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I am really struggling with fully grasping how to understand this question. Can someone please assist me with understanding how to do the following question? I would be very grateful, thank you in advance….
A Manufacturer produces a component for diesel engines and a similar component for petrol engines. In the course of production, both components must pass through a machine centre and a testing centre. Diesel engine components spend 4 hours in the machine centre and 2 hours in the testing centre. Petrol engine components spend 2 hours in each centre. There are 16 hours available per day in the machine centre and 12 hours in the testing centre. A contract with a customer stipulates that at least 3 diesel engine components must be produced per day. Each diesel component that is produced earns a contribution of £60 and each petrol component earns £45. How many units of each component should be manufactured each day in order to maximise contribution?
For linear programming it is best you start from the top, what am I trying to achieve it is either to minimize cost or maximize contribution and in this scenario it appears to be contribution,
So my first thing to write down is contribution is maximized at the point where; lets makes diesel as x and petrol as y
So for contribution to be maximized it would be at the point where 60x+45y
So contribution = 60x + 45y
Step two is to identify the constraints, they are items not allowing us produce more, so for machine centre the constraint is 4x+2y is less than or equal to 16. As there are just 16 hrs available per day.
The second constraint is in the testing centre the constraint here is 2x+2y is less than or equal to 12hrs
The next item we are to write is x has to be greater than or equal to 3 as at least three must be produced each day
And last of course is the non-negativity formula which is x,y is greater than or equal to zeroSo the solution to plot in the graph is:
Our first constraint which is 4x+2y is less than or equal to 16, we need to make certain assumptions here to get a figure for x and y, so if x=0, y would become ?
That would mean 4(0)+2y=16, therefore y would become 16/2=8 and if we apply the same logic into y being 0, x would become 4x+2(0)=16, therefore x would become 16/4=4, so we’ve already established that x=4,y=8
So all you need do is plot a graph with this points in it
Our second constraint apply the same ideology from the first and it should give you x=6,y=6
Third line to draw is where equals 3
I can’t possibly draw a graph here apologies about that
The next step is to determine at what point is contribution the highestThe idea is to find the point your iso contribution line is towards the point of leaving the feasible region, this can be solved via a simultaneous equation,
Which is 4x+2y=16 equation 1 and 2x+2y=12 equation 2, once you subtract equation one from two you should get 2x=4, therefore x=2 putting that into equation 1, it becomes 4(2)+2y=16, therefore 2y = 16-8 and so y becomes 8/2=4
Putting these two into our contribution formula it becomes 60(2)+45(4)=300Hello D.O.A
Thank you so much for taking the time to explain this to me. upon checking what i had already done it seems that I have done exactly the same as what you have describe above. However, my tutor said that it is wrong Stating that X cannot =2 because of the stipulation of a minimum of 3 X
He did not give me more than that. do you have any idea why he would say that its wrong?
hello frankwilliams977
i can try to do that, the whole idea around the minimum 3 is this, take for example prior to us commencing production, a certain customer Mr Weng has already made an order for three components of X, what this means is that we have to consider a point in our graph where x is at least 3 and y is at any other point, which point y is we have to ensure that whichever point we decide on it has to maximize contribution where X is at least 3
Can I suggest that you watch our free lectures on linear programming?
Our lectures are a complete course for Paper F5 and cover everything you need to be able to pass the exam well.
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