Chapter 4 The Simplex Algorithm – Part IIise.tamu.edu/INEN420/INEN420_2005Spring/SLIDES/Chapter...

L. Ntaimo (c) 2005 INEN420 TAMU1

Chapter 4The Simplex Algorithm – Part II

Based onIntroduction to Mathematical Programming: Operations Research, Volume 1

4th edition, by Wayne L. Winston and Munirpallam Venkataramanan

Lewis Ntaimo


So Far …

• Modeling LPs• Solving 2-variable LPs graphically• Simplex method - Max LPs• Simplex method - Min LPs

– Method 1: Multiply objective function by -1– Method 2: Modify Step 3 of simplex algorithm for Max LPs

• Simplex method – Special cases:– Alternative optimal solutions– Unbounded LPs


Degeneracy and Convergence of the Simplex Algorithm


4.11 Convergence of the Simplex Method

0 s.t.

zMax

≥=

=

xbAx

xcTRecall:

Where,

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

mnmm

n

n

aaa

aaaaaa

A

............

...

...

21

22221

11211

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nx

xx

x

.

.

.2

1

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nc

cc

c

.

.

.2

1

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

mb

bb

b

.

.

.2

1

Assumption: Constraint matrix A has n columns and m linearly independent rows. So we can form an m x m square matrix called the basis by setting (n - m) variables to zero (nonbasic variables or nbv’s). The remaining variables are the basic variables or bv’s.


4.11 Convergence of the Simplex MethodRecall that for any LP with m constraints, two bfs are said to be adjacent if their sets of bv’s have (m - 1) bv’s in common.

e.g. bfs of two immediate simplex tableaus are adjacent bfs.

• If an LP in standard form has m constraints and n variables, then there may be a basic solution for each choice of nonbasic variables.

• From n variables, in how many different ways can a set of (n - m)nonbasic variables (or equivalently, m basic variables) be chosen?

!)!(!

mmnn

mn

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

Thus an LP can have at most basic solutions!⎟⎟⎠

⎞⎜⎜⎝

⎛mn


4.11 Convergence of the Simplex Method

• For n = 20 and m = 10 we have:

?!10)!1020(

!201020

=−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ 184,756 bfs!

Assuming that no bfs is repeated the simplex algorithm will find the optimal bfs after a finite number of iterations.

• Vast experience with the simplex algorithm indicates that an optimal solution is found after examining fewer than 3m bfs (e.g. 3(10) = 30).

• Compared with examining 184,756 bfs, the simplex algorithm is quite efficient!


4.11 Degeneracy and Convergence of the Simplex Method

Definition: An LP is degenerate if it has at least one bfs in which a basic variable is equal to zero.

• We say that an LP is nondegenerate if all bfs are positive.

• Theoretically, the simplex algorithm (as you have learned it so far) can fail to find the optimal solution to an LP.

• However, LPs arising from actual applications seldom exhibit this behavior.



• Consider the following relationship for a max LP:

z-value for new bfs = z-value for current bfs – (value of entering variable in new bfs)*(coefficient of entering variable in row 0 of current bfs).

Recall: (coefficient of entering variable in row 0) < 0 and (value of entering variable in new bfs) >= 0.

Two Facts:(1) If (value of entering variable in new bfs) > 0), then

(z-value for new bfs) > (z-value for current bfs).

(2) If (value of entering variable in new bfs) = 0), then(z-value for new bfs) = (z-value for current bfs).



Fact 1 implies that each iteration of the simplex algorithm will increase z for a nondegenerate LP. This means that it is impossible to encounter the same bfs twice. So if we use the simplex algorithm to solve a nondegenerate LP, we are guaranteed to find the optimal solution in a finite number of iterations.

Fact 2 implies that the simplex algorithm may fail for a degenerate LP. In this case the algorithm will encounter the same bfs at least twice. This occurrence is called cycling. If cycling occurs then the algorithm can loop, or cycle, forever among a set of bfs and never get to the optimal solution!



Example: Solve the following LP using the simplex tableau method.

0, 0 6 s.t.

25Max

21

21

21

21

≥≤−≤+

+=

xxxxxx

xxz

Converting to standard form:

0,,, 0 6 s.t.

