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Simplex

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Simplex

1. 1. Several complications can occur while solving the LPP. Suchproblems are: Tie for the key row(degeneracy) Tie for the key column Unbounded problems Multiple optimal solutions Infeasible problems Redundant constraints Unrestricted Variables
2. 2.  Degeneracy occurs when there is tie for the minimum ratio(MR) for choosing the departing variable Basic Solution X1 X2 S1 S2 Min. variables variables ratio S1 20 4 9 1 0 5 S2 10 2 7 0 1 5 Zj 0 0 0 0 0 Tie of key Cj-Zj 5 3 0 0 row Key column
3. 3.  Find the coefficient of the slack variables and divide each coefficient by the corresponding positive numbers of the key column in the row, starting from left to right in order to break the tie X1 is replaced by S1 S1 S2 Lowest element 1/4= 0.25 0/4= 0 and S1 row is a key row 0/2= 0 1/2= 0.5
4. 4.  If the ratio do not break the tie, find the similar ratio for the coefficient of decision variables Compare the resulting ratio, column by column Select the row which contains smallest ratio.This row becomes the key row After resolving of this tie, simplex method is applied to obtain the optimum solution
5. 5.  This problem can arise in case of tie between identical Cj-Zj values In such a situation selection for key column can be made arbitrarily There is no wrong choice, although selection of one variable may result in more iteration Regardless of which variable column is chosen the optimal solution will eventually be found
6. 6. Basic Solution X1 X2 S1 S2 Min. ratiovariable values S1 2 3 2 1 0 S2 10 4 6 0 1 Zj 0 0 0 0 0 Cj-Zj 4 4 0 0 Tie of key column Any one of the decision variable is selected
7. 7.  It can be stated that a key row cannot be selected because minimum ratio column contains negative or infinity(∞) the solution is unbounded Basic Solution X1 X2 S1 S2 Min. variable variable ratio X1 7 1 0 1 0 ∞ S2 1 0 -1 -1 1 -1 Zj 35 5 0 5 0 Unbounded solution Cj-Zj 0 4 -5 0 Key column
8. 8.  If the index row indicates the value of Cj-Zj for a non basic variable to be zero, there exists an alternative optimum solution. To find the alternative optimal solution, the non basic variable with the Cj-Zj value of zero, should be selected as an entering variable and the simplex steps continued.
9. 9. Basic Solution X1 X2 X3 S1 S2 Min. variable variable ratio X1 10 4 5 6 1 0 X3 12 9 8 2 0 1X2 is not Zj 56 35 34 18 2 3there i.e.multiple Cj-Zj -33 -31 2 -2 -3optimalsolution
10. 10.  This condition occurs when the problem has incompatible constraints In final simplex table, all Cj-Zj elements +ve or zero in case of minimization and –ve or zero in case of maximization And if the basic variable include artificial variable, then LPP got an infeasible solution
11. 11. Basic var. Sol. Value X1 X2 S1 S2 S3 A1 X2 10 0 1 3 0 1 0 A1 20 0 0 -4 -1 -1 1 X1 40 1 0 -2 0 -1 0 Zj 190M-20M 4 3 1+4M M M-1 -M Cj-Zj 0 0 -1-4M -M 1-M 0 Since Cj-Zj row contains all elements –ve or zero , we are having optimum solution. Since artificial variable is present as a basic variable the given problem has infeasible solution.
12. 12. Consider the constraints, 3X1 + 4X2 < 7 neglected 3X1 + 4X2 < 15The second constraint is less restrictive(because both theconstraints have same co-efficient and variable) than first one,and is not required. Normally redundant constraint does not poseany problem except the computational work is unnecessarilyincreased
13. 13.  It is that decision variable which does not carry any value. To solve this problem, the variable can take two values i.e. one +ve & one –ve because difference between these two same +ve and –ve value is zero All variables become non-negative in the system and problem is solved.
14. 14.  Example: Max Z = 8x1 – 4x2 4x1 + 5x2 ≤ 20 -x1 + 3x2 ≥ -23 when x1 ≥ 0 , x2 is unrestricted in signSolution : Firstly replace the unrestricted variable x2 x2 = x3 – x4
15. 15.  After replacing the x2 , Max Z = 8x1 – 4x3 + 4x4 4x1 + 5x3 – 5x4 ≤ 20 x1 – 3x3 + 3x4 ≤ 23; x1,x3,x4 ≥ 0 After that slack variables are added to the constraints , Max Z = 8x1 – 4x3 + 4x4 + 0S1 + 0S2 4x1 + 5x3 – 5x4 + S1 + 0S2 = 20 x1 – 3x3 + 3x4 + 0S1 + S2 = 23; x1,x3,x4,S1,S2 ≥ 0
16. 16. Cj 8 -4 4 0 0 Basic Solution X1 X3 X4 S1 S2 Min. variable variable ratio 0 S1 20 4 5 -5 1 0 5 Key 0 S2 23 1 -3 3 0 1 23 row Zj 0 0 0 0 0 0 Cj-Zj 8 -4 4 0 0 Key column
17. 17. Cj 8 -4 4 0 0 Basic Solution X1 X3 X4 S1 S2 Min. variable variable ratio8 X1 5 1 5/4 -5/4 1/4 0 -ve S2 18 0 -17/4 17/4* -1/4 1 72/170 Zj 40 8 10 -10 2 0 Cj-Zj 0 -14 14 -2 0
18. 18. Basic Solution X1 X3 X4 S1 S2 variable variable X1 175/17 1 0 0 3/17 5/17 X4 72/17 0 -1 1 -1/17 4/17 Zj 1688/17 8 -4 4 20/17 56/17 Cj-Zj 0 0 0 -20/17 -56/17Since Cj-Zj ≤ 0 . The optimal solution is obtained.Ans. x1 = 175/17 x2 = x3 – x4 = 0 – 72/17 = -72/17 Z = 1688/17