[PDF] Finding feasible solutions to a LP PDF

Definition 3. A basic feasible solution is degenerate if there are more than n tight constraints. We say that a linear programming problem is degenerate if it

Chapter 9 Linear programming

has feasible solutions. But none of them is optimal (See Exercise 9.3). As a matter of fact for every number M

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not viable anymore Reverse mortgage as a feasible solution in Belgium? Part I: international analysis. (FR UK and USA) 5 factors influence the reverse

Methods for Initial Basic Feasible Solution Lecture 16 Transportation

Methods for Initial Basic Feasible Solution. Lecture 16. Transportation problem : (Vogal's Approximation method ). For each row of the table identify the

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4 mars 2019 on the most relevant feasible solutions and an infeasible local search ... n is the number of vertices of G. Notice that a feasible solution.

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SIGNMENT PROBLEM 1. The transportation problem 2. The matrix

Theorem 3 A balanced transportation problem always has a basic feasible solution. Such a solution consists of m + n ? 1 positive variables at most.

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https://faculty.math.illinois.edu/~mlavrov/docs/482-fall-2019/lecture8.pdf

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for any µ > 0 where P? is a feasible solution of (12.1) achieving the optimal value opt?. That is

Methods for Initial Basic Feasible Solution Lecture 15 Transportation

Some simple methods to obtain the initial basic feasible solution are. 1. North-West Corner Rule. 2. Lowest Cost Entry Method (Matrix Minima Method).

[PDF] Basic Feasible Solutions

Assume an LP in the following form Maximize cTx Subject to: Ax ? b x ? 0 • N Variables M constraints • U = Set of all feasible solutions

[PDF] Finding feasible solutions to a LP

We need to introduce artificial variables to help get an initial feasible solution We also negate the objective function and convert to a maximization problem

[PDF] 43 Basic feasible solutions and vertices of polyhedra

Due to the fundamental theorem of Linear Programming to solve any LP it 'suffices' to consider the vertices (finitely many) of the polyhedron P of the feasible

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THEOREM: For a feasible linear program in its standard form the optimum value of the objective over its nonempty feasible region is (a) either unbounded or

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Recall the definition of a basic feasible solution: Definition 1 Let P be a polyhedron defined by linear equality and inequality constraints and consider

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6 mar 2014 · We will start with discussing basic solutions and then show how this applies to the simplex algorithm 2 Basic Feasible Solutions Definition 1

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Basic feasible solutions: A basic solution which is nonnegative Basic solution: For a canonical form linear program (see below) a basic solution is a

[PDF] Methods for Initial Basic Feasible Solution Lecture 15 Transportation

15 1 Methods for Initial Basic Feasible Solution Some simple methods to obtain the initial basic feasible solution are 1 North-West Corner Rule

[PDF] Let to be a basic feasible solution to the LPP - Deshbandhu College

2 Such that Ax=b x>0 N has an optimat feasible solution then atleast one basic feasible solution must be optimal Proof- Let Zo= EB XB with x0 = Bb be a

[PDF] Solutions to Review Questions Exam 1 - Whitman People

A unique solution (either with or without an unbounded feasible set) • An unbounded solution - The feasible set is unbounded • An infinite number of solutions

Finding feasible solutions to a LPIn all the examples we have seen until now, there was an "easy" initial

basic feasible solution: put the slack variables on the left hand side. How- ever, this is not always the case, especially for minimization problems, or problems with equality constraints in the original model.

Consider the following simple LP.

minimizex s.tx≥5 x≥0 Forget for a minute that the solution is obvious. If we try to use simplex, we will first convert it to a max problem and negate the objective function. maximize-x s.tx≥5 x≥0 When we are finished, we will remember to negate the answer.

Converting to Standard Formmaximize-x

s.tx≥5 x≥0 Let"s convert the constraint to an equality by adding an excess variable. maximize-x s.tx-e= 5 x,e≥0 Notice that if we try to put the excess variable on the left hand side, we get the equation e=-5 +x , which would giveea value of-5and would not yield a bfs.

Continuedmaximize-x

s.tx-e= 5

x,e≥0•We deal with this situation, by using anartifical variable.•The artificial variable will allow us to easily find a bfs.•Problem:The artificial variable may allow us to find "solutions" that

are not really solutions to the original LP.•To compensate, we will add the artificial variable to the objective func-

tion with avery large negative coefficient (big M).Thus, if we make the artificial variable positive, the objective function will

be extremely negative. But we are maximizing, so we have every incentive to avoid making the objective function value negative, i.e. we want to avoid makingapositive. When we are done solving the LP, if the artificial variable is zero, we have solved the original LP. If it is positive, then we conclude that the original

LP was infeasible.

