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simplex method-maths 4 mumbai university

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The document explains the simplex method for solving linear programming problems when there are more than two decision variables, detailing the roles of slack and surplus variables, as well as basic solutions and basic feasible solutions. It outlines the steps to convert inequalities into equalities, find initial feasible solutions, and perform iterations to reach optimal solutions through a series of examples. Additionally, it introduces the concept of artificial variables in cases where certain constraints lead to infeasibility.

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simplex method-maths 4 mumbai university
Simplex Method •When decision variables are more than 2, we always use Simplex Method •Slack Variable: Variable added to a  constraint to convert it to an equation (=). • A slack variable represents unused resources • A slack variable contributes nothing to the objective function value. •Surplus Variable: Variable subtracted from a  constraint to convert it to an equation (=). • A surplus variable represents an excess above a constraint requirement level. • Surplus variables contribute nothing to the calculated value of the objective function.
Cont…. Basic Solution(BS) : This solution is obtained by setting any n variables (among m+n variables) equal to zero and solving for remaining m variables, provided the determinant of the coefficients of these variables is non-zero. Such m variables are called basic variables and remaining n zero valued variables are called non basic variables. Basic Feasible Solution(BFS) : It is a basic solution which also satisfies the non negativity restrictions.
Cont….. BFS are of two types: ◦ Degenerate BFS: If one or more basic variables are zero. ◦ Non-Degenerate BFS: All basic variables are non-zero. Optimal BFS: BFS which optimizes the objective function.
Example Max. Z = 13x1+11x2 Subject to constraints: 4x1+5x2 < 1500 5x1+3x2 < 1575 x1+2x2 < 420 x1, x2 > 0
Solution : Step 1: Convert all the inequality constraints into equalities by the use of slack variables. Let S1, S2 , S3 be three slack variables. Introducing these slack variables into the inequality constraints and rewriting the objective function such that all variables are on the left-hand side of the equation. Model can rewritten as: Z - 13x1 -11x2 = 0 Subject to constraints: 4x1+5x2 + S1 = 1500 5x1+3x2 +S2= 1575 x1+2x2 +S3 = 420 x1, x2, S1, S2, S3 > 0
Cont… Step II: Find the Initial BFS. One Feasible solution that satisfies all the constraints is: x1= 0, x2= 0, S1= 1500, S2= 1575, S3= 420 and Z=0. Now, S1, S2, S3 are Basic variables. Step III: Set up an initial table as:
Cont… Row NO. Basic Variable Coefficients of: Sol. Rati o Z x1 x2 S1 S2 S3 A1 Z 1 -13 -11 0 0 0 0 B1 S1 0 4 5 1 0 0 1500 375 C1 S2 0 5 3 0 1 0 1575 315 D1 S3 0 1 2 0 0 1 420 420 Step IV: a) Choose the most negative number from row A1(i.e Z row). Therefore, x1 is a entering variable. b) Calculate Ratio = Sol col. / x1 col. (x1 > 0) c) Choose minimum Ratio. That variable(i.e S2) is a departing variable.
Cont…. Step V: x1 becomes basic variable and S2 becomes non basic variable. New table is: Row NO. Basic Varia ble Coefficients of: Sol. Ratio Z x1 x2 S1 S2 S3 A1 Z 1 0 -16/5 0 13/5 0 4095 B1 S1 0 0 13/5 1 -4/5 0 240 92.3 C1 x1 0 1 3/5 0 1/5 0 315 525 D1 S3 0 0 7/5 0 -1/5 1 105 75
Cont… Next Table is : Row NO. Basic Variab le Coefficients of: Sol. Z x1 x2 S1 S2 S3 A1 Z 1 0 0 0 15/7 16/7 4335 B1 S1 0 0 0 1 -3/7 -13/7 45 C1 x1 0 1 0 0 2/7 -3/7 270 D1 x2 0 0 1 0 -1/7 5/7 75 Optimal Solution is : x1= 270, x2= 75, Z= 4335
Example Max. Z = 3x1+5x2+4x3 Subject to constraints: 2x1+3x2 < 8 2x2+5x3 < 10 3x1+2x2+4x3 < 15 x1, x2, x3 > 0
Cont… Let S1, S2, S3 be the three slack variables. Modified form is: Z - 3x1-5x2-4x3 =0 2x1+3x2 +S1= 8 2x2+5x3 +S2= 10 3x1+2x2+4x3+S3= 15 x1, x2, x3, S1, S2, S3 > 0 Initial BFS is : x1= 0, x2= 0, x3=0, S1= 8, S2= 10, S3= 15 and Z=0.
Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 x3 S1 S2 S3 Z 1 -3 -5 -4 0 0 0 0 S1 0 2 3 0 1 0 0 8 8/3 S2 0 0 2 5 0 1 0 10 5 S3 0 3 2 4 0 0 1 15 15/2 Therefore, x2 is the entering variable and S1 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 x3 S1 S2 S3 Z 1 1/3 0 -4 5/3 0 0 40/3 x2 0 2/3 1 0 1/3 0 0 8/3 - S2 0 -4/3 0 5 -2/3 1 0 14/3 14/15 S3 0 5/3 0 4 -2/3 0 1 29/3 29/12 Therefore, x3 is the entering variable and S2 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 x3 S1 S2 S3 Z 1 -11/15 0 0 17/15 4/5 0 256/15 x2 0 2/3 1 0 1/3 0 0 8/3 4 x3 0 -4/15 0 1 -2/15 1/5 0 14/15 - S3 0 41/15 0 0 2/15 -4/5 1 89/15 89/41 Therefore, x1 is the entering variable and S3 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Z x1 x2 x3 S1 S2 S3 Z 1 0 0 0 45/41 24/41 11/41 765/41 x2 0 0 1 0 15/41 8/41 -10/41 50/41 x3 0 0 0 1 -6/41 5/41 4/41 62/41 x1 0 1 0 0 -2/41 -12/41 15/41 89/41 Optimal Solution is : x1= 89/41, x2= 50/41, x3=62/41, Z= 765/41
Example Min.. Z = x1 - 3x2 + 2x3 Subject to constraints: 3x1 - x2 + 3x3 < 7 -2x1 + 4x2 < 12 -4x1 + 3x2 + 8x3 < 10 x1, x2, x3 > 0
Cont… Convert the problem into maximization problem Max.. Z’ = -x1 + 3x2 - 2x3 where Z’= -Z Subject to constraints: 3x1 - x2 + 3x3 < 7 -2x1 + 4x2 < 12 -4x1 + 3x2 + 8x3 < 10 x1, x2, x3 > 0
Cont… Let S1, S2 and S3 be three slack variables. Modified form is: Z’ + x1 - 3x2 + 2x3 = 0 3x1 - x2 + 3x3 +S1 = 7 -2x1 + 4x2 + S2 = 12 -4x1 + 3x2 + 8x3 +S3 = 10 x1, x2, x3 > 0 Initial BFS is : x1= 0, x2= 0, x3=0, S1= 7, S2= 12, S3 = 10 and Z=0.
Cont… Basic Variable Coefficients of: Sol. Ratio Z’ x1 x2 x3 S1 S2 S3 Z’ 1 1 -3 2 0 0 0 0 S1 0 3 -1 3 1 0 0 7 - S2 0 -2 4 0 0 1 0 12 3 S3 0 -4 3 8 0 0 1 10 10/3 Therefore, x2 is the entering variable and S2 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Ratio Z’ x1 x2 x3 S1 S2 S3 Z’ 1 -1/2 0 2 0 3/4 0 9 S1 0 5/2 0 3 1 1/4 0 10 4 x2 0 -1/2 1 0 0 1/4 0 3 - S3 0 -5/2 0 8 0 -3/4 1 1 - Therefore, x1 is the entering variable and S1 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Z’ x1 x2 x3 S1 S2 S3 Z’ 1 0 0 13/5 1/5 8/10 0 11 x1 0 1 0 6/5 2/5 1/10 0 4 x2 0 0 1 3/5 1/5 3/10 0 5 S3 0 0 0 11 1 -1/2 1 11 Optimal Solution is : x1= 4, x2= 5, x3= 0, Z’ = 11 Z = -11
Example Max.. Z = 3x1 + 4x2 Subject to constraints: x1 - x2 < 1 -x1 + x2 < 2 x1, x2 > 0
Cont… Let S1 and S2 be two slack variables . Modified form is: Z -3x1 - 4x2 = 0 x1 - x2 +S1 = 1 -x1 + x2 +S2 = 2 x1, x2, S1, S2 > 0 Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2 and Z=0.
Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 S1 S2 Z 1 -3 -4 0 0 0 S1 0 1 -1 1 0 1 - S2 0 -1 1 0 1 2 2 Therefore, x2 is the entering variable and S2 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 S1 S2 Z 1 -7 0 0 4 8 S1 0 0 0 1 1 3 - x2 0 -1 1 0 1 2 - x1 is the entering variable, but as in x1 column every no. is less than equal to zero, ratio cannot be calculated. Therefore given problem is having a unbounded solution.
simplex method-maths 4 mumbai university
Introduction LPP, in which constraints may also have > and = signs, we introduce a new type of variable , called the artificial variable. These variables are fictitious and cannot have any physical meaning. Two Phase Simplex Method is used to solve a problem in which some artificial variables are involved. The solution is obtained in two phases.
Example Min.. Z = 15/2 x1 - 3x2 Subject to constraints: 3x1 - x2 - x3 > 3 x1 - x2 + x3 > 2 x1, x2, x3 > 0
Cont… Convert the objective function into the maximization form Max. Z’ = -15/2 x1 + 3x2 where Z’= -Z Subject to constraints: 3x1 - x2 - x3 > 3 x1 - x2 + x3 > 2 x1, x2, x3 > 0
Cont… Modified form is : Introduce surplus variables S1 and S2, and artificial variables a1 and a2 Z’ + 15/2 x1 - 3x2 = 0 Subject to constraints: 3x1 - x2 - x3 –S1 + a1 = 3 x1 - x2 + x3 –S2 + a2 = 2 x1, x2, x3 , S1, S2, a1, a2 > 0
Cont… Phase I : Simplex method is applied to a specially constructed Auxiliary LPP leading to a final simplex table containing a BFS to the original problem. •Step 1: Assign a cost –1 to each artificial variable and a cost 0 to all other variables in the objective function. •Step 2: Construct the auxiliary LPP in which the new objective function Z* is to be maximized subject to the given set of constraints.