25Max

2121

221

121

21

≥=+−=++

+=

ssxxsxx

sxxxxz


4.11 Degeneracy and Convergence of the Simplex MethodExample: Solve using the simplex tableau methodInitial Tableau Solution

Row z x1 x2 s1 s2 RHS BV MRT0 1 -5 -2 0 0 0 z = 0 z = 01 0 1 1 1 0 6 s1 = 6 6 x1 = 02 0 1 -1 0 1 0 s2 = 0 0* x2 = 0

Second TableauRow z x1 x2 s1 s2 RHS BV MRT

0 1 0 -7 0 5 0 z = 0 z = 01 0 0 2 1 -1 6 s1 = 6 3* x1 = 02 0 1 -1 0 1 0 x1 = 0 None x2 = 0

Cycling: We encounter same bfs twice!

Third (Optimal) TableauRow z x1 x2 s1 s2 RHS BV MRT

0 1 0 0 3.5 1.5 21 z = 21 z = 211 0 0 1 0.5 -0.5 3 x2 = 3 x1 = 32 0 1 0 0.5 0.5 3 x1 = 3 x2 = 3

Optimal Solution: z = 21, x1 = 3, x2 =3



If an LP has many degenerate bfs (or a bfs with many bv’s equal to zero), then the simplex algorithm is often very inefficient

Example: Solve the previous example using the graphical methodThe extreme points of the feasible region are B,C and D. Three sets of basic variables correspond to extreme point C.

021 ≤− xx

621 ≤+ xx

102510

21 =+⇒=xx

z

B

D

AC x1

x2

2 4 6

2

6

4

21=z

BV's BFS Extreme Pointx1, x2 x1 = x2 = 3, s1 = s2 = 0 Dx1, s1 x1 = 0, s1 = 6, x2 = s2 = 0 Cx1, s2 x1 = 6, s2 = -6, x2 = s1 = 0 Infeasiblex2, s1 x2 = 0, s1 = 6, x1 = s2 = 0 Cx2, s2 x2 = 6, s2 = 6, s1 = x1 = 0 Bs1, s2 s1 = 6, s2 = 0, x1 = x2 = 0 C



For an LP with n decision variables to be degenerate, (n+1) or more of the LP’s constraints (including the sign restrictions xi ≥ 0 as constraints) must be binding at an extreme point.

Fortunately, the simplex method can be modified to ensure that cycling will never occur.

Apply the so called anti-cycling rules.



Bland’s Rules for Anti-Cycling: Smallest Subscript Pivoting Rule

(Assume that slack and excess variables are numbered xn+1, xn+2, ...).

Step 1. Choose as the entering variable (in a max problem) the variable with a negative coefficient in row 0 that has the smallest subscript.

Step 2. If there is a tie in the minimum ratio test, then break the tie by choosing the winner of the ratio test so that the variable leaving the basis has the smallest subscript.



Additional homework problems have been posted on the course website on degeneracy and convergence of the simplex method.


The Big M Method


4.12 The Big M Method

• Recall that the simplex method algorithm requires a starting bfs. In all the LPs we have considered so far, we have found starting bfs by using the slack variables as our basic variables.

• If an LP has ≥ or = constraints, however, a starting bfs may not be readily apparent. In such a case, we need another method to solve the problem.



Consider the following problem:

Bevco manufactures an orange-flavored soft drink called Oranj by combining orange soda and orange juice. Each orange soda contains 0.5 oz of sugar and 1 mg of vitamin C. Each ounce of orange juice contains 0.25 oz of sugar and 3 mg of vitamin C. It costs Bevco 2¢ to produce an ounce of orange soda and 3¢ to produce an ounce of orange juice. Bevco’s marketing department has decided that each 10-oz bottle of Oranj must contain at least 20 mg of vitamin C and at most 4 oz of sugar. Use linear programming to determine how Bevco canmeet the marketing department’s requirements at minimum cost.



