Lesson 32: Simplex Method II

Lesson 32 (KH, Section 11.2)The Simplex Method

Math 20

December 7, 2007

Announcements

I Pset 11 due December 11. Pset 12 due December 18.

I next PS Sunday 6-7 in SC B-10

We are going to solve the linear programming problem ofmaximizing

z = 2x1 − 4x2 + 5x3

subject to constraints

3x1 + 2x2 + x3 ≤ 6

3x1 − 6x2 + 7x3 ≤ 9

and x1, x2, x3 ≥ 0.

Introduce slack variables u1 and u2 and rewrite the LP problem toreplace the inequalities with equalities.

Solution

3x1+2x3+ x3+u1 =6

3x1−6x2+7x3 +u2 =9

The new constraints are x1, x2, x3, u1, u2 ≥ 0.

Introduce slack variables u1 and u2 and rewrite the LP problem toreplace the inequalities with equalities.

Solution

3x1+2x3+ x3+u1 =6

3x1−6x2+7x3 +u2 =9

The new constraints are x1, x2, x3, u1, u2 ≥ 0.

How many basic solutions are there to this problem? (Don’t findthem, just count them)

SolutionThere are 3 + 2 = 5 variables and 3 of them are decision variables,so the number of basic solutions is(

5

3

)= 10

How many basic solutions are there to this problem? (Don’t findthem, just count them)

SolutionThere are 3 + 2 = 5 variables and 3 of them are decision variables,so the number of basic solutions is(

5

3

)= 10

Let’s choose x1 = x2 = x3 = 0, u1 = 6, u2 = 9 for our initial basicsolution. Since it’s feasible (all x and u variables are nonnegative),we know it’s a corner of the feasible set. The geometric ideabehind the simplex method is to traverse the feasible set, movingat each set to the adjacent vertex which increases the objectivefunction the most. So let’s find the corners adjacent to (0, 0, 0).

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

Traversing the feasible set

•

(0, 0, 0, 6, 9)

• (0, 0, 6, 0,−6)

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

•(0,−3/2, 0, 3, 0)

• (2, 0, 0, 0, 3)• (3, 0, 0,−3, 0)

• (0, 0, 6, 0,−6)

•(0,−3/2, 0, 3, 0)

• (3, 0, 0,−3, 0)

•

• (0, 0, 9/7, 33/7, 0)

•(0, 3, 0, 0, 18)

• (2, 0, 0, 0, 3)

z = 45/7

z = −18

z = 6

•(0,−3/2, 0, 3, 0)

• (0, 0, 6, 0,−6)

• (3, 0, 0,−3, 0)

• (0, 0, 9/7, 33/7, 0)

•

(0, 0, 0, 6, 9)

•(0, 33/20, 27/10, 0, 0)

•(11/6, 0, 1/2, 0, 0)

z = 69/10

z = 37/6

•(0, 33/20, 27/10, 0, 0)

z = 69/10

(x1, x2, x3, u1, u2)

I Find all basic solutions which are either on the x1 axis(x2 = x3 = 0), the x2 axis (x1 = x3 = 0), or the x3 axis(x1 = x2 = 0).

I The ones that “count” are the basic solutions which are alsofeasible. Find them. You should get three; one on each axis.

I Find the value of the objective function on each of thesecorners and find which one was biggest. What nonbasicvariable went from zero to positive? What basic variable wentfrom positive to zero?

Now let’s look at the numerical algorithm for the simplex method.The initial tableau is

x1 x2 x3 u1 u2 z valueu1 3 2 1 1 0 0 6u2 3 −6 7 0 1 0 9z −2 4 −5 0 0 1 0

Because the variables are x1, x2 and x3, we can mentally block outthose columns of the tableau. What’s left are the equations u1 = 6,u2 = 9, and z = 0.

Now let’s look at the numerical algorithm for the simplex method.The initial tableau is

x1 x2 x3 u1 u2 z valueu1 3 2 1 1 0 0 6u2 3 −6 7 0 1 0 9z −2 4 −5 0 0 1 0

To go to the next tableau, we need to decide which variable becomesbasic (nonzero) and basic variable becomes zero (nonbasic). The en-tering variable is the variable in the column with the largest nega-tive entry in the bottom row. The departing variable is the variablein the row which has the smallest positive θ-ratio: the “value” col-umn divided by the column of the new entering variable.

