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Linear programming with absolute value objective function 
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#1
Mar512, 07:59 PM

P: 76

1. The problem statement, all variables and given/known data
Minimize 2x_{1}3x_{2} subject to x_{1}+x_{2}≤5 x_{1}+x_{2}≥1 x_{1}≥0, x_{2}≥0 (a) Solve the problem graphically. (b) Formulate a linear program that could be used to solve the problem. Use software to solve your LP and show how to reconstruct a solution to the original problem. 2. Relevant equations 3. The attempt at a solution I started graphing the problem and posted an attachment of what I have. I usually solve these by way of the corner point method (just to check that my graphical method is correct), but once I found my corner points for this problem I found that 2 of them have the same value but are not adjacent CPF solutions (0,0)=0 (1,0)=1 (3,2)=0 (0,5)=15 This is what's confusing me... also, how would I show the graphical method when absolute value is involved? I started to try it, but I would like to know if I'm doing it correctly before going further. 


#2
Mar512, 09:44 PM

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RGV 


#3
Mar612, 03:18 PM

P: 76

Yes that's basically what I'm trying to ask about here. I tried looking in my book but I can't seem to find absolute value objective functions.
I remember my teacher talking about this but her example isn't about absolute value objective let x_{i} = x_{i}^{+}  x_{i}^{} x_{i} = x_{i}^{+} + x_{i}^{} But I'm still confused as to how to solve this... 


#4
Mar612, 05:21 PM

P: 76

Linear programming with absolute value objective function
Alright, here is my graphical analysis:
optimal value = 0, occurs at (0,0) and (3,2) is this correct? 


#5
Mar612, 08:21 PM

P: 76

This is what I have after doing some research online:
Let X = 2x_{1}  3x_{2} > Minimize X subject to X ≤ X' X ≤ X' > Minimize X' subject to 2x_{1}  3x_{2} ≤ X' 2x_{1} + 3x_{2} ≤ X' Question, what is X'? 


#6
Mar612, 08:27 PM

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#7
Mar612, 08:58 PM

P: 76

But I'm trying to solve it using IOR and I can't have a function with absolute values I thought?



#8
Mar612, 10:07 PM

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Watch me compute 3 without using absolute values: min z s.t. z >= 3 z >= 3 Solution: z = 3. RGV 


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