Cont… Max. Z* = -a1 –a2 Z* + a1 + a2 = 0 Subject to constraints: 3x1 - x2 - x3 –S1 + a1 = 3 x1 - x2 + x3 –S2 + a2 = 2 x1, x2, x3 , S1, S2, a1, a2 > 0 Auxiliary LPP is: Initial solution is a1 = 3, a2 = 2 and Z* = 0
Cont… •Step 3: Solve the auxiliary problem by simplex method until either of the following three possibilities arise: •Max Z* < 0 and at least one artificial variable appear in the optimum basis at a positive level. In this case given problem does not have any feasible solution. •Max Z* = 0 and at least one artificial variable appear in the optimum basis at a zero level. In this case proceed to Phase II. •Max Z* = 0 and no artificial variable appear in the optimum basis. In this case also proceed to Phase II.
Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 x3 S1 S2 a1 a2 Z* 1 0 0 0 0 0 1 1 0 a1 0 3 -1 -1 -1 0 1 0 3 a2 0 1 -1 1 0 -1 0 1 2 This table is not feasible as a1 and a2 has non zero coefficients in Z* row. Therefore next step is to make the table feasible.
Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 x3 S1 S2 a1 a2 Z* 1 -4 2 0 1 1 0 0 -5 a1 0 3 -1 -1 -1 0 1 0 3 1 a2 0 1 -1 1 0 -1 0 1 2 2 Therefore, x1 is the entering variable and a1 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 x3 S1 S2 a2 Z* 1 0 2/3 -4/3 -1/3 1 0 -1 x1 0 1 -1/3 -1/3 -1/3 0 0 1 - a2 0 0 -2/3 4/3 1/3 -1 1 1 3/4 Therefore, x3 is the entering variable and a2 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 x3 S1 S2 Z* 1 0 0 0 0 0 0 x1 0 1 -1/2 0 -1/4 -1/4 5/4 x3 0 0 -1/2 1 1/4 -3/4 3/4 As there is no variable to be entered in the basis, this table is optimum for Phase I. In this table Max. Z* = 0 and no artificial variable appears in the optimum basis, therefore we can proceed to Phase II.
Cont… Phase II: The artificial variables which are non basic at the end of Phase I are removed from the table and as well as from the objective function and constraints. Now assign the actual costs to the variables in the Objective function. That is, Simplex method is applied to the modified simplex table obtained at the Phase I. Basic Variable Coefficients of: Sol. Z’ x1 x2 x3 S1 S2 Z’ 1 15/2 -3 0 0 0 0 x1 0 1 -1/2 0 -1/4 -1/4 5/4 x3 0 0 -1/2 1 1/4 -3/4 3/4 Again this table is not feasible as basic variable x1 has a non zero coefficient in Z’ row. So make the table feasible.
Cont… Basic Variable Coefficients of: Sol. Z’ x1 x2 x3 S1 S2 Z’ 1 0 3/4 0 15/8 15/8 -75/8 x1 0 1 -1/2 0 -1/4 -1/4 5/4 x3 0 0 -1/2 1 1/4 -3/4 3/4 Optimal Solution is : x1= 5/4, x2= 0, x3= 3/4, Z’ = -75/8 Z = 75/8
Example Min.. Z = x1 - 2x2 –3x3 Subject to constraints: -2x1 + x2 + 3x3 = 2 2x1 + 3x2 + 4x3 = 1 x1, x2, x3 > 0
Cont… Phase I: Introducing artificial variables a1 and a2 Auxiliary LPP is: Max. Z* = -a1 - a2 Z* + a1 + a2 = 0 Subject to constraints: -2x1 + x2 + 3x3 + a1 = 2 2x1 + 3x2 + 4x3 + a2 = 1 x1, x2, x3 > 0
Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 x3 a1 a2 Z* 1 0 0 0 1 1 0 a1 0 -2 1 3 1 0 2 a2 0 2 3 4 0 1 1 This table is not feasible as a1 and a2 has non zero coefficients in Z* row. Therefore next step is to make the table feasible.
Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 x3 a1 a2 Z* 1 0 -4 -7 0 0 -3 a1 0 -2 1 3 1 0 2 2/3 a2 0 2 3 4 0 1 1 1/4 Therefore, x3 is the entering variable and a2 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 x3 a1 Z* 1 7/4 5/4 0 0 -5/4 a1 0 -7/2 -5/4 0 1 5/4 x3 0 1/2 3/4 1 0 1/4 As there is no variable to be entered in the basis, therefore Phase I ends here. But one artificial variable is present in the basis and Z* < 0. Therefore we cannot proceed to Phase II. Given problem is having a non-feasible solution.
