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Linear Programming
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12-06-2011 3:24pm#1
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Looking at last year's paper then - assuming you have that one.
11a
i
The equation of a line when you have two points (x1,y1) and (x2,y2) is
y-y1=((y2-y1)/(x2-x1)) * (x-x1)
In our case x1=0 y1=2 x2=4 y2=0.
So
y-2=((0-2)/(4-0))*(x-0)
y-2=-1/2x
y=-x/2 + 2
ii
The three inequalities which must be satisfied are given by
> Everything in the shaded region is above the x-axis, so y >= 0.
> Everything is to the right of the y-axis so x >=0
and everything is underneath the line k so
y <= -x/2 + 2.
b
i
x is the number of heavy containers
y is the number of light containers
Heavy container weighs 160
Light containers weigh 40
Truck can carry 2080
So you need the combined weight of x heavy containers and y light containers to be less than or equal to 2080. Each of the x heavy containers weighs 160 and each of the y light containers weighs 40, so the total weight is going to be 160x + 40y.
We need this to be less than or equal to 2080, so 160x + 40y <= 2080.
Similarly, you should be able to see that the time constraint gives us the inequality:
3x+2y<=54.
These are the two inequalities they are looking for.
(They neglect to mention the inequalities x >= 0, y >=0.)
ii
Heavy containers are worth 48e.
Light containers are worth 36e.
So you have 48e for each of the x containers, and 36e for each of the y containers.
You will want to sub the values into the four vertices of your shaded region on your graph (which you are asked to draw, but I am not going to do here...) into 48x + 36y and see which gives the maximum value.
It should be easy for you to read the vertices when you've constructed the graph.
Q: 160x + 40y <= 2080.
QQ: 3x+2y<=54.
Q crosses the x-axis at (13,0) and the y-axis at (0,52) (found by subbing in 0 for y and x respectively to the equation)
QQ cross the x-axis at (18,0) and the y-axis at (0,27).
The vertex (18,0) is outside the region (as (13,0) is smaller).
Similarly (0,52) is outside the region.
So two vertices are (13,0) and (0,27).
To find the vertex where Q and QQ cross each other, solve the simultaneous equations:
160x + 40y = 2080
3x+ 2y =54
160x + 40 y = 2080
-60x - 40y = -1080
so
100x = 1000
x=10. Gives y =12.
The third vertex is (10,12).
(0,0) is also a vertex, as you can't have negative x or y, but this one is trivial and can probably be safely ignored.
So you sub your three vertices: (10,12), (13, 0) and (0,27) into
48x + 36y.
48(10)+36(12)=912
48(13)+36(0)=624
48(0)+36(27)=972
So to maximise profit you produce 27 of type y (the light containers) and 0 of the heavy containers.
edit: Jesus... miles too slow. Stupid old brain and fingers slowing down
Why, back in my day... 0 -
Mysteriouschic wrote: »Try this it's really good at explaining it http://www.studentxpress.ie/ordinary/ordlinprog.htm
This one 2010 linear programming http://www.studentxpress.ie/ordpapers/ordlpsoln2010.pdf solutionLooking at last year's paper then - assuming you have that one.
11a
i
The equation of a line when you have two points (x1,y1) and (x2,y2) is
y-y1=((y2-y1)/(x2-x1)) * (x-x1)
In our case x1=0 y1=2 x2=4 y2=0.
So
y-2=((0-2)/(4-0))*(x-0)
y-2=-1/2x
y=-x/2 + 2
ii
The three inequalities which must be satisfied are given by
> Everything in the shaded region is above the x-axis, so y >= 0.
> Everything is to the right of the y-axis so x >=0
and everything is underneath the line k so
y <= -x/2 + 2.
b
i
x is the number of heavy containers
y is the number of light containers
Heavy container weighs 160
Light containers weigh 40
Truck can carry 2080
So you need the combined weight of x heavy containers and y light containers to be less than or equal to 2080. Each of the x heavy containers weighs 160 and each of the y light containers weighs 40, so the total weight is going to be 160x + 40y.
We need this to be less than or equal to 2080, so 160x + 40y <= 2080.
Similarly, you should be able to see that the time constraint gives us the inequality:
3x+2y<=54.
These are the two inequalities they are looking for.
(They neglect to mention the inequalities x >= 0, y >=0.)
ii
Heavy containers are worth 48e.
Light containers are worth 36e.
So you have 48e for each of the x containers, and 36e for each of the y containers.
You will want to sub the values into the four vertices of your shaded region on your graph (which you are asked to draw, but I am not going to do here...) into 48x + 36y and see which gives the maximum value.
It should be easy for you to read the vertices when you've constructed the graph.
Q: 160x + 40y <= 2080.
QQ: 3x+2y<=54.
Q crosses the x-axis at (13,0) and the y-axis at (0,52) (found by subbing in 0 for y and x respectively to the equation)
QQ cross the x-axis at (18,0) and the y-axis at (0,27).
The vertex (18,0) is outside the region (as (13,0) is smaller).
Similarly (0,52) is outside the region.
So two vertices are (13,0) and (0,27).
To find the vertex where Q and QQ cross each other, solve the simultaneous equations:
160x + 40y = 2080
3x+ 2y =54
160x + 40 y = 2080
-60x - 40y = -1080
so
100x = 1000
x=10. Gives y =12.
The third vertex is (10,12).
(0,0) is also a vertex, as you can't have negative x or y, but this one is trivial and can probably be safely ignored.
So you sub your three vertices: (10,12), (13, 0) and (0,27) into
48x + 36y.
48(10)+36(12)=912
48(13)+36(0)=624
48(0)+36(27)=972
So to maximise profit you produce 27 of type y (the light containers) and 0 of the heavy containers.
edit: Jesus... miles too slow. Stupid old brain and fingers slowing down Why, back in my day...
thank you so much
i actually understand it now 0 -
i got 6/50 for it in the mocks 
This discussion has been closed.
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