具有二进制调度变量的R:成本函数优化？
下面详细介绍了一个优化问题的简化版本，我很难解决这个问题 目标是使通过卡车运送水的组织的成本函数最小化，并使用该等式生成最小化成本的卡车运送计划 该组织全年向约10000个家用水箱供水 储罐的最大容量为300加仑，最低要求限值为100加仑——也就是说，储罐在低于100加仑之前应加满300加仑 例如，如果第2周的油箱容量为115加仑，第3周预计使用20加仑，则需要在第3周重新加注 成本包括：具有二进制调度变量的R:成本函数优化？,r,optimization,linear-programming,nonlinear-optimization,lpsolve,R,Optimization,Linear Programming,Nonlinear Optimization,Lpsolve,下面详细介绍了一个优化问题的简化版本，我很难解决这个问题 目标是使通过卡车运送水的组织的成本函数最小化，并使用该等式生成最小化成本的卡车运送计划 该组织全年向约10000个家用水箱供水 储罐的最大容量为300加仑，最低要求限值为100加仑——也就是说，储罐在低于100加仑之前应加满300加仑 例如，如果第2周的油箱容量为115加仑，第3周预计使用20加仑，则需要在第3周重新加注 成本包括： 每次送货费10美元 每周卡车的费用。一辆卡车每周的费用是1000美元。因此，如果在一周内交付200次，成本
（200*10+1000*1）（201*10+1000*2）weekly_cost_function <- function(i){
cost <- (ceiling(sum(i)/200)) * 1600 + (sum(i) * 10)
cost
}
**example cost for one week with i = 199 deliveries:**
weekly_cost_function(i = 199)
[1] 3590
重要的是，
重新填充水平保持在100以上的任何值。上限x>=0y = ceiling(x)
我不担心这一点。对于爬山优化器来说，使用天花板功能似乎是一个难题。我认为遗传算法更合适。每个家庭每周是否交付的矩阵构成了一个很好的基因组
library(dplyr)
# Original given sample input data.
df.usage <- structure(list(reduction.group = c(1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5,
5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8), week = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12), water_usage = c(46, 50, 42, 47, 43, 39,
38, 32, 42, 36, 42, 30, 46, 50, 42, 47, 43, 39, 38, 32, 42, 36,
42, 30, 46, 50, 43, 47, 43, 39, 38, 32, 42, 36, 42, 30, 46, 50,
43, 47, 43, 39, 38, 32, 42, 36, 42, 30, 29, 32, 27, 30, 27, 25,
24, 20, 26, 23, 27, 19, 29, 32, 27, 30, 27, 25, 24, 20, 26, 23,
27, 19, 29, 32, 27, 30, 28, 25, 25, 21, 27, 23, 27, 19, 29, 32,
27, 30, 28, 25, 25, 21, 27, 23, 27, 20), tank.level.start = c(115,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 165, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, 200, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, 215, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, 225, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 230,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 235, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, 240, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA)), row.names = c(NA, 96L), class = "data.frame")
# Orginal given delivery cost function.
weekly_cost_function <- function(i){
cost <- (ceiling(sum(i)/200)) * 1600 + (sum(i) * 10)
cost
}
# Calculate the list of houses (reduction.groups) and number of delivery weeks (weeks).
reduction.groups <- unique(df.usage$reduction.group)
temp <- df.usage %>% filter(reduction.group == 1)
weeks <- nrow(temp)
# The genome consists of a matrix representing deliver-or-not to each house each week.
create_random_delivery_schedule <- function(number_of_houses, number_of_weeks, prob = NULL) {
matrix(sample(c(0, 1), number_of_houses * number_of_weeks, replace = TRUE, prob = prob), number_of_houses)
}
# Generate a population of random genes.
population_size <- 100
schedules <- replicate(population_size, create_random_delivery_schedule(length(reduction.groups), weeks), simplify = FALSE)
# Calculate fitness of an individual.
fitness <- function(schedule) {
# Fitness is related to delivery cost.
delivery_cost <- sum(apply(schedule, 2, weekly_cost_function))
