Question

Use the simplex method to solve the problem. Maximize subject to P = 4X2 + 12x2 2x1 + x2 58 Xy + 8x2 58 X1, X220 when X1 = The maximum value is P= (Simplify your answers.) and x2 =

Please solve this step by step I do not understand the pivoting in the table and I would like to learn how to do it thx!

Answer 1

Solution:
The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '≤' we should add slack variable S₁

2. As the constraint-2 is of type '≤' we should add slack variable S₂

After introducing slack variables

Max P = 4x₁ + 12x₂ + 0S₁ + 0S₂

Subject to

2x₁ + x₂ + S₁ = 8

x₁ + 8x₂ + S₂ = 8

and x₁, x₂, S₁,S₂ ≥ 0

Iteration-1		Cj	4	12	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XB / x2
S1	0	8	2	1	1	0	8 / 1=8
S2	0	8	1	(8)	0	1	8 / 8=1→
P=0		Pj	0	0	0	0
		Pj - Cj	-4	-12↑	0	0

Negative minimum P_j - C_j is -12 and its column index is 2. So, the entering variable is x₂.

Minimum ratio is 1 and its row index is 2. So, the leaving basis variable is S₂.

∴ The pivot element is 8.

Entering =x₂, Departing =S₂, Key Element =8

R₂(new)=R₂(old) ÷ 8

R₁(new)=R₁(old) - R₂(new)

Iteration-2		Cj	4	12	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XB / x1
S1	0	7	(1.875)	0	1	-0.125	7 / 1.875=3.7333→
x2	12	1	0.125	1	0	0.125	1 / 0.125=8
P=12		Pj	1.5	12	0	1.5
		Pj - Cj	-2.5↑	0	0	1.5

Negative minimum P_j - C_j is -2.5 and its column index is 1. So, the entering variable is x₁.

Minimum ratio is 3.7333 and its row index is 1. So, the leaving basis variable is S₁.

∴ The pivot element is 1.875.

Entering =x1, Departing =S1, Key Element =1.875

R₁(new) = R₁(old) ÷ 1.875

R₂(new) = R₂(old) - 0.125R₁(new)

Iteration-3		Cj	4	12	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio
x1	4	3.7333	1	0	0.5333	-0.0667
x2	12	0.5333	0	1	-0.0667	0.1333
P=21.3333		Pj	4	12	1.3333	1.3333
		Pj - Cj	0	0	1.3333	1.3333

Since all P_j - C_j ≥ 0

Hence, optimal solution is arrived with value of variables as :

x₁ = 3.7333,   x₂ = 0.5333

Max P = 21.3333