# benefitcost: Benefit-Cost Ratio (Engineering Economics) In iemisc: Irucka Embry's Miscellaneous Functions

## Description

Compute the benefit-cost ratio between two alternatives

## Usage

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 benefitcost( ic1, n1, ac1, ab1, i1, salvage1, ic2, n2, ac2, ab2, i2, salvage2, option1, option2, table = c("ptable", "rtable", "both") ) 

## 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≤ft[\frac{≤ft(1 + i\right)^n - 1}{i≤ft(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≤ft[\frac{≤ft(1 + i\right)^n - 1}{i≤ft(1 + i\right)^n}\right] - S≤ft[\frac{1}{≤ft(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} ≥q 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

## 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

## 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

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 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 

iemisc documentation built on Aug. 2, 2020, 9:07 a.m.