Example Problem 3- Charging a Blast Furnace You are the manager of an iron found

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Question

Example Problem 3- Charging a Blast Furnace You are the manager of an iron foundry. Your company has a firm order to produce 1000 lb of castings containing at least 0.45% Mn and between 3.25 and 5.50% Si. These are specialty castings, so you have to make them from scratch. Your foundry has abundant amounts of three types of pig iron available, with the following properties Pig Iron Si% Mn % 0.6 0.45 0.5 0.4 Also, you can add pure Mn directly to the melt. The castings sell for $ 0.45 per pound. The costs of the various inputs are: Pig A @ $21 per 1000 lb; Pig B @ $25 per 1009 lb; Pig C @ $15 per thousand lb; Mn @ $8 per lb. It costs 0.5 cents to melt one lb of pig iron. How would maximize your foundry's profits? Construct a LP model. Can you solve this problem graphically? 4-Dec-17 rm l View Slide 11 of 14 99%

Explanation / Answer

Solution:

According to the problem;

x1 = Thousands of pounds of pig iron A,

x2 = Thousands of pounds of pig iron B,

x3 = Thousands of pounds of pig iron C,

x4 = Pounds of pure manganese.

Formulation of the problem (objective function);

We want to maximize the total profit resulting from the production of 1000 pounds of castings. Since we are producing exactly 1000 pounds of castings, the total income will be the selling price per pound times 1000 pounds. That is:

Total income = 0.45 × 1000 = 450.

Further Add melting cost; $0.005/pound

Pig iron A 21 + 5 = 26,

Pig iron B 25 + 5 = 30,

Pig iron C 15 + 5 = 20.

Therefore, the total cost becomes:

Total cost = 26x1 + 30x2 + 20x3 + 8x4,

and the total profit we want to maximize is determined by the expression:

Total profit = Total income Total cost

Thus,

Total profit = 450 26x1 30x2 20x3 8x4

Now, let’s define constraints;

Since the producer does not want to keep any supply of the castings on hand, we should make the total amount to be produced exactly equal to 1000 pounds; that is,

1000x1 + 1000x2 + 1000x3 + x4 = 1000

The castings should contain at least 0.45 percent manganese, or 4.5 pounds in the 1000 pounds of castings to be produced. This restriction can be expressed as follows:

4.5x1 + 5.0x2 + 4.0x3 + x4 4.5

The restriction regarding the silicon content can be represented by the following inequalities:

40x1 + 10x2 + 6x3 32.5, (5) 40x1 + 10x2 + 6x3 55.0

Finally,

x1 0, x2 0, x3 0, x4 0

Hence Finally the formulation of this linear programming is as follows:

Minimize z = 26x1 + 30x2 + 20x3 + 8x4

subject to:

1000x1 + 1000x2 + 1000x3 + x4 = 1000

4.5x1 + 5.0x2 + 4.0x3 + x4 4.5

40x1 + 10x2 + 6x3 32.5

40x1 + 10x2 + 6x3 55.0

x1 0, x2 0, x3 0, x4 0

This problem can not be solevd graphically, as two variables are neeed in graphical solution.

Although following is the optimal soultion by simplex method

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 artificial variable A1

2. As the constraint 2 is of type '' we should subtract surplus variable S1 and add artificial variable A2

3. As the constraint 3 is of type '' we should subtract surplus variable S2 and add artificial variable A3

4. As the constraint 4 is of type '' we should add slack variable S3

After introducing slack,surplus,artificial variables

Min z

subject to

1000

4.5

32.5

and x1,x2,x3,x4,S1,S2,S3,A1,A2,A30

Iteration-1

x1 Entering variable

MinRatio
XB/x1

1000

1000/1000=1

4.5

-1

4.5/4.5=1

A3 Leaving variable

32.5

(40) (pivot element)

-1

32.5/40=0.8125

55/40=1.375

z=0 0=
Zj=CBXB

Zj Zj=CBxj

1044.5M 1044.5M=M×1000+M×4.5+M×40+0×40
Zj=CBx1

1015M 1015M=M×1000+M×5+M×10+0×10
Zj=CBx2

1010M 1010M=M×1000+M×4+M×6+0×6
Zj=CBx3

2M 2M=M×1+M×1+M×0+0×0
Zj=CBx4

-M -M=M×0+M×(-1)+M×0+0×0
Zj=CBS1

-M -M=M×0+M×0+M×(-1)+0×0
Zj=CBS2

0 0=M×0+M×0+M×0+0×1
Zj=CBS3

M M=M×1+M×0+M×0+0×0
Zj=CBA1

M M=M×0+M×1+M×0+0×0
Zj=CBA2

M M=M×0+M×0+M×1+0×0
Zj=CBA3

Cj-Zj

-1044.5M+26 -1044.5M+26=26-1044.5M

-1015M+30 -1015M+30=30-1015M

-1010M+20 -1010M+20=20-1010M

-2M+8 -2M+8=8-2M

M M=0-(-M)

0 0=0-0

0 0=M-M

Negative minimum Cj-Zj is -1044.5M+26 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 0.8125 and its row index is 3. So, the leaving basis variable is A3.

