Homework #7Engineering Economics and Pollution Prevention
1. You are working for a factory and have been asked to assess the economic impacts of
implementing a new pollution prevention program at the factory. The factory
managers required you to design the program so that the production from the factory
is not reduced by the pollution prevention program. Currently, the factory has an
annual operation cost of $480,000/yr. It currently costs the factory $75,000 per year
to dispose of its waste. The plan that you developed will have an initial equipment
cost of $100,000. However, the result of your plan will reduce the annual operating
cost to $400,000/yr and the waste disposal cost to $30,000/yr.
a. Assuming a constant discount rate of 7 percent, how long will it take the factory
to break even on their investment in your pollution prevention program?
2. A metal plating operator is considering installing an ion exchange unit to recover and
reuse metals currently lost in the rinse waters from the plating line. The company
presently pays a sewer usage fee of $4.00 per kg of metal sent to the sewage
treatment plant. The metal plater is currently discharging 1,000 kg/yr of metal to the
sewer. The metal costs the company $120 per kilogram to purchase. The ion
exchanger will recover 98 percent of this metal, and all of this recovered metal can be
reused in the metal plating operation. The ion exchange unit will cost $50,000 to
purchase and install and $12,000 per year to operate.
a. What is the net cost or benefit of the project, assuming a discount rate of 10
percent over 10 years (based on an expected equipment life of 10 years)?
b. What is the payback period for the ion exchange unit?
c. Based on your economic analysis, would you recommend that the ion exchange
unit should be installed to the metal plater? Why?
d. Discuss how two other factors besides economics might influence the decision.
Engineering Economics
I. Introduction
A. Engineering Economics Background
1. Costs
• Capital costs
• Operation and maintenance (O&M) costs
2. Benefits
3. Present Worth
PW =
F
(1 + i) n
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 (i.e. what is the present worth) for
a discount rate of 7%?
1
4. 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:
[(1 + 𝑖)𝑛 − 1]
𝑃𝑊 = 𝐴
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:
PW = (A )(PWF)
(1 + i) n − 1
where PWF = present worth factor =
i(1 + i) n
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 (i.e. what is the present worth) for a discount rate of 7%?
2
B. 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
[(1 + 𝑖)𝑛 − 1]
𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 = (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
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 )
−(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
=
⌊𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡) − (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )⌋
⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 ) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋
(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
𝑙𝑜𝑔[(1 + 𝑖)𝑛 ] = 𝑙𝑜𝑔 {
}
⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 ) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋
(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
𝑛[𝑙𝑜𝑔(1 + 𝑖)] = 𝑙𝑜𝑔 {
}
⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 ) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋
(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
𝑙𝑜𝑔 {
}
⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 ) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋
𝑛=
𝑙𝑜𝑔(1 + 𝑖)
3
D. 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.
1. 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?
2. How long will it take for the pollution prevention program to pay for
itself?
4
C. Environmental Cost Accounting
1. Benefits
a) Accuracy
b) Improved Decision Making
2. Environmental cost
5
U.S. EPA. 1995. An Introduction to Environmental Accounting as a Business Management Tool: Key Concepts and
Terms. EPA 742-R-95-001.
6
Pollution Prevention (P2)
I. Introduction
• Pollution Prevention Act 1990
A. 3M: Pollution Prevention Pays (3P) Program
• 3M is a huge company
• The program has some important goals
• The environmental results of the program have been dramatic:
• The program has also saved lots of money
• “We believe that environmental interests and our business interests
have merged.” – VP of 3M
• Example – Reduce solvent use in making tape
• Facility in Brazil. Replace solvent-based
manufacturing with water-based method.
• Reduced solvent emissions by 45 tonnes/year
• No need to treat waste for solvents, which
saves money and energy
• Reduced CO2 emissions by 145 tonnes per year
• Saves $850,000 per year.
1
II. Defining Pollution Prevention
• The U.S. EPA does not include recycling or recovery of a waste in its
definition, whereas many states in the U.S. and members of the European Union
include recycling and recovery.
A. Goals of Pollution Prevention
Source Reduction
Pollution
Prevention
Recycling and Recovery
Treatment
NOT
Pollution
Prevention
Disposal
2
B. Definition
• “The use of materials, processes and practices that reduce or eliminate
the creation of pollutants or wastes at the source. It includes practices
that reduce the use of hazardous materials, energy, water, or other
resources AND practices that protect natural resources through
conservation or more efficient use.” – U.S. EPA definition
Bishop, P.L. 2000. Pollution Prevention: Fundamentals and Practice. McGraw Hill: New York.
