Using the Simplex Method in Mixed Integer Linear Programming

Transcription

1 Integer Using the Simplex Method in Mixed Integer UTFSM Nancy, 17 december 2015 Using the Simplex Method in Mixed Integer

2 Outline Mathematical Programming Integer 1 Mathematical Programming Optimisation Problems Solving Techniques 2 3 Integer Using the Simplex Method in Mixed Integer

3 Integer Optimisation Problems Solving Techniques 1 Mathematical Programming Optimisation Problems Solving Techniques 2 3 Integer Using the Simplex Method in Mixed Integer

5 Integer Optimisation Problems Solving Techniques Constraint Satisfaction Problems A Constraint Satisfaction Problem (CSP) consists of: Variables: The unknown, what we can modify Parameters: The known, what we cannot modify Constraints: Relationships among variables and parameters Using the Simplex Method in Mixed Integer

6 Integer Optimisation Problems Solving Techniques N-Queens Problem Using the Simplex Method in Mixed Integer

7 Integer Optimisation Problems Solving Techniques Map Colouring Problem Using the Simplex Method in Mixed Integer

8 Integer Optimisation Problems Solving Techniques Constraint Satisfaction Problems Given a CSP we can define: Solution: An assignment of values to variables Feasible solution: A solution satisfying all constraints A CSP can be Satisfiable: If there exists at least one feasible solution Unsatisfiable: If there does not exist a feasible solution Solving a CSP means to search for one or all feasible solutions, or to prove that the CSP is unsatisfiable. Using the Simplex Method in Mixed Integer

9 Integer Optimisation Problems Solving Techniques Constraint Satisfaction Optimisation Problems A Constraint Satisfaction Optimisation Problem (CSOP) consists of: CSP: Variables, parameters, and constraints Objective function(s): Function(s) to be optimised satisfying all constraints Using the Simplex Method in Mixed Integer

10 Integer Optimisation Problems Solving Techniques Knapsack Problem Using the Simplex Method in Mixed Integer

11 Integer Optimisation Problems Solving Techniques Transportation Problems Using the Simplex Method in Mixed Integer

12 Integer Optimisation Problems Solving Techniques Traveling Salesperson Problem (TSP) Using the Simplex Method in Mixed Integer

13 Integer Optimisation Problems Solving Techniques Vehicle Routing Problems (VRP) Using the Simplex Method in Mixed Integer

14 Integer Optimisation Problems Solving Techniques Scheduling Problem Using the Simplex Method in Mixed Integer

15 Integer Optimisation Problems Solving Techniques Given a CSOP we can define: Optimal solution: A feasible solution better than or equal to all other feasible solutions Depending on the set of constraints, a CSOP can be Feasible: If there exists at least one feasible solution Unfeasible: If there does not exist a feasible solution Depending on the set of constraints and the objective function, a feasible CSOP can have Unbounded Feasible Solutions: Feasible solutions can grow indefinitely Unique Optimal Solution: Just one optimal solution Multiple Optimal Solutions: Several optimal solutions Solving a CSOP means to search for optimal solution(s) or to prove that the CSOP is unfeasible. Using the Simplex Method in Mixed Integer

17 Integer Solving Techniques Optimisation Problems Solving Techniques Both CSPs and CSOPs can be tackled using two approaches: Global or Complete Techniques: Search the global optimal solution, the best one among all feasible solutions Local or Incomplete Techniques: Search for a local optimal solution, the best one among the considered feasible solutions Using the Simplex Method in Mixed Integer

18 Integer 1 Mathematical Programming Optimisation Problems Solving Techniques 2 3 Integer Using the Simplex Method in Mixed Integer

19 Integer Linear: All mathematical functions are linear Programming: In the sense of planning activities Industrial Applications: Blending, transportation, etc. Origin: George B. Dantzig, 1947 Using the Simplex Method in Mixed Integer

