Part A Given Three production facilities Types

Part A Given Three production facilities Types

Part A

Given;

Three production facilities:

Types of pollutants:

New technology at facility 1 descriptions;

Cost =

Amount of pollutant reduced:

New technology at facility 2 descriptions;

Cost =

Amount of pollutant reduced:

New technology at facility 3 descriptions;

Cost

Amount of pollutant reduced:

Targets on the amount of pollutants;

Required:

  1. We start by taking the following;

Unit of facility 1 = x

Unit of facility 2 = y

Unit of facility 3 = z

From the data above, we will formulate our linear programming model as below.

Linear programming model

Subject to

We move on to solving our linear programming model.

Subject to

We convert our problem to canonical form by adding slack, surplus and artificial variables appropriately.

  1. As the constraint-1 is of type ‘’, we should subtract surplus variableand add artificial variable

  2. As the constraint-2 is of type ‘’, we should subtract surplus variableand add variable

After introducing surplus, artificial variables;

Subject to

Iteration-1


Cj

30

20

40

0

0

M

M


B

XB

x

y

z

S1

S2

A1

A2

MinRatio

M

25

0.1

0.2

(0.4)

-1

0

1

0

M

35

0.45

0.25

0.3

0

-1

0

1

Z = 60M


Zj

0.55M

0.45M

0.7M

-M

-M

M

M




Zj-Cj

0.55M-30

0.45M-20

-M

-M

0

0




Positive maximumisand its column index is 3. So, the entering variable is.

Minimum ratio is 62.5 and its row index is 1. So, the leaving basis variable is A1.

Therefore, the pivot element is 0.4

Entering =, departing =key element = 0.4

R1(new) = R2(old) * 2.5

R1(old)

25

0.1

0.2

0.4

-1

0

0

R1(new)=R1(old)*2.5

62.5

0.25

0.5

1

-2.5

0

0

R2(new) = R2(old)-0.3R1(new)

R2(old)=

35

0.45

0.25

0.3

0

-1

1

R1(new)=

62.5

0.25

0.5

1

-2.5

0

0

0.3*R1(new)=

18.75

0.075

0.15

0.3

-0.75

0

0

R2(new)=R2(old)-0.3R1(new)

16.25

0.375

0.1

0

0.75

-1

1



Iteration-2


Cj

30

20

40

0

0

M


B

CB

XB

x

y

z

S1

S2

A2

MinRatio

z

40

62.5

0.25

0.5

1

-2.5

0

0

A2

M

16.25

0.375

0.1

0

(0.75)

-1

1


Z=16.25M+2500


Zj

0.375M+10

0.1M+20

40

0.75M-100

-M

M




Zj-Cj

0.375M-20

0.1M

0

0.75M-100

-M

0




Positive maximum Zj-Cj is 0.75M-100 and its column index is 4. So, the entering variable is S1.

Minimum ratio is 21.6667 and its row index is 2. So, the leaving basis variable is A2.

Therefore, the pivot element is 0.75

Entering = S1, Departing = A2, key element = 0.75

R2(new) = R2(old)*1.3333

R2(old)=

16.25

0.375

0.1

0

0.75

-1

R2(new)=R2(old)*1.3333

21.6667

0.5

0.1333

0

1

-1.333



R1(new) = R1(old)+2.5R2(new)

R1(old)=

62.5

0.25

0.5

1

-2.5

0

R1(new)=R1(old)+2.5R2(new)

116.6667

1.5

0.8333

1

0

-3.3333



Iteration-3


Cj

30

20

40

0

0


B

CB

XB

x

y

z

S1

S2

MinRatio

z

40

116.667

1.5

0.8333

1

0

-3.3333

S1

0

21.667

(0.5)

0.1333

0

1

-1.3333

Z=4666.6667


Zj

60

33.3333

40

0

-133.333




Zj-Cj

30

13.3333

0

0

-133.333




Positive maximum Zj-Cjis 30 and its column index is 1. So, the entering variable is x.

Minimum ratio is 43.3333 and its row index is 2. So, the leaving basis variable is S1.

Therefore, the pivot element is 0.5

Entering = x, departing = S1, Key element = 0.5

R2(new) = R2(old)*2

R2(old)=

21.6667

0.5

0.1333

0

1

-1.3333

R2(new)=R2(old)*2

43.333

1

0.2667

0

2

-2.6667



R1(new)=R1(old)-1.5R2(new)

R1(old)=

116.6667

1.5

0.8333

1

0

-3.3333

R1(new)=R1(old)-1.5R2(new)


51.6667

0

0.4333

1

-3

0.6667



Iteration-4


Cj

30

20

40

0

0


B

CB

XB

x

y

z

S1

S2

MinRatio

z

40

51.6667

0

(0.4333)

1

-3

0.6667

x

30

43.3333

1

0.2667

0

2

-2.6667

Z=3366.6667


Zj

30

25.3333

40

-60

-53.3333




Zj-Cj

0

5.3333

0

-60

-53.3333




Positive maximum Zj-Cjis 5.3333 and its column index is 2. So, the entering variable is y.

Minimum ratio is 119.2308 and its row index is 1. So, the leaving basis variable is z.

