# 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) 116.667 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)= 116.667 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 119.231 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

Charnes, A. and Cooper, W.W., 1962. Programming with linear fractional functionals. Naval Research logistics quarterly, 9(3‐4), pp.181-186.

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.