21 Mathematical Programming Integer 1 Assign variables x 1 and x 2 to axes x and y of plane, respectively. 2 Identify the feasible area: Applying non-negativity constraints, select positive side of both axes. Draw functional constraints. 3 Identify the feasible solution in the feasible area that optimise the value of the objective function. Using the Simplex Method in Mixed Integer

22 Example Mathematical Programming Integer Considering the following problem: Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1, x 2 0 Using the Simplex Method in Mixed Integer

23 Integer Variable representation x 1 and x 2 are assigned to axes x and y of a plane, respectively. Each point (x 1, x 2 ) represents a solution. The solution where x 1 = 0 x 2 = 0 is called the origin. Using the Simplex Method in Mixed Integer

24 Integer Set of feasible solutions Non-negativity constraints, x 1 0 y x 2 0, impose (x 1, x 2 ) will be in the positive side of both axes. Using the Simplex Method in Mixed Integer

25 Integer Set of feasible solutions Constraint 1x 1 + 1x 2 17 allows solutions (x 1, x 2 ) under line 1x 1 + 1x 2 = 170: Doing the same for all constraints: Using the Simplex Method in Mixed Integer

26 Integer Set of feasible solutions The set of feasible solutions is called feasible area. Using the Simplex Method in Mixed Integer

27 Optimising Mathematical Programming Integer Assigning arbitrary values to the objective function 11x x 2 : 11x x 2 = 100: there exist some feasible solutions 11x x 2 = 135: there exist some feasible solutions 11x x 2 = 170: there does not exist any feasible solution Using the Simplex Method in Mixed Integer

28 Optimising Mathematical Programming Integer Different values of the objective function generate different parallel lines, the higher the value, the farther the line from the origin. Feasible solution(s) on the farthest line correspond to the optimal solution(s). Optimal solution is in the intersection of constraints c3 and c4: 3x 1 + 5x 2 = 48 3x 1 + 1x 2 = 30 solving the system of two equations we get: (x 1, x 2 ) = (17/2, 9/2); z = 11 17/ /2 = 313/2 Using the Simplex Method in Mixed Integer

29 Integer Properties of feasible solutions at vertices 1 Location of optimal solutions: If there exists an optimal solution, it has to be a feasible solution at a vertex. If there exist multiple optimal solutions, at least two of them have to be feasible solutions at adjacent vertices. 2 There exists a finite number of feasible solutions at the vertices. 3 If a feasible solution at a vertex is equal or better than all the feasible solutions at adjacent vertices, then it is equal or better than all the feasible solutions at the vertices, that is, it is optimal. Using the Simplex Method in Mixed Integer

30 Standard form Mathematical Programming Integer Subject to Min z = 1x 1 + 1x 2 1x 1 + 2x x 1 + 1x x 1 2x 2 = 10 x 1, x 2 0 The standard form of this problem is the following: Subject to Min z = 1x 1 + 1x 2 + 0s 1 + 0s 2 1x 1 + 2x 2 + 1s 1 = 30 2x 1 + 1x 2 1s 2 = 28 1x 1 2x 2 = 10 x 1, x 2, s 1, s 2 0 Using the Simplex Method in Mixed Integer

32 Integer Iterative Optimisation Algorithm Structure Initialisation procedure Iterative process Stopping criterion Optimisation Algorithm Structure Search for an initial solution Iteratively search for a better solution Optimality test Using the Simplex Method in Mixed Integer

33 Integer Overview of the Simplex Method 1 Initial step: Start in a feasible basic solution at a vertex. 2 Iterative step: Move to a better feasible basic solution at an adjacent vertex 3 Optimality test: A feasible basic solution at a vertex is optimal when it is equal or better than feasible basic solutions at all adjacent vertices. Using the Simplex Method in Mixed Integer

