Question

Find the optimal solution of the following linear program. Write down your optimal solution and optimal objective value. max -5z2-3r3-2x4 T1 T1,...,T4 20

Answer 1

Max Z-_512-3r3-2x4

subject to

$- x_1 + 2 x_3 + x_4 \leq 5$

.

.

After introducing surplus,artificial variables

subject to

Iteration-1		Cj	0	-5	-3	-2	0	-M	-M
B	CB	XB	x1	x2	x3	x4	S1	A1	A2	MinRatio XB/x3
A1	-M	2	-1	1	(1)	0	0	1	0	2/1=2→
A2	-M	5	-1	0	2	1	-1	0	1	5/2=2.5
Z=-7M		Zj	2M	-M	-3M	-M	M	-M	-M
		Zj-Cj	2M	-M+5	-3M+3↑	-M+2	M	0	0

Negative minimum Zj-Cj is -3M+3 and its column index is 3.

Minimum ratio is 2 and its row index is 1

The pivot element is 1.

Entering =x3, Departing =A1,

$R_2 \leftarrow R_2-2R_1$

Iteration-2		Cj	0	-5	-3	-2	0	-M
B	CB	XB	x1	x2	x3	x4	S1	A2	MinRatio XB/x4
x3	-3	2	-1	1	1	0	0	0	---
A2	-M	1	1	-2	0	(1)	-1	1	1/1=1→
Z=-M-6		Zj	-M+3	2M-3	-3	-M	M	-M
		Zj-Cj	-M+3	2M+2	0	-M+2↑	M	0

Negative minimum Zj-Cj is -M+2 and its column index is 4

Minimum ratio is 1 and its row index is 2

The pivot element is 1.

Entering =x4, Departing =A2,

Iteration-3		Cj	0	-5	-3	-2	0
B	CB	XB	x1	x2	x3	x4	S1	MinRatio
x3	-3	2	-1	1	1	0	0
x4	-2	1	1	-2	0	1	-1
Z=-8		Zj	1	1	-3	-2	2
		Zj-Cj	1	6	0	0	2

Since all $Zj-Cj \geq 0$

Hence, optimal solution is arrived

${\color{Red} x_1=0, \:\:\:x_2=0, \:\:\:x_3=2, \:\:\:x_4=1}$

${\color{Red} Max \:\:\: Z=-8}$