LPSolve - 为多列之和设置约束

Question

我对使用LPSolve 解决线性优化问题有很好的了解，但有一个方面被难住了。我想为多列的总和创建一个约束。例如，我有一个约束，不允许四个特定列中的任何一个大于 3。但是，我要求四个列中的任何一个都等于 3。

工作示例

在此示例中，我正在制作餐点以优化“价值”，同时保持低于 5 个单独的项目和 40 美元的成本。我还有四种不同的食物组 - 肉类、蔬菜、水果、淀粉 - 我要求任何一组食物中的食物不得超过四项，但任何一组食物必须包含 3 项（这就是我的我被难住了）。

除了最后一个约束之外，下面是得到我想要的结果的代码：

## Choose 5 food items remaining under $40 and maximizing Value ##
## There can be no more than 3 items from the same group chosen, but **there must be 3 items from at least one group**(??) ##

library(dplyr)
library(lpSolve)

# Constraints
totalItems <- 5
totalCost <- 40
maxAllGroups <- 3

# Setup problem
food <- c('Chicken', 'Beef', 'Lamb', 'Fish', 'Pork', 'Carrot', 'Lettuce', 'Asparagus', 'Beats', 'Broccoli', 'Orange', 'Apple', 'Pear', 'Banana', 'Watermelon', 'Potato', 'Corn', 'Beans', 'Bread', 'Pasta')
group <- c('Meat', 'Meat', 'Meat', 'Meat', 'Meat', 'Veggie', 'Veggie', 'Veggie', 'Veggie', 'Veggie', 'Fruit', 'Fruit', 'Fruit', 'Fruit', 'Fruit', 'Starch', 'Starch', 'Starch', 'Starch', 'Starch')
cost <- round(runif(length(food), 1, 20), 0)
value <- round(runif(length(food), 20, 60), 0)
df <- data.frame(food, group, cost, value, stringsAsFactors = FALSE) %>% 
  mutate(Total = 1)

# Value to be maximized
Value <- df$value

# Create constraint vectors
ConVec_Cost <- df$cost
ConVec_Items <- df$Total
# Make `Group` dummy variables
  groups <- unique(df$group)
  ConVec_Groups <- data.frame(row.names = 1:nrow(df))
  for(i in 1:length(groups)){
    currGroup <- groups[i]

    vec <- df %>% 
      mutate(isGroup = (group == currGroup)*1) %>% 
      select(isGroup)
    colnames(vec) <- currGroup

    ConVec_Groups <- cbind(ConVec_Groups, vec)
  }

# ConVec_AnyGroupEqual3 <- ???

ConVec_All <- t(cbind(ConVec_Cost, ConVec_Items, ConVec_Groups))

# Create constraint directions
ConDir_Cost <- "<="
ConDir_Items <- "=="
ConDir_Groups <- rep("<=", ncol(ConVec_Groups))
# ConDir_AnyGroupEqual3 <- "=="
ConDir_All <- c(ConDir_Cost, ConDir_Items, ConDir_Groups)

# Create constraint values
ConVal_Cost <- totalCost
ConVal_Items <- totalItems
ConVal_Groups <- rep(maxAllGroups, ncol(ConVec_Groups))
# ConVal_AnyGroupEqual3 <- 1 #1 group should have 3
ConVal_All <- c(ConVal_Cost, ConVal_Items, ConVal_Groups)

# Solve
sol <- lpSolve::lp("max",
                   objective.in = Value,
                   const.mat    = ConVec_All,
                   const.dir    = ConDir_All,
                   const.rhs    = ConVal_All,
                   all.bin      = TRUE
)

# Solution
df[sol$solution == 1,]

如果我需要一个特定的食物组来拥有 3 个，那么这很容易，但我需要任何一个组为 3 个的事实却让这变得困难。有没有办法在不诉诸LPSolveAPI（我承认对此知之甚少）的情况下做到这一点？

