Metode Kuantitatif Bisnis. Week 4 Linear Programming Simplex Method - Minimize

Transcription

1 Metode Kuantitatif Bisnis Week 4 Linear Programming Simplex Method - Minimize

2 Outlines Solve Linear Programming Model Using Graphic Solution Solve Linear Programming Model Using Simplex Method (Maximize) Solve Linear Programming Model Using Simplex Method (Minimize)

3 Linear Programming Problems Standard Maximization Standard Minimization Non- Standard Objective: Maximize Objective: Minimize Objective: Max or Min Constraints: ALL with Constraints: All with Constraints: Mixed, can be, or = Standard simplex Dual Two Phase

4 STANDARD MINIMIZATION PROBLEM

5 Holiday Meal Turkey Ranch Minimize cost = 2X 1 + 3X 2 subject to 5X X X 1 + 3X X X 1, X 2 0

6 The Dual Minimize cost = 2X 1 + 3X 2 subject to 5X X X 1 + 3X X X 1, X 2 0 A = A T =

7 The Dual A T = Maximize Z = 90Y Y Y 3 subject to 5Y 1 + 4Y Y Y 1 + 3Y 2 3 Y 1, Y 2, Y 3 0 Continue with the simplex method to solve standard maximization problems!!

8 Slack Variables When using the dual, slack variables are variables in the former problem formulation. Therefore, slack variables in this problem are X 1 and X 2. Constraints: 5Y 1 + 4Y Y Y 1 + 3Y 2 3 5Y 1 + 4Y Y 3 + X Y 1 + 3Y 2 + X 2 3 Z = 90Y Y Y 3-90Y 1-48Y 2-1.5Y 3 + Z = 0

9 First Simplex Tableu 5Y 1 + 4Y Y 3 + X 1 2 row 1 Basic Variable Y 1 Y 2 Y 3 X 1 X 2 Z RHS X X Z Y 1 + 3Y 2 + X 2 3 row 2-90Y 1-48Y 2-1.5Y 3 + Z = 0 row 3

10 Simplex Tableu Pivot Number Change the pivot number to 1 and other number in the same column to 0 Basic Variable Y 1 Y 2 Y 3 X 1 X 2 Z RHS X X Z = 2/5 = 0.4 = 3/20 = 0.3 Most negative, therefore variable Y 1 enter the solution mix Smallest result of the ratio, therefore variable X 2 leave the solution mix

11 Second Simplex Tableu Basic Var. Y 1 Y 2 Y 3 X 1 X 2 Z RHS X X Z = Row 2 * (-5) + Row 1 = Row 2 / 10 = Row 2 * 90 + Row 3 Basic Var. Y 1 Y 2 Y 3 X 1 X 2 Z RHS X Y Z Row P still contain negative value

12 Second Simplex Tableu Pivot Number Change the pivot number to 1 and other number in the same column to 0 Basic Var. Y 1 Y 2 Y 3 X 1 X 2 Z RHS X Y Z /2.5 = /0.3 = 1 Most negative, therefore variable Y 2 enter the solution mix Smallest result of the ratio, therefore variable X 1 leave the solution mix

13 Third Simplex Tableu Basic Var. Y 1 Y 2 Y 3 X 1 X 2 Z RHS X Y Z = Row 1 = Row 1 * (-1/2) + Row 2 = Row 1 * 15 + Row 3 Basic Var. Y 1 Y 2 Y 3 X 1 X 2 Z RHS Y Y Z OPTIMAL!!

14 Optimal Solution Because this solution is the solution of the dual, use row Z and column X 1, X 2 as the optimal solution. Basic Var. Y 1 Y 2 Y 3 X 1 X 2 Z RHS Y Y Z X 1 = 8.4 X 2 = 4.8 Minimal cost = 31.2

15 EXERCISE

16 M7-22 Solve the following LP problem first graphically and then by the simplex algorithm: Minimize cost = 4X 1 + 5X 2 subject to X 1 + 2X X 1 + X 2 75 X 1, X 2 0

17 NON-STANDARD LINEAR PROGRAMMING PROBLEM

18 The Muddy River Chemical Company Minimize cost = $5X 1 + $6X 2 subject to X 1 + X 2 = 1000 lb X lb X lb X 1, X 2 0

19 Surplus Variable Greater-than-or-equal-to ( ) constraints involve the subtraction of a surplus variable rather than the addition of a slack variable. Surplus is sometimes simply called negative slack X X 2 S 2 = 150

20 Artificial Variable Artificial variables are usually added to greater-than-or-equal-to ( ) constraints to avoid violating the non-negativity constraint. When a constraint is already an equality (=), artificial variable is also used to provide an automatic initial solution

21 Artificial Variable Equality X 1 + X 2 = 1000 Equality X 1 + X 2 +A 1 = 1000 Greater than or equal X Greater than or equal X 2 S 2 + A 2 = 150 Before the final simplex solution has been reached, all artificial variables must be gone from the solution mix

22 Two Phase Method Phase I Objective: Minimize all artificial variables Phase II Objective: LP original objective

23 Conversions Constraints: X 1 + X 2 = 1000 X X Objective function: Minimize C = 5X 1 + 6X 2 X 1 + X 2 + A 1 = 1000 X 1 + S 1 = 300 X 2 S 2 + A 2 = 150 Phase I : Minimize Z = A 1 + A 2 Phase II: Minimize C = 5X 1 + 6X 2