Decision variables:

x1 = number of ounces of orange soda in a bottle of Oranj

x2 = number of ounces of orange juice in a bottle of Oranj

LP:

Min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 ≤ 4 (sugar constraint)

x1 + 3x2 ≥ 20 (vitamin C constraint)

x1 + x2 = 10 (10 oz in 1 bottle of Oranj)

x1, x2 ≥ 0



The LP in standard:

Min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 + s1 = 4

x1 + 3x2 - e2 = 20

x1 + x2 = 10x1, x2, s1, e2 ≥ 0

The LP in standard form has z and s1 which could be used for BVs but row 2 would violate sign restrictions and row 3 has no readily apparent basic variable.



In order to use the simplex method, a bfs is needed. The idea of the Big M method is to create artificial variables to provide an initial bfs and then eliminate them from the final solution. The variables will be labeled aiaccording to the row i in which they are used as follows.

Row 0: z - 2x1 - 3x2

Row 1: 0.5x1 + 0.25x2 + s1 = 4

Row 2: x1 + 3x2 - e2 + a2 = 20

Row 3: x1 + x2 + a3 = 10


4.12 The Big M MethodIn the optimal solution, all artificial variables must be set equal to zero. To accomplish this, in a min LP, a term Mai is added to the objective function for each artificial variable ai. For a max LP, the term –Mai is added to the objective function for each ai. M represents some very large number. The modified Bevco LP in standard form then becomes:

Min z = 2x1 + 3x2 + Ma2 + Ma3

s.t. 0.5x1 + 0.25x2 + s1 = 4

x1 + 3x2 - e2 + a2 = 20

x1 + x2 + a3 = 10x1, x2, s1, e2, a2, a3 ≥ 0

Modifying the objective function this way makes it extremely costly for an artificial variable to be positive. The optimal solution should force a2 = a3 =0.


4.12 The Big M MethodThe rows of the simplex tableau would then be:

Row 0: z - 2x1 - 3x2 - Ma2 - Ma3 = 0Row 1: 0.5x1 + 0.25x2 + s1 = 4Row 2: x1 + 3x2 - e2 + a2 = 20Row 3: x1 + x2 + a3 = 10

But observe that to have canonical form we need to eliminate a2 and a3from row 0! Simply set new Row 0 to Row 0 + M(Row 2) + M(Row 3)

z - 2x1 - 3x2 - Ma2 - Ma3 = 0M( x1 + 3x2 - e2 + a2 = 20)

+ M( x1 + x2 + a3 = 10)z + (2M-2)x1 +(4M-3)x2 – Me2 = 30M



Description of the Big M Method1. Modify the constraints so that the rhs of each constraint is

nonnegative. Identify each constraint that is now an = or ≥ constraint.

2. Convert each inequality constraint to standard form (add a slackvariable for ≤ constraints, add an excess variable for ≥ constraints).

3. For each ≥ or = constraint, add artificial variables. Add sign restriction ai ≥ 0.

4. Let M denote a very large positive number. Add (for each artificial variable) Mai to min problem objective functions or -Mai to max problem objective functions.

5. Since each artificial variable will be in the starting basis, all artificial variables must be eliminated from row 0 before beginning the simplex. Remembering M represents a very large number, solve the transformed problem by the simplex.



If all artificial variables in the optimal solution equal zero, the solution is optimal.

If any artificial variables are positive in the optimal solution, the problem is infeasible.



Iteration 1 z x1 x2 s1 e2 a2 a3 rhs ratio ero

0 1.00 2M - 2 4M -3 -M 30M Row 0 + M(Row 2) + M(Row 3)

1 0.50 0.25 1.00 4.00 162 1.00 3 -1.00 1.00 20.00 6.673 1.00 1.00 1.00 10.00 10

ero 1 z x1 x2 s1 e2 a2 a3 rhs ero0 1.00 2M - 2 4M -3 -M 30M 1 0.50 0.25 1.00 4.002 0.33 1 -0.33 0.33 6.67 Row 2 x 1/33 1.00 1.00 1.00 10.00

ero 2 z x1 x2 s1 e2 a2 a3 rhs ero0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/3 Row 0 - (4M-3)*(Row 2)1 0.50 0.25 1.00 4.002 0.33 1 -0.33 0.33 6.673 1.00 1.00 1.00 10.00

ero 3 z x1 x2 s1 e2 a2 a3 rhs ero0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/3 1 0.42 1.00 0.08 -0.08 2.33 Row 1 - 0.25*(Row 2)2 0.33 1 -0.33 0.33 6.673 1.00 1.00 1.00 10.00

ero 4 z x1 x2 s1 e2 a2 a3 rhs ero0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/31 0.42 1.00 0.08 -0.08 2.332 0.33 1 -0.33 0.33 6.673 0.67 0.33 -0.33 1.00 3.33 Row 3 - Row 2

