pulp

问题 I've been using PuLP to solve a particular mixed integer linear program (MIP) I am interested in. However, as the problem size grows, PuLP is taking too long. I want to be able to run the solver for some time and terminate it prematurely if its taking to long and obtain the best feasible solution so-far computed. I have tried manually timing the solver out with signal, but the variables are all "None". I've looked at the documentation and PuLP does not seem to support this, though as I

问题 I have a problem that I don't know how to add indicator constraints in pulp . Can anyone help me? For example: I have a decision variable x[(i,j)] , LpBinary and a continuous variable u[i] When x[(i,j)] equals 1 , then u[i] + q[j] == u[j] ( q is just the demand of customers) Thank you for your helping. 回答1: Welcome to SO! My interpretation of your question is that you have binary variables x[(i,j)] , and continuos variables u[i] . When x[(i,j)]==1 then you want to enforce a constraint as

问题 I have a problem that I don't know how to add indicator constraints in pulp . Can anyone help me? For example: I have a decision variable x[(i,j)] , LpBinary and a continuous variable u[i] When x[(i,j)] equals 1 , then u[i] + q[j] == u[j] ( q is just the demand of customers) Thank you for your helping. 回答1: Welcome to SO! My interpretation of your question is that you have binary variables x[(i,j)] , and continuos variables u[i] . When x[(i,j)]==1 then you want to enforce a constraint as

问题 I've initialised my LpVariable like so: x = LpVariable('x', None, None) At this point, my variable has upper and lower bounds as float('inf') and float('-inf') . Now based on some parameters of my logic, I want to bound the upper limit of this variable to, say, any x < 20 . Can I only do this by adding in an LpProblem and modifying the Variable using the problem parameters? y = LpProblem('Minimizing Problem', LpMinimize) y += x < 20 Or is there another way to manipulate the variable? Changing

问题 I've initialised my LpVariable like so: x = LpVariable('x', None, None) At this point, my variable has upper and lower bounds as float('inf') and float('-inf') . Now based on some parameters of my logic, I want to bound the upper limit of this variable to, say, any x < 20 . Can I only do this by adding in an LpProblem and modifying the Variable using the problem parameters? y = LpProblem('Minimizing Problem', LpMinimize) y += x < 20 Or is there another way to manipulate the variable? Changing

问题 I have some Pyomo code (see below) which produces one possible solution to the linear equation: 3a + 2b + 2c = 15 Where: -20 <= a <= 0 0 <= b <= 20 20 <= c <= 50 a, b, c are real numbers How could I rewrite the program to produce a random solution every time it is run? I am open to using Pulp as well. My Code: # Coefficients coeffs = [3,2,2] y_target = 15 # Define model and variables model = ConcreteModel() model.a = Var(bounds=(-20,0)) model.b = Var(bounds=(0,20)) model.c = Var(bounds=(20,50

问题 I want to solve an LPP which has more than one optimal solution. How can I do that? For Example: Maximize 2000x1 + 3000x2 subject to 6x1 + 9x2 ≤ 100 2x1 + x2 ≤ 20 x1, x2 ≥ 0 In this LPP problem, there is more than one optimal solution i.e (0,100/9) and (20/3,20/3). When I solve this problem using the pulp library, it gives me only a (0,100/9) solution. I want all the possible solutions. 回答1: There is a good discusion on this here. There are two questions here: (i) How to find multiple optimal

问题 I have written the following Python Code using the PuLP Library for solving the Knapsack Problem using the Integer Programming formulation. I am using strings to generate the LpVariable commands and add the constraints and then executing them with eval. Is there a way to do this without using eval? from pulp import * #Knapsack problem items = input ('Enter the number of items :') items = int(items) #print('Enter %d items one by one') print ('Enter {0} items profit one by one'.format(items))

问题 I am trying to generate an equation via lpDot() , Such as PulpVar = [x1,x2] Constants = [5,6] then doing dot product as: model += lpDot(PulpVar, Constants) Form what I understand this should generate an equation as x1*5+x2*6 but I am getting lpAffineExpression as output and the lp file so generated is empty 回答1: lpDot() – given two lists of the form [a1, a2, …, an] and [ x1, x2, …, xn] will construct a linear epression to be used as a constraint or variable ref So, if you use with constants,

问题 I am still very new to Python, after years and years of Matlab. I am trying to use Pulp to set up an integer linear program. Given an array of numbers: {P[i]:i=1...N} I want to maximize: sum( x_i P_i ) subject to the constraints A x <= b A_eq x = b_eq and with bounds (vector based bounds) LB <= x <= UB In pulp however, I don't see how to do vector declarations properly. I was using: RANGE = range(numpy.size(P)) x = pulp.LpVariable.dicts("x", LB_ind, UB_ind, "Integer") where I can only enter