投资决策:净现值、内部收益率、净现值计算

Question

我试图计算不同数量项目的净现值(NPV)、内部收益率(IRR)和投资回收期(PB)，以评估哪个投资项目提供最佳回报。

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到目前为止，我可以在几行代码中分别为每个项目计算它。但我要做的是:编写一个函数，该函数接受一个包含许多不同项目及其现金流的矩阵，并自动返回NPV、IRR和PB。该函数应接受用于贴现现金流的无风险利率(利率)作为附加输入参数。

有谁有办法解决这个问题吗？关于如何编写一个函数来返回一个包含结果的矩阵？

计算每个度量本身的代码如下所示：

library(FinCal)

#NPV Project 1 & Project 2
NPV.Project1 <- NPV(-1000, c(1250, 10, 10, 20, 20), c(1:5), 0.06 ) 
NPV.Project2 <- NPV(-1000, c(-10, 0, 10, 20, 2000), c(1:5), 0.06 )

#Solution: NPV2 with 509 > NPV1 with 227 -> Pick Project 2

#IRR Project 1 & Project 2
IRR.Project1 <- IRR(-1000, c(1250, 10, 10, 20, 20), c(1:5))
IRR.Project2 <- IRR(-1000, c(-10, 0, 10, 20, 2000), c(1:5))
#Solution: IRR.Project1 with 28% > IRR.Project2 with 15% -> Pick Project 1

#PB Project 1 & Project 2
#PB Project 1
cf0 <- -1000
cf <- c(1250, 10, 10, 20, 20)
t <- 5

temp <- 0

#Calculating the Payback period
for (i in 1:5){
  temp[i]=sum(cf[1:i])
}
for (i in 1:5){
  if ((temp[i]+cf0) > 0){
    payback.Project1 <- i
    break
  }
  print(payback.Project1)
}

#PB Project 2
cf0 <- -1000
cf <- c(-10, 0, 10, 20, 2000)
t <- 5

temp <- 0

#Calculating the Payback period
for (i in 1:5){
  temp[i]=sum(cf[1:i])
}
for (i in 1:5){
  if ((temp[i]+cf0) > 0){
    payback.Project2 <- i
    break
  } 
  print(payback.Project2)
}
#Solution: Period for PB with Project 1 is way smaller (1 year) than with  Project 2 (5 years)

Answer 1

您的数据：

cf_df <- data.frame(
  Project1 = c(-1000, 1250, 10, 10, 20, 20),
  Project2 = c(-1000, -10, 0, 10, 20, 2000)
)

以下是一些有用的函数，您可以考虑保留它们来处理资本预算问题：

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))
 }

pbp <- function(x, ...){
  # calculates payback period (PBP)
  #
  # x - cash flows vector
  # ... - ignored
  i <- match(1, sign(cumsum(x)))
  i-2+(-cumsum(x)[i-1]/x[i])
 }

dpbp <- function(x, r, t0=FALSE){
  # calculates discounted payback period (DPBP) 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
  pbp(dcf(x, r, t0))
 }

irr <- function(x, t0=FALSE, ...){
  # calculates internal rate of return (IRR) given cash flow 
  #
  # x - cash flows vector
  # t0 - cash flow starts in year 0, default is FALSE
tryCatch(uniroot(f=function(i){sum(dcf(x, i, t0))}, 
                 interval=c(0,1))$root,
         error=function(e) return(NA)
        )
  }

将这些功能与dplyr的强大功能相结合，您将拥有一个能够与首席财务官最喜欢的电子表格竞争的工具箱：

library(dplyr)
library(tidyr)
library(forcats)

cf_df %>%
  summarise_all(funs(NPV=npv,PBP=pbp, DPBP=dpbp, IRR=irr), r=0.06, t0=TRUE) %>% 
  gather(key=key, value = value) %>% 
  separate(key, into = c("Project", "Metric")) %>% 
  spread(key=Project, value=value) %>% 
  mutate(Metric=fct_relevel(Metric, "NPV", "IRR", "PBP", "DPBP"),
         Metric=fct_recode(Metric, 
                           `Net Present Value (NPV), USD mln`="NPV", 
                           `Internal Rate of Return (IRR), %`="IRR",
                           `Payback Period, years` = "PBP",
                           `Discounted Payback Period, years`="DPBP")) %>% 
  arrange(as.numeric(Metric))

#                            Metric    Project1    Project2
#1 Net Present Value (NPV), USD mln 227.3284770 509.3204496
#2 Internal Rate of Return (IRR), %   0.2808485   0.1508404
#3            Payback Period, years   0.8000000   4.4900000
#4 Discounted Payback Period, years   0.8480000   4.6592072

Answer 2

 W=function(data,rate){
   NPV=apply(data,1,npv,r=rate)
   IRR=apply(data,1,irr)
   PB=apply(data,1,function(x)min(which(cumsum(x)[-1]>0)))
   data.frame(NPV=NPV,IRR=IRR,PB=PB)
 }

 funfun=function(data,r){
 mapply(apply,list(data),1,c(function(x)npv(r,x),irr,function(x)min(which(cumsum(x)[-1]>0))))
 }

W(dat,0.06)
       NPV       IRR PB
A 227.3285 0.2808516  1
B 509.3204 0.1508564  5

 funfun(dat,0.06)
      [,1]      [,2] [,3]
A 227.3285 0.2808516    1
B 509.3204 0.1508564    5

当然，您可以给出最终的矩阵列名。希望这能有所帮助

下面是使用的矩阵数据：

dat= structure(c(-1000, -1000, 1250, -10, 10, 0, 10, 10, 20, 20, 20, 
 2000), .Dim = c(2L, 6L), .Dimnames = list(c("A", "B"), NULL))