Simplex Method for LP (II) - Xiaoxi Li's homepageUnbounded objective z Multiple optimal solutions "=" or " " constraint: the artiﬁcial-variable technique and the Big-M method Negative

Simplex Method for LP (II)

Xiaoxi Li

Wuhan University

Sept. 27, 2017 (week 4)

Operations Research (Li, X.) Simplex Method for LP (II) Sept. 27, 2017 (week 4) 1 / 31

Organization of this lecture

Contents:

Tie for entering/leaving variableUnbounded objective z

Multiple optimal solutions"=" or "≥" constraint:the artificial-variable technique and the Big-M methodNegative variables allowedThe Two-Phase method for the artificial-variable technique


Tie for the entering or leaving variable

When there is a tie for determining the entering or the leaving variable,just choose one arbitrarily.

For an example:

Eq. Basic variables z x1 x2 s1 s2 s3 RHS(0) z 1 -3 -3 0 0 0 0(1) s1 0 1 0 1 0 0 4(2) s2 0 0 2 0 1 0 12(3) s3 0 3 2 0 0 1 18

It is possible to choose either x1 or x2 as the entering variable. Theywill lead to different optimal solution paths.


No leaving variable: unbounded objective z

For an example:

Eq. Basic variables z x1 x2 s1 s2 RHS(0) z 1 -3 -5 0 0 0(1) s1 0 1 0 1 0 0(2) s2 0 3 -2 0 1 6

Every coefficient in the pivot column is either zero or negative. Thenthe entering variable x2 can be increased infinitely large (while x1 = 0):{

x1 +0x2 + s1 = 4 =⇒ s1 = 4≥ 03x1−2x2 + s2 = 6 =⇒ 2x2 = s2−6≥−6.

As x2 increases while keeping x1 = 0, the objective z tends to infinity.


Multiple optimal solutions

.

It is possible for us to use the simplex method to identify all the CPFsolutions if there are multiple. Then, by taking convex combinations ofthe CPF solutions, we obtain all the optimal solutions.

When there are multiple optimal solutions, at least one of thenonbasic variables si has a zero coefficient in the final Eq. (0).Increasing such variable si will not increase the objective z, thus itis possible to leave the supporting hyperplane "si = 0" to arrive atan adjacent CPF optimal solution (having the same z value).


For an example: Max z = 3x1 +2x2,

Initialization:

Eq. Basic variables z x1 x2 s1 s2 s3 RHS Optimal?(0) z 1 -3 -2 0 0 0 0 NO(1) s1 0 1 0 1 0 0 4(2) s2 0 0 2 0 1 0 12(3) s3 0 3 2 0 0 1 18

The pivot number: 1.


Iteration 1:

Eq. Basic variables z x1 x2 s1 s2 s3 RHS Optimal?(0) z 1 0 -2 3 0 0 12 NO(1) x1 0 1 0 1 0 0 4(2) s2 0 0 2 0 1 0 12(3) s3 0 0 2 -3 0 1 6

Iteration 2:

Eq. Basic variables z x1 x2 s1 s2 s3 RHS Optimal?(0) z 1 0 0 0 0 1 18 Yes(1) x1 0 1 0 1 0 0 4(2) s2 0 0 0 3 1 -1 6(3) x2 0 0 1 -3/2 0 1/2 3

Optimality Test : after Iteration 2, s1,s3 are nonbasic variables, and theircoefficients in Eq. (0) are respectively 0 and 1. =⇒ (4,3,0,6,0) is anoptimal solution.


The coefficient for the nonbasic variable s1 in Eq. (0) is 0, thus it ispossible to find another BF optimal solution following an

Extra Iteration:

Eq. Basic variables z x1 x2 s1 s2 s3 RHS Optimal?(0) z 1 0 0 0 0 1 18 Yes(1) x1 0 1 0 0 -1/3 1/3 2(2) s1 0 0 0 1 1/3 -1/3 2(3) x2 0 0 1 0 1/2 0 6

We obtain another optimal BF solution (2,6,2,0,0). The set of optimalsolutions is {α(4,3,0,6,0)+(1−α)(2,6,2,0,0)| for some α ∈ [0,1]}.

Check. Now s2 is a nonbasic variable with 0 its coefficient in Eq. (0).Re-do an extra iteration by increasing s2 leads to the previous simplextable.


The artificial-variable technique for LP in non-standardforms

So far, we only studied the simplex method for LP in standard form. Inthis section we aim at solving the LP in non-standard forms, which aresorted into one of the following situations:

z is minimized.some RHS constant "bi" is negative.some functional constraints with "=" or "≥".some nonnegative constraint "xj ≥ 0" is withdrew.