The Bevco example continued (Using Method 2 for Min LP): Details


4.12 The Big M MethodIteration

2 z x1 x2 s1 e2 a2 a3 rhs ratio

0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/31 0.42 1.00 0.08 -0.08 2.33 5.602 0.33 1 -0.33 0.33 6.67 20.003 0.67 0.33 -0.33 1.00 3.33 5.00

ero 1 z x1 x2 s1 e2 a2 a3 rhs ero0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/31 0.42 1.00 0.08 -0.08 2.332 0.33 1 -0.33 0.33 6.673 1.00 0.50 -0.50 1.50 5.00 (Row 3)*(3/2)

ero 2 z x1 x2 s1 e2 a2 a3 rhs ero0 1.00 -0.50 (1-2M)/2 (3-2M)/2 25.00 Row 0 + (3-2M)*(Row 3)/31 0.42 1.00 0.08 -0.08 2.332 0.33 1.00 -0.33 0.33 6.673 1.00 0.50 -0.50 1.50 5.00

ero 3 z x1 x2 s1 e2 a2 a3 rhs ero0 1.00 -0.50 (1-2M)/2 (3-2M)/2 25.001 1.00 -0.13 0.13 -0.63 0.25 Row 1 - (5/12)*Row 3)2 0.33 1.00 -0.33 0.33 6.673 1.00 0.50 -0.50 1.50 5.00

ero 4 z x1 x2 s1 e2 a2 a3 rhs ero0 1.00 -0.50 (1-2M)/2 (3-2M)/2 25.001 1.00 -0.13 0.13 -0.63 0.252 1.00 -0.50 0.50 -0.50 5.00 Row 2 -(1/3)*Row 33 1.00 0.50 -0.50 1.50 5.00



4.12 The Big M MethodThe Bevco example continued (Using Method 2 for Min LP):

Initial TableauRow z x1 x2 s1 e2 a2 a3 RHS BV MRT

0 1 2M-2 4M-3 0 -M 0 0 30M z = 30M1 0 0.5 0.25 1 0 0 0 4 s1 = 4 162 0 1 3 0 -1 1 0 20 a2 = 20 6.67*3 0 1 1 0 0 0 1 10 a3 = 10 10

Second TableauRow z x1 x2 s1 e2 a2 a3 RHS BV MRT

0 1 (2M-3)/3 0 0 (M-3)/3 (3-4M)/3 0 (60+10M)/3 z = (60+10M)/31 0 5/12 0 1 1/12 -1/12 0 7/3 s1 = 7/3 5.62 0 1/3 1 0 -1/3 1/3 0 20/3 x2 = 20/3 203 0 2/3 0 0 1/3 -1/3 1 10/3 a3 = 10/3 5*

Third (Optimal) TableauRow z x1 x2 s1 e2 a2 a3 RHS BV MRT

0 1 0 0 0 -1/2 (1-2M)/2 (3-2M)/2 25 z = 251 0 0 0 1 -1/8 1/8 -5/8 1/4 s1 = 1/42 0 0 1 0 -1/2 1/2 -1/2 5 x2 = 53 0 1 0 0 1/2 -1/2 3/2 5 x1 = 5




Issues with the Big M Method

1) It is difficult to determine how large M should be

Generally chosen to be at least 100 times the largest

coefficient in the objective function.

2) Introduction of such large numbers cause round off errors and other computational difficulties (numerical instability). Read section on Scaling of LPs on page 167.

This motivates the need for yet another method!



Problem (10 pts). Get with your team and solve the following LP usingthe Big M method :

0, 32

6 2 s.t.3Min

21

21

21

1

≥=+−≤−−

=

xxxxxx

xz


4.12 The Big M MethodSolution to Problem (10 pts):

LP is standard form:

0,, 3 2 6 2 s.t.

