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MEPM 531 Operation Research - American University of Ras Al Khaimah

Solve the problems using LP Modeling and the Simplex Algorithm.

Problem 1: Reddy Mikks wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit.

Solution:

Reddy Mikks produces both interior and exterior paints from two raw materials, ??1 and ??2. The following table provides the basic data of the problem:

 

Exterior Paint

Interior Paint

Maximum daily availability(tons)

Raw Material,M1

6

4

24

Raw Material,M2

1

2

6

Profit per ton(1000$)

5

4

 

A market survey indicates that the daily demand for interior paint cannot exceed that of exterior paint by more than 1 ton. Also, the maximum daily demand of interior paint is 2 tons. Reddy Mikks wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit.

Constraints

6M1+4M2<=24

M1+2M2<=6

M2-M1<=1

M2<=2

Objective function(Maximize Profit)

Z=5M1+4M2

6M1+4M2=24

M1

0

4

M2

6

0

Points

(0,6)

(4,0)

M1+2M2=6

M1

0

6

M2

3

0

Points

(0,3)

(6,0)

M2-M1<=1

M1

0

-1

M2

1

0

Points

(0,1)

(-1,0)

Constraints

6M1+4M2+ S1=24

M1+2M2+ S2=6

M2-M1+ S3=1

M2+ S4=2

Objective function(Maximize Profit)

      Z=5M1+4M2+0 S1+ 0S2+0 S3+0 S4

CB

 

5

4

0

0

0

0

Solution

B.v

M1

M2

S1

S2

S3

S4

0

S1

6

4

1

0

0

0

24

0

S2

1

2

0

1

0

0

6

0

S3

-1

1

0

0

1

0

1

0

S4

0

1

0

0

0

1

2

 

zj

0

0

0

0

0

0

 

 

Cj-- zj

5

4

0

0

0

0

 

For optimal solution, Cj-- zj<=0           

Maximum positive value of Cj-- zj=5

CB

 

5

4

0

0

0

0

Solution

B.v

M1

M2

S1

S2

S3

S4

0

S1

6

4

1

0

0

0

24

0

S2

1

2

0

1

0

0

6

0

S3

-1

1

0

0

1

0

1

0

S4

0

1

0

0

0

1

2

 

zj

0

0

0

0

0

0

 

 

Cj-- zj

5

4

0

0

0

0

 

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

0

S1

6

4

1

0

0

0

24

24/6=4(MIN)

0

S2

1

2

0

1

0

0

6

6/1=6

0

S3

-1

1

0

0

1

0

1

1/-1=-1(REJECT)

0

S4

0

1

0

0

0

1

2

0/2=0(REJECT)

zj

0

0

0

0

0

0

0

 

Cj-- zj

5

4

0

0

0

0

0

 

New first row= old first row/pivot key

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

6/6=1

4/6=2/3

1/6

0/6=0

0/6=0

0/6=0

4

 

0

S2

 

 

 

 

 

 

 

 

0

S3

 

 

 

 

 

 

 

 

0

S4

 

 

 

 

 

 

 

 

zj

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

New second row =old second row -(pivot key for second row * new line)

                               =(1  2  0  1 0 0  6) -(1) * ( 1 2/3  1/6  0 0 0 4)

                             = (0 4/3 -1/6  1 0 0 2)

c

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

6/6=1

4/6=2/3

1/6

0/6=0

0/6=0

0/6=0

4

 

0

S2

0

4/3

-1/6

1

0

0

2

 

0

S3

 

 

 

 

 

 

 

 

0

S4

 

 

 

 

 

 

 

 

zj

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

New third row =old third row -(pivot key for third row * new line)

                               =(-1  1  0  0 1 0  1) -(-1) * ( 1 2/3  1/6  0 0 0 4)

                             = (0 5/3  1/6  0 1 0 5)

c

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

6/6=1

4/6=2/3

1/6

0/6=0

0/6=0

0/6=0

4

 

0

S2

0

4/3

-1/6

1

0

0

2

 

0

S3

0

5/3

1/6

0

1

0

5

 

0

S4

 

 

 

 

 

 

 

 

zj

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

New third row =old third row -(pivot key for third row * new line)

                               =(0  1  0  0  0 1 2) -(0) * ( 1 2/3  1/6  0 0 0 4)

                             = (0  1  0  0  0 1 2)

