Linear programming - incorrect constraints

In summary, the question involves calculating the maximum profit available for a company that supplies electronic components to three customers with varying prices and transportation costs. The maximum profit with no restrictions on capacity is £451,200, with a total of 7200 units produced in Manchester and 5300 units produced in London. However, with the given restrictions, the total maximum output for both factories is 8000 units, and the transportation costs need to be included in the constraints to find the optimal solution. Using a spreadsheet or software is recommended to solve this problem.
  • #1
Foilist
10
1
Hi,

I have been struggling with Part 3 of this question for some time:

Computico Limited, currently operating in the UK, assembles electronic components at its two factories, located at Manchester and London, and sells these components to three major customers. Next month the customers, in units, orders are:

Customer A|Customer B|Customer C
3000|4200|5300

As Computicos three customers are in different industries it allows them to charge slightly different prices to different customers. For each component, Customer A pays £108, Customer B pays £105 and Customer C pays £116. The variable costs of assembling the components at the two factories vary because of different labour costs and power costs; these are £28 per unit at Manchester and £36 per unit at London.

The transportation costs per unit between each factory and customer are:

From To Customer Cost (£)
Manchester A 45
B 40
C 55
London A 42
B 43
C 44

1. Set up a table showing the contribution to profit of supplying one unit to each customer.

Table 2.1 below shows the contributions available and the profit gained when one unit is supplied to each customer. It can be seen that this contribution is different dependent upon the factory the goods are supplied from:

Factory Customer Transport Cost
(per Unit) Labour Cost
(per Unit) Price to Customer
(per Unit) Profit
(per Unit)
Manchester A -£28 -£45 £108 £35
B -£40 £105 £37
C -£55 £116 £33
London A -£36 -£42 £108 £30
B -£43 £105 £26
C -£44 £116 £36

Table 2.1 – Contribution to Profit

2. Assuming no restrictions on capacity at the two factories calculate the maximum total contribution from next months orders and the associated capacities for the two factories.

Table 2.2 below shows the profits available from each customer if they are supplied from either factory:

Factory Profit (£) Units Ordered Total Profit (£)
Customer A Manchester 35 3000 105,000
London 30 90,000
Customer B Manchester 37 4200 155,400
London 26 109,200
Customer C Manchester 33 5300 174,900
London 36 190,800

From this it can be seen that should there be no limit on productions, the maximum profit available for the company would be to supply:

Customer A – Manchester - £105,000 profit
Customer B – Manchester - £155,400 profit
Customer C – London - £190,800

Maximum Profit available with NO restrictions = £451,200

This would require a total of:

7200 units to be produced in Manchester and 5300 units to be produced in London.

Next months orders from the three customers does in fact exceed the total capacity for the two factories. The maximum output for Manchester is 4000 units and for London it is 6000 units.

3. Calculate the total contribution and indicate which customer will have unsatisfied demand.

With the restrictions stated above, it is clear that the company will be unable to meet the needs of all customers, however the company will still wish to maximise profit. Let Customer A = X1, Customer B = X2 and Customer C = X3.

The objective function becomes:

MAXIMISE 35X1 + 37X2 + 36X3

Subject to:

3000X1¬ + 4200X2 + 5300X3 ≤ 10,000 - Maximum units which can be produced.

3000X1 + 4200X2 ≤ 4000 - Units produced in Manchester

5300X3 ≤ 6000 - Units produced in London

X1, X2 and X3 ≥ 0

With 3 variables, the Simplex Method will be used to calculate the optimal solution, the first step of which requires the introduction of Slack Variables. The constraints therefore become:

3000X1¬ + 4200X2 + 5300X3 + S1 = 10,000

3000X1 + 4200X2 + S2 = 4000

5300X3 + S3 = 6000

These are then entered into Table 2.3 below:

Products Slack Variables Solution
Quantity
Row Solution Variable X1 X2 X3 S1 S2 S3
1 S1 3000 4200 5300 1 0 0 10000
2 S2 3000 4200 0 0 1 0 4000
3 S3 0 0 5300 0 0 1 6000
4 Z 35 37 36 0 0 0 0

When this is progressed, using the simplex method, through the iterations the answer achieved is meaningless. I have claculated it using both pen and paper and some software, both of which arrive at the same answer. This would indicate to me that my constraints are wrong - any help appreciated!
 
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  • #2


Hello,

Thank you for sharing your question and progress with us. It seems like you have a good understanding of the problem and have made some progress in solving it. However, I would like to offer some suggestions and clarifications to help guide you towards the correct solution.

Firstly, in part 2 of the question, you have correctly calculated the maximum profit available if there were no restrictions on capacity. However, in part 3, you have assumed that the maximum output for Manchester is 4000 units and for London it is 6000 units. This is incorrect as the question states that the maximum output for Manchester is 4000 units and for London it is 4000 units. This means that the total maximum output for both factories is 8000 units, not 10000 as you have used in your constraints.

Secondly, in your constraints, you have not taken into account the transportation costs per unit between each factory and customer. These costs need to be included in your constraints as they will affect the total contribution and therefore the optimal solution. For example, for Customer A, if they are supplied from Manchester, the total contribution would be £108 - £28 - £45 = £35 per unit. However, if they are supplied from London, the total contribution would be £108 - £36 - £42 = £30 per unit. This difference needs to be reflected in your constraints.

Lastly, I would suggest using a spreadsheet or software to solve this problem as it involves multiple constraints and variables. This will make it easier to keep track of your calculations and to make changes if needed.

I hope this helps and good luck with finding the optimal solution!
 

What is linear programming?

Linear programming is a mathematical method used to optimize a system with linear constraints and a linear objective function. It involves finding the maximum or minimum value of a linear function, subject to a set of linear constraints.

What are constraints in linear programming?

Constraints in linear programming are restrictions that must be satisfied in order to find the optimal solution. They can be in the form of equations or inequalities and limit the values that the decision variables can take.

What are incorrect constraints in linear programming?

Incorrect constraints in linear programming refer to constraints that do not accurately represent the problem or the limitations of the system. This can lead to an incorrect or invalid solution.

What are the consequences of incorrect constraints in linear programming?

The consequences of incorrect constraints in linear programming can include an invalid solution, an infeasible problem, or an incorrect optimal solution. It can also lead to wasted time and resources if the problem needs to be re-evaluated with correct constraints.

How can incorrect constraints in linear programming be avoided?

Incorrect constraints in linear programming can be avoided by thoroughly understanding the problem and its limitations before formulating the constraints. It is also important to carefully review the constraints to ensure they accurately represent the problem and do not contradict each other.

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