iemisc (version 0.9.9)

benefitcost: Benefit-Cost Ratio (Engineering Economics)

Description

Compute the benefit-cost ratio between two alternatives

Usage

benefitcost(
  ic1,
  n1,
  ac1,
  ab1,
  i1,
  salvage1,
  ic2,
  n2,
  ac2,
  ab2,
  i2,
  salvage2,
  option1,
  option2,
  table = c("ptable", "rtable", "both")
)

Value

data.table with character vectors with the monetary values having thousands separator in a pretty table (ptable) & message with the best option, data.frame with numeric vectors without the thousands separator in regular table (rtable) & a message with the best option, or both options combined in a list

Arguments

ic1

numeric vector that contains the initial cost for option 1

n1

numeric vector that contains the useful life (years) for option 1

ac1

numeric vector that contains the annual cost [operations & maintenance (O&M)] for option 1

ab1

numeric vector that contains the annual benefits for option 1

i1

numeric vector that contains the effective interest rate per period as a percent for option 1

salvage1

numeric vector that contains the salvage value for option 1

ic2

numeric vector that contains the initial cost for option 2

n2

numeric vector that contains the useful life (years) for option 2

ac2

numeric vector that contains the annual cost [operations & maintenance (O&M)] for option 2

ab2

numeric vector that contains the annual benefits for option 2

i2

numeric vector that contains the effective interest rate per period as a percent for option 2

salvage2

numeric vector that contains the salvage value for option 2

option1

character vector that contains the name of option for option 1

option2

character vector that contains the name of option for option 2

table

character vector that contains the table output format (ptable, rtable, or both)

Details

Benefit is expressed as

$$Benefit = AB\left[\frac{\left(1 + i\right)^n - 1}{i\left(1 + i\right)^n}\right]$$

Benefit

the present equivalent benefit

AB

the annual benefit

i

the "effective interest rate" per year

n

the number of years

Cost is expressed as

$$Cost = PC + OM\left[\frac{\left(1 + i\right)^n - 1}{i\left(1 + i\right)^n}\right] - S\left[\frac{1}{\left(1 + i\right)^n}\right]$$

Cost

the present equivalent cost

PC

the present or initial cost

OM

the annual operations & maintenance cost

S

the salvage value

i

the "effective interest rate" per year

n

the number of years

Benefit-Cost ratio is expressed as

$$BC = \frac{B_2 - B_1}{C_2 - C_1} \geq 1$$

BC

the present equivalent cost

\(B_1\)

the benefit for alternative 1

\(B_2\)

the benefit for alternative 2

\(C_1\)

the cost for alternative 1

\(C_2\)

the cost for alternative 2

References

1. Michael R. Lindeburg, PE, EIT Review Manual, Belmont, California: Professional Publications, Inc., 1996, page 14-2, 14-4.

2. William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling, Engineering Economy, Fourteenth Edition, Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2009, page 133, 142, 442-443, 452-453.

Examples

library("iemisc")
# Example from Lindeburg Reference text (page 14-4)
benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000, i1 = 10,
salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000, i2 = 10,
salvage2 = 10000, option1 = "A", option2 = "B", table = "rtable")


# This is useful for saving the results as the named data.frame rtable
rtable <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "rtable")

rtable


# This is useful for saving the results as the named data.frame ptable
ptable <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "ptable")

ptable


# This is useful for saving the results as the named list of 2 data.frames
# called both
both <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "both")


both


# Example 10-8 from the Sullivan Reference text (page 452-453)
project <- benefitcost(ic1 = 750000, n1 = 35, ac1 = 120000, ab1 = 245000,
i1 = 9, salvage1 = 0, ic2 = 625000, n2 = 25, ac2 = 110000, ab2 = 230000,
i2 = 9, salvage2 = 0, option1 = "Project I", option2 = "Project II",
table = "rtable")

project