Passing set of variable data and optimise based on fitted curve

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Passing set of variable data and optimise based on fitted curve

by jeromecjl :: Rate this Message:

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Hello,

How do I pass a set of variable data into lpSolve for the following optimization?

Obj function: (max) revenue = price * volume

Constraints: price and volume pair must be from the following variable data set:

Variable data set:
#  price   volume
1  10      500
2  20      450
3  30      330
4  40      250
5  50      190

Result: 10,000 (variable row#4)

Could it be possible for lpSolve to generate a profit curve and find the optimum point (max revenue) in which the result of price could be based on the max point in the curve-fitted profit function (say price between 30 and 40)?

Example:
Fitted to: revenue = -700price^2 + 5200price + 780
Subject to: price range between 10 to 50

Thanks,
Jerome


RE: Passing set of variable data and optimise based on fitted curve

by William H. Patton :: Rate this Message:

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Neither of your sample objectives are linear (Price*volume) or price^2.

 

With some cleverness and thought about calculus, you might manage a
piecewise

Linear approximation. That is: you know the slopes between the breakpoints

See SOS2 examples in the docs.

http://lpsolve.sourceforge.net/5.5/SOS.htm

http://yetanothermathprogrammingconsultant.blogspot.com/2009/06/gams-piecewi
se-linear-functions-with.html

http://www.pitt.edu/~jvielma/presentations/LOG_ISMP_09.pdf

 

William

 

  _____  

From: lp_solve@... [mailto:lp_solve@...] On Behalf
Of jeromecjl
Sent: Friday, November 06, 2009 12:11 AM
To: lp_solve@...
Subject: [lp_solve] Passing set of variable data and optimise based on
fitted curve

 

 

Hello,

How do I pass a set of variable data into lpSolve for the following
optimization?

Obj function: (max) revenue = price * volume

Constraints: price and volume pair must be from the following variable data
set:

Variable data set:
# price volume
1 10 500
2 20 450
3 30 330
4 40 250
5 50 190

Result: 10,000 (variable row#4)

Could it be possible for lpSolve to generate a profit curve and find the
optimum point (max revenue) in which the result of price could be based on
the max point in the curve-fitted profit function (say price between 30 and
40)?

Example:
Fitted to: revenue = -700price^2 + 5200price + 780
Subject to: price range between 10 to 50

Thanks,
Jerome