Question

+ Use the simplex method to solve the linear Maximize z = xy + 2x2 + x3 +8X4 programming problem. subject to: xy + 2x2 + x3 + x4 = 48 4x4 + X2 + 2X2 + x4 S 105 with Xy 20, X220, X3 20, X4 20. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. X2 = A. The maximum is when X1 = (Simplify your answers.) , X3 =, and X4 = B. There is no maximum.

Answer 1

Sol:

Problem is

Max Z = x1 + 2 x2 + x3 + 8 x4

subject to

x1 + 2 x2 + x3 + x4 ≤ 48 4 x1 + x2 + 2 x3 + x4 ≤ 105

and x1,x2,x3,x4≥0;

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 Z = x1 + 2 x2 + x3 + 8 x4 + 0 S1 + 0 S2

subject to

x1 + 2 x2 + x3 + x4 + S1 = 48 4 x1 + x2 + 2 x3 + x4 + S2 = 105

and x1,x2,x3,x4,S1,S2≥0

Iteration-1		Cj	1	2	1	8	0	0
B	CB	XB	x1	x2	x3	x4	S1	S2	MinRatio XB/x4
S1	0	48	1	2	1	(1)	1	0	48/1=48→
S2	0	105	4	1	2	1	0	1	105/1=105
Z=0		Zj	0	0	0	0	0	0
		Zj-Cj	-1	-2	-1	-8↑	0	0

Negative minimum Zj-Cj is -8 and its column index is 4. So, the entering variable is x4.

Minimum ratio is 48 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 1.

Entering =x4, Departing =S1, Key Element =1

R1(new)=R1(old)

R2(new)=R2(old) - R1(new)

Iteration-2		Cj	1	2	1	8	0	0
B	CB	XB	x1	x2	x3	x4	S1	S2	MinRatio
x4	8	48	1	2	1	1	1	0
S2	0	105	3	-6	4	0	-1	1
Z=105		Zj	8	56	8	8	8	0
		Zj-Cj	7	54	7	0	8	0

Since all Zj-Cj≥0

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

x1=0,x2=0,x3=0,x4=20

Max Z=105z