Example Min.. Z = 4x1 + x2 Subject to constraints: 3x1 + x2 = 3 4x1 + 3x2 > 6 x1 +2x2 < 4 x1, x2 > 0
Cont… Phase I: Introducing artificial variable a1 and a2, surplus variable S1 and slack variable S2 Auxiliary LPP is: Max. Z* = -a1 - a2 Z* + a1 + a2 = 0 Subject to constraints: 3x1 + x2 +a1 = 3 4x1 + 3x2 –S1 + a2 = 6 x1 +2x2 +S2 = 4 x1, x2 > 0
Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 S1 S2 a1 a2 Z* 1 0 0 0 0 1 1 0 a1 0 3 1 0 0 1 0 3 a2 0 4 3 -1 0 0 1 6 S2 0 1 2 0 1 0 0 4 This table is not feasible as a1 and a2 has non zero coefficients in Z* row. Therefore next step is to make the table feasible.
Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 S1 S2 a1 a2 Z* 1 -7 -4 1 0 0 0 -9 a1 0 3 1 0 0 1 0 3 1 a2 0 4 3 -1 0 0 1 6 3/2 S2 0 1 2 0 1 0 0 4 4 Therefore, x1 is the entering variable and a1 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 S1 S2 a2 Z* 1 0 -5/3 1 0 0 -2 x1 0 1 1/3 0 0 0 1 3 a2 0 0 5/3 -1 0 1 2 6/5 S2 0 0 5/3 0 1 0 3 9/5 Therefore, x2 is the entering variable and a2 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 S1 S2 Z* 1 0 0 0 0 0 x1 0 1 0 1/5 0 3/5 x2 0 0 1 -3/5 0 6/5 S2 0 0 0 1 1 1 As there is no variable to be entered in the basis, this table is optimum for Phase I. In this table Max. Z* = 0 and no artificial variable appears in the optimum basis, therefore we can proceed to Phase II.
Cont… Phase II: Original Objective function is: Min.. Z = 4x1 + x2 Convert the objective function into the maximization form Max. Z’ = -4 x1 - x2 where Z’= -Z Basic Variable Coefficients of: Sol. Z* x1 x2 S1 S2 Z* 1 4 1 0 0 0 x1 0 1 0 1/5 0 3/5 x2 0 0 1 -3/5 0 6/5 S2 0 0 0 1 1 1 Again this table is not feasible as basic variable x1 and x2 has a non zero coefficient in Z’ row. So make the table feasible.
Cont… Basic Variable Coefficients of: Sol. Ratio Z’ x1 x2 S1 S2 Z’ 1 0 0 -1/5 0 -18/5 x1 0 1 0 1/5 0 3/5 3 x2 0 0 1 -3/5 0 6/5 - S2 0 0 0 1 1 1 1 Therefore, S1 is the entering variable and S2 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Z’ x1 x2 S1 S2 Z’ 1 0 0 0 1/5 -17/5 x1 0 1 0 0 -1/5 2/5 x2 0 0 1 0 3/5 9/5 S2 0 0 0 1 1 1 Optimal Solution is : x1= 2/5, x2= 9/5, Z’ = -17/5 Z = 17/5
simplex method-maths 4 mumbai university
Introduction At the stage of improving the solution during Simplex procedure, if a tie for the minimum ratio occurs at least one basic variable becomes equal to zero in the next iteration and the new solution is said to be Degenerate.
Example Max.. Z = 3x1 + 9x2 Subject to constraints: x1 + 4x2 < 8 x1 + 2x2 < 4 x1, x2 > 0
Cont… Let S1 and S2 be two slack variables. Modified form is: Z - 3x1 - 9x2 = 0 x1 + 4x2 + S1 = 8 x1 + 2x2 +S2 = 4 x1, x2, S1, S2 > 0 Initial BFS is : x1= 0, x2= 0, S1= 8, S2= 4 and Z=0.
Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 S1 S2 Z 1 -3 -9 0 0 0 S1 0 1 4 1 0 8 2 S2 0 1 2 0 1 4 2 In this table S1 and S2 tie for the leaving variable. So any one can be considered as leaving variable. Therefore, x2 is the entering variable and S1 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 S1 S2 Z 1 -3/4 0 9/4 0 18 x2 0 1/4 1 1/4 0 2 8 S2 0 1/2 0 -1/2 1 0 0 Therefore, x1 is the entering variable and S2 is the departing variable.
Cont… Basic Variable Coefficients of: Sol. Z x1 x2 S1 S2 Z 1 0 0 3/2 3/2 18 x2 0 0 1 1/2 -1/2 2 x1 0 1 0 -1 2 0 Optimal Solution is : x1= 0, x2= 2 Z = 18 It results in a Degenerate Basic Solution.

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simplex method-maths 4 mumbai university

  • 2. Simplex Method •When decision variables are more than 2, we always use Simplex Method •Slack Variable: Variable added to a  constraint to convert it to an equation (=). • A slack variable represents unused resources • A slack variable contributes nothing to the objective function value. •Surplus Variable: Variable subtracted from a  constraint to convert it to an equation (=). • A surplus variable represents an excess above a constraint requirement level. • Surplus variables contribute nothing to the calculated value of the objective function.