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

6/6=1

4/6=2/3

1/6

0/6=0

0/6=0

0/6=0

4

 

0

S2

0

4/3

-1/6

1

0

0

2

 

0

S3

0

5/3

1/6

0

1

0

5

 

0

S4

0

1

0

0

0

1

2

 

Zj

5

10/3

5/6

0

0

0

20

 

Cj-- Zj

0

2/3

-5/6

0

0

0

 

 

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

6/6=1

4/6=2/3

1/6

0/6=0

0/6=0

0/6=0

4

6

0

S2

0

4/3

-1/6

1

0

0

2

3/2(MIN)

0

S3

0

5/3

1/6

0

1

0

5

3

0

S4

0

1

0

0

0

1

2

2

Zj

5

10/3

5/6

0

0

0

20

 

Cj-- Zj

0

2/3

-5/6

0

0

0

 

 

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

6/6=1

4/6=2/3

1/6

0/6=0

0/6=0

0/6=0

4

 

0

M2

0

4/3

-1/6

1

0

0

2

 

0

S3

0

5/3

1/6

0

1

0

5

 

0

S4

0

1

0

0

0

1

2

 

Zj

5

10/3

5/6

0

0

0

20

 

Cj-- Zj

0

2/3

-5/6

0

0

0

 

 

NEW SECOND ROW

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

 

 

 

 

 

 

 

 

4

M2

0

1

-1/8

3/4

0

0

3/2

 

0

S3

 

 

 

 

 

 

 

 

0

S4

 

 

 

 

 

 

 

 

Zj

 

 

 

 

 

 

 

 

Cj-- Zj

 

 

 

 

 

 

 

 

NEW FIRST ROW

New FIRST row =old FIRST row -(pivot key for FIRST row * new line)

                               =(1  2/3  1/6  0  0 0 4) -(2/3) * ( 0 1 - 1/8  3/4 0 0 3/2)

                             = (1  0  1/4  -1/2  0 0 3)

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

1

0

1/4

-1/2

0

0

3

 

4

M2

0

1

-1/8

3/4

0

0

3/2

 

0

S3

 

 

 

 

 

 

 

 

0

S4

 

 

 

 

 

 

 

 

Zj

 

 

 

 

 

 

 

 

Cj-- Zj

 

 

 

 

 

 

 

 

NEW THIRD ROW

New THIRD row =old THIRD row -(pivot key for THIRD row * new line)

                               =(0  5/3  1/6  0  1 0 5) -(5/3) * ( 0 1 - 1/8  3/4 0 0 3/2)

                             = (0  0  9/24  -5/4  1 0 5/2)

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

1

0

1/4

-1/2

0

0

3

 

4

M2

0

1

-1/8

3/4

0

0

3/2

 

0

S3

0

0

9/24

-5/4

1

0

5/2

 

0

S4

 

 

 

 

 

 

 

 

Zj

 

 

 

 

 

 

 

 

Cj-- Zj

 

 

 

 

 

 

 

 

NEW FOURTH ROW

New FOURTH row =old FOURTH row -(pivot key for THIRD row * new line)

                               =(0  1  0  0  0  1 2) -(1) * ( 0 1 - 1/8  3/4 0 0 3/2)

                             = (0  0  1/8  -3/4  0 1 1/2)

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

1

0

1/4

-1/2

0

0

3

 

4

M2

0

1

-1/8

3/4

0

0

3/2

 

0

S3

0

0

9/24

-5/4

1

0

5/2

 

0

S4

0

0

1/8

-3/4

0

1

1/2

 

Zj

 

 

 

 

 

 

 

 

Cj-- Zj

 

 

 

 

 

 

 

 

 

CB

 

5

4

0

0

0

0

Solution

Ratio

B.v

M1

M2

S1

S2

S3

S4

5

M1

1

0

1/4

-1/2

0

0

3

 

4

M2

0

1

-1/8

3/4

0

0

3/2

 

0

S3

0

0

9/24

-5/4

1

0

5/2

 

0

S4

0

0

1/8

-3/4

0

1

1/2

 

Zj

5

4

3/4

1/2

0

0

21

 

Cj-- Zj

0

0

-3/4

-1/2

0

0

 

 

This is optimal solution as  Cj-- Zj <=0

M1=3, M2=3/2

Z=5(3)+4(3/2)=21

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Problem 2

A firm manufactures two types of products A and B and sells them at a profit of $2 on type A and $3 on type B. Each product is processed on two machines G and H. Type A requires one minute of processing time on G and two minutes on H; type B requires one minute on G and one minute on H. The machine G is available for not more than 6 hour 40 minutes while machine H is available for not more than 10 hours during any working day. Formulate the problem as a linear programming problem and determine the optimum product mix of types A and B products.

