleSolutionSolution The The dual problemdual problem associated with the given associated with the given primal primal

problemproblem is a is a standard maximization problemstandard maximization problem.. Thus, we can proceed with the Thus, we can proceed with the simplex methodsimplex method.. First, we First, we introduceintroduce to the system of equations the to the system of equations the slack slack

variablesvariables xx and and yy, and , and restaterestate the the inequalitiesinequalities as as equationsequations, , obtainingobtaining

40 10 5 6

10 15 15 8

2400 2100 1500 0

u v w x

u v w y

u v w P

40 10 5 6

10 15 15 8

2400 2100 1500 0

u v w x

u v w y

u v w P

Example 3, page 229

ExampleExampleSolutionSolution Next, we transcribe the Next, we transcribe the coefficientscoefficients of the system of of the system of

equations equations

into an into an initial simplex tableauinitial simplex tableau::

40 10 5 6

10 15 15 8

2400 2100 1500 0

u v w x

u v w y

u v w P

40 10 5 6

10 15 15 8

2400 2100 1500 0

u v w x

u v w y

u v w P

uu vv ww xx yy PP ConstantConstant

4040 1010 55 11 00 00 66

1010 1515 1515 00 11 00 88

––2400 2400 ––2100 2100 ––15001500 00 00 11 00

Example 3, page 229

ExampleExampleSolutionSolution Continue with the Continue with the simplex iterative methodsimplex iterative method until a until a final final

tableautableau is obtained with the is obtained with the solutionsolution for the problem: for the problem:

The The fundamental theorem of dualityfundamental theorem of duality tells us that the tells us that the solutionsolution to the to the primal problemprimal problem is is xx = 30 = 30 and and yy = 120 = 120, with a , with a minimum valueminimum value for for CC of of 11401140..

uu vv ww xx yy PP ConstantConstant

11 00 ––3/203/20 3/1003/100 ––1/501/50 00 1/501/50

00 11 11/1011/10 ––1/501/50 2/252/25 00 13/2513/25

0 0 00 450450 3030 120120 11 11401140

SolutionSolution for thefor the primal problemprimal problem

Example 3, page 229

4.34.3The Simplex Method: The Simplex Method: Nonstandard ProblemsNonstandard Problems

10–1/301/30 1

0

0

1

v

1

0

0

P

105/34/300

41/32/31 0

42/37/300

Constantwuyx

10–1/301/30 1

0

0

1

v

1

0

0

P

105/34/300

41/32/31 0

42/37/300

Constantwuyx

Nonstandard ProblemsNonstandard Problems

A A nonstandard problemnonstandard problem is one that is one that does not fitdoes not fit any of the any of the two categoriestwo categories of problems we have studied so far: of problems we have studied so far:✦ Standard maximization problem:Standard maximization problem:

1.1. The The objective functionobjective function is to be is to be maximizedmaximized..2.2. All the All the variablesvariables involved in the problem are involved in the problem are nonnegativenonnegative..3.3. All other All other linear constraintslinear constraints may be written so that the may be written so that the

expression involving the variables is expression involving the variables is less than or equal toless than or equal to a nonnegative constanta nonnegative constant..

✦ Standard minimization problem:Standard minimization problem:1.1. The The objective functionobjective function is to be is to be minimizedminimized..2.2. All the All the variables involvedvariables involved are are nonnegativenonnegative..3.3. All other All other linear constraintslinear constraints may be written so that the may be written so that the

expression involving the variables is expression involving the variables is greatergreater than or equal than or equal to a constantto a constant..

4.4. AllAll coefficientscoefficients in the in the objective functionobjective function are are nonnegativenonnegative..

Simplex Method for Solving Nonstandard ProblemsSimplex Method for Solving Nonstandard Problems

1.1. If necessary, If necessary, rewriterewrite the problem as a the problem as a maximizationmaximization problem. problem.

2.2. If necessary, If necessary, rewriterewrite all all constraintsconstraints (except (except xx 0 0, , yy 0 0, , zz 0 0, …) using less than or equal to (, …) using less than or equal to () ) inequalities.inequalities.

