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#1 Re: Help Me ! » Linear Programing - word problem confusing me ! » 2014-11-11 06:30:36

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

#2 Re: Help Me ! » Linear Programing - word problem confusing me ! » 2014-10-31 00:04:55

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

#3 Re: Help Me ! » Linear Programing - word problem confusing me ! » 2014-10-30 05:17:10

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

#4 Help Me ! » Linear Programing - word problem confusing me ! » 2014-10-30 03:28:05

Miss_band
Replies: 7

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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