simplex meathod
TRANSCRIPT
PRESENTATION ON SPECIAL CASES OF SIMPLEX METHOD
SPECIAL CASES IN APPLYING SIMPLEX METHOD
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
TIE FOR THE KEY ROW
Degeneracy occurs when there is tie for the minimum ratio(MR) for choosing the departing variable
Basic variables
Solution variables
X1 X2 S1 S2 Min. ratio
S1 20 4 9 1 0 5
S2 10 2 7 0 1 5
Zj 0 0 0 0 0
Cj-Zj 5 3 0 0 Tie of key row
Key column
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
S1 S2
1/4= 0.25 0/4= 0
0/2= 0 1/2= 0.5
Lowest element and S1 row is a key row
X1 is replaced by S1
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
TIE FOR THE KEY COLUMN
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
Basic variable
Solution values
X1 X2 S1 S2 Min. ratio
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
UNBOUNDED PROBLEM
It can be stated that a key row cannot be selected because minimum ratio column contains negative or infinity(∞) the solution is unbounded
Basic variable
Solution variable
X1 X2 S1 S2 Min. ratio
X1 7 1 0 1 0 ∞
S2 1 0 -1 -1 1 -1
Zj 35 5 0 5 0
Cj-Zj 0 4 -5 0
Unbounded solution
Key column
MULTIPLE OPTIMAL SOLUTION
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.
Basic variable
Solution variable
X1 X2 X3 S1 S2 Min. ratio
X1 10 4 5 6 1 0
X3 12 9 8 2 0 1
Zj 56 35 34 18 2 3
Cj-Zj -33 -31 2 -2 -3X2 is not there i.e. multiple optimal solution
INFEASIBLE PROBLEMS
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
Basic var. Sol. Value X1 X2 S1 S2 S3 A1X2 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.
REDUNDANT CONSTRAINT
Consider the constraints, 3X1 + 4X2 < 7 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
neglected
UNRESTRICTED VARIABLES
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.
Example:Max Z = 8x1 – 4x24x1 + 5x2 ≤ 20-x1 + 3x2 ≥ -23
when x1 ≥ 0 , x2 is unrestricted in signSolution :
Firstly replace the unrestricted variable x2x2 = x3 – x4
After replacing the x2 , Max Z = 8x1 – 4x3 + 4x44x1 + 5x3 – 5x4 ≤ 20x1 – 3x3 + 3x4 ≤ 23; x1,x3,x4 ≥ 0
After that slack variables are added to the constraints ,Max Z = 8x1 – 4x3 + 4x4 + 0S1 + 0S24x1 + 5x3 – 5x4 + S1 + 0S2 = 20x1 – 3x3 + 3x4 + 0S1 + S2 = 23;x1,x3,x4,S1,S2 ≥ 0
Basic variable
Solution variable
X1 X3 X4 S1 S2 Min. ratio
S1 20 4 5 -5 1 0 5
S2 23 1 -3 3 0 1 23
Zj 0 0 0 0 0 0
Cj-Zj 8 -4 4 0 0
Cj 8 -4 4 0 0
0
0
Key column
Key row
Basic variable
Solution variable
X1 X3 X4 S1 S2 Min. ratio
X1 5 1 5/4 -5/4 1/4 0 -ve
S2 18 0 -17/4 17/4* -1/4 1 72/17
Zj 40 8 10 -10 2 0
Cj-Zj 0 -14 14 -2 0
Cj 8 -4 4 0 0
8
0
Basic variable
Solution variable
X1 X3 X4 S1 S2
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 -56/17
Since Cj-Zj ≤ 0 . The optimal solution is obtained.
Ans. x1 = 175/17x2 = x3 – x4 = 0 – 72/17 = -72/17 Z = 1688/17
THANKYOU
Presented by:-Astha Harpreet SinghShivaniShweta