The pivot element is 40.

Entering =x1, Departing =A3, Key Element =40

R3(new)=R3(old)÷40

R1(new)=R1(old)-1000R3(new)

R2(new)=R2(old)-4.5R3(new)

R4(new)=R4(old)-40R3(new)

Iteration-2

x3 Entering variable

MinRatio
XB/x3

A1 Leaving variable

187.5 187.5=1000-1000×0.8125
R1(new)=R1(old)-1000R3(new)

0 0=1000-1000×1
R1(new)=R1(old)-1000R3(new)

750 750=1000-1000×0.25
R1(new)=R1(old)-1000R3(new)

(850) 850=1000-1000×0.15 (pivot element)
R1(new)=R1(old)-1000R3(new)

1 1=1-1000×0
R1(new)=R1(old)-1000R3(new)

0 0=0-1000×0
R1(new)=R1(old)-1000R3(new)

25 25=0-1000×(-0.025)
R1(new)=R1(old)-1000R3(new)

0 0=0-1000×0
R1(new)=R1(old)-1000R3(new)

1 1=1-1000×0
R1(new)=R1(old)-1000R3(new)

0 0=0-1000×0
R1(new)=R1(old)-1000R3(new)

187.5/850=0.2206

0.8438 0.8438=4.5-4.5×0.8125
R2(new)=R2(old)-4.5R3(new)

0 0=4.5-4.5×1
R2(new)=R2(old)-4.5R3(new)

3.875 3.875=5-4.5×0.25
R2(new)=R2(old)-4.5R3(new)

3.325 3.325=4-4.5×0.15
R2(new)=R2(old)-4.5R3(new)

1 1=1-4.5×0
R2(new)=R2(old)-4.5R3(new)

-1 -1=(-1)-4.5×0
R2(new)=R2(old)-4.5R3(new)

0.1125 0.1125=0-4.5×(-0.025)
R2(new)=R2(old)-4.5R3(new)

0 0=0-4.5×0
R2(new)=R2(old)-4.5R3(new)

1 1=1-4.5×0
R2(new)=R2(old)-4.5R3(new)

0.8438/3.325=0.2538

0.8125 0.8125=32.5÷40
R3(new)=R3(old)÷40

1 1=40÷40
R3(new)=R3(old)÷40

0.25 0.25=10÷40
R3(new)=R3(old)÷40

0.15 0.15=6÷40
R3(new)=R3(old)÷40

0 0=0÷40
R3(new)=R3(old)÷40

-0.025 -0.025=(-1)÷40
R3(new)=R3(old)÷40

0 0=0÷40
R3(new)=R3(old)÷40

0.8125/0.15=5.4167

22.5 22.5=55-40×0.8125
R4(new)=R4(old)-40R3(new)

0 0=40-40×1
R4(new)=R4(old)-40R3(new)

0 0=10-40×0.25
R4(new)=R4(old)-40R3(new)

0 0=6-40×0.15
R4(new)=R4(old)-40R3(new)

0 0=0-40×0
R4(new)=R4(old)-40R3(new)

1 1=0-40×(-0.025)
R4(new)=R4(old)-40R3(new)

1 1=1-40×0
R4(new)=R4(old)-40R3(new)

0 0=0-40×0
R4(new)=R4(old)-40R3(new)

---

z=21.125 21.125=26×0.8125
Zj=CBXB

Zj Zj=CBxj

26 26=M×0+M×0+26×1+0×0
Zj=CBx1

753.875M+6.5 753.875M+6.5=M×750+M×3.875+26×0.25+0×0
Zj=CBx2

853.325M+3.9 853.325M+3.9=M×850+M×3.325+26×0.15+0×0
Zj=CBx3

2M 2M=M×1+M×1+26×0+0×0
Zj=CBx4

-M -M=M×0+M×(-1)+26×0+0×0
Zj=CBS1

25.1125M-0.65 25.1125M-0.65=M×25+M×0.1125+26×(-0.025)+0×1
Zj=CBS2

0 0=M×0+M×0+26×0+0×1
Zj=CBS3

M M=M×1+M×0+26×0+0×0
Zj=CBA1

M M=M×0+M×1+26×0+0×0
Zj=CBA2

Cj-Zj

0 0=26-26

-753.875M+23.5 -753.875M+23.5=30-(753.875M+6.5)

-853.325M+16.1 -853.325M+16.1=20-(853.325M+3.9)

-2M+8 -2M+8=8-2M

M M=0-(-M)

-25.1125M+0.65 -25.1125M+0.65=0-(25.1125M-0.65)

0 0=0-0

0 0=M-M

Negative minimum Cj-Zj is -853.325M+16.1 and its column index is 3. So, the entering variable is x3.