  • 3. Cont…. Basic Solution(BS) : This solution is obtained by setting any n variables (among m+n variables) equal to zero and solving for remaining m variables, provided the determinant of the coefficients of these variables is non-zero. Such m variables are called basic variables and remaining n zero valued variables are called non basic variables. Basic Feasible Solution(BFS) : It is a basic solution which also satisfies the non negativity restrictions.
  • 4. Cont….. BFS are of two types: ◦ Degenerate BFS: If one or more basic variables are zero. ◦ Non-Degenerate BFS: All basic variables are non-zero. Optimal BFS: BFS which optimizes the objective function.
  • 5. Example Max. Z = 13x1+11x2 Subject to constraints: 4x1+5x2 < 1500 5x1+3x2 < 1575 x1+2x2 < 420 x1, x2 > 0
  • 6. Solution : Step 1: Convert all the inequality constraints into equalities by the use of slack variables. Let S1, S2 , S3 be three slack variables. Introducing these slack variables into the inequality constraints and rewriting the objective function such that all variables are on the left-hand side of the equation. Model can rewritten as: Z - 13x1 -11x2 = 0 Subject to constraints: 4x1+5x2 + S1 = 1500 5x1+3x2 +S2= 1575 x1+2x2 +S3 = 420 x1, x2, S1, S2, S3 > 0
  • 7. Cont… Step II: Find the Initial BFS. One Feasible solution that satisfies all the constraints is: x1= 0, x2= 0, S1= 1500, S2= 1575, S3= 420 and Z=0. Now, S1, S2, S3 are Basic variables. Step III: Set up an initial table as:
  • 8. Cont… Row NO. Basic Variable Coefficients of: Sol. Rati o Z x1 x2 S1 S2 S3 A1 Z 1 -13 -11 0 0 0 0 B1 S1 0 4 5 1 0 0 1500 375 C1 S2 0 5 3 0 1 0 1575 315 D1 S3 0 1 2 0 0 1 420 420 Step IV: a) Choose the most negative number from row A1(i.e Z row). Therefore, x1 is a entering variable. b) Calculate Ratio = Sol col. / x1 col. (x1 > 0) c) Choose minimum Ratio. That variable(i.e S2) is a departing variable.
  • 9. Cont…. Step V: x1 becomes basic variable and S2 becomes non basic variable. New table is: Row NO. Basic Varia ble Coefficients of: Sol. Ratio Z x1 x2 S1 S2 S3 A1 Z 1 0 -16/5 0 13/5 0 4095 B1 S1 0 0 13/5 1 -4/5 0 240 92.3 C1 x1 0 1 3/5 0 1/5 0 315 525 D1 S3 0 0 7/5 0 -1/5 1 105 75
  • 10. Cont… Next Table is : Row NO. Basic Variab le Coefficients of: Sol. Z x1 x2 S1 S2 S3 A1 Z 1 0 0 0 15/7 16/7 4335 B1 S1 0 0 0 1 -3/7 -13/7 45 C1 x1 0 1 0 0 2/7 -3/7 270 D1 x2 0 0 1 0 -1/7 5/7 75 Optimal Solution is : x1= 270, x2= 75, Z= 4335
  • 11. Example Max. Z = 3x1+5x2+4x3 Subject to constraints: 2x1+3x2 < 8 2x2+5x3 < 10 3x1+2x2+4x3 < 15 x1, x2, x3 > 0
  • 12. Cont… Let S1, S2, S3 be the three slack variables. Modified form is: Z - 3x1-5x2-4x3 =0 2x1+3x2 +S1= 8 2x2+5x3 +S2= 10 3x1+2x2+4x3+S3= 15 x1, x2, x3, S1, S2, S3 > 0 Initial BFS is : x1= 0, x2= 0, x3=0, S1= 8, S2= 10, S3= 15 and Z=0.
  • 13. Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 x3 S1 S2 S3 Z 1 -3 -5 -4 0 0 0 0 S1 0 2 3 0 1 0 0 8 8/3 S2 0 0 2 5 0 1 0 10 5 S3 0 3 2 4 0 0 1 15 15/2 Therefore, x2 is the entering variable and S1 is the departing variable.
  • 14. Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 x3 S1 S2 S3 Z 1 1/3 0 -4 5/3 0 0 40/3 x2 0 2/3 1 0 1/3 0 0 8/3 - S2 0 -4/3 0 5 -2/3 1 0 14/3 14/15 S3 0 5/3 0 4 -2/3 0 1 29/3 29/12 Therefore, x3 is the entering variable and S2 is the departing variable.
  • 15. Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 x3 S1 S2 S3 Z 1 -11/15 0 0 17/15 4/5 0 256/15 x2 0 2/3 1 0 1/3 0 0 8/3 4 x3 0 -4/15 0 1 -2/15 1/5 0 14/15 - S3 0 41/15 0 0 2/15 -4/5 1 89/15 89/41 Therefore, x1 is the entering variable and S3 is the departing variable.