Solution:

 

Machine G(min)

Machine H(min)

Profit

Type A,X1

1

2

$2

Type B,X2

1

1

$3

Availability of each machine

400

600

 

Constraints

X1 +X2 <=400

2X1+X2<=600

Objective function

Z=2X1+3X2

Converting inequalities into equations

X1+X2+S1=400

2X1+X2+S2=600

Z=2X1+3X2+0S1+0S2

CB

 

2

3

0

0

Solution

 

B.v

X1

X2

S1

S2

 

0

S1

1

1

1

0

400

 

0

S2

2

1

0

1

600

 

zj

Cj-- zj

0

0

0

0

0+

 

2

3

0

0

 

 

CB

 

2

3

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

0

S1

1

1

1

0

400

400(min)

0

S2

2

1

0

1

600

600

zj

Cj-- zj

0

0

0

0

0

0

2

3

0

0

 

 

New table

First row

CB

 

2

3

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

2

X2

1

1

1

0

400

 

0

S2

 

 

 

 

 

 

zj

Cj-- zj

 

 

 

 

 

 

 

 

 

 

 

 

Second row

New second row =(2 1 0 1 600) -(1)*(1 1 1 0 400)

                            =(1 0 -1 1 200)

CB

 

2

3

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

3

X2

1

1

1

0

400

 

0

S2

1

0

-1

1

200

 

zj

Cj-- zj

3

3

3

0

1200

 

0

0

-2

0

 

 

This is optimal solution as Cj-- zj <=0

X1=0 ,X2=400 , z=1200

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Problem 3

A company produces two types of hats. Each hat of the first type requires twice as much labor time as the second type. The company can produce a total of 500 hats a day. The market limits daily sales of the first and second type to 150 and 250 hats respectively. Assuming that the profits per hat are $8 for type A and $5 for type B, formulate the problem as a linear programming model in order to determine the number of hats to be produced of each type so as to maximize the profit.

Solution:

Let say first type hats are X1 and second type hats are X2

2X1+X2<=500

X1<=150

X2<=250

X1<=2X2

Objective Function (Maximize)

Z=8X1+5X2

Converting inequalities into equations

2X1+X2+S1=500

X1+S2=150

X2+S3=250

Z=8X1+5X2+0S1+0S2+0S3

CB

 

8

5

0

0

0

Solution

B.v

X1

X2

S1

S2

S3

0

S1

2

1

1

0

0

500

0

S2

1

0

0

1

0

150

0

S3

0

1

0

0

1

250

 

zj

0

0

0

0

0

 

 

Cj-- zj

8

5

0

0

0

 

CB

 

8

5

0

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

S3

0

S1

2

1

1

0

0

500

250

0

S2

1

0

0

1

0

150

150(Min)

0

S3

0

1

0

0

1

250

infinite

 

zj

0

0

0

0

0

 

 

 

Cj-- zj

8

5

0

0

0

 

 

New second row

CB

 

8

5

0

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

S3

0

S1

 

 

 

 

 

 

 

8

X1

1

0

0

1

0

150

 

0

S3

 

 

 

 

 

 

 

 

zj

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

New first row=(2 1 1 0 0 250)-(2)(1 0 0 1 0 150)= (0 1 1-2 0 200)

CB

 

8

5

0

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

S3

0

S1

0

1

1

-2

0

-50

 

8

X1

1

0

0

1

0

150

 

0

S3

 

 

 

 

 

 

 

 

zj

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

New third row =(0 1 0 0 1 250)-(0)(1 0 0 1 0 150)= (0 1 0 0 1 250)

CB

 

8

5

0

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

S3

0

S1

0

1

1

-2

0

-50

 

8

X1

1

0

0

1

0

150

 

0

S3

0

1

0

0

1

100

 

 

zj

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

CB

 

8

5

0

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

S3

0

S1

0

1

1

-2

0

-50

 

