Linear programming - profit maximisation

In summary, Gordon Ltd is contract to supply 1000 tennis and 2000 badminton racquets. There are only 4000kg of material and 50000 hours of labour available in the period. The company wishes to maximise profit in the period.
  • #1
jb7
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Homework Statement



I can do most the question, but just get stuck on the final question. Here is the whole question



1. Gordon Ltd makes 2 products, Tennis racquets and badminton racquets, each using the same materials and the same skilled labour.

The costs of the products per unit of production are as follows:

......... ...Tennis...Badminton
......... ....£......£
Selling price.......... 140 .....120
Materials @ £18 per KG........9.... ...6
Labour @ £12 per hour........60..... ... 60
Other variable cost.......18... ...12
Allocation of fixed cost........ 33......22

Profit per unit.........20. ......20

The company is drawing up production plans for the 3 months to 31 March 2010. The anticipated demand in the period is for 6000 tennis racquets and 6000 badminton racquets.

There are only 4000kg of material and 50000 hours of labour available in the period.

The company has a contract to supply 1000 tennis and 2000 badminton racquets which must be satisfied. The company wishes to maximise profit in the period


(a) Formulate a linear programming model for this problem. (10 marks)

(b) Use the graphical method to determine how many doors and windows should be produced (20 marks)

(c) What are the shadow prices of materials and labour? What do these prices mean? (10 marks)

(d) If new supplies of materials became available at £15 per kg should they be purchased? If so how much extra material should be bought? (10 marks)

I can do up to part C, i.e calculate the shadow prices for materials and labour. However I cannot figure out how to do part D! - is it something to do with the shadow prices calculated in part C? My thoughts of how to do it was to do the WHOLE THING again, but with materials priced at £15 per kg, rather then at £18 per kg as they were initially...but something tells me that doing this incorrect. Anyone have any idea how to do this part D?



Homework Equations



Maximise C = 53x+42y

subject to these contraints

0.5x+0.33y<=4000
5x+5y<=5000
x>=1000
y>=2000
x<=6000
y<=6000

The Attempt at a Solution



no idea where to start with part D (don't need the solutions to the other parts.. can do them just fine :)) thanks
 
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  • #2
any ideas of where to start? only need help with the question in bold..
 
  • #3
profit was maximised where the materials and labour constraints crossed, which gave 4000 units of x (tennis rackets), and 6000 units of y (badminton rackets), which is indeed more then the contract, which is 1000 units of x and 2000 units of of y. However, I have just noticed that at the start of the question it specified that there is demand for 6000 x and 6000 y, although my profit maximisation only gives 4000x and 6000y. Is this "part D" something to do with buying enough materials so that those extra 2000 units of x are sold, so that demand for x is met at 6000 units ? (demand for y however seemed to be already met with the current maximisation).

Furthermore, the shadow price for materials was calculated as £66 (part C), although I don't know if this has anything to do with helping me solve part D?

Still not sure where to start though, if materials are now available at £15 per kg, this would change the materials constraint to 0.6x + 0.4y <=4000. Should I just solve the whole thing again using this constraint?
 
  • #4
Draw those constraints on a graph- that is, draw the lines with "=" rather than ">" or "<" to find the boundary of the "feasible region. The maximum and minimum values of the target function will always occur at a vertex of that region.
 
  • #5
HallsofIvy said:
Draw those constraints on a graph- that is, draw the lines with "=" rather than ">" or "<" to find the boundary of the "feasible region. The maximum and minimum values of the target function will always occur at a vertex of that region.

thanks for your reply on how to do part D, so basically I just have to do the whole thing again, but with the new constraint where materials are available at £15 per kg? (i.e plot 0.6x+0.4y=4000?) - and then maximise using the same objective function and method from parts A and B?

cheers
 

What is linear programming?

Linear programming is a mathematical method used to determine the optimal solution to a problem with linear relationships between variables. It is commonly used in business and economics to determine the best course of action for profit maximisation.

What is profit maximisation in linear programming?

Profit maximisation is the process of finding the combination of variables that will result in the highest possible profit. In linear programming, this is achieved by setting up a mathematical model with constraints and objective functions to determine the optimal values of decision variables.

What are the constraints in linear programming?

Constraints in linear programming are limitations or restrictions on the decision variables that must be taken into account when finding the optimal solution. These can include resource limitations, budget constraints, and operational constraints.

What is the objective function in linear programming?

The objective function in linear programming is the mathematical expression that represents the goal of the problem, usually to maximise or minimise a certain quantity. It is typically a linear combination of the decision variables and is used to determine the optimal solution.

What are some real-world applications of linear programming for profit maximisation?

Linear programming has many real-world applications in industries such as manufacturing, transportation, and finance. It can be used to determine the optimal production levels of different products, the most efficient transportation routes, and the best investment decisions for maximum profit.

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