C. Activities That Are Not Part of Pollution Prevention
3
III. Benefits of Pollution Prevention
A. Environmental benefits
B. National and Societal Benefits
C. Economic Benefits
1. Tangible cost savings
2. Intangible costs
4
D. Examples
Solvent Waste Generation (kg)
1. Commonwealth Edison
Non-Chlorinated Solvents
300000
Chlorinated Solvents and
Oils
Chlorinated Solvents
250000
200000
Naptha
150000
100000
50000
0
1992
1993
5
1994
2. Chrysler (automobile)
• Reduced mercury in product
• $18,000/yr savings
3. Bristol-Meyers Squibb (medical)
• Reduced waste (e.g. solvents)
• $675,000/yr savings
4. Precision Circuits (electronics)
• Reduced waste (metals, nitric acid, waste sludge)
• $19,000/yr savings
5. Island Dog Breweries (small microbrewery)
• Received ~ $10,000 in grants from State of Massachusetts
• New equipment results in savings of $2,900/year
• Also reduces water use and cleaning supplies
6. BAE Systems Norfolk Ship Repair
• New methods to save energy saves $135,000/yr and 2,500
MT of CO2 per year
• Reduced water use by 15%, saving $1,600,000/yr
• Hazardous waste recycling (mostly ship paint) increased by
35%, saving $17,000/yr
IV. Approaches to Pollution Prevention
A. Plans and Documentation
• Record keeping
• Understand Facility
• Ongoing Analysis
• Specific Things to Document
6
B. Inventory Management
1. Match raw material purchases with production output
2. Economic savings
3. Control
4. Example. PVC manufacturer in North Carolina
5. Expiration dates
B. Production Process Modification
1. Reactor Design
raw
materials
product
Reactor
energy
waste
7
2. Improved Energy Efficiency
3. Materials Substitution
Source: Bishop, P. 2000. Pollution Preevntion: Fundamentals and Practice. McGraw Hill:New York.
• Dry Cleaning example
▪ Gasoline and kerosene
▪ PCE
8
▪ More recently, other alternatives: liquid CO2, nonaromatic hydrocarbons (petroleum industry), D5 (Dow
Chemical)
• D5 (Decamethylcyclopentasiloxane)
SiO2, CO2, and H2O
4. Hazardous Materials Storage
Table 6-3
Most common deficiencies in storing hazardous wastes and hazardous materials
Deficiency
Percent of audited facilities
having this deficiency
Inadequate containment/drum/tank storage areas
68
Lack of integrity testing
65
Lack of or inadequate inspection program for storage
48
Improper Labeling
44
Inadequate training program
40
Lack of or inadequate spill prevention, control, and
34
contingency plan
Incomplete wastewater analysis
24
Lack of or inadequate hazardous waste contingency plan
18
Inadequate handling of materials containing PCBs
18
Storm drains in liquid storage and transfer areas
18
Inadequate handling of empty drums
17
Source: LaGrega, M. et al. 2001. Hazardous Waste Management. McGraw Hill:New York.
9
C. Volume Reduction
1. Waste segregation
2. Waste concentration
D. Waste Recovery and Recycling
E. Personnel Training
• Personnel training is essential
• Ongoing Training.
• Explanations.
• Examples.
10
Source: Bishop, P. 2000. Pollution Preevntion: Fundamentals and Practice. McGraw Hill:New York.
V. Economic Rationale for Pollution Prevention
U.S. EPA. 1995. An Introduction to Environmental Accounting as a Business Management Tool: Key Concepts and
Terms. EPA 742-R-95-001.
11
A. Overview of Various P2 Costs and Benefits to Companies
1. Reduce costs from current environmental performance
a) Perhaps the most obvious area to consider.
1) Regulatory penalties.
2) Treatment costs.
3) Industrial costs and societal costs.
• Netherlands flower industry example.
12
2. Reduce potential costs from future liability
a) Out of site, out of mind
b) Past environmental performance
c) Liability types
d) Liability costs
e) Reducing future costs
1) Current environmental performance may create future
liabilities.
2) Complying with current regulations may not be enough to
prevent future liability
3. Reduced capital costs
a) Interest rates.
b) Short-term versus long-term.
4. Improved market opportunities
a) Product phase out.
13
b) Opportunity.
• Example – Phase out and opportunity. Refrigerants.
• R-12 (a CFC, freon) by DuPont. Global warming potential
(GWP) = 10,200. Ozone depletion potential (ODP) = 1.0
(VERY high)
• R-22 (an HCFC) by DuPont. GWP = 810. ODP = 0.05
(much less, but still too high).
• R-134a. DuPont. GWP = 1430. ODP ~ 0. Phase out starting
2012.
• New alternatives such as fluoro-ethers and fluoro-olefins,
and fluoro-ethanes.
o Growing concern for global warming potential of
refrigerants. EU requires all cars to use refrigerants
with GWP less than 150 (others banned since 2012).
o Example: R1234yf by Honeywell. GWP = 4.
