Robust algorithms for the solution of the inventory problem in Life Cycle Impact Assessment Maurizio Cellura, Antonino Marvuglia – University of Palermo (Italy) Marcello Pucci – Institute for Studies on Intelligent Systems for Automation (I.S.S.I.A), National Research Council, Palermo (Italy) Intelligent Analysis of Environmental Data (S4 ENVISA Workshop 2009)

1. University of Palermo
S4 ENVISA Workshop 2009
Palermo, 18-20 June 2009
Robust algorithms for the solution of
the inventory problem in Life Cycle
Impact Assessment
Maurizio Cellura Marcello Pucci
Antonino Marvuglia
Institute for Studies on Intelligent
University of Palermo Systems for Automation (I.S.S.I.A), 1
Dep. of Energy and Environmental Researches (DREAM) National Research Council, Palermo (Italy)

2. Introduction S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Every product has a “life” (1. design/development of the product; 2. resource extraction;
3. production; 4. use/consumption; 5. end-of-life activities, like collection/sorting,
reuse, recycling, waste disposal).
All activities, or processes, in a product’s life result in environmental impacts due to
consumption of resources, emissions of substances into the natural environment, and
2
other environmental exchanges (e.g. radiation).

3. Introduction S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Life cycle assessment (LCA) is a methodological framework for estimating and assessing the
environmental impacts attributable to the life cycle of a product, such as climate change,
stratospheric ozone depletion, tropospheric ozone (smog) creation, eutrophication, acidification,
toxicological stress on human health and ecosystems, the depletion of resources, water use, land use,
noise and others.
LCA is traditionally divided into four distinct though
interdependent phases:
1. Goal and scope definition attempts to set the extent of the
inquiry as well as specify the methods used to conduct it in
later phases.
2. Life cycle inventory analysis defines and quantifies the flow
of material and energy into, through, and from a product
system.
3. Life cycle impact assessment converts inventory data into
environmental impacts using a two-step process of
classification and characterization.
4. Life cycle interpretation marks the point in an LCA when
one draws conclusions and formulates recommendations
based upon inventory and impact assessment data.
Despite LCA is nowadays a universally accepted methodology, each of these
3
phases still contains some unresolved problems.

4. Introduction S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
PHASE PROBLEM
Goal and scope definition • Functional unit definition
• Boundary selection
• Social and economic impacts
• Alternative scenario considerations
Life cycle inventory analysis • Allocation
• Negligible contribution (“cutoff”) criteria
• Local technical uniqueness
Life cycle impact assessment • Impact category and methodology selection
• Spatial variation
• Local environmental uniqueness
• Dynamics of the environment
• Time horizons
Life cycle interpretation • Weighting and evaluation
• Uncertainty in the decision process
All • Data availability and quality
4
From: J. Reap et al., A survey of unresolved problems in life cycle assessment. Int J Life Cycle Assess (2008) 13:290–300

5. Introduction S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Allocation refers to the procedure of appropriately allocating the environmental burdens
of a multi-functional process amongst its functions or products.
Co-generation of Production Production
electricity and heat of electricity of heat
litre of fuel  −2   λ1a × −2   λ1b × −2 
     
kWh of electricity  10   10   0 
MJ of heat  18   0   18 
p1 =  p1a =  p1b = 
Kg of CO2 1  µ1a ×1  µ1b × 1 
     
Kg of SO2  0.1  λ1a + λ1b = 1  µ2 a × 0.1   µ 2b × 0.1 
     
0   0   0 
litre of crude oil  µ1a + µ1b = 1
λ2 a + λ2b = 1 ALLOCATION FACTORS
Obviously, arbitrary allocations could lead to incorrect LCA results and less preferable
decisions based on those results. 5

6. Introduction S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
ISO 14044 recommends that, where allocation cannot be avoided, physical causality is to
be used as the basis for allocation where possible.
Physical properties used as a basis for allocation include mass, energy or exergy content.
If allocation based on physical, causal relationship is not feasible or does not provide a
full solution, ISO 14044 suggests that the exchanges between the products and functions
have to be partitioned “in a way which reflects other relationships between them. For
example, input and output data might be allocated between co-products in proportion to
the economic value of the products.”
Anyhow, it has to be remarked that, regardless the method used, allocation introduces
uncertainty and subjectivity elements into the computation and leads to biased solutions.
6

