我认为你的问题可以解决，但你将不得不做一些代数，因为它更适合无约束的优化。它可以相对地使用方法和 来处理边界，但没有比这更复杂的约束。optimoptimL-BFGS-BBrent

通过使用拉格朗日乘子，可以将简单的约束优化问题转换为无约束优化问题。从本质上讲，定义一个与原始函数相同的函数，加上约束时间的常数。

L = function(a,b,c,lambda) {objective_function(a,b,c)-lambda*(max_budget-sum(a,b,c))}

并最大化 L。

我对拉格朗日乘数有点生疏，但每次超出预算时，这都应该会惩罚目标。lambda 越高，您对超出预算的宽容就越少。

我认为不能在这里解决您的问题，因为该问题是具有约束的约束优化问题。一个有前途的候选人是.optimmax_budgetconstrOptim

您需要注意两件事：

1. 既然你已经指定了，你就不需要在你的 ;否则，它将返回最小化过程。fnscale=-1-tot_ROIobjective_function

2. 您应该有一个明确的语句，这在约束优化问题中至关重要。sum(spend) <= max_budget

法典

下面是一个实现（我想移出，因为如果您需要更改目标函数，这将更容易维护代码）objective_functionbudget_optimal_allocation

# Objective function to maximize total ROI
objective_function <- function(spend, a, b, c) {
    # Calculate total revenue and total spend
    revenue <- a / (1 + b * (spend)^c)
    tot_revenue <- sum(revenue)
    tot_spend <- sum(spend)

    # Calculate total ROI
    tot_revenue / tot_spend
}

budget_optimal_allocation <- function(data, max_budget) {
    # Extract channel names and parameters
    channel_names <- data$channel_name
    a <- data$a
    b <- data$b
    c <- data$c
    spend_min <- data$spend_min
    spend_max <- data$spend_max


    # Initial guess for spend values (equal allocation for each channel)
    init <- spend_min + sqrt(.Machine$double.eps)
    # Box constraints for spend
    k <- nrow(data)
    ui <- rbind(diag(k), -diag(k), -rep(1, k))
    ci <- c(spend_min, -spend_max, -max_budget)
    # Optimize spend allocation to maximize total ROI with box constraints
    result <- constrOptim(
        theta = init,
        f = objective_function,
        a = a,
        b = b,
        c = c,
        ui = ui,
        ci = ci,
        method = "Nelder-Mead",
        control = list(fnscale = -1)
    )

    # Combine channel names and optimal spend into a data frame
    return(list(
        optimal_allocation = data.frame(
            channel_name = channel_names,
            optimal_spend = result$par
        ),
        max_tot_ROI = result$value
    ))
}

输出

使用与您的问题中相同的数据

# Example data
example_data <- data.frame(
    channel_name = c("Channel 1", "Channel 2", "Channel 3"),
    a = c(5000, 8000, 6000),
    b = c(0.001, 0.002, 0.0015),
    c = c(-1.05, -1.2, -0.95),
    spend_min = c(100, 200, 150),
    spend_max = c(500, 800, 600)
)

# Max budget for optimization
max_budget <- 1000

# Run the optimization function
optimal_allocation <- budget_optimal_allocation(example_data, max_budget)

我们拭目以待

> optimal_allocation
$optimal_allocation
  channel_name optimal_spend
1    Channel 1           100
2    Channel 2           200
3    Channel 3           150

$max_tot_ROI
[1] 42.2219

如果你想用完所有的预算

您可以进行以下更改，因此，代码是init <- (spend_min + spend_max) / 2ui <- rbind(diag(k), -diag(k), rep(1, k))ci <- c(spend_min, -spend_max, max_budget)

budget_optimal_allocation <- function(data, max_budget) {
    # Extract channel names and parameters
    channel_names <- data$channel_name
    a <- data$a
    b <- data$b
    c <- data$c
    spend_min <- data$spend_min
    spend_max <- data$spend_max


    # Initial guess for spend values (equal allocation for each channel)
    eps <- sqrt(.Machine$double.eps)
    init <- (spend_min + spend_max) / 2
    # Box constraints for spend
    k <- nrow(data)
    ui <- rbind(diag(k), -diag(k), rep(1, k))
    ci <- c(spend_min, -spend_max, max_budget)
    # Optimize spend allocation to maximize total ROI with box constraints
    result <- constrOptim(
        theta = init,
        f = objective_function,
        a = a,
        b = b,
        c = c,
        ui = ui,
        ci = ci,
        method = "Nelder-Mead",
        control = list(fnscale = -1)
    )

    # Combine channel names and optimal spend into a data frame
    return(list(
        optimal_allocation = data.frame(
            channel_name = channel_names,
            optimal_spend = result$par
        ),
        max_tot_ROI = result$value
    ))
}

您将看到输出

> optimal_allocation
$optimal_allocation
  channel_name optimal_spend
1    Channel 1      252.1112
2    Channel 2      427.5322
3    Channel 3      320.3566

$max_tot_ROI
[1] 18.99994

并验证所有预算是否已用完

> sum(optimal_allocation$optimal_allocation$optimal_spend)
[1] 1000