lesson-31-the-simplex-method-i-1197655204190297-5

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Lesson 31 (KH, Section 11.2)The Simplex Method

Math 20

December 5, 2007

Announcements

Pset 11 due December 10. Pset 12 due December 17.

ML OH: Today 13 (SC 323)

SS OH: Tonight 8:309:30 in Quincy dining hall

HW coming

Midterm II review slides online

Midterm II: tomorrow 78:30pm in Hall A
http://find/http://goback/


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Outline

Setup

Illustrative ProblemSlack Variables

The Simplex Method, By ExampleThe Initial Basic Feasible SolutionCreating a New Tableau

Recap of Steps

Example


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Any vector x which satifies all the inequalities is called a feasiblesolution to the given problem, and a feasible solution maximizing

the objective function is called an optimal solution.


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Why are basic feasible solutions necessary?

TheoremIf a LP problem has an optimal solution, then it has a basic

optimal solution.This is just a restatement of the corner principle. So we only needto find the basic feasible solutions!


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Back to the Baker

2x

+y

=50

x+2y=

70

10 20 30 40 50 60 70

(0, 0) (25, 0) (70, 0)

(0, 35)

(0, 50)

(10, 30)


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Back to the Baker

u

=0

v=0

10 20 30 40 50 60 70

(0, 0) (25, 0) not feasible

(0, 35)

not feasible

(10, 30)


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How many basic feasible solutions are there? Out of the m + nvariables, we choose n to set equal to zero, and solve for the rest.This can be done

n + m

m

=

(n + m)!

m! n!

ways. Thats a lot!The simplex method is a way to arrive at an optimal solution bytraversing the vertices of the feasible set, in each step increasingthe objective function by as much as possible.


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Outline

Setup

Illustrative ProblemSlack Variables

The Simplex Method, By ExampleThe Initial Basic Feasible SolutionCreating a New Tableau

Recap of Steps

Example


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Well work with the illustrative problem. We can start with thebasic feasible solution x = 0, y = 0. Thus u = 50 and v = 70.

This is our initial basic solution.Well start writing everything in a table (or tableau), so lets alsowrite the objective function with a right-hand side of zero. Thus

8x 10y + z = 0.

We put this all together, forming what is called the initial tableau:

x y u v z valueu 2 1 1 0 0 50

v 1 2 0 1 0 70z 8 10 0 0 1 0


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x y u v z valueu 2 1 1 0 0 50v 1 2 0 1 0 70z 8 10 0 0 1 0

Is the solution u = 50, v = 70 (i.e., x = 0, y = 0 optimal?)


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x y u v z valueu 2 1 1 0 0 50v 1 2 0 1 0 70z 8 10 0 0 1 0

Is the solution u = 50, v = 70 (i.e., x = 0, y = 0 optimal?) No,increasing x or y would increase z.


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x y u v z value

u 2 1 1 0 0 50v 1 2 0 1 0 70z 8 10 0 0 1 0

Move from one basic solution to another

One of the zero (nonbasic) variables becomes nonzero andone of nonzero (basic) variables becomes zero

Do this as efficiently as possible


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x y u v z value

u 2 1 1 0 0 50v 1 2 0 1 0 70z 8 10 0 0 1 0

Move from one basic solution to another

One of the zero (nonbasic) variables becomes nonzero andone of nonzero (basic) variables becomes zero

Do this as efficiently as possible

Which of x or y would you most like to increase?


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How much can we increase y? Well, since x is still zero, the

equations of constraint can be written

u = 50 y

v = 70 2y


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How much can we increase y? Well, since x is still zero, the

equations of constraint can be written

u = 50 y

v = 70 2y

We still need u 0 and v 0, so the most we can increase y is to35. This is the smallest of the ratios 50

1= 50 and 70

2= 35. So

were going to increase y to 35. This will make v = 0. We call vthe departing variable.The new basic solution therefore has y = 35, v = 0, u = 15, andx = 0. The new value of the objective function is z = 10y = 350.

