我正在尝试解决 R 中的容量设施位置问题。示例数据:
n<- 500 #number of customers
m<- 20 #number of facility centers
set.seed(1234)
fixedcost <- round(runif(m, min=5000, max=10000))
warehouse_locations <- data.frame(
id=c(1:m),
y=runif(m, 22.4, 22.6),
x= runif(m, 88.3, 88.48)
)
customer_locations <- data.frame(
id=c(1:n),
y=runif(n, 22.27, 22.99),
x= runif(n, 88.12, 88.95)
)
capacity <- round(runif(m, 1000, 4000))
demand <- round(runif(n, 5, 50))
具有成本函数的模型:
library(geosphere)
transportcost <- function(i, j) {
customer <- customer_locations[i, ]
warehouse <- warehouse_locations[j, ]
(distm(c(customer$x, customer$y), c(warehouse$x, warehouse$y), fun = distHaversine)/1000)*20
}
library(ompr)
library(magrittr)
model <- MIPModel() %>%
# 1 iff i gets assigned to SC j
add_variable(x[i, j], i = 1:n, j = 1:m, type = "binary") %>%
# 1 if SC j is built
add_variable(y[j], j = 1:m, type = "binary") %>%
# Objective function
set_objective(sum_expr(transportcost(i, j) * x[i, j], i = 1:n, j = 1:m) +
sum_expr(fixedcost[j] * y[j], j = 1:m), "min") %>%
#Demand of customers shouldn't exceed total facility capacities
add_constraint(sum_expr(demand[i] * x[i, j], i = 1:n) <= capacity[j] * y[j], j = 1:m) %>%
# every customer needs to be assigned to a SC
add_constraint(sum_expr(x[i, j], j = 1:m) == 1, i = 1:n) %>%
# if a customer is assigned to a SC, then this SC must be built
add_constraint(x[i,j] <= y[j], i = 1:n, j = 1:m)
model
library(ompr.roi)
library(ROI.plugin.glpk)
result <- solve_model(model, with_ROI(solver = "glpk", verbose = TRUE))
有什么办法可以减少计算时间?如果我理解正确,那么 0.4% 是当前模型与预期结果之间的差异。即使差异远大于此,我也会很高兴,我可以获得合适的模型。有什么办法可以设置吗?就像 5-6% 的差异就足够了。