linear-programming

I am working on some code in R to optimize my fantasy football lineup but I am having some difficulty with one constraint. I basically have a list of players, their position, expected points, and cost. Roster must include: 1 QB 2 RB 2 WR 1 TE 1 FLEX (either a RB, WR, or TE) Total cost under $200 My issue is that my code wants to pick the FLEX position as player it already selected as a WR, RB or TE. Here is the code I am using, I have a table that I imported with columns for player, position, points, and cost. In the table, any RB, WR, or TE is duplicated with the position as FLEX. I tried to

I am trying to solve the Bin Packing Problem using the Integer Programming Formulation in Python PuLP. The model for the problem is as follows: I have written the following Python Code using the PuLP library from pulp import * #knapsack problem def knapsolve(bins, binweight, items, weight): prob = LpProblem('BinPacking', LpMinimize) y = [LpVariable("y{0}".format(i+1), cat="Binary") for i in range(bins)] xs = [LpVariable("x{0}{1}".format(i+1, j+1), cat="Binary") for i in range(items) for j in range(bins)] #minimize objective nbins = sum(y) prob += nbins print(nbins) #constraints prob += nbins >

Suppose I have a point cloud given in 4D space, which can be set up arbitrarily dense. These points will lie on the surface of a polytope, which is an unknown object at this point. (Just to provide some unhelpful visualization, here is a rather arbitrary projection of the point cloud which I have given - i.e. plotting the first 3 columns of the 100000x4 array): My goal is to obtain the vertices of this polytope, and one of my numerous attempts to get those was computing the convex hull of the point cloud. The problem here was that the number of resulting vertices differed with the specified

I need to find an exact real solution to a linear program (where all inputs are integers). It is important that the solver also outputs the solutions as rational numbers, ideally without doing any intermediate steps with floating point numbers. GLPK can do exact arithmetic, but cannot display the solutions as rational numbers (i.e. I get 0.3333 for 1/3). I could probably attempt to guess which number is meant by that, but that seems very fragile. I was unable to find an LP solver that can do this kind of thing. Is there one? Performance is not a huge issue; my problems are very small. (I did

I'm using scipy.optimize.minimize to solve a complex reservoir optimization model (SQSLP and COBYLA as the problem is constrained by both bounds and constraint equations). There is one decision variable per day (storage), and releases from the reservoir are calculated as a function of change in storage, within the objective function. Penalties based on releases and storage penalties are then applied with the goal of minimizing penalties (the objective function is a summation of all penalties). I've added some constraints within this model to limit the change in storage to the physical system

Reproducable Example: I described a simple 0/1-Knapsack problem with lpSolveAPI in R , which should return 2 solutions: library(lpSolveAPI) lp_model= make.lp(0, 3) set.objfn(lp_model, c(100, 100, 200)) add.constraint(lp_model, c(100,100,200), "<=", 350) lp.control(lp_model, sense= "max") set.type(lp_model, 1:3, "binary") lp_model solve(lp_model) get.variables(lp_model) get.objective(lp_model) get.constr.value((lp_model)) get.total.iter(lp_model) get.solutioncount(lp_model) Problem: But get.solutioncount(lp_model) shows that there's just 1 solution found: > lp_model Model name: C1 C2 C3

I have two variables: x>= 0 and y binary (either 0 or 1), and I have a constant z >= 0. How can I use linear constraints to describe the following condition: If x = z then y = 1 else y = 0. I tried to solve this problem by defining another binary variable i and a large-enough positive constant U and adding constraints y - U * i = 0; x - U * (1 - i) = z; Is this correct? Really there are two classes of constraints that you are asking about: If y=1 , then x=z . For some large constant M , you could add the following two constraints to achieve this: x-z <= M*(1-y) z-x <= M*(1-y) If y=1 then these