Linear programming simplex method explained in detail
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The document discusses the simplex method for solving linear programming problems. It explains that the simplex method is an iterative procedure developed by George Dantzig in 1946 that provides a systematic way to examine the vertices of the feasible region to determine the optimal value of the objective function. The document then provides step-by-step instructions for applying the simplex method, including preparing the initial tableau, selecting pivots, and checking for optimality. It also includes an example problem demonstrating the simplex method.
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Linear programming simplex method explained in detail
- 2. 4/5/2020 Dr. Abdulfatah Salem 2 For linear programming problems involving two variables, the graphical solution method introduced before is convenient. It is better to use solution methods that are adaptable to computers. One such method is called the simplex method, The simplex method is an iterative procedure developed by George Dantzig in 1946. It provides us with a systematic way of examining the vertices of the feasible region to determine the optimal value of the objective function. more than two variables problems involving a large number of constraints However, for problems involving
- 3. 4/5/2020 Dr. Abdulfatah Salem 3 Standard Minimization Form • The objective function is to be minimized. • All the RHS involved in the problem are nonnegative. • All other linear constraints may be written so that the expression involving the variables is greater than or equal to a nonnegative constant. Standard Maximization Form • The objective function is to be maximized. • All the RHS involved in the problem are nonnegative. • All other linear constraints may be written so that the expression involving the variables is less than or equal to a nonnegative constant.
- 4. 4/5/2020 Dr. Abdulfatah Salem 4 1) Be sure that the model is standard model 2) Be sure that the right hand side is +ve value 4X1 - 7X2 > -50 4X1 - 7X2 < 50 -8X1 - 6X2 ≤ -57 8X1 + 6X2 ≥ 57 3) Convert each inequality in the set of constraints to an equation as follows: • Slack Variable added to the LHS of a constraint contains ≤ separator to convert it to an equal separator (=). 4X1 - 7X2 < 50 4X1 - 7X2 + S = 50 or • Surplus Variable subtracted from the LHS and artificial variable added to the LHS of a constraint contains ≥ to convert it to an equal separator (=). 8X1 + 6X2 ≥ 57 8X1 + 6X2 - S + A = 57 Simplex Algorithm
- 5. 4/5/2020 Dr. Abdulfatah Salem 5 4) Prepare the equations to move all the unknown terms to the left hand side and the fixed values to the right hand side Z = 4X1 + 9X2 + 3X3 Z - 4X1 - 9X2 - 3X3 = 0 8X1 + 6X2 = 57 + 4X3 8X1 + 6X2 - 4X3 = 57 X1 + 5X2 = 70 - X3 X1 + 5X2 + X3 = 70 5) Create the initial simplex tableau Simplex Algorithm (cont.) Z - 4X1 - 9X2 - 3X3 = 0 X1 + 5X2 + X3 + S1 = 70 5X2 + S2 = 81 Basic Variables X1 X2 X3 S1 S2 Z RHS S1 1 5 1 1 0 0 70 S2 0 5 0 0 1 0 81 Z -4 -9 -3 0 0 1 0
- 6. 4/5/2020 Dr. Abdulfatah Salem 6 6) Select the pivot column. ( The column with the “most negative value” element in the last row. ) 7) Select the pivot row. (The row with the smallest non-negative result when the last element in the row is divided by the corresponding in the pivot column) Simplex Algorithm (cont.) Basic Variables X1 X2 X3 S1 S2 Z RHS S1 1 5 1 1 0 0 70 S2 0 5 0 0 1 0 81 Z -4 -9 -3 0 0 1 0 Basic Variables X1 X2 X3 S1 S2 Z RHS X2 1 5 1 1 0 0 70 S2 0 5 0 0 1 0 81 Z -4 -9 -3 0 0 1 0
- 7. 4/5/2020 Dr. Abdulfatah Salem 7 8) Use elementary row operations to make all numbers in the pivot column equal to 0 except for the pivot number equal to 1. 9) Check the optimality (entries in the bottom row are zero or positive, If so, this the final tableau (optimal solution) , If not, go back to step 5 10) The linear programming problem has been solved and maximum solution obtained, which is given by the entry in the lower-right corner of the tableau. Simplex Algorithm (cont.) Basic Variables X1 X2 X3 S1 S2 Z RHS S1 # 1 # # # # # S2 # 0 # # # # # Z # 0 # # # # # Basic Variables X1 X2 X3 S1 S2 Z RHS S1 # 1 # # # # # S2 # 0 # # # # # Z # 0 # # # # #
- 8. 4/5/2020 Dr. Abdulfatah Salem 8 = + Artificial (A) Constraint separator Action ≥ + Slack (s) ≤ - Surplus (s) + Artificial (A) Simplex Algorithm (cont.)