24 Converting The Objective Function In minimization problem, minimizing the cost objective is the same as maximizing the negative of the cost objective function Minimize C = 5X 1 + 6X 2 and Z = A 1 + A 2 Maximize (-C) = 5X 1 6X 2 and (-Z) = A 1 A 2

25 Conversions Constraints: X 1 + X 2 = 1000 X X Objective function: Phase I : ( Z) = A 1 A 2 Phase II: ( C) = 5X 1 6X 2 X 1 + X 2 + A 1 = 1000 X 1 + S 1 = 300 X 2 S 2 + A 2 = 150 Phase I : A 1 + A 2 Z = 0 Phase II: 5X 1 + 6X 2 C = 0

26 Simplex Tableu X 1 + X 2 + A 1 = 1000 row 1 X 1 + S row 2 X2 S 2 + A row 3 Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS A S A Z (-1) 0 0 C (-1) 0 A 1 + A 2 Z = 0 row 4 5X 1 + 6X 2 C = 0 row 5

27 Simplex Tableu A1 and A2 are supposed to be basic variable. However, in this tableu the condition is not satisfy. infeasible Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS A S A Z (-1) 0 0 C (-1) 0 row 1 * (-1) + row 3 * (-1) + row 4 These values need to be converted to zero (0)

28 Phase I: First Simplex Tableu Basic Variable condition is satisfied Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS A S A Z (-1) C (-1) 0 Continue with simplex procedure to eliminate the negative values in row Z first (Phase I)

29 Phase I: Second Simplex Tableu Pivot Number Change the pivot number to 1 and other number in the same column to 0 Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS A /1 0 = S /0 0 = ~ A /1 0 = Z (-1) C (-1) 0 Most negative, therefore variable X 2 enter the solution mix Smallest result of the ratio, therefore variable A 2 leave the solution mix

30 Phase I: Second Simplex Tableu Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS A = row 03 * (-1) 0 + row S = row A = row Z = row 03 * 2 (-1) + row C = row 03 * (-6) 0 + row (-1) 5 0 Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS A S X Z (-1) C (-1) -900 Row Z still contain negative values

31 Phase I: Third Simplex Tableu Pivot Number Change the pivot number to 1 and other number in the same column to 0 Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS A /1 0 = S /0 0 = ~ X Z (-1) C (-1) -900 Most negative, therefore variable S 2 enter the solution mix Smallest result of the ratio, therefore variable A 1 leave the solution mix

32 Phase I: Third Simplex Tableu Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS A = row S = row X = row 11 + row Z = row 21 + row (-1) C = row -61 * (-6) 0 + row (-1) Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS S S X Z (-1) 0 0 C (-1) Phase I OPTIMAL!!

33 Phase II: First Simplex Tableu Pivot Number Change the pivot number to 1 and other number in the same column to 0 Basic Var. X 1 X 2 S 1 S 2 C RHS S S X C (-1) /1 = /1 = /1 = 1000 Most negative, therefore variable X 1 enter the solution mix Smallest result of the ratio, therefore variable S 1 leave the solution mix

34 Third Simplex Tableu Basic Var. X 1 X 2 S 1 S 2 C RHS S S X C (-1) = Row 2 * (-1) + row 1 = Row 2 = Row 2 * (-1) + Row 3 = Row 2 + Row 4 Basic Var. X 1 X 2 S 1 S 2 C RHS S X X C (-1) OPTIMAL!!

35 Optimal Solution Remember that the objective function was converted. Thus: Max (-C) = Min C = 5700 Basic Var. X 1 X 2 S 1 S 2 C RHS S X X C (-1) X 1 = 300 X 2 = 700

36 EXERCISE

37 M7-27 A Pharmaceutical firm is about to begin production of three new drugs. An objective function designed to minimize ingredients costs and three production constraints are as follows: Minimize cost = 50X X X 3 subject to X 1 X 2 = X 2 + 2X 3 = X X 1, X 2, X 3 0

38 SPECIAL CASE IN SIMPLEX METHOD

39 1. Infeasibility No feasible solution is possible if an artificial variable remains in the solution mix Basic Var. X 1 X 2 S 1 S 2 A 1 A 2 Z C RHS X X A Z (-1) C (-1) -900 Phase I already optimal (no negative value)

40 2. Unbounded Solutions Unboundedness describes linear programs that do not have finite solutions Basic Var. X 1 X 2 S 1 S 2 P RHS X S P Since both pivot column numbers are negative, an unbounded solution is indicated

41 3. Degeneracy Degeneracy develops when three constraints pass through a single point Basic Var. X 1 X 2 X 3 S 1 S 2 S 3 C RHS X / = S /4 1 = S /2 0 = C Tie for the smallest ratio indicates degeneracy Degeneracy could lead to a situation known as cycling, in which the simplex algorithm alternates back and forth between the same non-optimal solutions

42 4. More Than One Optimal Solution If the value of P (objective function s row) is equal to 0 for a variable that is not in the solution mix, more than one optimal solution exists. Basic Var. X 1 X 2 S 1 S 2 P RHS X S P

43 THANK YOU