# If the schedule allows a tank level to drop below 100, apply a fitness penalty.
# Don't make the fitness penalty too large.
# If the fitness penalty is large enough to be catastrophic (essentially zero children)
# then solutions that are close to optimal will also be likely to generate children
# who fall off the catastropy cliff so there will be a selective pressure away from
# close to optimal solutions.
# However, if your optimizer generates a lot of infeasible solutions raise the penalty.
for (i in reduction.groups) {
temp <- df.usage %>% filter(reduction.group == i)
temp$level <- temp$tank.level.start
if (weeks > 1) for (j in 2:weeks) {
if (1 == schedule[i,j]) {
temp$level[j] <- 300
} else {
temp$level[j] <- ( temp$level[j-1] - temp$water_usage[j] )
if (100 > temp$level[j]) {
# Fitness penalty.
delivery_cost <- delivery_cost + 10 * (100 - temp$level[j])
}
}
}
}
# Return one over delivery cost so that lower cost is higher fitness.
1 / delivery_cost
}
# Generate a new schedule by combining two parents chosen randomly weighted by fitness.
make_baby <- function(population_fitness) {
# Choose some parents.
parents <- sample(length(schedules), 2, prob = population_fitness)
# Get DNA from mommy.
baby <- schedules[[parents[1]]]
# Figure out what part of the DNA to get from daddy.
house_range <- sort(sample(length(reduction.groups), 2))
week_range <- sort(sample(weeks, 2))
# Get DNA from daddy.
baby[house_range[1]:house_range[2],week_range[1]:week_range[2]] <- schedules[[parents[2]]][house_range[1]:house_range[2],week_range[1]:week_range[2]]
# Mutate, 1% chance of flipping each bit.
changes <- create_random_delivery_schedule(length(reduction.groups), weeks, c(0.99, 0.01))
baby <- apply(xor(baby, changes), c(1, 2), as.integer)
}
lowest_cost <<- Inf
# Loop creating and evaluating generations.
for (ii in 1:100) {
population_fitness <- lapply(schedules, fitness)
lowest_cost_this_generation <- 1 / max(unlist(population_fitness))
print(sprintf("lowest cost = %f", lowest_cost_this_generation))
if (lowest_cost_this_generation < lowest_cost) {
lowest_cost <<- lowest_cost_this_generation
best_baby <<- schedules[[which.max(unlist(population_fitness))]]
}
schedules <<- replicate(population_size, make_baby(population_fitness), simplify = FALSE)
}
库（dplyr）
#原始给定样本输入数据。
用法我看到很多问题，其中大多数是逻辑问题。首先写下数学模型。只有当你对数学模型的正确性有信心时才开始编码。我只是简单地回顾了这个答案，但它看起来很神奇。我现在就给悬赏金，悬赏金即将到期，如果我以后有问题，我会回来的！谢谢你，鲍勃。
weekly_cost_function <- function(i){
cost <- (ceiling(sum(i)/200)) * 1600 + (sum(i) * 10)
cost
}
**example cost for one week with i = 199 deliveries:**
weekly_cost_function(i = 199)
[1] 3590
num_groups <- length(unique(df.usage$reduction.group))
num_weeks <- length(unique(df.usage$week))
MIPModel() %>%
add_variable(x[i,w], # create decision variable: deliver or not by...
i = 1:num_groups, # group,
w = 1:num_weeks, # in week.
type = "integer", # Integers only
lb = 0, ub = 1) %>% # between 0 and 1, inclusive
set_objective(sum_expr( x[i,w]/200 * 1600 + x[i,w] * 10,
i = 1:num_groups,
w = 1:num_weeks),
sense = "min") %>%
# add constraint to achieve ceiling(x[i,w]/200), or should this be in the set_objective call?
add_constraint(???) %>%
solve_model(with_ROI("glpk"))
reduction.group week water.usage refill level
1 1 46 0 115
1 2 50 1 300
1 3 42 0 258
1 4 47 0 211
1 5 43 0 168
1 6 39 0 129
y = ceiling(x)
x <= y <= x+1
y integer
x+0.0001 <= y <= x+1
y integer
library(dplyr)