Now let’s look at the numerical algorithm for the simplex method.The initial tableau is

x1 x2 x3 u1 u2 z valueu1 3 2 1 1 0 0 6u2 3 −6 7 0 1 0 9z −2 4 −5 0 0 1 0

What is the entering variable? What is the departing variable?

Now let’s look at the numerical algorithm for the simplex method.The initial tableau is

x1 x2 x3 u1 u2 z valueu1 3 2 1 1 0 0 6u2 3 −6 7 0 1 0 9z −2 4 −5 0 0 1 0

What is the entering variable? What is the departing variable?

Solutionx3 is the entering variable and u2 is the departing variable.

Now that we’ve decided which variables to swap, we need to alterthe tableau. After mentally blocking out the columnscorresponding to the nonbasic variables, we would like the basicsolution to read off as easily as it did before. That is, we want thecolumn of the entering variable to be all zero except in oneposition where it’s one. Since there is a row being vacated by thedeparting variable, we can do row operations so that:

I The entry in the column of the entering variable and the rowof the departing variable is one.

I All the rest of the entries in the column of the enteringvariable are zero.

Do the necessary row operations.

Now that we’ve decided which variables to swap, we need to alterthe tableau. After mentally blocking out the columnscorresponding to the nonbasic variables, we would like the basicsolution to read off as easily as it did before. That is, we want thecolumn of the entering variable to be all zero except in oneposition where it’s one. Since there is a row being vacated by thedeparting variable, we can do row operations so that:

I The entry in the column of the entering variable and the rowof the departing variable is one.

I All the rest of the entries in the column of the enteringvariable are zero.

Do the necessary row operations.

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

u2 3 −6 7 0 1 0 9

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

u2 3 −6 7 0 1 0 9

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

u2 3 −6 7 0 1 0 9

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

u2 3 −6 7 0 1 0 9

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

u2 3 −6 7 0 1 0 9

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

u2 3 −6 7 0 1 0 9

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

u2 3 −6 7 0 1 0 9

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

u2 3 −6 7 0 1 0 9

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7

−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

x3 3/7 −6/7 1 0 1/7 0 9/7

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 3 2 1 1 0 0 6

x3 3/7 −6/7 1 0 1/7 0 9/7

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7

−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z −2 4 −5 0 0 1 0

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

u1 18/7 20/7 0 1 −1/7 0 33/7

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/20

6/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

x2 9/10 1 0 7/20−1/20 0 33/20

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

x2 9/10 1 0 7/20−1/20 0 33/20

x3 3/7 −6/7 1 0 1/7 0 9/7

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/20

6/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

x2 9/10 1 0 7/20−1/20 0 33/20

x3 6/5 0 1 3/10 1/10 0 27/10

z 1/7 −2/7 0 0 5/7 1 45/7

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

x2 9/10 1 0 7/20−1/20 0 33/20

x3 6/5 0 1 3/10 1/10 0 27/10

z 2/5 0 0 1/10 7/10 1 69/10

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

x2 9/10 1 0 7/20−1/20 0 33/20

x3 6/5 0 1 3/10 1/10 0 27/10

z 2/5 0 0 1/10 7/10 1 69/10

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10

Tableau

x1 x2 x3 u1 u2 z value

x2 9/10 1 0 7/20−1/20 0 33/20

x3 6/5 0 1 3/10 1/10 0 27/10

z 2/5 0 0 1/10 7/10 1 69/10

largest negativecoefficient in ob-jective row

entering variable

θ

69/7

smallest positiveθ-ratio

departing variable

make this entry 1and the rest of itscolumn 0

×1/7−1

5

largest negativecoefficient in ob-jective row

entering variable

θ33/20

−3/2

smallest positiveθ-ratiodeparting variable

make this entry 1and the rest of itscolumn 0

×7/206/7

2/7

No more negative coefficients.We are done!

solution is

x1 = 0

x2 = 33/20

x3 = 27/10

z = 69/10