3.3. IntroduceIntroduce slack variablesslack variables and set up the initial and set up the initial simplex tableau.simplex tableau.

4.4. ScanScan the upper part of the the upper part of the column of constantscolumn of constants of of the tableau the tableau for negative entriesfor negative entries..a.a. If there are If there are no negative entriesno negative entries, , complete the complete the

solution using the simplex methodsolution using the simplex method for problems for problems in standard form.in standard form.

b.b. If there are If there are negative entriesnegative entries, proceed to , proceed to step 5step 5..

Simplex Method for Solving Nonstandard ProblemsSimplex Method for Solving Nonstandard Problems

5.5. PivotPivot thethe tableau. tableau.a.a. Pick any Pick any negative entrynegative entry in a in a rowrow in which a in which a

negative entrynegative entry in in the the columncolumn ofof constants constants occurs. occurs. The column containing this entry is the The column containing this entry is the pivot pivot columncolumn. .

b.b. ComputeCompute the the positive ratiospositive ratios of the numbers in the of the numbers in the column of constantscolumn of constants to the corresponding to the corresponding numbers in the numbers in the pivot columnpivot column. . The The pivot rowpivot row corresponds to the corresponds to the smallest ratiosmallest ratio. . The The intersectionintersection of the pivot column and the of the pivot column and the pivot row determines the pivot row determines the pivot elementpivot element..

c.c. Pivot the tableauPivot the tableau about the about the pivot elementpivot element. . Then return to Then return to step 4step 4..

ExampleExample

Solve the linear programming problemSolve the linear programming problem

2 3Minimize C x y 2 3Minimize C x y 5

3 9

2 2

, 0

subject to

x y

x y

x y

x y

5

3 9

2 2

, 0

subject to

x y

x y

x y

x y

Example 4, page 246

ExampleExampleSolutionSolution We first We first rewrite the problemrewrite the problem as a as a maximizationmaximization problemproblem

with with constraintsconstraints using using , which gives the following , which gives the following equivalent problemequivalent problem::

2 3Maximize P C x y 2 3Maximize P C x y 5

3 9

2 2

, 0

subject to

x y

x y

x y

x y

5

3 9

2 2

, 0

subject to

x y

x y

x y

x y

Example 4, page 246

ExampleExampleSolutionSolution Introduce the Introduce the slack variablesslack variables uu and and vv, and set up the , and set up the initial initial

simplex tableausimplex tableau::

xx yy uu vv ww PP ConstantConstant

11 11 11 00 00 00 55

––11 ––3 3 00 11 00 00 ––99

––22 1 1 00 00 11 00 22

22 ––33 00 00 00 11 00

Example 4, page 246

ExampleExampleSolutionSolution Follow the Follow the procedureprocedure for solving for solving nonstandard problemsnonstandard problems

outlined previously:outlined previously:

Pivot Pivot ElementElement

51

93

21

5

3

2

51

93

21

5

3

2


11 11 11 00 00 00 55

––11 ––3 3 00 11 00 00 ––99

––22 1 1 00 00 11 00 22

22 ––33 00 00 00 11 00

Example 4, page 246



1 3

2 3

4 3

3

3

R R

R R

R R

1 3

2 3

4 3

3

3

R R

R R

R R


11 11 11 00 00 00 55

––11 ––3 3 00 11 00 00 ––99

––22 1 1 00 00 11 00 22

22 ––33 00 00 00 11 00

Example 4, page 246



1 3

2 3

4 3

3

3

R R

R R

R R

1 3

2 3

4 3

3

3

R R

R R

R R


33 00 11 00 ––11 00 33

––77 0 0 00 11 33 00 ––33

––22 1 1 00 00 11 00 22

––44 00 00 00 33 11 66

Example 4, page 246




33 00 11 00 ––11 00 33

––77 0 0 00 11 33 00 ––33

––22 1 1 00 00 11 00 22

––44 00 00 00 33 11 66

33

3 37 7

1

33

3 37 7

1


Example 4, page 246




33 00 11 00 ––11 00 33

––77 0 0 00 11 33 00 ––33

––22 1 1 00 00 11 00 22

––44 00 00 00 33 11 66

127 R 127 R

Example 4, page 246




33 00 11 00 ––11 00 33

11 0 0 00 ––1/71/7 ––3/73/7 00 3/73/7

––22 1 1 00 00 11 00 22

––44 00 00 00 33 11 66

127 R 127 R

Example 4, page 246



1 2

3 2

4 2

3

2

4

R R

R R

R R

1 2

3 2

4 2

3

2

4

R R

R R

R R


33 00 11 00 ––11 00 33

11 0 0 00 ––1/71/7 ––3/73/7 00 3/73/7

––22 1 1 00 00 11 00 22

––44 00 00 00 33 11 66

Example 4, page 246



1 2

3 2

4 2

3

2

4

R R

R R

R R

1 2

3 2

4 2

3

2

4

R R

R R

R R


00 00 11 3/73/7 2/72/7 00 12/712/7

11 0 0 00 ––1/71/7 ––3/73/7 00 3/73/7

00 1 1 00 ––2/72/7 1/71/7 00 20/720/7

00 00 00 ––4/74/7 9/79/7 11 54/754/7

Example 4, page 246

ExampleExampleSolutionSolution We now use the We now use the simplex methodsimplex method for problems in for problems in standard standard

formform to complete the problem. to complete the problem.


00 00 11 3/73/7 2/72/7 00 12/712/7

11 0 0 00 ––1/71/7 ––3/73/7 00 3/73/7

00 1 1 00 ––2/72/7 1/71/7 00 20/720/7

00 00 00 ––4/74/7 9/79/7 11 54/754/7


12/73/7 412/73/7 4

Example 4, page 246




00 00 11 3/73/7 2/72/7 00 12/712/7

11 0 0 00 ––1/71/7 ––3/73/7 00 3/73/7

00 1 1 00 ––2/72/7 1/71/7 00 20/720/7

00 00 00 ––4/74/7 9/79/7 11 54/754/7

713 R713 R

Example 4, page 246




00 00 7/37/3 11 2/32/3 00 44

11 0 0 00 ––1/71/7 ––3/73/7 00 3/73/7

00 1 1 00 ––2/72/7 1/71/7 00 20/720/7

00 00 00 ––4/74/7 9/79/7 11 54/754/7

Example 4, page 246

713 R713 R




00 00 7/37/3 11 2/32/3 00 44

11 0 0 00 ––1/71/7 ––3/73/7 00 3/73/7

00 1 1 00 ––2/72/7 1/71/7 00 20/720/7

00 00 00 ––4/74/7 9/79/7 11 54/754/7

12 17

23 17

44 17

R R

R R

R R

12 17

23 17

44 17

R R

R R

R R

Example 4, page 246




00 00 7/37/3 11 2/32/3 00 44

11 0 0 1/31/3 00 ––1/31/3 00 11

00 1 1 2/32/3 00 1/31/3 00 44

00 00 4/34/3 00 5/35/3 11 1010

12 17

23 17

44 17

R R

R R

R R

12 17

23 17

44 17

R R

R R

R R

Example 4, page 246



All the All the entries in the lastentries in the last row are row are nonnegativenonnegative and hence and hence the the tableautableau is is finalfinal..


00 00 7/37/3 11 2/32/3 00 44

11 0 0 1/31/3 00 ––1/31/3 00 11

00 1 1 2/32/3 00 1/31/3 00 44

00 00 4/34/3 00 5/35/3 11 1010

Example 4, page 246



Thus, the Thus, the optimal solutionoptimal solution is: is:

xx = 1 = 1 yy = 4 = 4 uu = 0 = 0 vv = 4 = 4 ww = 0 = 0 CC = – = – PP = –10 = –10


00 00 7/37/3 11 2/32/3 00 44

11 0 0 1/31/3 00 ––1/31/3 00 11

00 1 1 2/32/3 00 1/31/3 00 44

00 00 4/34/3 00 5/35/3 11 1010

Example 4, page 246

End of End of Chapter Chapter

4 The Simplex Method: Standard Maximization Problems The Simplex Method: Standard Minimization...

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