  • 16. Cont… Basic Variable Coefficients of: Sol. Z x1 x2 x3 S1 S2 S3 Z 1 0 0 0 45/41 24/41 11/41 765/41 x2 0 0 1 0 15/41 8/41 -10/41 50/41 x3 0 0 0 1 -6/41 5/41 4/41 62/41 x1 0 1 0 0 -2/41 -12/41 15/41 89/41 Optimal Solution is : x1= 89/41, x2= 50/41, x3=62/41, Z= 765/41
  • 17. Example Min.. Z = x1 - 3x2 + 2x3 Subject to constraints: 3x1 - x2 + 3x3 < 7 -2x1 + 4x2 < 12 -4x1 + 3x2 + 8x3 < 10 x1, x2, x3 > 0
  • 18. Cont… Convert the problem into maximization problem Max.. Z’ = -x1 + 3x2 - 2x3 where Z’= -Z Subject to constraints: 3x1 - x2 + 3x3 < 7 -2x1 + 4x2 < 12 -4x1 + 3x2 + 8x3 < 10 x1, x2, x3 > 0
  • 19. Cont… Let S1, S2 and S3 be three slack variables. Modified form is: Z’ + x1 - 3x2 + 2x3 = 0 3x1 - x2 + 3x3 +S1 = 7 -2x1 + 4x2 + S2 = 12 -4x1 + 3x2 + 8x3 +S3 = 10 x1, x2, x3 > 0 Initial BFS is : x1= 0, x2= 0, x3=0, S1= 7, S2= 12, S3 = 10 and Z=0.
  • 20. Cont… Basic Variable Coefficients of: Sol. Ratio Z’ x1 x2 x3 S1 S2 S3 Z’ 1 1 -3 2 0 0 0 0 S1 0 3 -1 3 1 0 0 7 - S2 0 -2 4 0 0 1 0 12 3 S3 0 -4 3 8 0 0 1 10 10/3 Therefore, x2 is the entering variable and S2 is the departing variable.
  • 21. Cont… Basic Variable Coefficients of: Sol. Ratio Z’ x1 x2 x3 S1 S2 S3 Z’ 1 -1/2 0 2 0 3/4 0 9 S1 0 5/2 0 3 1 1/4 0 10 4 x2 0 -1/2 1 0 0 1/4 0 3 - S3 0 -5/2 0 8 0 -3/4 1 1 - Therefore, x1 is the entering variable and S1 is the departing variable.
  • 22. Cont… Basic Variable Coefficients of: Sol. Z’ x1 x2 x3 S1 S2 S3 Z’ 1 0 0 13/5 1/5 8/10 0 11 x1 0 1 0 6/5 2/5 1/10 0 4 x2 0 0 1 3/5 1/5 3/10 0 5 S3 0 0 0 11 1 -1/2 1 11 Optimal Solution is : x1= 4, x2= 5, x3= 0, Z’ = 11 Z = -11
  • 23. Example Max.. Z = 3x1 + 4x2 Subject to constraints: x1 - x2 < 1 -x1 + x2 < 2 x1, x2 > 0
  • 24. Cont… Let S1 and S2 be two slack variables . Modified form is: Z -3x1 - 4x2 = 0 x1 - x2 +S1 = 1 -x1 + x2 +S2 = 2 x1, x2, S1, S2 > 0 Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2 and Z=0.
  • 25. Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 S1 S2 Z 1 -3 -4 0 0 0 S1 0 1 -1 1 0 1 - S2 0 -1 1 0 1 2 2 Therefore, x2 is the entering variable and S2 is the departing variable.
  • 26. Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 S1 S2 Z 1 -7 0 0 4 8 S1 0 0 0 1 1 3 - x2 0 -1 1 0 1 2 - x1 is the entering variable, but as in x1 column every no. is less than equal to zero, ratio cannot be calculated. Therefore given problem is having a unbounded solution.
  • 28. Introduction LPP, in which constraints may also have > and = signs, we introduce a new type of variable , called the artificial variable. These variables are fictitious and cannot have any physical meaning. Two Phase Simplex Method is used to solve a problem in which some artificial variables are involved. The solution is obtained in two phases.
  • 29. Example Min.. Z = 15/2 x1 - 3x2 Subject to constraints: 3x1 - x2 - x3 > 3 x1 - x2 + x3 > 2 x1, x2, x3 > 0
  • 30. Cont… Convert the objective function into the maximization form Max. Z’ = -15/2 x1 + 3x2 where Z’= -Z Subject to constraints: 3x1 - x2 - x3 > 3 x1 - x2 + x3 > 2 x1, x2, x3 > 0
  • 31. Cont… Modified form is : Introduce surplus variables S1 and S2, and artificial variables a1 and a2 Z’ + 15/2 x1 - 3x2 = 0 Subject to constraints: 3x1 - x2 - x3 –S1 + a1 = 3 x1 - x2 + x3 –S2 + a2 = 2 x1, x2, x3 , S1, S2, a1, a2 > 0
  • 32. Cont… Phase I : Simplex method is applied to a specially constructed Auxiliary LPP leading to a final simplex table containing a BFS to the original problem. •Step 1: Assign a cost –1 to each artificial variable and a cost 0 to all other variables in the objective function. •Step 2: Construct the auxiliary LPP in which the new objective function Z* is to be maximized subject to the given set of constraints.