Solving the LPmaximize-x

s.tx-e= 5 x,e≥0

Adding the artificial variable to the LP, we get

maximize-x-Ma(1)subject to x-e+a= 5(2)x,e,a≥0.(3)In "Standard form." z=-x-Ma(4)a= 5-x+e(5)

Standard Formz=-x-Ma(6)a= 5-x+e(7)Note that this is not really in standard form.ais on the left hand side

of a constraint, and the right hand side of the objective function. We can"t allow this, so we use the equation to get rid of theain the objective function. The objective function is now -x-Ma=-x-M(5-x+e) =-5M+ (M-1)x-Me

Rewriting, we obtain

z=-5M+ (M-1)x-Me(8)a= 5-x+e(9)Now we have a basic feasible solution(x,e,a) = (0,0,5)and can continue with the simplex algorithm.

Solving (cont)z=-5M+ (M-1)x-Me(10)a= 5-x+e(11)Remember thatMis a big number. We choosexas the entering variable,

andaas the leaving variable.

z=-5-e+ (1-M)a(12)x= 5 +e-a(13)All the coefficients in the objective function are negative, so we have an

optimal solution. Notice that the value ofais 0, which means that the originalLPis feasible. The value ofxis5and the objective function is-5. Negating that we get that the optimal objective function value is5, as we expected. The bevco ExampleHere is a more involved example, that comes from the Bevco problem in the book. In this case the LP is: minimize2x1+ 3x2(14)subject to excesses non zero) is not feasible. We need to introduce artificial variables to help get an initial feasible solution. We also negate the objective function and convert to a maximization problem.

Conversionminimize2x1+ 3x2(19)subject to

maximize-2x1-3x2-Ma1-Ma2(24)subject to .5x1+.25x2+s1= 4(25)x1+ 3x2-e1+a1= 20(26)x1+x2+a2= 10(27)x1,x2,s1,e1,a1,a2≥0.(28) Standard Formmaximize-2x1-3x2-Ma1-Ma2(29)subject to

.5x1+.25x2+s1= 4(30)x1+ 3x2-e1+a1= 20(31)x1+x2+a2= 10(32)x1,x2,s1,e1,a1,a2≥0.(33)Now we convert it to standard form. Either a slack variable or an artificial

variable goes on the left hand side. We also will substitute fora1anda2in the objective function. This yields: z=-30M+ (2M-2)x1+ (4M-3)x2-Me1(34)s1= 4-.5x1-.25x2(35)a1= 20-x1-3x2+e1(36)a2= 10-x1-x2(37)

Iteration 1z=-30M+ (2M-2)x1+ (4M-3)x2-Me1(38)s1= 4-.5x1-.25x2(39)a1= 20-x1-3x2+e1(40)a2= 10-x1-x2(41)We choosex2as the entering variable anda1as the leaving variable. After

an annoying amount of algebra, we obtain:

Iteration 2z=-60-10M3+2M-33x1+M-33e1+3-4M3a1(46)x2=203-x13+e13-a13(47)s1=73-512x1-e112+a112(48)a2=103-2x13-e13+a13(49)Now we pivot inx1and pivot outa2and obtain:

z= 25-e12+1-2M2a1+3-2M2a2(50)x1= 5-e12+a12-3a22(51)x2= 5 +e12-a12+a22(52)s1=14+e18-a18+5a28(53)Now all the coefficients in the objective row are negative, so we have an

optimal solution. Also,a1anda2are both non-basic, so the problem is feasible. The optimal solution, in the original variables, isx1= 5,x2= 5 with objective value25. An infeasible LPLet"s see what happens if our original LP is infeasible. Consider the LP: maximizex1(54)subject to x able to the second, and obtain: z=x1-Ma1(58)a1= 7-x1-x2+e1(59)s1= 6-x1-x2(60)

Eliminatinga1z=x1-Ma1(61)a1= 7-x1-x2+e1(62)s1= 6-x1-x2(63)Next we eliminatea1from the objective, yielding:

z=-7M+Mx1+Mx2-Me1(64)a1= 7-x1-x2+e1(65)s1= 6-x1-x2(66)Now, we choosex2to enter ands1to leave: z=-M-5Mx2-6Ms1-Me1(67)x1= 6-x2-s1(68)a1= 1 +s1+e1(69)

Final tableauxz=-M-5Mx2-6Ms1-Me1(70)x1= 6-x2-s1(71)a1= 1 +s1+e1(72)The simplex algorithm terminates because all the objective coefficients

are negative. But, the objective function value is-Manda1= 1. This tells us that the original problem isinfeasible.quotesdbs_dbs14.pdfusesText_20

[PDF] [PDF] Finding feasible solutions to a LP