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#1 2014-10-30 03:28:05

Miss_band
Member
Registered: 2014-10-30
Posts: 4

Linear Programing - word problem confusing me !

Hi,
I'm having a problem with the following word problem. The whole thing about the quantities of material available is throwing me because the numbers are huge.

A pig feed company produce three different qualities of pig feed. In each product there are a blend of three raw materials.
Feed 1 sells at $30 per 100kg bag and is made of 40% of material A, 40% of material B and 20% of material C.
Feed 2 sells at $20 per 100kg bag and is made of 20% of material A, 50% of material B and 30% of material C.
Feed 3 sells at $10 per 100kg bag and is made of 65% of material B and 35% of material C.

Material A costs $200 per 1000kg and there are 50,000kg available.
Material B costs $100 per 1000kg and there are 75,000kg available.
Material C costs $65 per 1000kg and there are 80,000kg available.

Re-write this as a as a linear programming problem showing the Max and constraints ? ?

What I have so far is this and I’m not sure if I now multiply the $200 by 50, and so on…. ? ?

          Material A        Material B        Material C        Price per 100kg
Feed 1     40                  40                  20                       30
Feed 2     20                  50                  30                       20
Feed 3                           65                  35                      10

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#2 2014-10-30 04:00:39

Bob
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Registered: 2010-06-20
Posts: 10,058

Re: Linear Programing - word problem confusing me !

hi Miss_band

Welcome to the forum.

What you have is ok so far.  It is a bit confusing that they keep jumping from 100s to 1000s to 10000s.  So I would stick to Kg and adjust everything as you make the equations.

Consider what are your variables.  The question wants us to decide how much of each feed to make, so call the amounts (in Kg) x, y and z.

Now you have to make the constraints.

If you make x Kg  of feed 1, then you'll need A =  0.40x Kg,     B = 0.40x Kg,        C =  0.20x Kg

Do similar for feed 2 and feed 3.

So the total of material A you'll need will be 0.40x + 0.20y  + 0z.  This must be less than 50000

Get two more constraints like this.

Now what about your profit.  The cost of x will be  $ x/1000 times 200 = $ x/5 = $ 0.20x

Get two more expressions like this.

Then work out your sales (assuming you sell it all)

For x you'll sell at $ 30x/100 = $ 0.30x.  Do the same for y and z.

Then get the profit expression which will be sales minus costs.  That's what you want to maximise.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2014-10-30 05:17:10

Miss_band
Member
Registered: 2014-10-30
Posts: 4

Re: Linear Programing - word problem confusing me !

Hi Bob,

Thank you for the reply and I think I'm starting to see this a little clearer.

I'm down as far as getting the constraints and I think that's fine as I have

0.40x + 0.20y  <= 50000
0.40x + 0.50y + 0.65z <= 75000
0.20x + 0.30y + 0.35z <= 80000

After that is maybe where I'm going wrong. I've calculated the cost as

$ x/1000 times 200 = $ x/5
$ y/1000 times 100 = $ y/10
$ z/1000 times 65 = $ 13z/200

I've calculated the sales as

For x you’ll sell at $ 30/100
For y you’ll sell at $ 20/100
For z you’ll sell at $ 10/100

Now you say the profit expression is sales minus the cost. When I do that I get

Max P = (30/100 – x/5) + (20/100 – y/10) + (10/100 – 13z/200)

This is where I think I might be gone wrong because the examples in my book seem to have a "neater" looking profit expression, usually something like Max P = 40x + 50y +30z

Thanks, Martha

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#4 2014-10-30 06:10:00

Bob
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Registered: 2010-06-20
Posts: 10,058

Re: Linear Programing - word problem confusing me !

hi Martha

Good effort.  smile

The sales need x, y and z in them.

And I think you'll find it easier to write the costs and sales as decimals  eg    $0.20x .  Then the 'sums' are easier.

The final expression will have 0.1x etc.

But don't worry.  There is a simple reason for this: I suggested working in Kg.  I actually started with 100s of Kg but it was getting too complicated when 1000s came up as well so I simplified.  But it wouldn't be wrong to work in 100s or even in 1000s.  The question doesn't say.  With a two variable problem you can plot the constraints as lines and shade areas.  The 'profit' expression can be made into an equation by choosing an arbitrary profit value, plotting that line, and then tracking up the graph with parallel lines until we find the last allowable point as the parallel lines run out of the solution space.  So you can take your profit line, multiply every term by 10 or 100 or 1000 and it will still be a profit line and you can still find the maximum.

With 3 variables you cannot draw a graph but the same principle applies.  Just scale up all the terms until you have the book values.  If you actually went all the way to a solution, one person might have x = 2500 Kg say and another might have x = 2.5 hundreds of Kg.  Same answer.

If you want to complete the solution of  this problem you could go to a site I found recently for someone else:

http://www.zweigmedia.com/RealWorld/simplex.html

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#5 2014-10-31 00:04:55

Miss_band
Member
Registered: 2014-10-30
Posts: 4

Re: Linear Programing - word problem confusing me !

Hi Bob,

Again thank you for the clear explanation.
I added the  x, y and z to the sales and converted them to decimals as you suggested and I was left with

Max P = 0.1x + 0.1y + 0.035z

If I multiply that line by 1000 I get Max P = 100x + 100y + 35z

Just wondering if I multiply the profit line by 1000, do I also have to multiply the constraints by 1000 ?

I'm guessing not because 50000 x 1000 = 50000000 which seems like a very big number to be dealing with.

Thanks for the tip about that site, it's very helpful

Thanks, Martha

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#6 2014-10-31 21:05:18

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Linear Programing - word problem confusing me !

hi Martha,

You cannot alter the constraints because they are fixed by the numbers given in the problem.  But the 'profit' expression is always a movable object because you don't know the value of P when you start the problem.

10 = 0.1x + 0.1y + 0.035z

0.1 = 0.1x + 0.1y + 0.035z

5 = 100x + 100y + 35z

are all possible profit lines.  Solving the problem means finding the biggest P satisfying the constraints.  So it won't matter which P equation you use.

Bob

ps.  btw I agree with your values for P


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#7 2014-11-11 06:30:36

Miss_band
Member
Registered: 2014-10-30
Posts: 4

Re: Linear Programing - word problem confusing me !

Hi,

I handed in my solution and it came back to me because I forgot to complete a part.
They want me to calculate the ranges for the objective co-efficients and I just can't see anywhere that explains how to do that....

I found the optimum point and the shadow prices, but what is actually meant by "the ranges for the objective co-efficients" ?

Thanks, Martha

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#8 2014-11-11 07:00:02

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Linear Programing - word problem confusing me !

hi Martha,

I'm not sure what is meant by this.  The objective function is the expression you are trying to maximise.  I've been calling it the 'profit' line.  So the coefficients are the numbers in front of x, y and z.  Why these would have 'ranges' is a mystery.  Please post your solution (also the exact wording of this part of the question),  and I'll give it some more thought.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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