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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.

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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

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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