Therefore, the pivot element is 0.4333

Entering = y, departing = z, key element = 0.4333

R1(new) = R1(old) * 2.3077

R1(old)=

51.6667

0

0.4333

1

-3

0.6667

R1(new)=R1(old)* 2.3077

119.2308

0

1

2.3077

-6.9231

1.5385



R2(new) = R2(old) – 0.2667R1(new)

R2(old)=

43.3333

1

0.2667

0

2

-2.6667

R2(new) = R2(old) –0.2667R1(new)


11.5385

1

0

-0.6154

3.8462

-3.0769





Iteration-5


Cj

30

20

40

0

0


B

CB

XB

x

y

z

S1

S2

MinRatio

y

20

119.2308

0

1

2.3077

-6.9231

1.5385


x

30

11.5385

1

0

-0.6154

3.8462

-3.0769


Z=2730.7692


Zj

30

20

27.6923

-23.0769

-61.5385




Zj- Cj

0

0

-12.3077

-23.0769

-61.5385




Since all Zj- Cj

Hence, optimal solution is arrived with value of variables as:



  1. The optimal solution obtained satisfies all constraints. The amount of pollutants obtained in the optimal solution are within the range required by the government.

For pollutant 1,

For pollutant 2,



  1. Reducing the pollution target mean that the values of x and y will decrease. This in return will result to a decrease in the minimum cost. When pollutant 1 is reduced to 20 tons, the optimal cost will be reduced to a value close to the ratio of reduction. When the pollutant is increased to 30 tons, the optimal cost will increase.

  2. The cost coefficients were used to obtain the optimal cost to be incurred. The pattern observed is that the lesser the cost coefficient, the high the obtained value. This means that a less costly technology will be preferred in order to lower the cost of production. If the cost at facility 3 is reduced from $40 to $30, the technology involved will be adopted to a value similar to facility 1.



































Part B



Provided data:


C1

C2

C3

Production cost

Sale

Supervision requirement in hr per barrel

Minimum demand quantities (barrels)

Octane rating

Sulfur content

Gasoline 1

$6

$140

0.02

6000

96

1.5%

Gasoline 2

$6

$150

0.02

8000

98

1%

Gasoline 3

$5

$110

0.01

4000

86

2%

Gasoline 4

$5

$125

0.01

4000

88

2%

Purchase price

$65

$70

$75







Octane rating

88

95

99







Sulfur content

2%

1%

0.2%









Profit

Subject to

Solving the linear programme;

Part C


Salmon obtained(ton per month)

Transport to F1

Transport to F2

Production capacity


Region R1

200

$14

$11

150


Region R2

200

$13

$16

150










We formulate our program as follows

We get

Subject to

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate.

1. As the constraint-1 is of type '' we shouldsubtractsurplus variableS1and add artificial variableA1

2. As the constraint-2 is of type '' we should subtract surplus variableS2and add artificial variableA2

After introducing surplus, artificial variables

Max Z

=


10950

x1

+

15840

x2

+

0

S1

+

0

S2

-

M

A1

-

M

A2


subject to


14

x1

+

11

X2

-


S1




+


A1




=

2200


13

x1

+

16

x2




-


S2




+


A2

=

2600


and x1, x2, S1, S2, A1, A2≥ 0



Negative minimum Zj-Cjis -27M – 15480 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 162.5 and its row index is 2. So, the leaving basis variable is A2

Therefore, the pivot element is 16.

Entering = x2, departing = A2, Key element = 16



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

Minimum ratio is 81.4815 and its row index is 1. So, the leaving basis variable is A1.

Therefore, the pivot element is

Entering = x1, departing = A1, key element =

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

minimum ratio is 600 and its row index is 1. so, the leaving basis variable is x1.

Therefore, the pivot element is

Entering = S2, Departing = x1, key element =

Variable S1should enter into the basis, but all the coefficients in the S1column are negative or zero. So, S1can not be entered into the basis. Hence, the solution to the problem is unbound.



2. an increase in demand results to increase in transportation cost.

By factoring in the penalty costs, we will have







Part D

Provided data:

Total pollution before the projects = S units

Set of projects = P

Units of money = y

Time period = t

Interest rate = r%

Money available after t+1 = (1+r)y

Total initial budget = I

Outsource = Bt

Share = f

Total budget = allocatedtotal budget allocated in the previous period

Balance = E



Linear program

Subject to





Part E

Q1

  1. The model is infeasible because some constraints are relatively high as compared to other constraints.

  2. The model can be unbounded. Reason behind this is that it is infeasible meaning that some constraints will not be satisfied.

Q2

  1. The model will remain feasible. Since the value of optimal objective value is very high, the model will still be feasible.

  2. The optimal objective value will not remain the same. It will reduce.

Q3. An increase in resources results to an increase in objective goal. Therefore, the objective goal will be greater than 999.

Q4. Slack variables are tight and bounding when they are zero. Therefore, when the slack is positive the result become lose and less bounded. The demand satisfaction will then be reduced.







References

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Karmarkar, N., 1984, December. A new polynomial-time algorithm for linear programming. In Proceedings of the sixteenth annual ACM symposium on Theory of computing (pp. 302-311). ACM.

Zimmermann, H.J., 1978. Fuzzy programming and linear programming with several objective functions. Fuzzy sets and systems, 1(1), pp.45-55.

Candes, E.J. and Tao, T., 2005. Decoding by linear programming. IEEE transactions on information theory, 51(12), pp.4203-4215.

Luenberger, D.G., 1973. Introduction to linear and nonlinear programming.

Nash, S.G. and Sofer, A., 1996. Linear and nonlinear programming.