Creating a New Tableau


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Creating a New TableauWe are exchanging the basic variable v for y. This means the objective row has to be replaced with one that has a zero in the ycolumns. We can do this by adding multiples of row 2. Since y is

an entering variable, we might as well normalize row 2 to have aone.So we scale the second row to have a one in the y column.

x y u v z valueu 2 1 1 0 0 50v 1 2 0 1 0 70

z 8 10 0 0 1 0

Creating a New Tableau


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Creating a New TableauNow we zero out the rest of this column by adding 10 times row 2to row 3, and subtracting row 2 from row 1.

x y u v z value2 1 1 0 0 50

1/2 1 0 1/2 0 35

z 8 10 0 0 1 0


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Rinse Lather Repeat


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Rinse, Lather, RepeatThe x column in the objective row has a negative entry, so increasingx will increase z.

x y u v z valueu 3/2 0 1 1/2 0 15y 1/2 1 0 1/2 0 35z 3 0 0 5 1 350

Rinse Lather Repeat


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Rinse, Lather, RepeatThe x column in the objective row has a negative entry, so increasingx will increase z. How much can we increase it?

x y u v z valueu 3/2 0 1 1/2 0 15y 1/2 1 0 1/2 0 35z 3 0 0 5 1 350


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Rinse Lather Repeat


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Rinse, Lather, RepeatThe x column in the objective row has a negative entry, so increasingx will increase z. How much can we increase it? The minimumof the two ratios 15

3/2 = 10 and35

1/2 = 70. So x in the entering

variable and u is the departing variable.

x y u v z value u 3/2 0 1 1/2 0 15

y 1/2 1 0 1/2 0 35z 3 0 0 5 1 350

Rinse, Lather, Repeat


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3/2 = 10 and35


variable and u is the departing variable.We scale row one by 2/3 to make it one in the basic column.

x y u v z valueu 3/2 0 1 1/2 0 15y 1/2 1 0 1/2 0 35z 3 0 0 5 1 350



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3/2 = 10 and35


variable and u is the departing variable.We scale row one by 2/3 to make it one in the basic column.And we zero out the rest of the column by subtracting half of row1 from row 2, and adding 3 times row 1 to row 3.

x y u v z valueu 1 0 2/3 1/3 0 10y 1/2 1 0 1/2 0 35

3 0 0 5 1 350



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, , pThe x column in the objective row has a negative entry, so increasingx will increase z. How much can we increase it? The minimumof the two ratios 15

3/2 = 10 and35


variable and u is the departing variable.We scale row one by 2/3 to make it one in the basic column.And we zero out the rest of the column by subtracting half of row1 from row 2, and adding 3 times row 1 to row 3.

x y u v z valueu 1 0 2/3 1/3 0 10y 1/2 1 0 1/2 0 35

3 0 0 5 1 350


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x y u v z valuex 1 0 2/3 1/3 0 10y 0 1 1/3 2/3 0 30

0 0 2 4 1 380Now any increase in the decision variables or slack variables wouldresult in a decrease of z. We are done!


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Recap of Steps


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1. Set up the initial tableau.

2. Apply the optimality test. If the objective row has no negative

entries in the columns labeled with variables, then theindicated solution is optimal; we can stop.

3. Choose a pivotal column by determining the column with themost negative entry in the objective row. If there are severalcandidates for a pivotal column, choose any one.

4. Choose a pivotal row. Form the ratios of the entries above theobjective row in the rightmost column by the correspondingentries of the pivotal column for those entries in the pivotalcolumn which are positive. The pivotal row is the row forwhich the smallest of these ratios occurs. If there is a tie,choose any one of the qualifying rows. If none of the entriesin the pivotal column above the objective row is positive, theproblem has no finite optimum. We stop.

5. Perform pivotal elimination to construct a new tableau and

return to Step 2.


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We insert slack variables u1, u2, and u3.The equations of constraint become

2x1+4x2+ x3+u1 4

2x12x2+3x3 +u2 4

2x1+ x2 x3 +u3 8

with all variables nonnegative.

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