使用 R 寻找最佳项目团队

Question

我刚开始学习 R 并且运行遇到了一些我不确定如何在代码中处理的问题。

我正在创建一个 data.frame，其中包含可以分配给项目的个人池。该项目需要一名 BA、一名 PM、两名 SA， 和一名可以是 SA 或 BA 的额外人员。每个人都有一个评级和与之相关的成本，我需要最高评级，同时将成本保持在特定阈值以下。

我不确定如何实现上述场景的粗体部分。下面的代码有效但没有考虑额外的 BA/SA。

(这是自学..没有布置作业)

EDIT-Desired output 其中最后一行可以是 SA 或 BA 位置。

name     position rating cost   BA PM SA  
Matt        SA     95    9500   0  0  1    
Aaron       BA     85    4700   1  0  0    
Stephanie   SA     95    9200   0  0  1    
Molly       PM     88    5500   0  1  0    
Jake        SA     74    5300   0  0  1

代码：

#load libraries
library(lpSolve)

# create data.frame
name = c("Steve", "Jeremy", "Matt", "Aaron", "Stephanie", "Molly", "Jake", "Tony", "Jay", "Katy", "Alison") 
position = c("BA", "PM", "SA", "BA", "SA", "PM", "SA", "SA", "PM", "BA", "SA")
rating = c(75, 90, 95, 85, 95, 88, 74, 81, 55, 65, 68) 
cost = c(5000, 8000, 9500, 4700, 9200, 5500, 5300, 7300, 3300, 4100, 4400)
df = data.frame(name, position, rating, cost)

# create restrictions
num_ba = 1
num_pm = 1
num_sa = 2
max_cost = 35000

# create vectors to constrain by position
df$BA = ifelse(df$position == "BA", 1, 0)
df$PM = ifelse(df$position == "PM", 1, 0)
df$SA = ifelse(df$position == "SA", 1, 0)

# vector to optimize against
objective = df$rating

# constraint directions
const_dir <- c("=", "=", "=", "<=")

# matrix
const_mat = matrix(c(df$BA, df$PM, df$SA, df$cost), 4, byrow=TRUE)
const_rhs = c(num_ba, num_pm, num_sa, max_cost)

#solve
x = lp("max", objective, const_mat, const_dir, const_rhs, all.bin=TRUE, all.int=TRUE)
print(df[which(x$solution==1), ])

Answer 1

如果我答对了问题，这可能有效：

 library(lpSolve)

 # create data.frame
 name = c("Steve", "Jeremy", "Matt", "Aaron", "Stephanie", "Molly", "Jake", "Tony", "Jay", "Katy", "Alison") 
position = c("BA", "PM", "SA", "BA", "SA", "PM", "SA", "SA", "PM", "BA", "SA")
rating = c(75, 90, 95, 85, 95, 88, 74, 81, 55, 65, 68) 
cost = c(5000, 8000, 9500, 4700, 9200, 5500, 5300, 7300, 3300, 4100, 4400)

df = data.frame(name, position, rating, cost)

# create restrictions
num_pm = 1
min_num_ba = 1
min_num_sa = 2
tot_saba   = 4
max_cost = 35000

# create vectors to constrain by position
df$PM = ifelse(df$position == "PM", 1, 0)
df$minBA = ifelse(df$position == "BA", 1, 0)
df$minSA = ifelse(df$position == "SA", 1, 0)
df$SABA = ifelse(df$position %in% c("SA","BA"), 1, 0)

# vector to optimize against
objective = df$rating

# constraint directions
const_dir <- c("==", ">=", "<=", "==", "<=")

# matrix
const_mat = matrix(c(df$PM, df$minBA, df$minSA, df$SABA, df$cost), 5, byrow=TRUE)
const_rhs = c(num_pm, min_num_ba,min_num_sa, tot_saba, max_cost)

#solve
x = lp("max", objective, const_mat, const_dir, const_rhs, all.bin=TRUE, all.int=TRUE)
print(df[which(x$solution==1), ])

我正在做的是修改一些约束并添加一个新的：BA的数量必须> = 1。SA的数量> = 2，和的总和BA 和 SA 必须是 4，这样你总是 select 5 个人。

然而，这给出了与 OP 所写不同的解决方案：

       name position rating cost PM minBA minSA SABA
1     Steve       BA     75 5000  0     1     0    1
3      Matt       SA     95 9500  0     0     1    1
4     Aaron       BA     85 4700  0     1     0    1
5 Stephanie       SA     95 9200  0     0     1    1
6     Molly       PM     88 5500  1     0     0    0

但是，将此解决方案的评分相加得出 438，而运算结果为 437，因此这应该是正确的。

HTH。

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