Linear programming with absolute value objective function

In summary: That's what I'm trying to ask about...I don't understand how to solve the problem using IOR without absolute values.You don't have any absolute values any more---that was the whole point of introducing the new variable X'.
  • #1
csc2iffy
76
0

Homework Statement



Minimize |2x1-3x2|
subject to
x1+x2≤5
-x1+x2≥-1
x1≥0, x2≥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.

Homework Equations





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.
 

Attachments

  • abs val minimization.png
    abs val minimization.png
    5.8 KB · Views: 979
Physics news on Phys.org
  • #2
csc2iffy said:

Homework Statement



Minimize |2x1-3x2|
subject to
x1+x2≤5
-x1+x2≥-1
x1≥0, x2≥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.

Homework Equations





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.

I assume the question wants you to give an LP formulation---that is, a formulation that does *not* have any absolute values in it. Formulating such problems is standard, and can be found in any introductory LP or Operations Research book. You can even look on line.

RGV
 
  • #3
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 xi = xi+ - xi-
|xi| = xi+ + xi-

But I'm still confused as to how to solve this...
 
  • #4
Alright, here is my graphical analysis:

optimal value = 0, occurs at (0,0) and (3,2)
is this correct?
 
Last edited:
  • #5
This is what I have after doing some research online:

Let X = 2x1 - 3x2

---->
Minimize |X|
subject to
X ≤ X'
-X ≤ X'

---->
Minimize X'
subject to
2x1 - 3x2 ≤ X'
-2x1 + 3x2 ≤ X'

Question, what is X'?
 
  • #6
csc2iffy said:
This is what I have after doing some research online:

Let X = 2x1 - 3x2

---->
Minimize |X|
subject to
X ≤ X'
-X ≤ X'

---->
Minimize X'
subject to
2x1 - 3x2 ≤ X'
-2x1 + 3x2 ≤ X'

Question, what is X'?

I don't understand your question. The optimal X' will be the minimal value of |2x1 - 3x2|; after all, it is >= both (2x1 - 3x2) and -(2x1 - 3x2), and you minimize it.

RGV
 
  • #7
But I'm trying to solve it using IOR and I can't have a function with absolute values I thought?
 
  • #8
csc2iffy said:
But I'm trying to solve it using IOR and I can't have a function with absolute values I thought?

You don't have any absolute values any more---that was the whole point of introducing the new variable X'.

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

Solution: z = 3.

RGV
 
Last edited:

FAQ: Linear programming with absolute value objective function

1. What is linear programming with absolute value objective function?

Linear programming with absolute value objective function is a mathematical optimization technique used to find the maximum or minimum value of an objective function subject to certain linear constraints. The objective function includes an absolute value expression, which makes the optimization problem more complex.

2. How is linear programming with absolute value objective function different from traditional linear programming?

Unlike traditional linear programming, which involves optimizing a linear objective function, linear programming with absolute value objective function involves optimizing an objective function containing absolute value expressions. This makes the problem more challenging and may require the use of more advanced techniques and algorithms.

3. What are some applications of linear programming with absolute value objective function?

Linear programming with absolute value objective function can be used in various fields such as economics, finance, engineering, and operations research. Some common applications include portfolio optimization, production planning, resource allocation, and transportation network design.

4. What are the steps involved in solving a linear programming problem with absolute value objective function?

The steps for solving a linear programming problem with absolute value objective function are similar to traditional linear programming. They include formulating the problem, constructing the objective function, setting up the constraints, solving the problem using appropriate algorithms, and interpreting the results to make decisions.

5. What are some challenges associated with linear programming with absolute value objective function?

Linear programming with absolute value objective function can be more challenging than traditional linear programming due to the complexity of the objective function. It may also require the use of more sophisticated algorithms and techniques, and the interpretation of the results may be more difficult. Additionally, the problem may have multiple optimal solutions, which can be challenging to handle.

Similar threads

Replies
3
Views
638
Replies
2
Views
1K
Replies
2
Views
4K
Replies
3
Views
677
Replies
4
Views
3K
Replies
2
Views
987
Back
Top