# Original given sample input data.
df.usage <- structure(list(reduction.group = c(1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5,
5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8,
8, 8, 8), week = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,
12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12), water_usage = c(46, 50, 42, 47, 43, 39,
38, 32, 42, 36, 42, 30, 46, 50, 42, 47, 43, 39, 38, 32, 42, 36,
42, 30, 46, 50, 43, 47, 43, 39, 38, 32, 42, 36, 42, 30, 46, 50,
43, 47, 43, 39, 38, 32, 42, 36, 42, 30, 29, 32, 27, 30, 27, 25,
24, 20, 26, 23, 27, 19, 29, 32, 27, 30, 27, 25, 24, 20, 26, 23,
27, 19, 29, 32, 27, 30, 28, 25, 25, 21, 27, 23, 27, 19, 29, 32,
27, 30, 28, 25, 25, 21, 27, 23, 27, 20), tank.level.start = c(115,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 165, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, 200, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, 215, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, 225, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 230,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 235, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, 240, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA)), row.names = c(NA, 96L), class = "data.frame")
# Orginal given delivery cost function.
weekly_cost_function <- function(i){
cost <- (ceiling(sum(i)/200)) * 1600 + (sum(i) * 10)
cost
}
# Calculate the list of houses (reduction.groups) and number of delivery weeks (weeks).
reduction.groups <- unique(df.usage$reduction.group)
temp <- df.usage %>% filter(reduction.group == 1)
weeks <- nrow(temp)
# The genome consists of a matrix representing deliver-or-not to each house each week.
create_random_delivery_schedule <- function(number_of_houses, number_of_weeks, prob = NULL) {
matrix(sample(c(0, 1), number_of_houses * number_of_weeks, replace = TRUE, prob = prob), number_of_houses)
}
# Generate a population of random genes.
population_size <- 100
schedules <- replicate(population_size, create_random_delivery_schedule(length(reduction.groups), weeks), simplify = FALSE)
# Calculate fitness of an individual.
fitness <- function(schedule) {
# Fitness is related to delivery cost.
delivery_cost <- sum(apply(schedule, 2, weekly_cost_function))
# If the schedule allows a tank level to drop below 100, apply a fitness penalty.
# Don't make the fitness penalty too large.
# If the fitness penalty is large enough to be catastrophic (essentially zero children)
# then solutions that are close to optimal will also be likely to generate children
# who fall off the catastropy cliff so there will be a selective pressure away from
# close to optimal solutions.
# However, if your optimizer generates a lot of infeasible solutions raise the penalty.
for (i in reduction.groups) {
temp <- df.usage %>% filter(reduction.group == i)
temp$level <- temp$tank.level.start
if (weeks > 1) for (j in 2:weeks) {
if (1 == schedule[i,j]) {
temp$level[j] <- 300
} else {
temp$level[j] <- ( temp$level[j-1] - temp$water_usage[j] )
if (100 > temp$level[j]) {
# Fitness penalty.
delivery_cost <- delivery_cost + 10 * (100 - temp$level[j])
}
}
}
}
# Return one over delivery cost so that lower cost is higher fitness.
1 / delivery_cost
}
# Generate a new schedule by combining two parents chosen randomly weighted by fitness.
make_baby <- function(population_fitness) {
# Choose some parents.
parents <- sample(length(schedules), 2, prob = population_fitness)
# Get DNA from mommy.
baby <- schedules[[parents[1]]]
# Figure out what part of the DNA to get from daddy.
house_range <- sort(sample(length(reduction.groups), 2))
week_range <- sort(sample(weeks, 2))
# Get DNA from daddy.
baby[house_range[1]:house_range[2],week_range[1]:week_range[2]] <- schedules[[parents[2]]][house_range[1]:house_range[2],week_range[1]:week_range[2]]
# Mutate, 1% chance of flipping each bit.
changes <- create_random_delivery_schedule(length(reduction.groups), weeks, c(0.99, 0.01))
baby <- apply(xor(baby, changes), c(1, 2), as.integer)
}
lowest_cost <<- Inf
# Loop creating and evaluating generations.
for (ii in 1:100) {
population_fitness <- lapply(schedules, fitness)
lowest_cost_this_generation <- 1 / max(unlist(population_fitness))
print(sprintf("lowest cost = %f", lowest_cost_this_generation))
if (lowest_cost_this_generation < lowest_cost) {
lowest_cost <<- lowest_cost_this_generation
best_baby <<- schedules[[which.max(unlist(population_fitness))]]
}
schedules <<- replicate(population_size, make_baby(population_fitness), simplify = FALSE)
}