  • 33. Cont… Max. Z* = -a1 –a2 Z* + a1 + a2 = 0 Subject to constraints: 3x1 - x2 - x3 –S1 + a1 = 3 x1 - x2 + x3 –S2 + a2 = 2 x1, x2, x3 , S1, S2, a1, a2 > 0 Auxiliary LPP is: Initial solution is a1 = 3, a2 = 2 and Z* = 0
  • 34. Cont… •Step 3: Solve the auxiliary problem by simplex method until either of the following three possibilities arise: •Max Z* < 0 and at least one artificial variable appear in the optimum basis at a positive level. In this case given problem does not have any feasible solution. •Max Z* = 0 and at least one artificial variable appear in the optimum basis at a zero level. In this case proceed to Phase II. •Max Z* = 0 and no artificial variable appear in the optimum basis. In this case also proceed to Phase II.
  • 35. Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 x3 S1 S2 a1 a2 Z* 1 0 0 0 0 0 1 1 0 a1 0 3 -1 -1 -1 0 1 0 3 a2 0 1 -1 1 0 -1 0 1 2 This table is not feasible as a1 and a2 has non zero coefficients in Z* row. Therefore next step is to make the table feasible.
  • 36. Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 x3 S1 S2 a1 a2 Z* 1 -4 2 0 1 1 0 0 -5 a1 0 3 -1 -1 -1 0 1 0 3 1 a2 0 1 -1 1 0 -1 0 1 2 2 Therefore, x1 is the entering variable and a1 is the departing variable.
  • 37. Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 x3 S1 S2 a2 Z* 1 0 2/3 -4/3 -1/3 1 0 -1 x1 0 1 -1/3 -1/3 -1/3 0 0 1 - a2 0 0 -2/3 4/3 1/3 -1 1 1 3/4 Therefore, x3 is the entering variable and a2 is the departing variable.
  • 38. Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 x3 S1 S2 Z* 1 0 0 0 0 0 0 x1 0 1 -1/2 0 -1/4 -1/4 5/4 x3 0 0 -1/2 1 1/4 -3/4 3/4 As there is no variable to be entered in the basis, this table is optimum for Phase I. In this table Max. Z* = 0 and no artificial variable appears in the optimum basis, therefore we can proceed to Phase II.
  • 39. Cont… Phase II: The artificial variables which are non basic at the end of Phase I are removed from the table and as well as from the objective function and constraints. Now assign the actual costs to the variables in the Objective function. That is, Simplex method is applied to the modified simplex table obtained at the Phase I. Basic Variable Coefficients of: Sol. Z’ x1 x2 x3 S1 S2 Z’ 1 15/2 -3 0 0 0 0 x1 0 1 -1/2 0 -1/4 -1/4 5/4 x3 0 0 -1/2 1 1/4 -3/4 3/4 Again this table is not feasible as basic variable x1 has a non zero coefficient in Z’ row. So make the table feasible.
  • 40. Cont… Basic Variable Coefficients of: Sol. Z’ x1 x2 x3 S1 S2 Z’ 1 0 3/4 0 15/8 15/8 -75/8 x1 0 1 -1/2 0 -1/4 -1/4 5/4 x3 0 0 -1/2 1 1/4 -3/4 3/4 Optimal Solution is : x1= 5/4, x2= 0, x3= 3/4, Z’ = -75/8 Z = 75/8
  • 41. Example Min.. Z = x1 - 2x2 –3x3 Subject to constraints: -2x1 + x2 + 3x3 = 2 2x1 + 3x2 + 4x3 = 1 x1, x2, x3 > 0
  • 42. Cont… Phase I: Introducing artificial variables a1 and a2 Auxiliary LPP is: Max. Z* = -a1 - a2 Z* + a1 + a2 = 0 Subject to constraints: -2x1 + x2 + 3x3 + a1 = 2 2x1 + 3x2 + 4x3 + a2 = 1 x1, x2, x3 > 0
  • 43. Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 x3 a1 a2 Z* 1 0 0 0 1 1 0 a1 0 -2 1 3 1 0 2 a2 0 2 3 4 0 1 1 This table is not feasible as a1 and a2 has non zero coefficients in Z* row. Therefore next step is to make the table feasible.
  • 44. Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 x3 a1 a2 Z* 1 0 -4 -7 0 0 -3 a1 0 -2 1 3 1 0 2 2/3 a2 0 2 3 4 0 1 1 1/4 Therefore, x3 is the entering variable and a2 is the departing variable.
  • 45. Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 x3 a1 Z* 1 7/4 5/4 0 0 -5/4 a1 0 -7/2 -5/4 0 1 5/4 x3 0 1/2 3/4 1 0 1/4 As there is no variable to be entered in the basis, therefore Phase I ends here. But one artificial variable is present in the basis and Z* < 0. Therefore we cannot proceed to Phase II. Given problem is having a non-feasible solution.