8

X1

1

0

0

1

0

150

 

0

S3

0

1

0

0

1

100

 

 

zj

8

0

0

8

0

1200

 

 

Cj-- zj

0

5

0

-8

0

 

 

 

CB

 

8

5

0

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

S3

0

S1

0

1

1

-2

0

-50

-50

8

X1

1

0

0

1

0

150

infinite

0

X2

0

1

0

0

1

100

100(min)

 

zj

8

0

0

8

0

1200

 

 

Cj-- zj

0

5

0

-8

0

 

 

 

CB

 

8

5

0

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

S3

0

S1

0

1

1

-2

0

-50

-50

8

X1

1

0

0

1

0

150

infinite

5

X2

0

1

0

0

1

100

100(min)

 

zj

8

0

0

8

0

1200

 

 

Cj-- zj

0

5

0

-8

0

 

 

 

CB

 

8

5

0

0

0

Solution

Ratio

B.v

X1

X2

S1

S2

S3

0

S1

0

0

1

-2

-1

-150

 

8

X1

1

0

0

1

0

150

 

5

X2

0

1

0

0

1

100

 

 

zj

8

5

0

8

5

1700

 

 

Cj-- zj

0

0

0

-8

-5

 

 

This is optimal solution

X1=150

X2=100

Z=1700

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Problem 4

The manufacturer of patent medicines is proposed to prepare a production plan for medicines A and B. There are sufficient ingredients available to make 20,000 bottles of medicine A and 40,000 bottles of medicine B, but there are only 45,000 bottles into which either of the medicines can be filled.

Solution:

1. Formulate this problem as L.P.P.

2. How the manufacturer schedule his production in order to maximize profit

Objective function

Z=8x+7y

Constraints

x+y<=.45000

x<=20000

y<=40000

3x+y<=66000

Converting into equations

x+y+S1=45000

x+S2=20000

y+S3=40000

3x+y+S4=66000

Z=8x+7y+0S1+0S2+0S3+0S4

CB

 

8

7

0

0

0

0

Solution

B.v

x

y

S1

S2

S3

S4

0

S1

1

1

1

0

0

0

45000

0

S2

1

0

0

1

0

0

20000

0

S3

0

1

0

0

1

0

40000

0

S4

3

1

0

0

0

1

66000

 

zj

0

0

0

0

0

0

0

 

Cj-- zj

8

7

0

0

0

0

 

 

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

1

1

1

0

0

0

45000

45000

0

S2

1

0

0

1

0

0

20000

20000(Min)

0

S3

0

1

0

0

1

0

40000

Infinity

0

S4

3

1

0

0

0

1

66000

22000

 

zj

0

0

0

0

0

0

0

 

 

Cj-- zj

8

7

0

0

0

0

 

 

New second row

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

 

 

 

 

 

 

 

 

8

x

1

0

0

1

0

0

20000

 

0

S3

 

 

 

 

 

 

 

 

0

S4

 

 

 

 

 

 

 

 

 

zj

 

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

New third row and first row

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

0

1

1

-1

0

0

25000

 

8

x

1

0

0

1

0

0

20000

 

0

S3

-1

1

0

-1

1

0

20000

 

0

S4

 

 

 

 

 

 

 

 

 

zj

 

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

New fourth row

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

0

1

1

-1

0

0

25000

 

8

x

1

0

0

1

0

0

20000

 

0

S3

0

1

0

0

1

0

40000

 

0

S4

0

1

0

-3

0

1

6000

 

 

zj

 

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

 

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

0

1

1

-1

0

0

25000

 

8

x

1

0

0

1

0

0

20000

 

0

S3

0

1

0

0

1

0

40000

 

0

S4

0

1

0

-3

0

1

6000

 

 

zj

8

0

0

8

0

0

160000

 

 

Cj-- zj

0

7

0

-8

0

0

 

 

 

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

0

1

1

-1

0

0

25000

25000

8

X

1

0

0

1

0

0

20000

infinite

0

S3

0

1

0

0

1

0

40000

40000

0

S4

0

1

0

-3

0

1

6000

6000(min)

 

zj

8

0

0

8

0

0

160000

 

 

Cj-- zj

0

7

0

-8

0

0

 

 

 

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

 

 

 

 

 

 

 

 

8

X

 

 

 

 

 

 

 

 

0

S3

 