34 Integer Creating the initial simplex tableau Given the following linear programming problem in standard form: Subject to Max z = 11x x 2 + 0s 1 + 0s 2 + 0s 3 + 0s 4 1x 1 + 1x 2 + 1s 1 = 17 3x 1 + 7x 2 + 1s 2 = 63 3x 1 + 5x 2 + 1s 3 = 48 3x 1 + 1x 2 + 1s 4 = 30 x 1, x 2, s 1, s 2, s 3, s 4 0 Let: c j : coefficient for variable j in the objective function a ij : coefficient for variable j in constraint i b i : right-hand side value of constraint i Using the Simplex Method in Mixed Integer

35 Integer A simplex tableau is the following: c 1... c n a a 1n b a m1... a mn b m Row c Matrix A Column b Using the Simplex Method in Mixed Integer

36 Integer Creating the initial simplex tableau Subject to Max z = 11x x 2 + 0s 1 + 0s 2 + 0s 3 + 0s 4 1x 1 + 1x 2 + 1s 1 = 17 3x 1 + 7x 2 + 1s 2 = 63 3x 1 + 5x 2 + 1s 3 = 48 3x 1 + 1x 2 + 1s 4 = 30 x 1, x 2, s 1, s 2, s 3, s 4 0 x 1 x 2 s 1 s 2 s 3 s b i Using the Simplex Method in Mixed Integer

37 Integer To read the solutions we add two columns: Basis: Current basic variables c j : Coefficients in the objective function of current basic variables x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i s s s s As all no-basic variables are assigned value zero, the value of basic variables is obtained from column b i at the corresponding row: (x 1, x 2, s 1, s 2, s 3, s 4 ) = (0, 0, 17, 63, 48, 30) The product of the column c j and the column b i turns in the value of the objective function (indicated under column b i ) Using the Simplex Method in Mixed Integer

38 Graphically Mathematical Programming Integer Using the Simplex Method in Mixed Integer

39 Integer Evaluating adjacent feasible basic solutions How about the quality of this basic solution with respect to adjacent basic solutions? Changing the value of a variable: Impacts directly the objective function through its coefficient in the objective function (c j ) Impacts indirectly the objective function through variations in other variables We need to take into account both impacts Using the Simplex Method in Mixed Integer

40 Integer Evaluating adjacent feasible basic solutions The direct impact in the objective function is easily computed by c j, but for the decrease and the net change we will add two rows into the tableau: z j : Decrease in the objective function when increasing variable j by one. c j z j : Net change in the objective function when increasing variable j by one. Using the Simplex Method in Mixed Integer

41 Integer Computing z j z 1 = = 0 z 2 = = 0 z 3 = = 0 z 4 = = 0 z 5 = = 0 z 6 = = 0 Adding this information into the tableau... x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i s s s s z j Entering x 1 or x 2 does not decrease the obj func Using the Simplex Method in Mixed Integer

42 Integer Computing c j z j c1 z1 = 11 0 = 11 c 2 z 2 = 14 0 = 14 c 3 z 3 = 0 0 = 0 c 4 z 4 = 0 0 = 0 c 5 z 5 = 0 0 = 0 c 6 z 6 = 0 0 = 0 Adding this information into the tableau... x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i s s s s z j c j z j It is not optimal, entering x 1 or x 2 allows to improve the obj func Using the Simplex Method in Mixed Integer

43 Integer Criterion to enter a non-basic variable into the basis The first criterion proposed to select the entering non-basic variable was to choose the variable with highest c j z j, i.e, with highest unit increase in the objective function. x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i s s s s z j c j z j Of course, we would like to increase the value of x 2 as much as possible, but we have to take into account all constraints! Using the Simplex Method in Mixed Integer

44 Integer Criterion to remove a basic variable from the basis The maximum value for x 2 is giving by the first current basic variable getting value zero when x 1 is increasing. Considering column j corresponding to the non-basic variable entering into the basis, we compute rate b i a ij (only for a ij > 0) for each row i, and we select the basic variable giving the minimum b i a ij. x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i b i s s s s z j c j z j a ij 1 = = 9 = = 30 1 Using the Simplex Method in Mixed Integer

