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# Give it a go: solution

Explore technology-assisted decision-making. Learn to optimize outcomes with advanced algorithms and techniques in different industries.

The company BioFuels Ltd manufactures fuels for the automotive transport system. All of their fuels are sustainably manufactured from biomass. BioFuel produces a E10 and E85 unleaded petrol. E10 unleaded petrol contains 90% traditional hydrocarbons, including octane, (typically from fossil fuels) and 10% bio-ethanol (from biomass). E85 unleaded petrol contains 15% traditional hydrocarbons, including octane, and 85% bio-ethanol.

E10 fuel is priced at £1.46 per litre and BioFuels Ltd makes £0.23 profit on each litre. The E85 fuel is prized at £1.97 per litre and BioFuels Ltd makes £0.46 profit on each litre.

## The solution

To find the solution to this problem you first have to formulate the problem as a linear program.

### Identify the decision variables

BioFuels Ltd is trying to determine the amount of E10 fuels and the amount of E85 fuel to manufacture. The decision variables are the amount of E10 fuel and the amount of E85 fuel that BioFuels Ltd should manufacture.

Let (x) be the amount of E10 fuel in litres that BioFuels Ltd should produce and let (y) be the amount of E85 fuel in litres that BioFuels Ltd should produce.

### Establish the constraints

There are five constraints in this formulation. Three of the constraints concern the quantities of bio-ethanol, octance and other hydrocarbons that go into each of the fuel types. All constraints will be formulated in litres.

Any solution to the problem can’t use more than 1500L of bio-ethanol as that is all that is available. The constraint can be modelling by the following inequality.

(0.1x + 0.85y leq1500)

Any solution to the problem can’t use more than 6000L of Octane as that is all that is available. The constraint can be modelling by the following inequality.

(0.81x + 0.12y leq 6000)

Any solution to the problem can’t use more than 4000L of other hydrocarbons as that is all that is available. The constraint can be modelling by the following inequality.

(0.09x + 0.03y leq4000)

There are also two implicit constraints which we need to model. It is assumed in the problem that the number of litres of E10 fuel must be greater than or equal to zero. That is, you can’t make a negative amount of E10 fuel. And likewise for E85, you can’t make a negative amount of E85 fuel so it can be assumed that the number of litres of E85 fuel must be greater than or equal to zero. These two constraints can be modelled by the following inequalities:

(x geq 0)

(ygeq 0)

Notice that any value of (x) and (y) which satisfy the five inequalities are a potential solution to BioFuels Ltd’s planning problem. The constraints are plotted on the graph below. If you observe closely, there is a small feasible region, which is bounded by a black lined polygon.

### Define the objective function

There are two objective functions. One for the profit, and one for the turnover.

The objective function for maximising the profit is:

(P = 0.23x + 0.46y)

The objective function for maximising the profit is:

(T = 1.46x + 1.97y)

Now that the formulation is complete, the questions can be answered:

• What is the best production plan for them to maximise their profit?
• How much E10 fuel should they make?
• How much E85 fuel should they make?
• What is their profit for the production plan?
• What is the best production plan for them to maximise their turnover?
• How much E10 fuel should they make?
• How much E85 fuel should they make?
• What is their turnover for the production plan?

## Maximising profit

Maximise (P = 0.23x + 0.46y)

Subject to (begin{cases} begin{align*} 0.1x + 0.85y &leq1500 \0.81x + 0.12y &leq 6000 \0.09x + 0.03y &leq4000\x &geq 0\y &geq 0.end{align*} end{cases})

The optimal solution which maximises the profit is when BioFuels Ltd produce (7272.73) L of E10 fuel, and (909.1) L of E85 fuel. The total profit will be £2090.91.

## Maximising turnover

Maximise (T = 1.46x + 1.97y)

Subject to (begin{cases} begin{align*} 0.1x + 0.85y &leq1500 \0.81x + 0.12y &leq 6000 \0.09x + 0.03y &leq4000\x &geq 0\y &geq 0.end{align*} end{cases})

The optimal solution which maximises the turnover is when BioFuels Ltd produce (7272.73) L of E10 fuel, and (909.1) L of E85 fuel. The total turnover will be £12409.10.

## Things to note

The constraint (0.09x +0.03y leq 4000)  is not a limiting factor in the decision about how much of each fuel to make. This can be observed by seeing that the constraint does not appear on the boundary of the feasible region.

Although in this example the solution that maximises the profit and the solution to maximises the turnover are the same, however, this is a coincidence. In general, if you are attempting to maximise two different quantities then you are likely to get two different solutions. When you use the graphical solution to solve this problem the order in which you encounter the vertices of the feasible region are the same in both the maximising the profit case and maximising the turnover case.

## Next step

In the next step, you will be invited to share this exercise with a family member or friend. This will help you consolidate your own learning and understanding – are you up for the challenge?