lpsolve 与 r 中的约束

Question

我想使用 R 使用 lpSolve 解决优化问题，它可以执行类似于 excel 中的求解器插件的过程。下面是一个简单的例子，我想专门使用 lpSolve 来最大化 npv 值。

df<-structure(list(id = c(1, 2, 3, 4, 5, 6, 7, 8), Revenue = c(109, 
111, 122, 139, 156, 140, 137, 167)), row.names = c(NA, 8L), class = "data.frame")

dcf <- function(x, r, t0=FALSE){
  # calculates discounted cash flows (DCF) given cash flow and discount rate
  #
  # x - cash flows vector
  # r - vector or discount rates, in decimals. Single values will be recycled
  # t0 - cash flow starts in year 0, default is FALSE, i.e. discount rate in first period is zero.
  if(length(r)==1){
    r <- rep(r, length(x))
    if(t0==TRUE){r[1]<-0}
  }
  x/cumprod(1+r)
}

npv <- function(x, r, t0=FALSE){
  # calculates net present value (NPV) given cash flow and discount rate
  #
  # x - cash flows vector
  # r - discount rate, in decimals
  # t0 - cash flow starts in year 0, default is FALSE
  sum(dcf(x, r, t0))
}
npv(df$Revenue,.2)
#Non optimized npv yields a value of 492.
#How can i use lpSolve to optimize my table? Said another way how can I rearrange the table to maximize npv using lpSolve?

更复杂的问题涉及具有以下规则的惩罚列： Id 代表项目。

如果 Id 项目不是开始期（第 1 行）。检查之前的 Id 是否在 2 的增量范围内（从其他之前的行中减去行 Id 的绝对值。如果为真，则将 Revenue 减去 20%。我认为这个问题仍然涉及解决正确的顺序。我该如何优化它功能？

#Randomize order to give base npv. Now i need to optimize the order to find max value
df<- df%>%mutate(random_sort= sample(nrow(df)))

x=function(i){
  df_fcn<- i
  df_fcn<- df_fcn%>%mutate(Penalty= if_else(abs(random_sort-lag(random_sort))>2,1,.8))%>%mutate(Penalty=ifelse(is.na(Penalty),1,Penalty))
  df_fcn<- df_fcn%>%mutate(Revenue_Penalized= Revenue*Penalty)
  
  npv(df_fcn$Revenue_Penalized,.2)
  }

Answer 1

我想到的最好办法是随机重新排列数据并找到最大值。

schedule_function=function(i){
  i<- i%>%mutate(random_sort=sample(random_sort))
  df_fcn<- i%>%mutate(Penalty= if_else(abs(random_sort-lag(random_sort))>2,1,.8))%>%mutate(Penalty=ifelse(is.na(Penalty),1,Penalty))
  df_fcn<- df_fcn%>%mutate(Revenue_Penalized= Revenue*Penalty)
  final_df<-print(df_fcn)
  npv(df_fcn$Revenue_Penalized,.2)
}


n <- 1:10000
MAX = -Inf    ## initialize maximum
for (i in 1:length(n)) {
  x <- schedule_function(df)
  if (x > MAX) MAX <- x

  }

lpsolve 与 r 中的约束

lpsolve with constraints in r

optimization

r

linear-programming

lpsolve