45 Pivoting Mathematical Programming Integer x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i b i a ij s s s s z j c j z j The adjacent basic solution is obtained replacing s 2 by x 2 in the basis, and operating rows to obtain in column x 2 the following vector: Using the Simplex Method in Mixed Integer

46 Pivoting Mathematical Programming Integer Using the Gauss Elimination Method, we can Multiply a constraint by a constant k (k 0) Add constraints In our case: Multiplying the second row by 1 7 we get the new second row. Multiplying the new second row by 1 and adding it to the first row we get the new first row. Multiplying the new second row by 5 and adding it to the third row we get the new third row. Multiplying the new second row by 1 and adding it to the fourth row we get the new fourth row. Using the Simplex Method in Mixed Integer

47 Pivoting Mathematical Programming Integer x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i b i a ij s s s s z j c j z j x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i 4 1 s x s s New feasible basic solution: (x 1, x 2, s 1, s 2, s 3, s 4 ) = (0, 9, 8, 0, 3, 21); z = 126 Using the Simplex Method in Mixed Integer

49 Integer Evaluating adjacent feasible basic solutions x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i 4 1 s x s s Computing z j and c j z j... x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i 4 1 s x s s z j c j z j c j z j > 0 so there exists a better adjacent basic solution and the current one is not optimal Using the Simplex Method in Mixed Integer

50 Pivoting Mathematical Programming Integer x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i b i s = x = s = s = z j c j z j x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i 1 2 s x x s New feasible basic solution: (x 1, x 2, s 1, s 2, s 3, s 4 ) = ( 7 2, 15 2 a ij, 6, 0, 0, 12); z = Using the Simplex Method in Mixed Integer

52 Integer Evaluating adjacent feasible basic solutions x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i 1 2 s x x s Computing z j and c j z j... x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i 1 2 s x x s z j c j z j c j z j > 0 so there exists a better adjacent basic solution and the current one is not optimal Using the Simplex Method in Mixed Integer

53 Pivoting Mathematical Programming Integer x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i b i s x a ij = x s z j c j z j x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i s x x s New feasible basic solution: (x 1, x 2, s 1, s 2, s 3, s 4 ) = ( 17 2, = 15 2 = 6, 4, 6, 0, 0); z = Using the Simplex Method in Mixed Integer

55 Integer Evaluating adjacent feasible basic solutions x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i 1 1 s x x s Computing z j and c j z j... x 1 x 2 s 1 s 2 s 3 s 4 Basis c j b i 1 1 s x x s z j c j z j All values in row c j z j are zero or negative so the Stopping Criterion is verified! Optimal Solution: (x, xcarlos, scastro, s, s Using, s the ) = Simplex ( 17 Method, 9, 4, in Mixed 6, 0, Integer 0)

57 Integer Integer Why? Many practical problems need variables to take integer values How to solve it? Exhaustive enumeration is only useful when dealing with few combinations of values In general, we deal with a combinatorial problem Solving techniques are based on divide and conquer approach Using the Simplex Method in Mixed Integer

59 Integer Solve linear programming relaxations to bound the objective function Create branches by adding constraints that eliminate non-integer values Using the Simplex Method in Mixed Integer

60 Example Mathematical Programming Integer Considering the following Integer Programming problem: Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1, x 2 0 x 1, x 2 Z Using the Simplex Method in Mixed Integer

61 Integer Solving the Relaxation Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1, x 2 0 x = (8.5, 4.5) z = Using the Simplex Method in Mixed Integer