Idea: transform the non-standard LP problem to some equivalent LP instandard form, mainly by introducing the artificial variable.


z = c1x1 + · · ·+ cnxn is minimized.Let z′ =−z be the new objective function, then subject to the same

constraints,Max − z =−c1x1−·· ·− cnxn

is equivalent toMin z = c1x1 + · · ·+ cnxn

Some RHS constant "bi" is negative.Multiplying both sides −1 to obtain a positive constant on the RHS.

For example:5x1 +6x2 ≥−5⇐⇒−5x1−6x2 ≤ 5, which is in standard form;5x1 +6x2 ≤−3⇐⇒−5x1−6x2 ≥ 3, with "≥", to be solved.5x1 +6x2 =−1⇐⇒−5x1−6x2 = 1, with "=", to be solved.


Some functional constraints with "=".For an example:

Maximize z = 3x1 +5x2

s.t.

x1 ≤ 4 1

2x2 ≤ 12 23x1 +2x2 =18 3x1,x2 ≥ 0 +

(0.1)

The problem is as before except that we have replaced in Eq. +

3x1 +2x2 ≤ 18 by 3x1 +2x2 = 18.


The feasible region for the LP problem with an equality constraint:


As before, we introduce slack variables to constraints 1 , 2 to obtain:


s.t.

x1 + s1 = 4 1

2x2 + s2 = 12 23x1 +2x2 = 18 3x1, x2, s1, s2 ≥ 0 +

(0.2)

The previous LP (0.1) is not in standard form, thus this LP (0.2) isnot in canonical form: there is no slack variable in constraint 3 .The crucial point is that the Initialization step of the simplexmethod can not be carried out directly since there is not anobvious BF solution

(as the origin (0,0,4,12,18) in standard LP (0.1)

).


To solve the problem, introduce an artificial variable s̄3 ≥ 0 in 3 ,

i.e. 3x1 +2x2 + s̄3 = 18 3 ,

to obtain an initial BF solution (x1,x2,s1,s2, s̄3) = (0,0,4,12,18),where {s1,s2, s̄3} are the basic variables.This is as to extend the feasible region to include the origin (0,0).However, in order to obtain an equivalent artificial problem, oneputs a heavy penalty in the objective function for s̄3 ≥ 0 beingstrictly positive, i.e.

Max z = 3x1 +5x2−Ms̄3, where M > 0 is sufficiently large.

The technique of introducing an artificial variable into the a LPconstraint is called the artificial-variable technique; the method ofrelaxing the constraint and introducing a huge penalty on infeasiblesolutions (outside the pre-relaxed region) is called the Big-M method.


Define the artificial problem as:

Define s̄3 = 18−3x1−2x2

Maximize z = 3x1 +5x2−Ms̄3

s.t.

x1 ≤ 4 12x2 ≤ 12 2

3x1 +2x2 ≤18 3(so 3x1 +2x2 + s̄3 = 18

)x1, x2, s̄3 ≥ 0 +

(0.3)


Graphical analysis for the artificial problem:


ClaimSuppose that LP (0.1) has a nonempty feasible region. Then for M > 0sufficiently largea, any optimal solution (x∗1,x

∗2, s̄∗3) to LP problem (0.3)

satisfies: (i). (x∗1,x∗2) is optimal to LP problem (0.1); (ii). s̄∗3 = 0.

aCompared with other coefficients/constants in LP (0.1).

To apply the simplex method to artificial problem (0.3) (in standardform!), we first write its associated LP problem in canonical form.

Step I. Introduce slack variables "s1,s2" to "≤" constraints 1 and 2 ,and put the objective equation into the equation system:

Eq.(0) z−3x1−5x2 +Ms̄3 = 0Eq.(1) x1 + s1 = 4Eq.(2) 2x2 + s2 = 12Eq.(3) 3x1 +2x2 + s̄3 = 18


Step II. Use elementary algebraic operations to write the objectivefunction in terms of the nonbasic variables {x1,x2} only.

{s1,s2, s̄3} are basic variables, and {x1,x2} are the nonbasic variables,so we make the operation

Eq.(0′) = Eq.(0)−M ∗Eq.(3)

to obtain the new equation system (ready to apply the simplexmethod):

Eq.(0) z− (3M+3)x1− (2M+5)x2 =−18MEq.(1) x1 + s1 = 4Eq.(2) 2x2 + s2 = 12Eq.(3) 3x1 +2x2 + s̄3 = 18


We now put the coefficients of the above system into the simplex tableto start solving the artificial problem with simplex method.