7. Mathematical background S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
The matrix method for the solution of the so called inventory problem in LCA generally
determines the inventory vector or eco-profile related to a specific process by solving the
system of linear equations:
economic
economic (or
technologic) matrix A ⋅s = f functional unit
scale vector
More in general, taking into account also the environmental part of the system:
A  fecn 
f =   ⋅s =   s = A −1 ⋅ fecn fenv = B ⋅ s
B  fenv 
From the mathematical point of view, the presence of multifunctional processes in the
investigated system makes the economic matrix rectangular and thus non invertible.
A possible strategy to deal with this problem without using allocation procedures is
based on the pseudo-inverse of the technology matrix.
The pseudo-inverse of a matrix A ∈ ℜ m×n (with m>n) is defined as:
A = ( A A) A
−1 7
† T T

8. Mathematical background S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Using the pseudo-inverse of the economic matrix A, it is possible to compute the scale
vector also when A is rectangular, through the expression:
s = A† ⋅ f
However, it is necessary to assess in every case whether the obtained solution is
satisfactory.
To accomplish this assessment it is necessary to substitute the computed value of s in
the original system, obtaining the so called discrepancy vector :
f = A ⋅ ( A† ⋅ f )
%
For the normal inverse A-1 we have: A ⋅ ( A −1 ⋅ f ) − f = 0
While for the pseudo-inverse we have: A ⋅ ( A† ⋅ f ) − f = f − f ≠ 0
%
If the Euclidean norm %
f −f is not too high (with respect to a fixed tolerance) the
solution obtained with the pseudo-inverse can be considered acceptable. 8

9. Problem description S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
One considers, for example, the system with the following inventory matrix:
Production of Production of Incineration of
electricity fuel waste
kWh of electricity 10 -500 -5
l of fuel -1 100 0
ECONOMIC
MATRIX
kg of organic waste 0 0 -1000
A
kg of chemical waste 0 0 -200
kg of CO2 1 10 1000
ENVIRON.
kg of SO2 0.1 2 30
MATRIX
kg of crude oil 0 -50 0 B
and let the functional unit be the following vector:
1000  Production of 1000 kWh of
electricity
 
0 
f =
 0 
  9
 0 

10. Problem description S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
In this case, using the pseudo-inverse of A, one obtains:
1000 
 200   
0 

A† ⋅ f =  2 
 f = A ⋅ ( A† ⋅ f ) = 
%
 0 
 0   
 
 0 
% (
As a consequence, the discrepancy vector d = f − f ) is:
0
 
% − f = 0
d=f d 2 =0
0
 
0
Thus the pseudo-inverse gives in this case (for this particular choice of the functional
unit vector) an exact solution to the problem.
10

11. Problem description S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
If we chose the following functional unit:
 0 
 
0 
f1 = 
 0 
 
 −1000 
Disposal of 1000 kg of chemical
waste
We obtain:
 0 
 0.1923   
0 
 
A † ⋅ f1 =  0.0019  f1 = A ⋅ ( A † ⋅ f1 ) = 
%
 −192.31
 0.1923   
  −38.46 

11

12. Problem description S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
The discrepancy vector is in this case equal to:
 0 
 
0  = 980.6
d1 = f − f = 
% d1 2
 −192.31
 
 961.54 
According to the goals of the study, this value could be considered too high and thus the
solution obtained through the pseudo-inverse should be in this case discarded.
In those cases in which the pseudo-inverse is not able to provide an acceptable solution,
the use of least squares techniques could be very useful.
12

13. Problem solution S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Given the system of equations A⋅x = f
(where A ∈ ℜ m×n (m > n) and f ∈ ℜ m×1 )
There is no exact solution if f ∉ range( A)
To solve this over-determined system we can use, for example, the Ordinary Least
Squares (OLS) technique, which means solving the system A ⋅ sOLS = f + ∆f
∆f = As-f
f
y=As
range(A)
13
∆f is the residual error vector corresponding to a perturbation in f .

14. Problem solution S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
A different approach that yields a consistent estimator is the Total Least Squares
(TLS), which is a linear parameter estimation technique that has been devised to
compensate for data errors.
It is a natural generalization of the OLS approximation method when the data both in
A and f are perturbed.
The classical TLS problem looks for the minimal corrections ∆A and ∆f on the given
data A and f that make solvable the corrected system of equations:
( A + ∆A ) sTLS = ( f + ∆f )
A particular case of TLS is the so called Data Least Squares (DLS) problem, in which
the error is assumed to lie only in the data matrix and thus the problem is converted
into the solution of the system:
( A + ∆A ) s DLS = f
14

15. Problem solution S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Classical TLS problem formulation
OLS approach TLS approach
• Find a matrix F' • Find a matrix Â
Frobenius norm
such that m×n such that
min F − F′ M ∈ℜ A − A F − F
ˆ ˆ
min
F ′∈ range( A ) F
m n F∈ range( A )
ˆ   F
∑∑ m
2
M = i, j
• Solve the system F
i =1 j =1
• Solve the system
A ⋅ SOLS = F′ ˆ
 ⋅ STLS = F
^
a2
f a2
f ^ ) a2
(A ^
n ge a1
ra
^
range(A) f
a1 range(A)
a1
f'
15
0 a 0 a

