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ISE 203 OR I

ISE 203 OR I . Chapter 4 Solving Linear Programming Problems: The Simplex Method Asst. Prof. Dr. Nerg iz Kasımbeyli. The SIMPLEX Method. Simplex method is an algebraic procedure However, its underlying concepts are geometric

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ISE 203 OR I

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  1. ISE 203 OR I Chapter 4 Solving Linear Programming Problems: The Simplex Method Asst. Prof. Dr. Nergiz Kasımbeyli

  2. The SIMPLEX Method • Simplex method is an algebraic procedure • However, its underlying concepts are geometric • Understanding these geometric concepts helps before going into their algebraic equivalents

  3. In this example, the point of intersection of twoconstraint boundaries are the corner-point solutions of the problem. • For a linear programming problem with n decision variables, each corner point solution lies at the intersection point of n constraints.

  4. The points that lie on the corners of the feasible region are the corner-point feasible (CPF) solutions or extreme points.

  5. Edges • For any LP problem with n decision variables, two extremepoints are adjacent to each other if they share n-1 constraint boundaries. The two adjacent extreme points are connected by a line segment that lies on these same shared constraint boundaries. Such a line segment is referred to as an edge of the feasible region.

  6. An edge of thefeasibleregion

  7. Optimality test Consider any LP problem that possesses at least one optimal solution. If a CPF solution has no adjacent CPF solutions that are better (as measured by Z), then it must be an optimal solution. (This is due to the fact that the feasible region is a convex set, we will see in Chapter 5) • (2,6) must be optimal since Z=36 is greater than Z=30 for (0,6) and Z=27 for (4,3). • This is the optimality test used by Simplex.

  8. The Simplex Method in a Nutshell Initialization (Find initial CPF solution) Is the current CPF solution optimal? Yes Stop No Move to a better adjacent CPF solution

  9. Introducing slack variables x3, x4 and x5 to convert inequalities into equalities – Convert LP to Augmented Form

  10. Slack variables • Slack variable = 0 in the current solutionThis solution lies on the constraint boundary for the corresponding functional constraint. Feasible solution, Binding constraint • If slack variable ≥ 0 the solution lies on the feasible side of this constraint boundary.  Feasible solution, Nonbinding constraint • If slack variable ≤ 0, the solution lies on the infeasible side of this constraint boundary.  Infeasible solution

  11. ‘Language’ of the Simplex Method

  12. Initial Assumptions • All constraints are of the form ≤ • All right-hand-side values (bj, j=1, …,m) are positive • We’ll learn how to address other forms later

  13. The Augmented Form Set up the method first: Convert inequality constraints to equality constraints byadding slack variables Augmented(Standard) Form Maximize Z = 3x1+ 5x2 subject to x1+s1= 4 2x2+s2= 12 3x1+ 2x2+s3= 18 x1,x2 , s1, s2, s3≥ 0 Original Form Maximize Z = 3x1+ 5x2 subject to x1 ≤ 4 2x2 ≤ 12 3x1+ 2x2 ≤ 18 x1,x2 ≥ 0

  14. Basic and Basic Feasible Solutions X2 Augmented Form Maximize Z = 3x1+ 5x2 subject to x1+s1= 4 2x2+s2= 12 3x1+ 2x2+s3= 18 x1,x2, s1, s2, s3 ≥ 0 (0,9,4,-6,0) (0,6,4,0,6) (2,6,2,0,0) (4,6,0,0,-6) • Augmented solution • Basic infeasible solution • Basic feasible solution (BFS) • Nonbasic feasible solution (2,3,2,6,6) (4,3,0,6,0) (0,2,4,8,14) (0,0,4,12,18) (4,0,0,12,6) (6,0,-2,12,0) X1

  15. Basic, Nonbasic Solutions and the Basis • In an LP, number of variables > number of equations • The difference is the degrees of freedom of the system • e.g. in Wyndor Glass, degrees of freedom (d.f.)= 5-3=2 • Can set some variables (# = d.f.) to an arbitrary value (simplex uses 0) • These variables (set to 0) are called nonbasic variables • The rest can be found by solving the remaining system: basis • The basis: the set of basic variables • If all basic variables are ≥ 0, we have a BFS • Between two basic solutions, if their bases are the same except for one variable, then they are adjacent

  16. Algebra of the Simplex MethodInitialization Maximize Z = 3x1+ 5x2 subject to x1+s1= 4 2x2+s2= 12 3x1+ 2x2+s3 = 18 x1,x2, s1, s2, s3 ≥ 0 • Find an initial basic feasible solution • If possible, use the origin as the initial CPF solution • Equivalent to:Choose original variables to be nonbasic (xi=0, i=1,…n) and let the slack variables be basic (sj=bj, j=1,…m))

  17. Algebra of the Simplex MethodOptimality Test Maximize Z = 3x1+ 5x2 subject to x1+s1= 4 2x2+s2= 12 3x1+ 2x2+s3 = 18 x1,x2, s1, s2, s3 ≥ 0 • Are any adjacent BF solutions better than the current one? • Rewrite Z in terms of nonbasic variables and investigate rate of improvement • Current nonbasic variables:x1,x2 • Corresponding Z= 0 • Optimal? No. If I increase the value of one of the nonbasic variables from 0 to a positive value, the objective function will increase.

