You will not these days be asked to draw a graph (because of computer based exams).
However you must be able to understand the graph because it is very likely that the examiner will give you a graph and expect you to calculate things from it.
(Solving without a graph will not work – just solving individual pairs of equations together will likely give solutions that break other constraints, and without the graph this will not be clear.)
Thanks John for the presentation. Solving for the optimum solution graphically involves plotting all of the constraints coordinates arrived at in formulating the problem.
Sir, I do checking for point A,C,D contribution. Is it correct as below? I understood that is no need to calculate in the exam. Thanks a lot. Point A,(E=0, S=36) Max C= (6×36)+0 = $216<= Point C, Mats; 2S+4E=80 Demds; E=10 Substitute: 2S+40=80 S=20 Max C=(6×20)+(9×10)= $210 <= Point D; (S=0,E=10) Max C=0+90 = $90 <= Therefore, The optimum point is B ($225).
You will not be expected to check the other corners in the exam – you are expected to understand the idea of using the iso-profit line to find the ‘best’ corner.
This one was excellent presentation.Even better than kaplan.
Thank you 馃檪
Hi John
Great lectures indeed, thank you Sir.
Thank you for your comment 馃檪
Dear Sir if examiner require us to what is feasible region what its proper answer should be?
The area on the graph representing all possible combinations of the two products that can be produced given the constraints.
I put E on the y axis and S x axis and 2x+4y=80
5x+6y=180
Does it matter the exam which way I do it
No, it doesnt matter (but appreciate that you cannot be asked to draw the graph in the exam).
Hi John, I wanted to ask how you got the Material and labour calculation.
Material: If S is 0, E= 20, If E= 0, S= 40
Labour If S =0, E= 30 & If E= 0, S =36.
I know it is simple calculation but didn’t get this.
Thank you so much.
For materials, 2S + 4E = 80
If you put S equal to 0, then 0 + 4E = 80
so 4E = 80. So E = 80/4 = 20
If you put E equal to 0, then 2S + 0 = 80
So 2S = 80. So S = 80/2 = 40
Do exactly the same for the labour equation 馃檪
Thank you so much! 馃檪
You are welcome 馃檪
Dear sir,
is it ok if we dont draw the graph in the exam ?
and just slove it with the help of equation ?
thanks
Krupali
You will not these days be asked to draw a graph (because of computer based exams).
However you must be able to understand the graph because it is very likely that the examiner will give you a graph and expect you to calculate things from it.
(Solving without a graph will not work – just solving individual pairs of equations together will likely give solutions that break other constraints, and without the graph this will not be clear.)
Sir,what about paper based exams? Is it possible a linear programming graph to be asked in section C of the paper
You can no longer be asked to draw a graph in the paper based exams.
(It is to make things fair to everyone, because it is impossible to ask you to draw graphs in the computer based exams 馃檪 )
Thanks John for the presentation. Solving for the optimum solution graphically involves plotting all of the constraints coordinates arrived at in formulating the problem.
Yes – and the graphical approach is the only approach in the syllabus for F5.
Noted sir
Sir,
I do checking for point A,C,D contribution. Is it correct as below? I understood that is no need to calculate in the exam. Thanks a lot.
Point A,(E=0, S=36)
Max C= (6×36)+0 = $216<=
Point C, Mats; 2S+4E=80
Demds; E=10
Substitute: 2S+40=80
S=20
Max C=(6×20)+(9×10)= $210 <=
Point D; (S=0,E=10)
Max C=0+90 = $90 <=
Therefore, The optimum point is B ($225).
Sorry Sir. I am sleepy. This checking should be next video 馃槢
You will not be expected to check the other corners in the exam – you are expected to understand the idea of using the iso-profit line to find the ‘best’ corner.