Initialization

Eq. Basic variables z x1 x2 s1 s2 s̄3 RHS(0) z 1 -3M-3 -2M-5 0 0 0 -18M(1) s1 0 1 0 1 0 0 4(2) s2 0 0 2 0 1 0 12(3) s̄3 0 3 2 0 0 1 18

Keep in mind: M > 0 is larger than any ordinary number youencounter during the solution process.

In view of this, we see that the initial BF solution (0,0,4,12,18) is notoptimal and we choose x1 as the entering variable since3M+3 > 2M+5. The minimum ratio test identifies s1 as the leavingvariable so "1" is the pivot number.


Iteration 1

Eq. Basic variables z x1 x2 s1 s2 s̄3 RHS(0) z 1 0 -2M-5 3M+3 0 0 -6M+12(1) x1 0 1 0 1 0 0 4(2) s2 0 0 2 0 1 0 12(3) s̄3 0 0 2 -3 0 1 6

Iteration 2

Eq. Basic variables z x1 x2 s1 s2 s̄3 RHS(0) z 1 0 0 -9/2 0 M+5/2 27(1) x1 0 1 0 1 0 0 4(2) s2 0 0 0 3 1 -1 6(3) x2 0 0 1 -3/2 0 1/2 3

Iteration 3 (after which it is optimal)

Eq. Basic variables z x1 x2 s1 s2 s̄3 RHS(0) z 1 0 0 0 3/2 M+1 36(1) x1 0 1 0 0 -1/3 1/3 2(2) s1 0 0 0 1 1/3 -1/3 2(3) x2 0 0 1 0 1/2 0 6


The optimal solution path is (0,0)→ (4,0)→ (4,3)→ (2,6);After Iteration 2, the artificial variable s̄3 leaves the basis, i.e. wereach at the supporting hyperplane "s̄3 = 0" ("3x1 +2x2 = 18"), thusthe feasible region of the original LP (0.1)Another iteration leads us to the optimal solution (2,6).


Some functional constraints with "≥".Adding one surplus variable first and then one artificial variable.

For an example:

Maximize z = 0.4x1 +0.5x2

s.t.

0.3x1 +0.1x2 ≤ 2.7 10.6x1 +0.4x2 ≥ 6 2x1, x2 ≥ 0 +

(0.4)

Let "x5 ≥ 0" be a surplus variable. The system (0.4) is equivalent to:0.3x1 +0.1x2 ≤ 2.7 10.6x1 +0.4x2−x5 = 6 2x1, x2, x5 ≥ 0 +

(0.5)

=⇒ It is then a problem of LP with an "=" constraint, which can besolved by introducing an artificial variable (same as before).


Remark.It is possible to have several constraints of type "≥", "=" or a mixof them. In this case, several artificial variables are needed (onefor each constraint), and we put a big M for each of them in theobjective function. At an optimal solution, all artificial variablesshould be zero. See pages. 120-125 in H-L for an example.It is possible to obtain at an optimal solution to an artificial LPproblem that the artificial variable is nonzero. In this case, wededuce that there is no feasibility for the original LP problem.The coefficient for nonbasic variable in Eq. (0) is in the form ofaM+b. In identifying the entering variable, when there is a tie onthe multiplication term "a", we look at the addition term "b".


Negative variables allowed

In some applications, the constraint xj ≥ 0 for some j may not berequired. A negative xj may be interpreted as a decrease in stock.

Two possibilities:

xj ≥ bj for some bj ∈ R. In this case, we define x′j = xj−bj ≥ 0 to bea decision variable, and substitute "xj = x′j +bj" in other equations.

xj ∈ R with no constraint. Introduce x+j ,x−j ≥ 0 two decision

variables with xj = x+j − x−j .


For an example.Maximize z = 3x1 +5x2

s.t.

x1 ≤ 4

2x2 ≤ 123x1 +2x2 ≤ 18x2 ≥ 0

is equivalent to the following (letting xj = x+j − x−j )

Maximize z = 3x+1 −3x−1 +5x2

s.t.

x+1 − x−1 ≤ 4

2x2 ≤ 123x+1 −3x−1 +2x2 ≤ 18x+1 , x−1 , x2 ≥ 0


The Two-Phase method

An easier way to treat the artificial problem involving no "Big M". Idea:first, start from the origin and arrive at the feasible region, byminimizing only the penalties;next, optimize the objective function within the feasible region.