63 Enumeration tree Mathematical Programming Integer Using the Simplex Method in Mixed Integer

64 Branching Mathematical Programming Integer Branching on x 1 : Using the Simplex Method in Mixed Integer

65 Integer Solving the Relaxation Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1 8 x 1, x 2 0 x = (8, 4.8) z = Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1 9 x 1, x 2 0 x = (9, 3) z = 141 Using the Simplex Method in Mixed Integer

67 Branching Mathematical Programming Integer Branching on x 2 : Using the Simplex Method in Mixed Integer

68 Integer Solving the Relaxation Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1 8 & x 2 4 x 1, x 2 0 x = (8, 4) z = 144 Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1 8 & x 2 5 x 1, x 2 0 x = (7.67, 5) z = Using the Simplex Method in Mixed Integer

70 Branching Mathematical Programming Integer Branching again on x 1 : Using the Simplex Method in Mixed Integer

71 Integer Solving the Relaxation Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1 8 & x 2 5 & x 1 7 x 1, x 2 0 x = (7, 5.4) z = Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1 8 & x 2 5 & x 1 8 x 1, x 2 0 Unfeasible Using the Simplex Method in Mixed Integer

73 Branching Mathematical Programming Integer Branching again on x 2 : Using the Simplex Method in Mixed Integer

74 Integer Solving the Relaxation Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1 8 & x 2 5 & x 1 7 & x 2 5 x 1, x 2 0 x = (7, 5) z = 147 Max z = 11x x 2 Subject to 1x 1 + 1x x 1 + 7x x 1 + 5x x 1 + 1x 2 30 x 1 8 & x 2 5 & x 1 7 & x 2 6 x 1, x 2 0 x = (6, 6) z = 150 Using the Simplex Method in Mixed Integer

77 Integer Solve linear programming relaxations to bound the objective function Create branches by adding constraints that fix variables Using the Simplex Method in Mixed Integer

78 Example Mathematical Programming Integer Max z = 9x 1 + 5x 2 + 6x 3 + 4x 4 Subject to 6x 1 + 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 1 + x 3 0 x 2 + x 4 0 x i {0, 1} i = 1,..., 4 Using the Simplex Method in Mixed Integer

79 Integer Constraints allowing values A constraint allowing values for variables: x i {0, 1} is equivalent to the following constraints: x i 0 x i 1 x i Z This correspond to an IP problem with upper bounds for all variables. Using the Simplex Method in Mixed Integer

80 Integer Solving the Relaxation Max z = 9x 1 + 5x 2 + 6x 3 + 4x 4 Subject to 6x 1 + 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 1 + x 3 0 x 2 + x 4 0 x i 1 i = 1,..., 4 x i 0 i = 1,..., 4 x = ( 5 6, 1, 0, 1); z = Using the Simplex Method in Mixed Integer

82 Branching Mathematical Programming Integer Remember that x i {0, 1} is equivalent to the following constraints: x i 0 x i 1 x i Z When branching we will add constraints x i 0 and x i 1: x i 0 x i 0 x i 1 x i 1 x i 0 x i 1 x i Z x i Z x i = 0 x i = 1 Using the Simplex Method in Mixed Integer

83 Branching Mathematical Programming Integer Subproblem 1 (original problem x 1 = 0): Max z = 9x 1 + 5x 2 + 6x 3 + 4x 4 Subject to 6x 1 + 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 1 + x 3 0 x 2 + x 4 0 x 1 = 0 x i {0, 1} i = 1,..., 4 Subproblem 2 (original problem x 1 = 1): Max z = 9x 1 + 5x 2 + 6x 3 + 4x 4 Subject to 6x 1 + 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 1 + x 3 0 x 2 + x 4 0 x 1 = 1 x i {0, 1} i = 1,..., 4 Using the Simplex Method in Mixed Integer