  18. Algebra of the Simplex MethodStep 1 of Iteration 1: Direction of Movement Maximize Z = 3x1+ 5x2 subject to x1+s1= 4 2x2+s2= 12 3x1+ 2x2+s3 = 18 x1,x2, s1, s2, s3 ≥ 0 • Which edge to move on? • Determine the direction of movement by selecting the entering variable (variable ‘entering’ the basis) • Choose the direction of steepest ascent (increase, since maximization) • x1: Rate of improvement in Z =3 • x2: Rate of improvement in Z =5 • Entering basic variable = x2(pivot column)

  19. Algebra of the Simplex MethodStep 2 of Iteration 1: Where to Stop Maximize Z = 3x1+ 5x2 subject to x1+s1= 4 (1) 2x2+s2= 12 (2) 3x1+ 2x2+s3 = 18 (3) x1,x2, s1, s2, s3 ≥ 0 • How far can we go? • Determine where to stop by selecting the leaving variable (variable ‘leaving’ the basis) • Increasing the value of x2 decreases the value of basic variables. • The minimum ratio test • Constraint (1):x1≤4 no bound on x2(s1= 4 - x1≥ 0) • Constraint (2):2x2+s2= 12  x2can be increased up to 6 before s2= 0. Min. Ratio • Constraint (3):3x1+2x2+s3= 18  x2can be increased up to 9 before s3= 0. • Leaving basic variable = s2(pivot row)

  20. Algebra of the Simplex MethodStep 3 of Iteration 1: Solving for the New BF Solution Z - 3x1- 5x2 = 0 (0) x1 +s1 = 4 (1) 2x2 +s2 = 12 (2) 3x1+ 2x2 +s3 = 18 (3) Z - 3x1+ + 5/2 s2 = 30 (0) x1 +s1 = 4 (1) x2 + 1/2 s2 = 6 (2) 3x1 - s2 + s3 = 6 (3) • Convert the system of equations to a more proper form for the new BF solution • Elementary algebraic operations: Gaussian elimination • Eliminate the entering basic variable (x2) from all but constraint 2 (pivot row) Since x1=0 and s2=0 we obtain (x1,x2,s1,s2,s3)= (0,6,4,0,6)

  21. Algebra of the Simplex MethodOptimality Test Z - 3x1+ + 5/2 s2 = 30 (0) x1 +s1 = 4 (1) x2 + 1/2 s2 = 6 (2) 3x1 - s2 + s3 = 6 (3) • Are any adjacent BF solutions better than the current one? • Rewrite Z in terms of nonbasic variables and investigate rate of improvement • Current nonbasic variables:x1, s2 • Corresponding Z= 30 • Optimal? No (increasing x1increases Z value)

  22. Algebra of the Simplex MethodStep 1 of Iteration 2: Direction of Movement Z - 3x1+ + 5/2 s2 = 30 (0) x1 +s1 = 4 (1) x2 + 1/2 s2 = 6 (2) 3x1 - s2 + s3 = 6 (3) • Which edge to move on? • Determine the direction of movement by selecting the entering variable (variable ‘entering’ the basis) • Choose the direction of steepest ascent • x1: Rate of improvement in Z = 3 • s2: Rate of improvement in Z = - 5/2 • Entering basic variable = x1

  23. Algebra of the Simplex MethodStep 2 of Iteration 2: Where to Stop Z - 3x1+ + 5/2 s2 = 30 (0) x1 +s1 = 4 (1) x2 + 1/2 s2 = 6 (2) 3x1 - s2 + s3 = 6 (3) • How far can we go? • Determine where to stop by selecting the leaving variable (variable ‘leaving’ the basis) • Increasing the value of x1 decreases the value of basic variables • The minimum ratio test • Constraint (1):x1 ≤ 4 • Constraint (2): no upper bound on x1 • Constraint (3):x1 ≤ 6/3= 2 • Leaving basic variable = s3

  24. Algebra of the Simplex MethodStep 3 of Iteration 2: Solving for the New BF Solution Z - 3x1+ + 5/2 s2 = 30 (0) x1 +s1 = 4 (1) x2 + 1/2 s2 = 6 (2) 3x1 - s2 + s3 = 6 (3) • Convert the system of equations to a more proper form for the new BF solution • Elementary algebraic operations: Gaussian elimination • Eliminate the entering basic variable (x1) from all but its equation The next BF solution is (x1,x2,s1,s2,s3)= (2,6,2,0,0)

  25. Algebra of the Simplex MethodOptimality Test Z + 3/2 s2 + s3 = 36 (0) +s1 + 1/3 s2 - 1/3 s3 = 2 (1) x2 + 1/2 s2 = 6 (2) x1 - 1/3 s2 + 1/3 s3 = 2 (3) • Are any adjacent BF solutions better than the current one? • Rewrite Z in terms of nonbasic variables and investigate rate of improvement • Current nonbasic variables: s2, s3 • Corresponding Z= 36 • Optimal? yes

  26. The Simplex Method in Tabular Form • For convenience in performing the required calculations • Record only the essential information of the (evolving) system of equations in tableaux • Coefficients of the variables • Constants on the right-hand-sides • Basic variables corresponding to equations

  27. Table 4.3

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