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Published byMadalyn Dolson Modified about 1 year ago

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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

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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 R i Given M is sufficiently large positive value, The variable R i is penalized in objective function using (-M R i ) in case of maximization and (+ M R i ) in case of minimization. ( Penalty Role)

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The M-Method ( Example) Minimized Subject to:

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The M-Method ( Solution) Minimized Subject to: By subtracting surplus x 3 in second constraint and adding slack x 4 in third constraint, thus we get:

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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 R 1, R 2 and x 4 as the starting basic feasible solution.

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Solution x4x4 R2R2 R1R1 x3x3 x2x2 x1x1 Basic 00-M 0-4z 3001013R1R1 601034R2R2 4100021x4 New z-row = Old z-row + M* R 1 -row + M*R 2 -row Solution x4x4 R2R2 R1R1 x3x3 x2x2 x1x1 Basic 9M000-M-1+4M-4+7Mz 3001013R1R1 601034R2R2 4100021x4 Artificial variables become zero

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Solution x4x4 R2R2 R1R1 x3x3 x2x2 x1x1 Basic 4+2M00(4-7M)/3-M(1+5M)/30z 1001/30 1x1x1 201-4/35/30R2R2 310-1/305/30x4 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: 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:

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Observation regarding the Use of M-method: 1.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.

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Example: Maximize Subject to: Using Computer solution, apply the simplex method M=10, and repeat it using M=999.999. the first M yields the correct solution x 1 =1 and x 2 =1.5, whereas the second gives the incorrect solution x 1 =4 and x 2 =0 Multiplying the objective function by 1000 to get z= 200x 1 + 500x 2 and solve the problem using M=10 and M=999.999 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.

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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. Two-phase method

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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 Note : Example Minimized Subject to:

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Solution: Phase I: Minimize: Subject to:

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Solution x4x4 R2R2 R1R1 x3x3 x2x2 x1x1 Basic 00 000r 3001013R1R1 6010 34R2R2 4100021x4 New r-row = Old r-row + 1* R 1 -row + R 2 -row Solution x4x4 R2R2 R1R1 x3x3 x2x2 x1x1 Basic 900047r 3001013R1R1 6010 34R2R2 4100021x4

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Solution x4x4 R2R2 R1R1 x3x3 x2x2 x1x1 Basic 00 000r 3/50-1/53/51/501x1x1 6/503/5-4/5-3/510x2x2 111100x4 By using new r-row, we solve Phase I of the problem which yields the following optimum tableau Because minimum r=0, Phase I produces the basic feasible solution:

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Phase II After eliminating artificial variables column, the original problem can be written as: Minimize: Subject to:

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Solution x4x4 x3x3 x2x2 x1x1 Basic 000-4z 3/501/501x1x1 6/50-3/510x2x2 11100x4 Again, because basic variables x 1 and x 2 have nonzero coefficient in he z row, they must be substituted out, using the following computation: New z-row = Old z-row + 4* x 1 -row + 1*x 2 -row

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Solution x4x4 x3x3 x2x2 x1x1 Basic 18/501/500z 3/501/501x1x1 6/50-3/510x2x2 11100x4 The initial tableau of Phase II is as the following:

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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. Remarks:

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