  • 46. Example Min.. Z = 4x1 + x2 Subject to constraints: 3x1 + x2 = 3 4x1 + 3x2 > 6 x1 +2x2 < 4 x1, x2 > 0
  • 47. Cont… Phase I: Introducing artificial variable a1 and a2, surplus variable S1 and slack variable S2 Auxiliary LPP is: Max. Z* = -a1 - a2 Z* + a1 + a2 = 0 Subject to constraints: 3x1 + x2 +a1 = 3 4x1 + 3x2 –S1 + a2 = 6 x1 +2x2 +S2 = 4 x1, x2 > 0
  • 48. Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 S1 S2 a1 a2 Z* 1 0 0 0 0 1 1 0 a1 0 3 1 0 0 1 0 3 a2 0 4 3 -1 0 0 1 6 S2 0 1 2 0 1 0 0 4 This table is not feasible as a1 and a2 has non zero coefficients in Z* row. Therefore next step is to make the table feasible.
  • 49. Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 S1 S2 a1 a2 Z* 1 -7 -4 1 0 0 0 -9 a1 0 3 1 0 0 1 0 3 1 a2 0 4 3 -1 0 0 1 6 3/2 S2 0 1 2 0 1 0 0 4 4 Therefore, x1 is the entering variable and a1 is the departing variable.
  • 50. Cont… Basic Variable Coefficients of: Sol. Ratio Z* x1 x2 S1 S2 a2 Z* 1 0 -5/3 1 0 0 -2 x1 0 1 1/3 0 0 0 1 3 a2 0 0 5/3 -1 0 1 2 6/5 S2 0 0 5/3 0 1 0 3 9/5 Therefore, x2 is the entering variable and a2 is the departing variable.
  • 51. Cont… Basic Variable Coefficients of: Sol. Z* x1 x2 S1 S2 Z* 1 0 0 0 0 0 x1 0 1 0 1/5 0 3/5 x2 0 0 1 -3/5 0 6/5 S2 0 0 0 1 1 1 As there is no variable to be entered in the basis, this table is optimum for Phase I. In this table Max. Z* = 0 and no artificial variable appears in the optimum basis, therefore we can proceed to Phase II.
  • 52. Cont… Phase II: Original Objective function is: Min.. Z = 4x1 + x2 Convert the objective function into the maximization form Max. Z’ = -4 x1 - x2 where Z’= -Z Basic Variable Coefficients of: Sol. Z* x1 x2 S1 S2 Z* 1 4 1 0 0 0 x1 0 1 0 1/5 0 3/5 x2 0 0 1 -3/5 0 6/5 S2 0 0 0 1 1 1 Again this table is not feasible as basic variable x1 and x2 has a non zero coefficient in Z’ row. So make the table feasible.
  • 53. Cont… Basic Variable Coefficients of: Sol. Ratio Z’ x1 x2 S1 S2 Z’ 1 0 0 -1/5 0 -18/5 x1 0 1 0 1/5 0 3/5 3 x2 0 0 1 -3/5 0 6/5 - S2 0 0 0 1 1 1 1 Therefore, S1 is the entering variable and S2 is the departing variable.
  • 54. Cont… Basic Variable Coefficients of: Sol. Z’ x1 x2 S1 S2 Z’ 1 0 0 0 1/5 -17/5 x1 0 1 0 0 -1/5 2/5 x2 0 0 1 0 3/5 9/5 S2 0 0 0 1 1 1 Optimal Solution is : x1= 2/5, x2= 9/5, Z’ = -17/5 Z = 17/5
  • 56. Introduction At the stage of improving the solution during Simplex procedure, if a tie for the minimum ratio occurs at least one basic variable becomes equal to zero in the next iteration and the new solution is said to be Degenerate.
  • 57. Example Max.. Z = 3x1 + 9x2 Subject to constraints: x1 + 4x2 < 8 x1 + 2x2 < 4 x1, x2 > 0
  • 58. Cont… Let S1 and S2 be two slack variables. Modified form is: Z - 3x1 - 9x2 = 0 x1 + 4x2 + S1 = 8 x1 + 2x2 +S2 = 4 x1, x2, S1, S2 > 0 Initial BFS is : x1= 0, x2= 0, S1= 8, S2= 4 and Z=0.
  • 59. Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 S1 S2 Z 1 -3 -9 0 0 0 S1 0 1 4 1 0 8 2 S2 0 1 2 0 1 4 2 In this table S1 and S2 tie for the leaving variable. So any one can be considered as leaving variable. Therefore, x2 is the entering variable and S1 is the departing variable.
  • 60. Cont… Basic Variable Coefficients of: Sol. Ratio Z x1 x2 S1 S2 Z 1 -3/4 0 9/4 0 18 x2 0 1/4 1 1/4 0 2 8 S2 0 1/2 0 -1/2 1 0 0 Therefore, x1 is the entering variable and S2 is the departing variable.
  • 61. Cont… Basic Variable Coefficients of: Sol. Z x1 x2 S1 S2 Z 1 0 0 3/2 3/2 18 x2 0 0 1 1/2 -1/2 2 x1 0 1 0 -1 2 0 Optimal Solution is : x1= 0, x2= 2 Z = 18 It results in a Degenerate Basic Solution.