how to use piecewise linear constraint in pulp?

[pancerZH]

For example, I have one constraints goes:
x + y >=2 or x + y ==2
there are many such constraints in my model, so how to express the or in pulp?

[naveenashwa77]

import pulp

#define whether you gonna minimize or maximize ur objective
lp_solve = pulp.LpProblem("My LP Problem", pulp.LpMaximize or pulp.LpMinimize)

#define your variables
x = pulp.LpVariable('x', lowBound=0, cat='Continuous')
y = pulp.LpVariable('y', lowBound=2, cat='Continuous')

define Objective function

lp_solve +=

set your Constraints

lp_solve +=

#call solver
lp_solve.solve()
for variable in lp_solve.variables():
print "{} = {}".format(variable.name, variable.varValue)
print pulp.value(lp_solve.objective)

[stumitchell]

your example seems to be strange as x+y >=2 includes x+y == 2 so you only need the first. however in general you will need to at binary integer decision variables to your model to use 'or' constraints Stu Stuart Mitchell PhD Engineering Science Extraordinary Freelance Programmer and Optimisation Guru www.stuartmitchell.com

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On Mon, Apr 30, 2018 at 3:50 AM, naveenashwa77 ***@***.***> wrote: import pulp #define whether you gonna minimize or maximize ur objective lp_solve = pulp.LpProblem("My LP Problem", pulp.LpMaximize or pulp.LpMinimize) #define your variables x = pulp.LpVariable('x', lowBound=0, cat='Continuous') y = pulp.LpVariable('y', lowBound=2, cat='Continuous') define Objective function lp_solve += set your Constraints lp_solve += (x+y) == 25 - x lp_solve += (x + y) >= 2 #call solver lp_solve.solve() for variable in lp_solve.variables(): print "{} = {}".format(variable.name, variable.varValue) print pulp.value(my_lp_problem.objective) — You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub <#173 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AAtEWeokYDe9UKpd8Z8qJsBofRjuQJ8Xks5tteEogaJpZM4TrY7C> .

[pancerZH]

Thank you for help but so sorry to find I made a mistake, the constraints should be:
x + y >= 2 or x + y == 0
I have tried to write them like this:
lp_solve += (x+y)>=2
lp_solve += (x+y)==0
It just did not work(obviously) and no result returned.
And I can't use constraint as lp_solve += (x+y)>=0, because I don't want the (0, 2) part.
Then how to deal with it?

[stumitchell]

have a look at this video https://www.youtube.com/watch?v=-3my1TkyFiM Stuart Mitchell PhD Engineering Science Extraordinary Freelance Programmer and Optimisation Guru www.stuartmitchell.com

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On Mon, Apr 30, 2018 at 1:08 PM, 张文喆 ***@***.***> wrote: Thank you for help but so sorry to find I made a mistake, the constraints should be: x + y >= 2 or x + y == 0 I have tried to write them like this: lp_solve += (x+y)>=2 lp_solve += (x+y)==0 It just did not work(obviously) and no result returned. And I can't use constraint as lp_solve += (x+y)>=0, because I don't want the (0, 2) part. Then how to deal with it? — You are receiving this because you commented. Reply to this email directly, view it on GitHub <#173 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AAtEWcAHMNp4oMyP5EZhtFoF7xzj-Agkks5ttmPwgaJpZM4TrY7C> .

[pancerZH]

Thank you for help! Finally I solved it with nonlinear programming : )