3Min

121

21

121

1

≥=+=−+

=

exxxx

exx xz

Big M Formulation:

0,,,, 3 2 6 2 s.t.

3Min

21121

221

1121

211

≥=++=+−+

++=

aaexxaxx

aexx MaMaxz



Eliminate a1 and a2 from row 0. Simply set new Row 0 to Row 0 + M(Row 2) + M(Row 3):

z - 3x1 - Ma1 - Ma2 = 0M( 2x1 + x2 - e1 + a1 = 6)

+ M( x1 + 2x2 + a2 = 3)z + (3M-3)x1 + (3M)x2 – Me1 = 9M


4.12 The Big M MethodUsing Method 2 of the simplex algorithm for min LPs:Initial Tableau

Row z x1 x2 e1 a1 a2 RHS BV MRT0 1 3M-3 3M -M 0 0 9M z = 9M1 0 2 1 -1 1 0 6 a1 = 6 62 0 1 2 0 0 1 3 a2 = 3 3/2*

2 pts

Second TableauRow z x1 x2 e1 a1 a2 RHS BV MRT

0 1 (3M-6)/2 0 -M 0 -3/2M 4.5M z = 4.5M1 0 1.5 0 -1 1 -1 4.5 a1 = 4.5 3*2 0 0.5 1 0 0 0.5 1.5 x2 = 1.5 3

3 pts

Third (Optimal) TableauRow z x1 x2 e1 a1 a2 RHS BV MRT

0 1 0 0 -2 2-M -(4+M)/2 9 z = 91 0 1 0 -2/3 2/3 -2/3 3 x1 = 32 0 0 1 0.5 -0.5 1 0 x2 = 0

3 pts

Optimal Solution: z = 9, x1 = 3, x2 = 0 2 pts


The Two-Phase Simplex Method


4.13 The Two-Phase Simplex Method

• When an initial basic feasible solution is not available, the two-phase simplex method can be used as an alternative to the Big M method.

• In this method we can add artificial variables, and then focus entirely on obtaining an initial basic feasible solution, any basic feasible solution (This is called Phase I).

• We can then start the simplex algorithm with the basic feasible solution we have found (This is called Phase II)



• Observe that if all we want is a basic feasible solution, then we can select any objective function.

• Once we find a basic feasible solution, we can reconsider the original cost coefficients.

• Also note that it is easy to bring cost coefficients into canonical form.

• It is desirable to choose an objective function such that solving the Phase I LP will force the artificial variables to be zero.

• Minimize the sum of artificial variables!



Consider solving the Bevco problem using the two-phase simplex method:

Decision variables:

x1 = number of ounces of orange soda in a bottle of Oranj

x2 = number of ounces of orange juice in a bottle of Oranj

LP:Min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 ≤ 4 (sugar constraint)

x1 + 3x2 ≥ 20 (vitamin C constraint)

x1 + x2 = 10 (10 oz in 1 bottle of Oranj)

x1, x2 ≥ 0



The LP in standard:

Min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 + s1 = 4

x1 + 3x2 - e2 = 20

x1 + x2 = 10x1, x2, s1, e2 ≥ 0

Note again that the LP in standard form has z and s1 which could be used for BVs but row 2 would violate sign restrictions and row 3 has no readily apparent basic variable.


4.13 The Two-Phase Simplex MethodAdding artificial variables we get the following:

Min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 + s1 = 4

x1 + 3x2 - e2 + a2 = 20

x1 + x2 + a3 = 10

x1, x2, s1, e2, a2, a3 ≥ 0

Then the Phase I LP is:

Min w’ = a2 + a3

s.t. 0.5x1 + 0.25x2 + s1 = 4

x1 + 3x2 - e2 + a2 = 20

x1 + x2 + a3 = 10

x1, x2, s1, e2, a2, a3 ≥ 0


4.13 The Two-Phase Simplex MethodNote that row 0 for this tableau contains the basic variables a2 and a3:

Row 0: w’ - a2 - a3 = 0Row 1: 0.5x1 + 0.25x2 + s1 = 4Row 2: x1 + 3x2 - e2 + a2 = 20Row 3: x1 + x2 + a3 = 10

As in the Big M method, we need to eliminate a2 and a3 from row 0 to get a canonical form.

