bilinear.R
# Copyright 2023, Gurobi Optimization, LLC # # This example formulates and solves the following simple bilinear model: # maximize # x # subject to # x + y + z <= 10 # x * y <= 2 (bilinear inequality) # x * z + y * z = 1 (bilinear equality) # x, y, z non-negative (x integral in second version) library(gurobi) library(Matrix) model <- list() # Linear constraint matrix model$A <- matrix(c(1, 1, 1), nrow=1, byrow=T) model$rhs <- c(10.0) model$sense <- c('<') # Variable names model$varnames <- c('x', 'y', 'z') # Objective function max 1.0 * x model$obj <- c(1, 0, 0) model$modelsense <- 'max' # Bilinear inequality constraint: x * y <= 2 qc1 <- list() qc1$Qc <- spMatrix(3, 3, c(1), c(2), c(1.0)) qc1$rhs <- 2.0 qc1$sense <- c('<') qc1$name <- 'bilinear0' # Bilinear equality constraint: x * z + y * z == 1 qc2 <- list() qc2$Qc <- spMatrix(3, 3, c(1, 2), c(3, 3), c(1.0, 1.0)) qc2$rhs <- 1.0 qc2$sense <- c('=') qc2$name <- 'bilinear1' model$quadcon <- list(qc1, qc2) # Solve bilinear model, display solution. The problem is non-convex, # we need to set the parameter 'NonConvex' in order to solve it. params <- list() params$NonConvex <- 2 result <- gurobi(model, params) print(result$x) # Constrain 'x' to be integral and solve again model$vtype <- c('I', 'C', 'C') result <- gurobi(model, params) print(result$x)