16. Problem solution S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Ordinary Least Squares Total Least Squares Data Least Squares
(OLS) (TLS) (DLS)
f f f
(fi) (fi) (fi)
arctan(s ) (ai)
OLS A arctan(s ) (ai)
TLS A arctan(s ) (ai) A
DLS
The OLS, TLS and DLS regression problems can be solved in several ways.
One way to find the solution is the minimization of a cost function:

( As − f ) ( As − f )
T
(OLS problem)

 ( As − f ) ( As − f )
T
E ( x) =  (TLS problem)
 1 + s Ts
 ( As − f ) T ( As − f ) 16
 (DLS problem)
 sTs

17. Problem solution S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
The approach here applied all these problems have been generalized by using a
parameterized formulation of an error function whose minimization yields the
corresponding solution. This error is given by the expression:
ξ = 0 OLS
1 ( A ⋅s - f ) ( A ⋅s - f )
T

E ( s) = ξ = 0.5 TLS
2 ( 1 − ξ ) + ξ sT s ξ = 1 DLS

This approach comes from a work by G. Cirrincione, M. Cirrincione and S. Van Huffel
("The GeTLS EXIN Neuron for Linear Regression", 2000)
The above function can be regarded as the cost function of a linear neuron (the GeTLS
EXIN linear neuron) whose weight vector is s(t).
In particular is:
( )
2
m
1 ai s - fi
EGeTLS ( s ) = ∑ E ( i)
( s) E ( s) =
( i)
i =1 2 ( 1 − ξ ) + ξ sT s
17
where ai is the i-th row of A and f is the i-th element of f.

18. Problem solution S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
( ) ( )
2
Hence: ( i) ai s - fi ai a i s - fi ξ s
dE
= −
ds ( 1 − ξ ) + ξ s ( t ) s ( t )  ( 1 − ξ ) + ξ sT ( t ) s ( t )  2
T
 
The iterative algorithm (learning law) exploiting the gradient (steepest descent) is
given by:
s ( t + 1) = s ( t ) − α ( t ) γ ( t ) aT + ξα ( t ) γ 2 ( t )  s ( t )
i  
sT ( t ) aT − fi
where: γ ( k) = i
; α (t) = learning rate
( 1 − ξ ) + ξ sT ( t ) s ( t )
The GeTLS EXIN neuron is a linear unit with:
• n inputs (vector ai);
• n weights (vector s);
• one output (scalar yi = sT ai );
(
• a training error (scalar ai s - fi ) )
In the application showed in this work, the parameter ς is made variable according to
18
a predefined scheduling (in general monotonically from 0 to 1).

19. Case study S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Illustrative case study: bricks production
Rectangular technology matrix A ∈ℜ14×8
Production Production Production Production Supply Production Final
Production of Production of
of of of of of of demand
oil-derivatives natural gas
electricity clay sand crude oil biomass bricks vector f
1 MJ of electricity 1 -7.20E-03 -1.80E-02 0 -3.20E-02 0 0 -3.69E+02 0
1 MJ of heat 2.48E+00 0 0 -4.87E-02 -1.13E+00 -4.87E-02 0 -1.01E+03 0
1 kg of white clay 0 1 0 0 0 0 0 -1.37E+03 0
1 kg of red clay 0 1 0 0 0 0 0 -8.51E+02 0
1 kg of recycled inerts 0 0 0 0 0 0 0 6.90E+01 0
1 kg of sand 0 0 1 0 0 0 0 -5.00E+02 0
1 kg of gravel 0 0 1 0 0 0 0 -3.47E+02 0
1 kg of olive cake 0 0 0 0 0 0 1 -1.53E+02 0
1 kg of straw 0 0 0 0 0 0 1 -1.92E+01 0
1 MJ of crude oil 0 0 0 1 -2.34E+00 0 0 0 0
1 MJ of diesel oil 0 -4.54E-03 -3.60E-02 -5.29E-03 1 -3.61E+01 -4.06E-01 -1.41E+03 0
1 MJ of fuel oil 0 0 0 0 1 0 0 -1.11E+03 0
1 MJ of natural gas -4.27E+00 0 0 0 0 1 0 -5.52E+03 0
1 t of bricks 0 0 0 0 0 0 0 1 1
Substitution Least
and Squares
Allocation solutions
19