 

 

 

 

 

 

 

7

Y

0

1

0

-3

0

1

6000

6000(min)

 

zj

 

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

New first row

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

0

0

1

2

0

-1

19000

 

8

x

 

 

 

 

 

 

 

 

0

S3

 

 

 

 

 

 

 

 

7

y

0

1

0

-3

0

1

6000

6000(min)

 

zj

 

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

New second row

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

0

1

1

-1

0

0

25000

 

8

x

1

0

0

1

0

0

20000

 

0

S3

 

 

 

 

 

 

 

 

7

y

0

1

0

-3

0

1

6000

6000(min)

 

zj

 

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

New third row

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

0

1

1

-1

0

0

25000

 

8

x

1

-1

0

4

0

-1

14000

 

0

S3

0

0

0

3

1

-1

34000

 

7

y

0

1

0

-3

0

1

6000

6000(min)

 

zj

 

 

 

 

 

 

 

 

 

Cj-- zj

 

 

 

 

 

 

 

 

 

CB

 

8

7

0

0

0

0

Solution

Ratio

B.v

x

Y

S1

S2

S3

S4

0

S1

0

1

1

-1

0

0

25000

 

8

x

1

-1

0

4

0

-1

14000

 

0

S3

0

1

0

0

1

0

40000

 

7

y

0

1

0

-3

0

1

6000

 

 

zj

8

-1

0

11

0

-1

140000

 

 

Cj-- zj

0

8

0

-53

0

1

 

 

This has multiple optimal solutions.

24/7 AVAILABILITY OF TRUSTED MEPM 531 OPERATION RESEARCH ASSIGNMENT WRITERS! ORDER ASSIGNMENTS FOR BETTER RESULTS!

Problem 5

A toy company manufactures two types of doll, a basic version - doll A and a deluxe version - doll B. Each doll of type B takes twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day. The supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined). The deluxe version requires a fancy dress of which there are only 600 per day available. If the company makes a profit of $3 and $5 per doll, respectively on doll A and B, then how many of each doll should be produced per day in order to maximize the total profit. Formulate this problem in order to determine the number of dolls to be produced of each type so as to maximize the profit.

Solution:

Maximize

Z=3x +5y

x+2y<=2000

x+y<=1500

Y<=600

Adding slack variables

Z=3x+5y+0S1+0S2+0S3

X+2y+S1=2000

X+y+S2=1500

y+S3=600

CB

 

3

5

0

0

0

Solution

B.v

x

y

S1

S2

S3

0

S1

1

2

1

0

0

2000

0

S2

1

1

0

1

0

1500

0

S3

0

1

0

0

1

600

 

zj

0

0

0

0

0

0

 

Cj-- zj

3

5

0

0

0

 

 

CB

 

3

5

0

0

0

Solution

Ratio

B.v

x

y

S1

S2

S3

0

S1

1

2

1

0

0

2000

1000

0

S2

1

1

0

1

0

1500

1500

0

S3

0

1

0

0

1

600

600(min)

 

zj

0

0

0

0

0

0

 

 

Cj-- zj

3

5

0

0

0

 

 

NEW table

CB

 

3

5

0

0

0

Solution

Ratio

B.v

x

y

S1

S2

S3

0

S1

1

1

1

0

0

1400

 

0

S2

1

0

0

1

-1

900

 

5

y

0

1

0

0

1

600

 

 

zj

0

5

0

0

5

3000

 

 

Cj-- zj

3

0

0

0

-5

 

 

 

CB

 

3

5

0

0

0

Solution

Ratio

B.v

x

y

S1

S2

S3

0

S1

1

1

1

0

0

1400

1400

0

S2

1

0

0

1

-1

900

900(min)

5

y

0

1

0

0

1

600

infinite

 

zj

0

5

0

0

5

3000

 

 

Cj-- zj

3

0

0

0

-5

 

 

New table

CB

 

3

5

0

0

0

Solution

Ratio

B.v

x

y

S1

S2

S3

0

S1

0

1

1

-1

1

500

 

3

x

1

0

0

1

-1

900

 

5

y

0

1

0

0

1

600

 

 

zj

3

5

0

3

2

5700

 

 

Cj-- zj

0

0

0

-3

-2

 

 

This is optimal solution as Cj-- zj<=0

X=900

Y=600

Z=5700

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