What is the net present worth from all costs and benefits for the new pollution prevention program over 20 yrs (use a 7% discount rate) in terms of today’s dollars?

Engineering Economics

Introduction

Engineering Economics Background

Costs

Capital costs

Operation and maintenance (O&M) costs

Benefits

Present Worthn

)i1(

PW +

where PW = present worth of the future benefit or cost

F = future benefit or cost

i = discount rate (annual)

n = number of years in the future

Example

A company will experience a savings of $15,000 in raw materials three years from now if they implement a pollution prevention program. How much is that savings worth today’s dollars for a discount rate of 7%?

Compound Present Worth for a period of Multiple Years

If we estimate that a cost or benefit is the same every year, then we can estimate the total cost or benefit for a multiple-year time frame:

𝑃𝑊 = 𝐴 [(1 + 𝑖)𝑛 − 1]

𝑖(1 + 𝑖)𝑛

PW = present worth of the future benefit or cost A = annual amount of future benefit or cost

i = interest rate (annual)

n = number of years into the future

This equation is also often written as:( )( )

PWFAPW =

where PWF = present worth factor =

)i1(i

1)i1(

−+

A = annual amount of future benefit or cost

i = interest rate (annual)

n = number of years into the future

Example

A company will experience a savings of $15,000 in raw materials every year. How much of that savings will the company realize after three years in today’s dollars for a discount rate of 7%?

Payback Period

It is important to know how long it will take before the financial benefits equal the financial costs.

Use equation for present worth and set the present worth of a New proposed changed method equal to the present the Old current method. Solve for ‘n’, the number of years it will take before they are equal. At that point, your project reaches payback (after that, it is all

profit).

New = 𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 + 𝐴𝑛𝑒𝑤

[(1+𝑖)𝑛−1]

𝑖(1+𝑖)𝑛 = 𝐴𝑜𝑙𝑑

[(1+𝑖)𝑛−1]

𝑖(1+𝑖)𝑛 = Old

𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 = (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤) [(1 + 𝑖)𝑛 − 1]

𝑖(1 + 𝑖)𝑛

𝑖(1 + 𝑖)𝑛(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡) = (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)[(1 + 𝑖)𝑛 − 1]

𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)(1 + 𝑖)𝑛 = (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)(1 + 𝑖)𝑛 − (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)

(1 + 𝑖)𝑛⌊𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡) − (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)⌋ = −(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)

(1 + 𝑖)𝑛 = −(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)

⌊𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡) − (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)⌋ = (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)

⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋

𝑙𝑜𝑔[(1 + 𝑖)𝑛] = 𝑙𝑜𝑔 { (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)

⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋}

𝑛[𝑙𝑜𝑔(1 + 𝑖)] = 𝑙𝑜𝑔 { (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)

⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋}

𝑛 =

𝑙𝑜𝑔 { (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤)

⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋}

𝑙𝑜𝑔(1 + 𝑖)

Present Worth Analysis of a Pollution Prevention Program

Example

A company currently pays $90,000/yr in fines for violating environmental regulations, $8,000/yr in insurance, and $70,000/yr to treat its waste. A pollution prevention program would cost $500,000 to install new equipment, $5,000/yr in employee training, and $20,000/yr in operation and maintenance costs. It would reduce the fines to $10,000/yr, insurance to $5,000/yr, and treatment costs to $60,000/yr.