Question

Min 8X1+3X2

S.T.

50X1 + 100X2 ≤ 1200

5X1 + 4X2 ≥ 60

X2 ≥ 3

X1,X2 ≥0

Using two-phase simplex method to solve it.

Answer 1

*****Please please please LIKE THIS ANSWER, so that I can get a small benefit, Please*****

Solution:
Problem is

Max Z = 8 x1 + 3 x2

subject to

50 x1 + 100 x2 ≤ 1200 5 x1 + 4 x2 ≥ 60 x2 ≥ 3

and x1,x2≥0;

-->Phase-1<--

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 subtract surplus variable S2 and add artificial variable A1

3. As the constraint-3 is of type '≥' we should subtract surplus variable S3 and add artificial variable A2

After introducing slack,surplus,artificial variables

Max Z = - A1 - A2

subject to

50 x1 + 100 x2 + S1 = 1200 5 x1 + 4 x2 - S2 + A1 = 60 x2 - S3 + A2 = 3

and x1,x2,S1,S2,S3,A1,A2≥0

Iteration-1		Cj	0	0	0	0	0	-1	-1
B	CB	XB	x1	x2	S1	S2	S3	A1	A2	MinRatio XBx1
S1	0	1200	50	100	1	0	0	0	0	1200/50=24
A1	-1	60	(5)	4	0	-1	0	1	0	60/5=12→
A2	-1	3	0	1	0	0	-1	0	1	---
Z=-63		Zj	-5	-5	0	1	1	-1	-1
		Zj-Cj	-5↑	-5	0	1	1	0	0

Negative minimum Zj-Cj is -5 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 12 and its row index is 2. So, the leaving basis variable is A1.

∴ The pivot element is 5.

Entering =x1, Departing =A1, Key Element =5

R2(new)=R2(old)÷5

R2(old) =	60	5	4	0	-1	0	0
R2(new)=R2(old)÷5	12	1	0.8	0	-0.2	0	0

R1(new)=R1(old) - 50R2(new)

R1(old) =	1200	50	100	1	0	0	0
R2(new) =	12	1	0.8	0	-0.2	0	0
50×R2(new) =	600	50	40	0	-10	0	0
R1(new)=R1(old) - 50R2(new)	600	0	60	1	10	0	0

R3(new)=R3(old)

R3(old) =	3	0	1	0	0	-1	1
R3(new)=R3(old)	3	0	1	0	0	-1	1

Iteration-2		Cj	0	0	0	0	0	-1
B	CB	XB	x1	x2	S1	S2	S3	A2	MinRatio XBx2
S1	0	600	0	60	1	10	0	0	600/60=10
x1	0	12	1	0.8	0	-0.2	0	0	12/0.8=15
A2	-1	3	0	(1)	0	0	-1	1	3/1=3→
Z=-3		Zj	0	-1	0	0	1	-1
		Zj-Cj	0	-1↑	0	0	1	0

Negative minimum Zj-Cj is -1 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 3 and its row index is 3. So, the leaving basis variable is A2.

∴ The pivot element is 1.

Entering =x2, Departing =A2, Key Element =1

R3(new)=R3(old)

R3(old) =	3	0	1	0	0	-1
R3(new)=R3(old)	3	0	1	0	0	-1

R1(new)=R1(old) - 60R3(new)

R1(old) =	600	0	60	1	10	0
R3(new) =	3	0	1	0	0	-1
60×R3(new) =	180	0	60	0	0	-60
R1(new)=R1(old) - 60R3(new)	420	0	0	1	10	60

R2(new)=R2(old) - 0.8R3(new)

R2(old) =	12	1	0.8	0	-0.2	0
R3(new) =	3	0	1	0	0	-1
0.8×R3(new) =	2.4	0	0.8	0	0	-0.8
R2(new)=R2(old) - 0.8R3(new)	9.6	1	0	0	-0.2	0.8

Iteration-3		Cj	0	0	0	0	0
B	CB	XB	x1	x2	S1	S2	S3	MinRatio
S1	0	420	0	0	1	10	60
x1	0	9.6	1	0	0	-0.2	0.8
x2	0	3	0	1	0	0	-1
Z=0		Zj	0	0	0	0	0
		Zj-Cj	0	0	0	0	0

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=9.6,x2=3

Max Z=0

-->Phase-2<--

we eliminate the artificial variables and change the objective function for the original,
Max Z=8x1+3x2+0S1+0S2+0S3

Iteration-1		Cj	8	3	0	0	0
B	CB	XB	x1	x2	S1	S2	S3	MinRatio XBS2
S1	0	420	0	0	1	(10)	60	420/10=42→
x1	8	9.6	1	0	0	-0.2	0.8	---
x2	3	3	0	1	0	0	-1	---
Z=85.8		Zj	8	3	0	-1.6	3.4
		Zj-Cj	0	0	0	-1.6↑	3.4

Negative minimum Zj-Cj is -1.6 and its column index is 4. So, the entering variable is S2.

Minimum ratio is 42 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 10.

Entering =S2, Departing =S1, Key Element =10

R1(new)=R1(old)÷10

R1(old) =	420	0	0	1	10	60
R1(new)=R1(old)÷10	42	0	0	0.1	1	6

R2(new)=R2(old) + 0.2R1(new)

R2(old) =	9.6	1	0	0	-0.2	0.8
R1(new) =	42	0	0	0.1	1	6
0.2×R1(new) =	8.4	0	0	0.02	0.2	1.2
R2(new)=R2(old) + 0.2R1(new)	18	1	0	0.02	0	2

R3(new)=R3(old)

R3(old) =	3	0	1	0	0	-1
R3(new)=R3(old)	3	0	1	0	0	-1

Iteration-2		Cj	8	3	0	0	0
B	CB	XB	x1	x2	S1	S2	S3	MinRatio
S2	0	42	0	0	0.1	1	6
x1	8	18	1	0	0.02	0	2
x2	3	3	0	1	0	0	-1
Z=153		Zj	8	3	0.16	0	13
		Zj-Cj	0	0	0.16	0	13

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=18,x2=3

Max Z=153

*****Please please please LIKE THIS ANSWER, so that I can get a small benefit, Please*****