Minimum ratio is 0.2206 and its row index is 1. So, the leaving basis variable is A1.

The pivot element is 850.

Entering =x3, Departing =A1, Key Element =850

R1(new)=R1(old)÷850

R2(new)=R2(old)-3.325R1(new)

R3(new)=R3(old)-0.15R1(new)

R4(new)=R4(old)

Iteration-3

x4 Entering variable

MinRatio
XB/x4

0.2206 0.2206=187.5÷850
R1(new)=R1(old)÷850

0 0=0÷850
R1(new)=R1(old)÷850

0.8824 0.8824=750÷850
R1(new)=R1(old)÷850

1 1=850÷850
R1(new)=R1(old)÷850

0.0012 0.0012=1÷850
R1(new)=R1(old)÷850

0 0=0÷850
R1(new)=R1(old)÷850

0.0294 0.0294=25÷850
R1(new)=R1(old)÷850

0 0=0÷850
R1(new)=R1(old)÷850

0.2206/0.0012=187.5

A2 Leaving variable

0.1103 0.1103=0.8438-3.325×0.2206
R2(new)=R2(old)-3.325R1(new)

0 0=0-3.325×0
R2(new)=R2(old)-3.325R1(new)

0.9412 0.9412=3.875-3.325×0.8824
R2(new)=R2(old)-3.325R1(new)

0 0=3.325-3.325×1
R2(new)=R2(old)-3.325R1(new)

(0.9961) 0.9961=1-3.325×0.0012 (pivot element)
R2(new)=R2(old)-3.325R1(new)

-1 -1=(-1)-3.325×0
R2(new)=R2(old)-3.325R1(new)

0.0147 0.0147=0.1125-3.325×0.0294
R2(new)=R2(old)-3.325R1(new)

0 0=0-3.325×0
R2(new)=R2(old)-3.325R1(new)

1 1=1-3.325×0
R2(new)=R2(old)-3.325R1(new)

0.1103/0.9961=0.1107

0.7794 0.7794=0.8125-0.15×0.2206
R3(new)=R3(old)-0.15R1(new)

1 1=1-0.15×0
R3(new)=R3(old)-0.15R1(new)

0.1176 0.1176=0.25-0.15×0.8824
R3(new)=R3(old)-0.15R1(new)

0 0=0.15-0.15×1
R3(new)=R3(old)-0.15R1(new)

-0.0002 -0.0002=0-0.15×0.0012
R3(new)=R3(old)-0.15R1(new)

0 0=0-0.15×0
R3(new)=R3(old)-0.15R1(new)

-0.0294 -0.0294=(-0.025)-0.15×0.0294
R3(new)=R3(old)-0.15R1(new)

0 0=0-0.15×0
R3(new)=R3(old)-0.15R1(new)

---

22.5 22.5=22.5
R4(new)=R4(old)

0 0=0
R4(new)=R4(old)

1 1=1
R4(new)=R4(old)

0 0=0
R4(new)=R4(old)

---

z=24.6765 24.6765=20×0.2206+26×0.7794
Zj=CBXB

Zj Zj=CBxj

26 26=20×0+M×0+26×1+0×0
Zj=CBx1

0.9412M+20.7059 0.9412M+20.7059=20×0.8824+M×0.9412+26×0.1176+0×0
Zj=CBx2

20 20=20×1+M×0+26×0+0×0
Zj=CBx3

0.9961M+0.0189 0.9961M+0.0189=20×0.0012+M×0.9961+26×(-0.0002)+0×0
Zj=CBx4

-M -M=20×0+M×(-1)+26×0+0×0
Zj=CBS1

0.0147M-0.1765 0.0147M-0.1765=20×0.0294+M×0.0147+26×(-0.0294)+0×1
Zj=CBS2

0 0=20×0+M×0+26×0+0×1
Zj=CBS3

M M=20×0+M×1+26×0+0×0
Zj=CBA2

Cj-Zj

0 0=26-26

-0.9412M+9.2941 -0.9412M+9.2941=30-(0.9412M+20.7059)

0 0=20-20

-0.9961M+7.9811 -0.9961M+7.9811=8-(0.9961M+0.0189)

M M=0-(-M)

-0.0147M+0.1765 -0.0147M+0.1765=0-(0.0147M-0.1765)

0 0=0-0

0 0=M-M

Negative minimum Cj-Zj is -0.9961M+7.9811 and its column index is 4. So, the entering variable is x4.