Simply add Row 2+ Row 3 to Row 0:

w’ - a2 - a3 = 0x1 + 3x2 - e2 + a2 = 20

+ x1 + x2 + a3 = 10w’ + 2x1 + 4x2 – e2 = 30



Description of the Two-Phase Simplex Method

1. Modify the constraints so that the rhs of each constraint is nonnegative. Identify each constraint that is now an = or ≥ constraint.

2. Convert each inequality constraint to standard form (add a slack variable for ≤ constraints, add an excess variable for ≥ constraints).

3. For each ≥ or = constraint, add artificial variables. Add sign restriction ai ≥0.

4. For now, ignore the original LP’s objective function. Instead solve an LP whose objective function is min w’ = (sum of all artificial variables). This is the Phase I LP. Solving the Phase I LP will force the artificial variables to be zero.



Description of the Two-Phase Simplex Method cont…

Since each ai ≥ 0, solving the Phase I LP will result in one of the following three cases:

Case 1: The optimal value of w’ > 0. In this case the original LP is infeasible.

Case 2: The optimal value of w’ = 0 and no artificial variables are in the optimal Phase I basis. Drop all columns in the optimal Phase I tableau that correspond to the artificial variables. In this case, combine the original objective function with the constraints from the optimal Phase I tableau. This yields the Phase II LP. The optimal solution to the Phase II LP is the optimal solution to the original LP.

Case 3: The optimal value of w’ = 0 and at least one artificial variable is in the optimal Phase I basis. In this case, we can find the optimal solution to the original LP if at the end of Phase I we drop from the optimal Phase I tableau all nonbasic artificial variables and any variable from the original problem that has a negative coefficient in row 0 of the optimal Phase I tableau.


4.13 The Two-Phase Simplex MethodThe Bevco example continued (Using Method 2 for Min LP):

Second Phase I TableauRow w' x1 x2 s1 e2 a2 a3 RHS BV MRT

0 1 2/3 0 0 1/3 -4/3 0 10/3 w' = 10/31 0 5/12 0 1 1/12 -1/12 0 7/3 s1 = 7/3 28/52 0 1/3 1 0 -1/3 1/3 0 20/3 x2 = 20/3 20.003 0 2/3 0 0 1/3 -1/3 1 10/3 a3 = 10/3 5*

Initial Phase I TableauRow w' x1 x2 s1 e2 a2 a3 RHS BV MRT

0 1 2 4 0 -1 0 0 30 w' = 301 0 0.5 0.25 1 0 0 0 4 s1 = 4 162 0 1 3 0 -1 1 0 20 a2 = 20 6.67*3 0 1 1 0 0 0 1 10 a3 = 10 10

Third (Optimal) Phase I TableauRow w' x1 x2 s1 e2 a2 a3 RHS BV MRT

0 1 0 0 0 0 -1 -1 0 w' = 01 0 0 0 1 -1/8 1/8 -5/8 1/4 s1 = 1/42 0 0 1 0 -1/2 1/2 -1/2 5 x2 = 53 0 1 0 0 1/2 -1/2 3/2 5 x1 = 5

We will now drop columns a2 and a3.


4.13 The Two-Phase Simplex MethodWe begin Phase II with the following set of equations:

Row 0: z - 2x1 – 3x2 = 0Row 1: s1 - 1/8e2 = 1/4Row 2: x2 - 1/2e2 = 5Row 3: x1 + 1/2e2 = 5

We need to eliminate x1 and x2 from row 0 to get a canonical form.

Add 3(Row 2) + 2(Row 3) to Row 0:

z - 2x1 – 3x2 = 03( x2 - 1/2e2 = 5)

+ 2( x1 + 1/2e2 = 5)

z - 1/2e2 = 25



Initial (Optimal) Phase II TableauRow z x1 x2 s1 e2 RHS BV MRT

0 1 0 0 0 -1/2 0 z = 251 0 0 0 1 -1/8 1/4 s1 = 1/42 0 0 1 0 -1/2 5 x2 = 53 0 1 0 0 1/2 5 x1 = 5

In this case Phase II requires no pivots to find an optimal solution. If the Phase II row 0 does not indicate an optimal tableau, then simply continue with the simplex until an optimal row 0 is obtained.

