Linear Programming word problem confusing me

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Miss_band

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Oct 30, 2014
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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
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
 
You will also need the costs of the material and, for the constraints, the amount of material available. It will possibly be better to write the costs of the material as costs per kg, i.e. material A cost $0.200/kg, B is $0.100/kg, and C is $0.065/kg.

Now just start writing down what you need:
Let F1 (F2, F3) be the number of bags of Feed 1 (2, 3) sold. Then
Income from Feed 1 = 30 F1
Cost for Feed 1 = (0.200 * 40 + 0.100 * 40 + 0.065 * 20) F1
Use of material A for Feed 1 = 40 F1
Use of material B for Feed 1 = 40 F1
Use of material C for Feed 1 = 20 F1
Similarly for Feed 2 and Feed 3. Take it from there.
 
Thanks for the reply Ishuda, I am still trying to figure this out
What I have so far is
Income from Feed 1 = 30F1
Income from Feed 2 = 20F2
Income from Feed 3 = 10F3
Cost for Feed 1 = (0.2 * 40 + 0.1 * 40 + 0.065 * 20) F1 = 10.3F1
Cost for Feed 2 = (0.2 * 20 + 0.1 * 50 + 0.065 * 30) F2 = 10.95F2
Cost for Feed 3 = (0.1 * 65 + 0.065 * 35) F31 = 8.775F3
I presume Profit = Income – Cost
Profit for Feed 1 = 19.7
Profit for Feed 1 = 9.05
Profit for Feed 1 = 1.225
I now have the following table
Material A
Material B
Material C
Price per 100kg
Profit
Feed 1 (F1)

40​

40​

20​

30​

19.7​
Feed 2 (F2)

20​

50​

30​

20​

9.05​
Feed 3 (F3)

65​

35​

10​

1.225​
Cost/kg

0.2​

0.1​

0.0065​

So the answer I have is
Max P = 19.7F1 + 9.05F2 + 1.225F3
Subject to
40F1 + 20F2 <= 0.2
40F1 + 50F2 + 65F3 <= 0.1
20F1 + 30F2 + 35F3 <= 0.065

Am I on the right track?
Thanks
 
Almost there but I would recompute the cost of Feed 1 if I were you and those last three equations (the constraints) are incorrect. What does 40 F1 + 20 F2 represent? It isn't the cost of material A
 
Ah yes, the cost of Feed 1 is 13.3, therefore the profit is 16.7
so the Max P = 16.7F1 + 9.05F2 + 1.225F3
Subject to
40F1 + 20F2 <= 10000
40F1 + 50F2 + 65F3 <= 7500
20F1 + 30F2 + 35F3 <= 5200
Am I close?
 
Miss_band said:
Ah yes, the cost of Feed 1 is 13.3, therefore the profit is 16.7
so the Max P = 16.7F1 + 9.05F2 + 1.225F3
Subject to
40F1 + 20F2 <= 10000
40F1 + 50F2 + 65F3 <= 7500
20F1 + 30F2 + 35F3 <= 5200
Am I close?
Click to expand...
40 F1 + 20 F2 represent how much of material A you will use (40 kilos for each bag of feed 1 and 20 kilos for each bag of feed 2). So how much do you have in total of material A? What about material B and material C?
 
Hi, Also, as part of this question I need to find "the ranges for the objective co-efficients" ? Can anyone help me with that? Thanks, Martha
 
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