- 9. 4/5/2020 Dr. Abdulfatah Salem 9 The national co. for assembling computer systems. assembles laptops and printers, Each laptop takes four hours of labor from the hardware department and two hours of labor from the software department. Each printer requires three hours of hardware and one hour of software. During the current week, 240 hours of hardware time are available and 100 hours of software time. Each laptop assembled gives a profit of $70 and each printer a profit of $50. How many printers and laptops should be assembled in order to maximize the profit? Example Solution ConstraintsprinterlaptopResource 24034hardware 10012software 5070Unit profit $ Max Z = 70x1 + 50x2 4x1 + 3x2 ≥ 240 2x1 + x2 ≥ 100 Z - 70x1 - 50x2 = 0 4x1 + 3x2 + s1 = 240 2x1 + x2 + s2 = 100
- 10. 4/5/2020 Dr. Abdulfatah Salem 10 4x1 + 3x2 + s1 = 240 2x1 + x2 + s2 = 100 Z - 70x1 - 50x2 = 0 Basic Variable X1 X2 S1 S2 Z Rhs S1 4 3 1 0 0 240 S2 2 1 0 1 0 100 Z -70 -50 0 0 1 0
- 11. 4/5/2020 Dr. Abdulfatah Salem 11 Basic Variable X1 X2 S1 S2 Z Rhs S1 4 3 1 0 0 240 S2 2 1 0 1 0 100 Z -70 -50 0 0 1 0 Basic Variable X1 X2 S1 S2 Z Rhs S1 4 3 1 0 0 240 S2 2 1 0 1 0 100 Z -70 -50 0 0 1 0 Ratio 60 50 0 Basic Variable X1 X2 S1 S2 Z Rhs S1 4 3 1 0 0 240 S2 2 1 0 1 0 100 Z -70 -50 0 0 1 0 X1 1 1/2 0 1/2 0 50 /2 X1 4 2 -70
- 12. 4/5/2020 Dr. Abdulfatah Salem 12 Basic Variable X1 X2 S1 S2 Z Rhs S1 4 3 1 0 0 240 X1 1 1/2 0 1/2 0 50 Z -70 -50 0 0 1 0 4 1 -70 ❶ ❷ 1 1/2 0 1/2 0 50 4 2 0 2 0 200 1 ½ 0 1/2 0 50 -70 -35 0 -35 0 -3500 Basic Variable X1 X2 S1 S2 Z Rhs S1 0 1 1 -2 0 40 X1 1 1/2 0 1/2 0 50 Z 0 -15 0 35 1 3500 ❸ First iteration
- 13. 4/5/2020 Dr. Abdulfatah Salem 13 1 1/2 -15 ❶ ❷ Basic Variable X1 X2 S1 S2 Z Rhs X2 0 1 1 -2 0 40 X1 1 1/2 0 1/2 0 50 Z 0 -15 0 35 1 3500 0 1 1 -2 0 40 0 1 1 -2 0 40 0 1/2 1/2 -1 0 20 0 -15 -15 30 0 -600 Basic Variable X1 X2 S1 S2 Z Rhs X2 0 1 1 -2 0 40 X1 1 0 -1/2 3/2 0 30 Z 0 0 15 5 1 4100 ❸ Second iteration
- 14. 4/5/2020 Dr. Abdulfatah Salem 14 BV X1 X2 S1 S2 Z Rhs X2 0 1 1 -2 0 40 X1 1 0 -1/2 3/2 0 30 Z 0 0 15 5 1 4100 X1 = 30 No of laptops = 30 X2 = 40 No of printers = 40 Z = 4100 The optimal profit = $4100
Editor's Notes
- In the last two lectures we discussed about the graphical method for solving linear programming problems depending on graphical representation of the constraints to form a feasible area and testing the values of objective function at each corner of the area. Although the graphical method is an invaluable aid to understand the properties of linear programming models, it provides very little help in handling practical problems. For linear programming problems the graphical solution method is convenient when the problem involving two variables and a few number of constraints. In real-world most applications have more than two variables and many constraints, which is too complex for graphical solution , it is better to use a new method that is adaptable to the real-world applications. Many different methods have been proposed to solve linear programming problems. One such method is called the Dantzig Simplex method, in honor of George Dantzig, the mathematician who developed the approach in 1946. The simplex method is an iterative systematic process based on algebraically examining the vertices or what we call corners of the polygon representing the feasible solutions to determine the optimal value of the objective function. Simplex Method is applicable to any problem that can be formulated in terms of linear objective function, subject to a set of linear constraints. The Simplex algorithm is an iterative procedure for solving LP model in a finite number of steps. It consists of Having a trial basic feasible solution to constraint-equations then testing whether it is an optimal solution or not, If not an improvement to the trial by a set of mathematical rules and repeating the process till an optimal solution is obtained.