The optimal Phase II tableau shows that the optimal solution to the Bevco problem is:

Optimal Solution: z = 25, x1 = 5, x2 = 5

Note that we obtained the same solution using the Big M Method.



• For examples of Case 1 and Case 3 of the two-phase simplex method read Example 6 and 7 of Chapter 4.

• Important note: Example 7 should be a max problem and not a min problem!



Problem (10 pts). Get with your team and solve the following LP usingthe two-phase simplex method :

0, 32

6 2 s.t.3Min

21

21

21

1

≥=+−≤−−

=

xxxxxx

xz


4.13 The Two-Phase Simplex MethodSolution to Problem (10 pts):

0, 3 2 6 2 s.t.

3Min

21

21

121

1

≥=+=−+

=

xxxx

exx xzLP is standard form:

0, 32 6 2 s.t.

3Min

21

21

21

1

≥=+≥+

=

xxxxxx

xzMaintaining nonegative RHS:

0, 3 2 6 2 s.t.

'Min

21

221

1121

21

≥=++=+−+

+=

xxaxx

aexx aawPhase I Formulation:



Eliminate a1 and a2 from row 0. Add Row 2 + Row 3 to Row 0:

w’ - a1 - a2 = 02x1 + x2 - e1 + a1 = 6

+ x1 + 2x2 + a2 = 3w’ + 3x1 + 3x2 - e1 = 9


4.13 The Two-Phase Simplex MethodUsing Method 2 of the simplex algorithm for min LPs:

Initial Phase I TableauRow w' x1 x2 e1 a1 a2 RHS BV MRT

0 1 3 3 -1 0 0 9 w' = 91 0 2 1 -1 1 0 6 a1 = 6 3*2 0 1 2 0 0 1 3 a2 = 3 3

2 pts

Second Phase I TableauRow w' x1 x2 e1 a1 a2 RHS BV MRT

0 1 0 3/2 1/2 -3/2 0 0 w' = 01 0 1 1/2 -1/2 1/2 0 3 x1 = 3 62 0 0 3/2 1/2 -1/2 1 0 a2 = 0 0*

2 pts

Third (Optimal) Phase I TableauRow w' x1 x2 e1 a1 a2 RHS BV MRT

0 1 0 0 0 -1/2 -1 0 w' = 01 0 1 0 -5/6 7/12 -1/3 3 x1 = 32 0 0 1 1/3 -1/6 2/3 0 x2 = 0

2 pts

We will now drop columns a1 and a2.


4.13 The Two-Phase Simplex MethodWe begin Phase II with the following set of equations:

Row 0: z - 3x1 = 0Row 1: x1 - 5/6e1 = 3Row 2: x2 + 1/3e2 = 0

We need to eliminate x1 from row 0 to get a canonical form.

Add 3(Row 1) to Row 0:

z - 3x1 = 0+ 3( x1 - 5/6e1 = 3)

z - 5/2e1 = 9



Initial Phase II TableauRow z x1 x2 e1 RHS BV MRT

0 1 0 0 -5/2 0 z = 91 0 1 0 -5/6 3 x1 = 32 0 0 1 1/3 0 x2 = 0

2 pts

In this case Phase II requires no pivots to find an optimal solution. If the Phase II row 0 does not indicate an optimal tableau, then simply continue with the simplex until an optimal row 0 is obtained.

The optimal Phase II tableau shows that the optimal solution to the problem is:

Optimal Solution: z = 9, x1 = 3, x2 = 0 2 pts

Note that we obtained the same solution using the Big M Method.


4.13 The Two-Phase Simplex MethodSummary

• To get started with the simplex method, add an artificial basis,but ensure that the artificial variables do not occur in an optimal solution.

• Big M method: place a large cost (penalty) on each of the artificial variables

• Two-phase method: In phase I minimize the sum of the artificial variables. At the end of phase I, we will have a basic feasible solution to the original problem. Use this as a starting point for Phase II, which solves the original problem.

Chapter 4 The Simplex Algorithm – Part IIise.tamu.edu/INEN420/INEN420_2005Spring/SLIDES/Chapter...

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