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Operation Research Chapter 3 Simplex Method.

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Presentation on theme: "Operation Research Chapter 3 Simplex Method."— Presentation transcript:

1 Operation Research Chapter 3 Simplex Method

2 Operation Research: An Introduction By Dr: Alaa Sagheer
Chapter 2 Outline: Part II Introduction The Linear Programming in Standard Form Simplex Method Artificial starting Solution The M-Method The Two-Phase Method Operation Research: An Introduction By Dr: Alaa Sagheer

3 Operation Research: An Introduction By Dr: Alaa Sagheer
Artificial Starting Solution (1): In our presentation of the Simplex method we have used the slack variables as the starting solution. These were coming from the standardized form of constraints that are type of “” However, if the original constraint is a “≥” or “=” type of constraint, we no longer have an easy starting solution. Therefore, Artificial Variables are used in such cases. An artificial variable is a variable introduced into each equation that has a surplus variable. To ensure that we consider only basic feasible solutions, an artificial variable is required to satisfy the nonnegative constraint. The two method used are: The M- method The Two-phase method Operation Research: An Introduction By Dr: Alaa Sagheer

4 Operation Research: An Introduction By Dr: Alaa Sagheer
Artificial Starting Solution (2): The M-Method The M-method starts with the LP in the standard form For any equation (i) that does not have slack, we augment an artificial variable Ri Given M is sufficiently large positive value, The variable Ri is penalized in objective function using (-M Ri ) in case of maximization and (+ M Ri ) in case of minimization ( Penalty Role) Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (3): The M-Method ( Example) Minimized Subject to: Operation Research: An Introduction By Dr: Alaa Sagheer

6 Operation Research: An Introduction By Dr: Alaa Sagheer
Artificial Starting Solution (4): The M-Method ( Solution) Minimized By subtracting surplus x3 in second constraint and adding slack x4 in third constraint, thus we get: Subject to: Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (5): By using artificial variables in equations that haven’t slack variables and penalized them in objective function, we got: Minimized Subject to: Then, we can use R1 , R2 and x4 as the starting basic feasible solution. Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (6): Solution x4 R2 R1 x3 x2 x1 Basic -M -1 -4 z 3 1 6 4 2 New z-row = Old z-row + M* R1-row + M*R2 -row Solution x4 R2 R1 x3 x2 x1 Basic 9M -M -1+4M -4+7M z 3 1 6 -1 4 2 Artificial variables become zero Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (7): Thus, the entering value is (-4+7M) because it the most positive coefficient in the z-row. The leaving variable will be R1 by using the ratios of the feasibility condition After determining the entering and leaving variable, the new tableau can be computed by Gauss-Jordan operations as follow: Solution x4 R2 R1 x3 x2 x1 Basic 4+2M (4-7M)/3 -M (1+5M)/3 z 1 1/3 2 -4/3 -1 5/3 3 -1/3 The last tableau shows that x2 is the entering variable and R2 is the leaving variable. The simplex computation must thus continued for two more iteration to satisfy the optimally condition. The results for optimality are: Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (8): Observation regarding the Use of M-method: The use of penalty M may not force the artificial variable to zero level in the final simplex iteration. Then the final simplex iteration include at least one artificial variable at positive level. This indication that the problem has no feasible condition. 2.( M )should be large enough to act as penalty, but it should not be too large to impair the accuracy of the simplex computations. Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (9): Example: Maximize Subject to: Using Computer solution, apply the simplex method M=10, and repeat it using M= the first M yields the correct solution x1 =1 and x2 =1.5, whereas the second gives the incorrect solution x1 =4 and x2 =0 Multiplying the objective function by 1000 to get z= 200x x2 and solve the problem using M=10 and M= and observe the second value is the one that yields the correct solution in this case The conclusion from two experiments is that the correct choice of the value of M is data dependent. Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (10): Two-phase method When a basic feasible solution is not readily available, the two-phase simplex method may be used as an alternative to the big M method. In the two-phase simplex method, we add artificial variables to the same constraints as we did in big M method. Then we find a basic feasible solution to the original LP by solving the Phase I LP. In the Phase I LP, the objective function is to minimize the sum of all artificial variables. At the completion of Phase I, we use Phase II and reintroduce the original LP’s objective function and determine the optimal solution to the original LP. Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (11): Note: In Phase I, If the optimal value of sum of the artificial variables are greater than zero, the original LP has no feasible solution which ends the solution process. Other wise, We move to Phase II Example Minimized Subject to: Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (12): Solution: Phase I: Minimize: Subject to: Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (13): Solution x4 R2 R1 x3 x2 x1 Basic -1 r 3 1 6 4 2 New r-row = Old r-row + 1* R1-row + R2 -row Solution x4 R2 R1 x3 x2 x1 Basic 9 -1 4 7 r 3 1 6 2 Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (14): By using new r-row, we solve Phase I of the problem which yields the following optimum tableau Solution x4 R2 R1 x3 x2 x1 Basic -1 r 3/5 -1/5 1/5 1 6/5 -4/5 -3/5 Because minimum r=0, Phase I produces the basic feasible solution: Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (15): Phase II After eliminating artificial variables column, the original problem can be written as: Minimize: Subject to: Operation Research: An Introduction By Dr: Alaa Sagheer

18 Artificial Starting Solution (16):
x4 x3 x2 x1 Basic -1 -4 z 3/5 1/5 1 6/5 -3/5 Again, because basic variables x1 and x2 have nonzero coefficient in he z row, they must be substituted out, using the following computation: New z-row = Old z-row + 4* x1-row + 1*x2 -row Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (17): The initial tableau of Phase II is as the following: Solution x4 x3 x2 x1 Basic 18/5 1/5 z 3/5 1 6/5 -3/5 Operation Research: An Introduction By Dr: Alaa Sagheer

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Artificial Starting Solution (18): Remarks: The removal of artificial variables and their column at the end of Phase I can take place only when they are all nonbasic. If one or more artificial variables are basic ( at zero level) at the end of Phase I, then the following additional steps must be under taken to remove them prior to start Phase II Step 1. Select a zero artificial variable to leave the basic solution and designate its row as pivot row. The entering variable can be any nonbasic (nonartificial) variable with nonzero (positive or negative) coefficient in the pivot row. Perform the associated simplex iteration. Step 2. Remove the column of the (Just-leaving) artificial from the tableau. If all the zero artificial variables have been removed , go to Phase II. Otherwise, go back to Step I. Operation Research: An Introduction By Dr: Alaa Sagheer

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