- The main function in a LP problem is that you try to maximize something like profit or to minimize something like cost, waste, loss or risks for example then you organize your other mathematical relationship even equations or inequations which have variables who affect this. The first step in using the simplex method is to classify the LP model to two types, maximization model or minimization model. The simplex method in its simple form is called the simple algorithm, that algorithm require having the LP model in a standard Form. So, you have to formulate the model to either standard maximization model or standard minimization model. The standard maximization model require: The objective function is to be maximized, that mean the objective function start with the word “Maximize Z =“ for example. All the RHS involved in the problem are nonnegative, that mean all the right hand side are positive values. All other linear constraints may be written so that the expression involving the variables is less than or equal to a nonnegative constant. On the other hand, The standard minimization model require: The objective function is to be minimized, that mean the objective function start with the word “Minimize Z =“ for example. All the RHS involved in the problem are nonnegative, that mean all the right hand side are positive values. All other linear constraints may be written so that the expression involving the variables is greater than or equal to a nonnegative constant.
- After preparing the model by converting it into standard model and become ready to solve using the simple algorithm, follow the following steps: Be sure that the model is standard model by applying the rules of the standard forms explained in details before. Be sure that the right hand side is +ve value Convert each inequality in the set of constraints to an equation as follows: Slack Variable added to the LHS of a constraint contains ≤ separator to convert it to an equal separator (=). or Surplus Variable subtracted from the LHS of a constraint contains ≥ to convert it to an equal separator (=). A slack variable represents unused resources A slack variable contributes nothing to the objective function value. A surplus variable represents an excess above a constraint requirement level. Surplus variables contribute nothing to the calculated value of the objective function.
- Write the objective function as an equation in the form "left hand side"= 0 where terms involving variables are negative Prepare the equations to move all the unknown terms to the left hand side and the fixed values to the right hand side Create the initial simplex tableau consisting of all the coefficients of the decision variables, slack variables, surplus variables and artificial variables added in the 3rd step. Each variable assign its separate column. the 1st. Column assigned to the basic variables while the last column assigned to the LHS values. each constraint assign a separate row while the revised objective function assign the last row.
- Select the pivot column. ( The column with the “most negative value” element in the last row. ) Select the pivot row. (The row with the smallest non-negative ratio resulting from dividing the value of when the last element in the row “under the RHS column” by the corresponding value under the pivot column)
- Use elementary row operations to make all numbers in the pivot column equal to 0 except for the pivot number equal to 1. Check the optimality case (entries in the bottom row are non-negative values, If so, this the final tableau (optimal solution) , If not, go back to step 6 The linear programming problem has been solved and maximum solution obtained, which is given by the entry in the lower-right corner of the tableau.