Apply the simplex method in two phases:Phase I. Minimize a new objective function as the sum of all theartificial variables.

Conclude that the problem is infeasible if the resulting optimalsolution contains a strictly positive artificial variable;Proceed to Phase II when all artificial variables leave the basicvariables, thus equal zero, at an optimal solution.

Phase II. Take the optimal solution in Phase I as an initial BFsolution, and apply the simplex method to the original LP problem.


For an example.


s.t.

x1 ≤ 4 1

2x2 ≤ 12 23x1 +2x2 = 18 3x1,x2 ≥ 0 +

Prepare as in the Big-M method: introducing slack variables (or evensurplus variable if needed) and artificial variables to obtain anequivalent artificial problem in canonical form:

Maximize z = 3x1 +5x2 +Ms̄3

s.t.

x1 + s1 = 4 1

2x2 + s2 = 12 23x1 +2x2 + s̄3 = 18 3x1, x2, s1, s2, s̄3 ≥ 0 +

(0.6)


To solve the above artificial LP (0.6), we use the two-phase methodinstead of the Big-M method.

Phase I.Minimize y = s̄3

s.t.

x1 + s1 = 4 1

2x2 + s2 = 12 23x1 +2x2 + s̄3 = 18 3x1, x2, s1, s2, s̄3 ≥ 0 +

{s1,s2, s̄3} are the initial basic variables, so we need first to replace s̄3by x1 and x2 using equation "s̄3 = 18−3x1−2x2". Further, this is aminimization problem, defining y′ =−y, one obtains the

Eq.(0) : −y−3x1−2x2 =−18.


Initialization

Eq. Basic variables -y x1 x2 s1 s2 s̄3 RHS(0) -y 1 -3 -2 0 0 0 -18(1) s1 0 1 0 1 0 0 4(2) s2 0 0 2 0 1 0 12(3) s̄3 0 3 2 0 0 1 18

Iteration 1

Eq. Basic variables -y x1 x2 s1 s2 s̄3 RHS(0) -y 1 0 -2 3 0 0 -6(1) x1 0 1 0 1 0 0 4(2) s2 0 0 2 0 1 0 12(3) s̄3 0 0 2 -3 0 1 6

Iteration 2

Eq. Basic variables -y x1 x2 s1 s2 s̄3 RHS(0) -y 1 0 0 0 0 1 0(1) x1 0 1 0 1 0 0 4(2) s2 0 0 0 3 1 -1 6(3) x2 0 0 1 -3/2 0 1/2 3


After Iteration 2, it is optimal for the Phase I problem, where theartificial variable s̄3 leaves the basis. The optimal BF solution is suchthat (x1,s2,x2) are basic variables equal to (4,6,3).

Phase II.Maximize z = 3x1 +5x2

s.t.

x1 + s1 = 4 1

3s1 + s2 − s̄3 = 6 2x2− 3

2 s1 + 12 s̄3 = 3 3

x1, x2, s1, s2, s̄3 ≥ 0 +

It’s ready to apply the simplex method to the above LP in canonicalform except that the objective function needs to be revised to containonly nonbasic variables {s1, s̄3}: 1 : x1 = 4− s1; 3 : x2 = 3+ 3

2 s1− 12 s̄3,

we obtain the revised objective function:

z = 3(4− s1)+5(3+32

s1−12

s̄3) = 27+92

s1−52

s̄3.


Initialization (of Phase II, compare with Iteration in Big-M method)

Eq. Basic variables z x1 x2 s1 s2 s̄3 RHS(0) z 1 0 0 -9/2 0 x 27(1) x1 0 1 0 1 0 x 4(2) s2 0 0 0 3 1 x 6(3) x2 0 0 1 -3/2 0 x 3

Iteration 1 (of Phase II)

Eq. Basic variables z x1 x2 s1 s2 s̄3 RHS(0) z 1 0 0 0 3/2 x 36(1) x1 0 1 0 0 -1/3 x 2(2) s1 0 0 0 1 1/3 x 2(3) x2 0 0 1 0 1/2 x 6

It is optimal the BF solution: (x1,s1,x2) = (2,2,6) basic variables and(s2, s̄3) = (0,0) the nonbasic variables. The optimal BF solution to theoriginal LP problem is then (x1,x2) = (1,6) and the optimal value is 36.


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Simplex Method for LP (II) - Xiaoxi Li's homepageUnbounded objective z Multiple optimal solutions "=" or " " constraint: the artiﬁcial-variable technique and the Big-M method Negative