Answer 1

不久前，我通过剖析某人对类似问题的回答来弄清楚这一点。基本上，您必须添加 5 个“虚拟行”（每个食物组一个），对角线值为 -3。然后，您必须添加另一列，除了最后 5 行为 1（您刚刚添加到矩阵中的行）之外，所有值都为 0。您强制新列至少为 1，这意味着必须选择最后 5 行之一。

由于将选择其中一个行，并且该行将 -3 添加到它所在的任何食物组，并且您有强制每个食物组至少为 0 的约束，然后选择的 -3 行将强制该食物组列选择 3，以便总和 >= 0。

新代码

## Choose 5 food items remaining under $40 and maximizing Value ##
## There can be no more than 3 items from the same group chosen, but **there must be 3 items from at least one group**(??) ##

library(dplyr)
library(lpSolve)

# Constraints
totalItems <- 5
totalCost <- 40
minAllGroups <- 0
atLeastNFrom1 <- 3

# Setup problem
food <- c('Chicken', 'Beef', 'Lamb', 'Fish', 'Pork', 'Carrot', 'Lettuce', 'Asparagus', 'Beats', 'Broccoli', 'Orange', 'Apple', 'Pear', 'Banana', 'Watermelon', 'Potato', 'Corn', 'Beans', 'Bread', 'Pasta')
group <- c('Meat', 'Meat', 'Meat', 'Meat', 'Meat', 'Veggie', 'Veggie', 'Veggie', 'Veggie', 'Veggie', 'Fruit', 'Fruit', 'Fruit', 'Fruit', 'Fruit', 'Starch', 'Starch', 'Starch', 'Starch', 'Starch')
cost <- round(runif(length(food), 1, 20), 0)
value <- round(runif(length(food), 20, 60), 0)
df <- data.frame(food, group, cost, value, stringsAsFactors = FALSE) %>% 
  mutate(Total = 1)


# Create constraint vectors
ConVec_Cost <- df$cost
ConVec_Items <- df$Total
# Make `Group` dummy variables
  groups <- unique(df$group)
  ConVec_Groups <- data.frame(row.names = 1:nrow(df))
  for(i in 1:length(groups)){
    currGroup <- groups[i]

    vec <- df %>% 
      mutate(isGroup = (group == currGroup)*1) %>% 
      select(isGroup)
    colnames(vec) <- currGroup

    ConVec_Groups <- cbind(ConVec_Groups, vec)
  }

# New vector for atleast
ConVec_AtLeastN <- 0

ConVec_All <- cbind(ConVec_Cost, ConVec_Items, ConVec_Groups, ConVec_AtLeastN)

# Add the negative values to the matrix
ConVec_All <- rbind(ConVec_All, c(0,0,-atLeastNFrom1,0,0,0,1))
ConVec_All <- rbind(ConVec_All, c(0,0,0,-atLeastNFrom1,0,0,1))
ConVec_All <- rbind(ConVec_All, c(0,0,0,0,-atLeastNFrom1,0,1))
ConVec_All <- rbind(ConVec_All, c(0,0,0,0,0,-atLeastNFrom1,1))

# Create constraint directions
ConDir_Cost <- "<="
ConDir_Items <- "=="
ConDir_Groups <- rep(">=", ncol(ConVec_Groups))
ConDir_AtLeastN <- ">="
ConDir_All <- c(ConDir_Cost, ConDir_Items, ConDir_Groups, ConDir_AtLeastN)

# Create constraint values
ConVal_Cost <- totalCost
ConVal_Items <- totalItems
ConVal_Groups <- rep(minAllGroups, ncol(ConVec_Groups))
ConVal_AtLeastN <- 1
ConVal_All <- c(ConVal_Cost, ConVal_Items, ConVal_Groups, ConVal_AtLeastN)


# Value to be maximized
Value <- c(df$value, rep(0, nrow(ConVec_All) - length(df$value)))

# Solve
sol <- lpSolve::lp("max",
                   objective.in = Value,
                   const.mat    = t(ConVec_All),
                   const.dir    = ConDir_All,
                   const.rhs    = ConVal_All,
                   all.bin      = TRUE
)

# Solution
df[sol$solution[1:nrow(df)] == 1,]