##### HI6006 Competitive Strategy Editing Service

Delivery in day(s): 4

**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:

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.

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

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

After introducing surplus, artificial variables;

Subject to

Iteration-1 | C | 30 | 20 | 40 | 0 | 0 | M | M | ||

B | X | x | y | z | S | S | A | A | 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 | Z | 0.55M | 0.45M | 0.7M | -M | -M | M | M | ||

Z | 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 A_{1}.

Therefore, the pivot element is 0.4

Entering =, departing =key element = 0.4

R_{1}(new) = R_{2}(old) * 2.5

R | 25 | 0.1 | 0.2 | 0.4 | -1 | 0 | 0 |

R | 62.5 | 0.25 | 0.5 | 1 | -2.5 | 0 | 0 |

R_{2}(new) = R_{2}(old)-0.3R_{1}(new)

R | 35 | 0.45 | 0.25 | 0.3 | 0 | -1 | 1 |

R | 62.5 | 0.25 | 0.5 | 1 | -2.5 | 0 | 0 |

0.3*R | 18.75 | 0.075 | 0.15 | 0.3 | -0.75 | 0 | 0 |

R | 16.25 | 0.375 | 0.1 | 0 | 0.75 | -1 | 1 |

Iteration-2 | C | 30 | 20 | 40 | 0 | 0 | M | ||

B | C | X | x | y | z | S | S | A | MinRatio |

z | 40 | 62.5 | 0.25 | 0.5 | 1 | -2.5 | 0 | 0 | |

A | M | 16.25 | 0.375 | 0.1 | 0 | (0.75) | -1 | 1 | |

Z=16.25M+2500 | Z | 0.375M+10 | 0.1M+20 | 40 | 0.75M-100 | -M | M | ||

Z | 0.375M-20 | 0.1M | 0 | 0.75M-100 | -M | 0 |

Positive maximum Z_{j}-Cj is 0.75M-100 and its column index is 4. So, the entering variable is S_{1}.

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

Therefore, the pivot element is 0.75

Entering = S_{1}, Departing = A_{2}, key element = 0.75

R_{2}(new) = R_{2}(old)*1.3333

R | 16.25 | 0.375 | 0.1 | 0 | 0.75 | -1 |

R | 21.6667 | 0.5 | 0.1333 | 0 | 1 | -1.333 |

R_{1}(new) = R_{1}(old)+2.5R_{2}(new)

R | 62.5 | 0.25 | 0.5 | 1 | -2.5 | 0 |

R | 116.6667 | 1.5 | 0.8333 | 1 | 0 | -3.3333 |

Iteration-3 | C | 30 | 20 | 40 | 0 | 0 | ||

B | C | X | x | y | z | S | S | MinRatio |

z | 40 | 116.667 | 1.5 | 0.8333 | 1 | 0 | -3.3333 | |

S | 0 | 21.667 | (0.5) | 0.1333 | 0 | 1 | -1.3333 | |

Z=4666.6667 | Zj | 60 | 33.3333 | 40 | 0 | -133.333 | ||

Z | 30 | 13.3333 | 0 | 0 | -133.333 |

Positive maximum Z_{j}-C_{j}is 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 S_{1}.

Therefore, the pivot element is 0.5

Entering = x, departing = S_{1}, Key element = 0.5

R_{2}(new) = R_{2}(old)*2

R | 21.6667 | 0.5 | 0.1333 | 0 | 1 | -1.3333 |

R | 43.333 | 1 | 0.2667 | 0 | 2 | -2.6667 |

R_{1}(new)=R_{1}(old)-1.5R_{2}(new)

R | 116.6667 | 1.5 | 0.8333 | 1 | 0 | -3.3333 |

R | 51.6667 | 0 | 0.4333 | 1 | -3 | 0.6667 |

Iteration-4 | C | 30 | 20 | 40 | 0 | 0 | ||

B | C | X | x | y | z | S | S | 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 | Z | 30 | 25.3333 | 40 | -60 | -53.3333 | ||

Z | 0 | 5.3333 | 0 | -60 | -53.3333 |

Positive maximum Z_{j}-C_{j}is 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

R_{1}(new) = R_{1}(old) * 2.3077

R | 51.6667 | 0 | 0.4333 | 1 | -3 | 0.6667 |

R | 119.2308 | 0 | 1 | 2.3077 | -6.9231 | 1.5385 |

R_{2}(new) = R_{2}(old) – 0.2667R_{1}(new)

R | 43.3333 | 1 | 0.2667 | 0 | 2 | -2.6667 |

R | 11.5385 | 1 | 0 | -0.6154 | 3.8462 | -3.0769 |

Iteration-5 | C | 30 | 20 | 40 | 0 | 0 | ||

B | C | X | x | y | z | S | S | 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 | Z | 30 | 20 | 27.6923 | -23.0769 | -61.5385 | ||

Z | 0 | 0 | -12.3077 | -23.0769 | -61.5385 |

Since all Z_{j}- C_{j}

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

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,

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.

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 variableS_{1}and add artificial variableA_{1}

2. As the constraint-2 is of type '≥' we should subtract surplus variableS_{2}and add artificial variableA_{2}

**After introducing surplus, artificial variables**

| ||||||||||||||||||||||||||||||||||||||||

subject to | ||||||||||||||||||||||||||||||||||||||||

| ||||||||||||||||||||||||||||||||||||||||

and x |

Negative minimum Z_{j}-C_{j}is -27M – 15480 and its column index is 2. So, the entering variable is x_{2}.

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

Therefore, the pivot element is 16.

Entering = x_{2}, departing = A_{2}, Key element = 16

Negative minimum Z_{j}-C_{j}isand its column index is 1. So, the entering variable is x_{1}.

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

Therefore, the pivot element is

Entering = x_{1}, departing = A_{1}, key element =

Negative minimum Z_{j}-C_{j}isand its column index is 4. So, the entering variable is S_{2}

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

Therefore, the pivot element is

Entering = S_{2}, Departing = x_{1}, key element =

Variable S_{1}should enter into the basis, but all the coefficients in the S_{1}column are negative or zero. So, S_{1}can 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 = B_{t}

Share = f

Total budget = allocatedtotal budget allocated in the previous period

Balance = E

Linear program

Subject to

**Part E**

**Q1**

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

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

**Q2**

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

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.