Minimum ratio is 0.1107 and its row index is 2. So, the leaving basis variable is A2.

The pivot element is 0.9961.

Entering =x4, Departing =A2, Key Element =0.9961

R2(new)=R2(old)×1.0039

R1(new)=R1(old)-0.0012R2(new)

R3(new)=R3(old)+0.0002R2(new)

R4(new)=R4(old)

Iteration-4

MinRatio

0.2205 0.2205=0.2206-0.0012×0.1107
R1(new)=R1(old)-0.0012R2(new)

0 0=0-0.0012×0
R1(new)=R1(old)-0.0012R2(new)

0.8812 0.8812=0.8824-0.0012×0.9449
R1(new)=R1(old)-0.0012R2(new)

1 1=1-0.0012×0
R1(new)=R1(old)-0.0012R2(new)

0 0=0.0012-0.0012×1
R1(new)=R1(old)-0.0012R2(new)

0.0012 0.0012=0-0.0012×(-1.0039)
R1(new)=R1(old)-0.0012R2(new)

0.0294 0.0294=0.0294-0.0012×0.0148
R1(new)=R1(old)-0.0012R2(new)

0 0=0-0.0012×0
R1(new)=R1(old)-0.0012R2(new)

0.1107 0.1107=0.1103×1.0039
R2(new)=R2(old)×1.0039

0 0=0×1.0039
R2(new)=R2(old)×1.0039

0.9449 0.9449=0.9412×1.0039
R2(new)=R2(old)×1.0039

0 0=0×1.0039
R2(new)=R2(old)×1.0039

1 1=0.9961×1.0039
R2(new)=R2(old)×1.0039

-1.0039 -1.0039=(-1)×1.0039
R2(new)=R2(old)×1.0039

0.0148 0.0148=0.0147×1.0039
R2(new)=R2(old)×1.0039

0 0=0×1.0039
R2(new)=R2(old)×1.0039

0.7794 0.7794=0.7794+0.0002×0.1107
R3(new)=R3(old)+0.0002R2(new)

1 1=1+0.0002×0
R3(new)=R3(old)+0.0002R2(new)

0.1178 0.1178=0.1176+0.0002×0.9449
R3(new)=R3(old)+0.0002R2(new)

0 0=0+0.0002×0
R3(new)=R3(old)+0.0002R2(new)

0 0=(-0.0002)+0.0002×1
R3(new)=R3(old)+0.0002R2(new)

-0.0002 -0.0002=0+0.0002×(-1.0039)
R3(new)=R3(old)+0.0002R2(new)

-0.0294 -0.0294=(-0.0294)+0.0002×0.0148
R3(new)=R3(old)+0.0002R2(new)

0 0=0+0.0002×0
R3(new)=R3(old)+0.0002R2(new)

22.5 22.5=22.5
R4(new)=R4(old)

0 0=0
R4(new)=R4(old)

1 1=1
R4(new)=R4(old)

z=25.5602 25.5602=20×0.2205+8×0.1107+26×0.7794
Zj=CBXB

Zj Zj=CBxj

26 26=20×0+8×0+26×1+0×0
Zj=CBx1

28.247 28.247=20×0.8812+8×0.9449+26×0.1178+0×0
Zj=CBx2

20 20=20×1+8×0+26×0+0×0
Zj=CBx3

8 8=20×0+8×1+26×0+0×0
Zj=CBx4

-8.0124 -8.0124=20×0.0012+8×(-1.0039)+26×(-0.0002)+0×0
Zj=CBS1

-0.0586 -0.0586=20×0.0294+8×0.0148+26×(-0.0294)+0×1
Zj=CBS2

0 0=20×0+8×0+26×0+0×1
Zj=CBS3

Cj-Zj

0 0=26-26

1.753 1.753=30-28.247

0 0=20-20

0 0=8-8

8.0124 8.0124=0-(-8.0124)

0.0586 0.0586=0-(-0.0586)

0 0=0-0

Since all Cj-Zj0

Hence, optimal solution is arrived with value of variables as :
x1=0.7794,x2=0,x3=0.2205,x4=0.1107

Min z=25.5602

Min z

subject to

1000

4.5

32.5

and x1,x2,x3,x4,S1,S2,S3,A1,A2,A30

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