ODP ~ 0.
c) Access to markets
14
5. Improved image
6. More Examples
a) 3M Reusable Packaging Systems
b) Dow – SafeChem System
15
c) Volkswagen
d) American Reinsurance
16
Homework #7
Engineering Economics and Pollution Prevention
1.) You are working for a factory and have been asked to assess the economic impacts of
implementing a new pollution prevention program at the factory. The factory
managers required you to design the program so that the production from the factory
is not reduced by the pollution prevention program. Currently, the factory has an
annual operation cost of $480,000/yr. It currently costs the factory $75,000 per year
to dispose of its waste. The plan that you developed will have an initial equipment
cost of $100,000. However, the result of your plan will reduce the annual operating
cost to $400,000/yr and the waste disposal cost to $30,000/yr.
A.) Assuming a constant discount rate of 7 percent, how long will it take the factory to
break even on their investment in your pollution prevention program?
Answer:
A.) The break-even point can be calculated using the following formula:
Break-even point (years) = Initial Investment / (Annual Savings x (1 – Discount Rate))
Break-even point (years) = $100,000 / ($45,000 x (1 – 0.07))
Break-even point (years) = 2.2 years
2.) A metal plating operator is considering installing an ion exchange unit to recover and
reuse metals currently lost in the rinse waters from the plating line. The company
presently pays a sewer usage fee of $4.00 per kg of metal sent to the sewage
treatment plant. The metal plater is currently discharging 1,000 kg/yr of metal to the
sewer. The metal costs the company $120 per kilogram to purchase. The ion
exchanger will recover 98 percent of this metal, and all of this recovered metal can be
reused in the metal plating operation. The ion exchange unit will cost $50,000 to
purchase and install and $12,000 per year to operate.
A.) What is the net cost or benefit of the project, assuming a discount rate of 10
percent over 10 years (based on an expected equipment life of 10 years)?
B.) What is the payback period for the ion exchange unit?
C.) Based on your economic analysis, would you recommend that the ion exchange
unit should be installed to the metal plater? Why?
D.) Discuss how two other factors besides economics might influence the decision.
Answer:
A.) The net cost or benefit of the project can be calculated using the following formula:
Net Cost/Benefit = (Initial Investment x (1 – Discount Rate)) – (Annual Savings x (1 – Discount
Rate))
Net Cost/Benefit = ($50,000 x (1 – 0.1)) – ($12,000 x (1 – 0.1))
Net Cost/Benefit = $31,800
B.) The payback period for the ion exchange unit can be calculated using the following
formula:
Payback Period (years) = Initial Investment / Annual Savings
Payback Period (years) = $50,000 / $12,000
Payback Period (years) = 4.2 years
C.) Based on my economic analysis, I would recommend that the ion exchange unit
should be installed to the metal plater. The initial investment is offset by the annual savings over
the 10-year life of the equipment, so the net cost/benefit of the project is a positive $31,800.
Additionally, the payback period of 4.2 years is relatively short.
D.) Two other factors that might influence the decision are the environmental impacts of
the ion exchange unit and the operational complexity of the unit. The environmental impacts
should be considered in order to determine if the unit will have any negative impacts on the
surrounding environment. Additionally, the operational complexity should be considered to
ensure that the unit can be operated safely and efficiently.
Engineering Economics
I. Introduction
A. Engineering Economics Background
1. Costs
• Capital costs
• Operation and maintenance (O&M) costs
2. Benefits
3. Present Worth
PW =
F
(1 + i) n
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 (i.e. what is the present worth) for
a discount rate of 7%?
1
4. 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:
[(1 + 𝑖)𝑛 − 1]
𝑃𝑊 = 𝐴
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:
PW = (A )(PWF)
(1 + i) n − 1
where PWF = present worth factor =
i(1 + i) n
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 (i.e. what is the present worth) for a discount rate of 7%?
2
B. 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
[(1 + 𝑖)𝑛 − 1]
𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 = (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
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 )
−(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
=
⌊𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡) − (𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )⌋
⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 ) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋
(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
𝑙𝑜𝑔[(1 + 𝑖)𝑛 ] = 𝑙𝑜𝑔 {
}
⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 ) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋
(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
𝑛[𝑙𝑜𝑔(1 + 𝑖)] = 𝑙𝑜𝑔 {
}
⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 ) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋
(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 )
𝑙𝑜𝑔 {
}
⌊(𝐴𝑜𝑙𝑑 − 𝐴𝑛𝑒𝑤 ) − 𝑖(𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑐𝑜𝑠𝑡)⌋
𝑛=
𝑙𝑜𝑔(1 + 𝑖)
3
D. 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.
1. 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?
2. How long will it take for the pollution prevention program to pay for
itself?
4
C. Environmental Cost Accounting
1. Benefits
a) Accuracy
b) Improved Decision Making
2. Environmental cost
5
U.S. EPA. 1995. An Introduction to Environmental Accounting as a Business Management Tool: Key Concepts and
Terms. EPA 742-R-95-001.
6
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