84 Integer Solving the Relaxation Subproblem 1: Max z = 5x 2 + 6x 3 + 4x 4 Subject to 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 3 0 x 2 + x 4 0 x i 1 i = 2,..., 4 x i 0 i = 2,..., 4 x = (0, 1, 0, 1); z = 9 Subproblem 2: Max z = 9 + 5x 2 + 6x 3 + 4x 4 Subject to 3x 2 + 5x 3 + 2x 4 4 x 3 + x 4 1 x 3 1 x 2 + x 4 0 x i 1 i = 2,..., 4 x i 0 i = 2,..., 4 x = (1, 4 5, 0, 4 5 ); z = Using the Simplex Method in Mixed Integer

86 Branching Mathematical Programming Integer Subproblem 3 (subproblem 2 x 2 = 0): Subject to Max z = 9 + 5x 2 + 6x 3 + 4x 4 3x 2 + 5x 3 + 2x 4 4 x 3 + x 4 1 x 3 1 x 2 + x 4 0 x 2 = 0 x i {0, 1} i = 2,..., 4 Subproblem 4 (subproblem 2 x 2 = 1): Max z = 9 + 5x 2 + 6x 3 + 4x 4 Subject to 3x 2 + 5x 3 + 2x 4 4 x 3 + x 4 1 x 3 1 x 2 + x 4 0 x 2 = 1 x i {0, 1} i = 2,..., 4 Using the Simplex Method in Mixed Integer

87 Integer Solving the Relaxation Subproblem 3: Max z = 9 + 6x 3 + 4x 4 Subject to 5x3 + 2x 4 4 x 3 + x 4 1 x 3 1 x 4 0 x i 1 i = 3, 4 x i 0 i = 3, 4 x = (1, 0, 4 5, 0); z = Subproblem 4: Max z = x 3 + 4x 4 Subject to 5x3 + 2x 4 1 x 3 + x 4 1 x 3 1 x 4 1 x i 1 i = 3, 4 x i 0 i = 3, 4 x = (1, 1, 0, 1 2 ); z = 16 Using the Simplex Method in Mixed Integer

89 Branching Mathematical Programming Integer Subproblem 5 (subproblem 4 x 4 = 0): Max z = x 3 + 4x 4 Subject to 5x 3 + 2x 4 1 x 3 + x 4 1 x 3 1 x 4 1 x 4 = 0 x i {0, 1} i = 3, 4 Subproblem 6 (subproblem 4 x 4 = 1): Max z = x 3 + 4x 4 Subject to 5x 3 + 2x 4 1 x 3 + x 4 1 x 3 1 x 4 1 x 4 = 1 x i {0, 1} i = 3, 4 Using the Simplex Method in Mixed Integer

90 Integer Solving the Relaxation Subproblem 5: Subject to Max z = x 3 5x 3 1 x 3 1 x 3 1 x 3 0 x = (1, 1, 1 5, 0); z = Subproblem 6: Subject to Max z = x 3 5x 3 1 x 3 0 x 3 1 x 3 1 x 3 0 Unfeasible Using the Simplex Method in Mixed Integer

92 Branching Mathematical Programming Integer Subproblem 7 (subproblem 5 x 3 = 0): Max z = x 3 Subject to 5x 3 1 x 3 1 x 3 1 x 3 0 x 3 = 0 x 3 {0, 1} Subproblem 8 (subproblem 5 x 3 = 1): Max z = x 3 Subject to 5x 3 1 x 3 1 x 3 1 x 3 0 x 3 = 1 x 3 {0, 1} Using the Simplex Method in Mixed Integer

93 Integer Solving the Relaxation Subproblem 7: Variable enumeration: x = (1, 1, 0, 0) Feasible solution z = 14 Subproblem 8: Variable enumeration: x = (1, 1, 1, 0) Unfeasible solution Using the Simplex Method in Mixed Integer

95 Integer Solving Mixed Integer Programming Problems Mixed Integer Programming Problems involve real, integer, and binary variables: Real variables: Simplex Method Integer variables: Simplex Method + constraints and Binary variables: Simplex Method + constraints = Using the Simplex Method in Mixed Integer