Introduction to Simplex Method
It was invented by G. Danztig in 1947. The simplex method gives an algorithm, that is, a rule of procedure generally involving repetitive application of a prescribed operation, which is based on the basic theorem of linear programming.
The Simplex algorithm is an iterative method for resolving LP problems in a finite number of steps. It contains
Benefits
Computational Procedure of Simplex Method
Take an example
Maximize Z = 3x1 + 2x2
Subject to
x1 + x2 ≤ 4
x1 - x2 ≤ 2
& x1 ≥ 0, x2 ≥ 0
Answer
Step 1 - Note down or write the given GLPP in the form of SLPP
Maximize Z = 3x1 + 2x2 + 0s1 + 0s2
x1 + x2+ s1= 4
x1 - x2 + s2= 2
x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0
Step 2 - Then present the constraints in the form of matrix
Step 3 - Now construct the starting simplex table with the use of notations
Cj → 3 2 0 0
Basic Variables
CB XB
X1 X2 S1 S2
Min ratio
XB /Xk
s1
s2
0 4
0 2
1 1 1 0
1 -1 0 1
Z= CB XB
Δj
Step 4 - Calculation of Z and Δj and check the basic feasible solution for optimality by the rules known.
= 0 *4 + 0 * 2 = 0
Δj = Zj - Cj
= CB Xj - Cj
Δ1 = CB X1 - Cj = 0 * 1 + 0 * 1 - 3 = -3
Δ2 = CB X2 - Cj = 0 * 1 + 0 * -1 - 2 = -2
Δ3 = CB X3 - Cj = 0 * 1 + 0 * 0 - 0 = 0
Δ4 = CB X4 - Cj = 0 * 0 + 0 * 1 - 0 = 0
Process to check the basic feasible solution for optimality by the rules known
Rule 1 - If all Δj ≥ 0, then the solution under the test will be optimal. Other optimal solution will exist if any non-basic Δj is zero as well.
Rule 2 - If at least one Δj is negative, then the solution is not optimal and one can proceed to improve the solution in the further step.
Rule 3 - If corresponding to any negative Δj, eacjh and every elements of the column Xj are negative or zero, then the solution under examination will be unbounded.
In this problem it is seem that Δ1 and Δ2 are negative. Therefore proceed to enhance or improve this solution
Step 5 - To enhance the basic feasible solution, the vector entering the basis matrix and the vector to be removed from the basis matrix are to be find out.
The incoming vector Xk is always chosen parallel to the most negative value of Δj. It is signified through (↑).
The outgoing vector is chosen parallel to the minimum positive value of minimum ratio. It is symbolized through (→).
Step 6 - Now mark the key element or pivot element through '1''.The element at the intersection of incoming vector and outgoing vector is the pivot element.
(Xk)
4 / 1 = 4
2 / 1 = 2 → outgoing
Z= CB XB = 0
↑incoming
Δ1= -3 Δ2= -2 Δ3=0 Δ4=0
x1
3 2
(R1=R1 - R2)
0 2 1 -1
2 / 2 = 1 → outgoing
2 / -1 = -2 (neglect in case of negative)
Z=0*2+3*2= 6
Δ1=0 Δ2= -5 Δ3=0 Δ4=3
Step 7 - Then repeat step 4 through step 6 until an optimal solution is attained.
x2
2 1
3 3
(R1=R1 / 2)
0 1 1/2 -1/2
(R2=R2 + R1)
1 0 1/2 1/2
Z = 11
Δ1=0 Δ2=0 Δ3=5/2 Δ4=1/2
As all Δj ≥ 0, optimal basic feasible solution is achieved
Thus the solution is Max Z = 11, x1 = 3 and x2 = 1
Worked Examples
Solve with the help of simplex method
Example 1
Maximize Z = 80x1 + 55x2
4x1 + 2x2 ≤ 40
2x1 + 4x2 ≤ 32
SLPP
Maximize Z = 80x1 + 55x2 + 0s1 + 0s2
4x1 + 2x2+ s1= 40
2x1 + 4x2 + s2= 32
Cj → 80 55 0 0
0 40
0 32
4 2 1 0
2 4 0 1
40 / 4 = 10→ outgoing
32 / 2 = 16
Δ1= -80 Δ2= -55 Δ3=0 Δ4=0
80 10
0 12
(R1=R1 / 4)
1 1/2 1/4 0
(R2=R2- 2R1)
0 3 -1/2 1
10/1/2 = 20
12/3 = 4→ outgoing
Z = 800
Δ1=0 Δ2= -15 Δ3=40 Δ4=0
80 8
55 4
(R1=R1- 1/2R2)
1 0 1/3 -1/6
(R2=R2 / 3)
0 1 -1/6 1/3
Z = 860
Δ1=0 Δ2=0 Δ3=35/2 Δ4=5
As all Δj ≥ 0, optimal basic feasible solution is achieved. Hence the solution is Max Z = 860, x1 = 8 and x2 = 4
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