不良规划R中的模拟数据

Question

我在试着估计糟糕的规划对企业的影响。我有两个不同的仓库，一个比另一个贵得多。在一个完美的情况下，我们总是选择最便宜的价格，然而，这个过程并不完美，有时有些订单是从错误的仓库购买的。

为了测试我们可能会损失多少，我正在开发一个仿真模型来测试3种场景。这些建议如下：

1. 每次选择最便宜的仓库
2. 10%的订单是以错误的价格购买的
3. 25%的订单是以错误的价格购买的

我现在不知道该从哪里开始，我知道我的数据需要符合正态分布，我需要多次模拟第2和第3场景，因为每个场景都会给出不同的结果。

structure(list(Customer = c("a", "b", "c", "d", "e"), `Option 1` = c(5, 
100, 107, 400, 30), `Option 2` = c(19, 200, 50, 300, 70), `Probability Selecting the more expensive option` = c(0.1, 
0.1, 0.1, 0.1, 0.1)), row.names = c(NA, -5L), class = c("tbl_df", 
"tbl", "data.frame"))
# A tibble: 5 x 4
  Customer `Option 1` `Option 2` `Probability Selecting the more expensive option`
  <chr>         <dbl>      <dbl>                                             <dbl>
1 a                 5         19                                               0.1
2 b               100        200                                               0.1
3 c               107         50                                               0.1
4 d               400        300                                               0.1
5 e                30         70                                               0.1

任何有关如何解决这个问题的帮助都将不胜感激。

谢谢

Answer 1

好的，如果我理解正确，我可能会这样做：

library(dplyr)

#I've renamed the columns to be easier to work with
df <- structure(list(customer = c("a", "b", "c", "d", "e"),
                     op1 = c(5, 100, 107, 400, 30),
                     op2 = c(19, 200, 50, 300, 70), 
                     probs = c(0.1,  0.1, 0.1, 0.1, 0.1)), row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"))
df


# Here is a simple dplyr solution to getting your best and worst selection to chose from 
df <- df %>% 
  rowwise() %>% 
  mutate(best = min(c(op1, op2)),
         worst = max(c(op1, op2)))


# this function repeatedly samples from the best and worst option with a probability of the argument probs
# it is deendent on integer indexing so if you move or add columns the function will need adjusting
# it can also be adjusted to use the prob of each row if that were to change through time
# The output is a matrix with the number of rwplicates as columns
# order of probability is best worst so probs = c(0.4,0.6) means it is less likely to choose best
resample_func <- function(df, reps = 5, probs = c(0.4,0.6)) {
  mat <- matrix(nrow = nrow(df), ncol = reps)
  nc <- df[, c(5,6)]
  # print(nc)
  
  for (j in 1:ncol(mat)) {
    for (i in 1:nrow(df)) {
      # nc <- df[,c(2,3)]
      mat[i, j] <- t(sample(nc[i,], 1, replace = F, prob = probs))
    }
  }
  rownames(mat) <- df$customer
  colnames(mat) <- paste0("rep_", 1:ncol(mat))
  return(mat)
}

resample_func(df)

请仔细检查输出！请让我知道这是否是您所追求的:)还请参见@r2evans的评论