20. Case study S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
The intervention matrix of the system at hand (used for the solutions obtained with the least
squares techniques) is:
21×8
B ∈ℜ
Production Poduction Production Production Production Production Supply Production
of of of of of of natural of of
electricity clay sand crude oil oil-derivatives gas biomass bricks
Resources and raw materials
MJ of Coal 1.49E-03 5.16E-03 1.36E-02 7.46E-07 5.77E-02 7.46E-07 1.40E-03 0
MJ of Lignite 2.74E-04 6.46E-03 1.63E-02 1.77E-09 2.68E-04 1.77E-09 8.32E-04 0
MJ of Hydropower 0 3.82E-04 1.04E-03 9.58E-08 7.66E-03 9.58E-08 4.54E-04 0
MJ of Geothermal Energy 0 2.80E-08 1.17E-07 6.08E-15 1.05E-04 6.08E-15 0 0
kg of Water 0 7.80E-03 2.40E-02 1.49E-02 7.26E-03 1.49E-02 0 1.03E+03
kg of Ores (sand, gravel, etc.) 2.20E-05 2.00E+00 2.00E+00 3.51E-05 4.55E-04 3.51E-05 0 0.00E+00
MJ of Crude Oil 1.21E-02 1.82E-02 1.43E-01 1.02E+00 2.34E+00 1.02E+00 4.15E-01 1.27E+03
kg of other Ores (iron, copper, etc.) 1.25E-04 7.37E-06 3.95E-05 5.60E-10 2.40E-04 5.60E-10 0 0.00E+00
Emissions to air
kg of CO2 8.32E-02 2.52E-03 1.33E-02 4.17E-03 2.88E-02 4.17E-03 4.93E+00 6.02E+02
kg of CO 1.63E-04 3.83E-06 2.79E-05 1.14E-05 3.34E-05 1.14E-05 2.70E-02 1.10E+00
kg of CH4 3.68E-04 2.97E-06 1.09E-05 7.13E-05 1.02E-04 7.13E-05 6.00E-03 1.18E+00
kg of SO2 2.96E-05 2.61E-06 1.64E-05 5.31E-06 2.69E-04 5.31E-06 7.43E-03 2.20E+00
kg of NMVOC 3.70E-05 4.20E-07 2.87E-06 1.93E-05 4.23E-05 1.93E-05 3.07E-02 1.13E+00
Emissions to water
kg of COD 6.25E-07 2.00E-07 1.11E-06 1.57E-11 6.75E-06 1.57E-11 2.21E-04 2.38E-01
kg of BOD 4.36E-08 6.11E-09 3.51E-08 4.41E-13 1.89E-07 4.41E-13 6.75E-06 2.25E-01
kg of P 3.27E-10 4.95E-11 3.91E-10 3.73E-18 9.84E-13 3.73E-18 0.00E+00 1.26E-03
kg of N 3.11E-08 2.91E-09 2.29E-08 2.19E-16 1.08E-10 2.19E-16 1.61E-04 6.00E-03
kg of AOX 5.78E-11 3.69E-12 2.90E-11 4.64E-18 2.01E-12 4.64E-18 2.95E-07 1.43E-05
Solid wastes
kg of Ash 0 1.11E-04 2.83E-04 1.48E-08 2.56E-04 1.48E-08 0 0
kg of Sludge 0 2.46E-07 1.93E-06 5.30E-14 2.10E-05 5.30E-14 0 0
kg of Nuclear Waste 0 2.54E-09 6.49E-09 8.82E-17 7.70E-10 8.82E-17 0 0
20

21. Case study S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Flow chart of the case study on bricks production
Production of c ru
de
o il
oil-derivatives
Production of
oil
natural gas diesel diesel oil
Production of
ity crude oil
tric
ec
s
el
ga
t
hea
al
die
at
tur
se
die
na
na
he l oi
na
es
l
se
tur
u
lo
diesel oil
a
all g
iil
Production of
as
s
electricity and
electr ic
heat ity Supply of
elec
tric it
biomass
y Production of Production of
he ele red and white sand and
at ctr
ic i clay gravel w
ty ra
st
ke
white clay
ca
red clay
v e
gravel oli
fuel oil
sand
Production of
bricks
21
bricks

22. Case study S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Square technology matrix A′ ∈ℜ13×13 (physical allocation)
Production Production Production Production Production Production Production Production Production Production Supply Production
Supply of
of of of of of of of of of of of of
straw
electricity heat white clay red clay sand gravel crude oil fuel oil diesel oil natural gas olive cake bricks
1 MJ of electricity 1 0 -3.60E-03 -3.60E-03 -9.00E-03 -9.00E-03 0 -1.60E-02 -1.60E-02 0 0 0 -3.69E+02
1 MJ of heat 0 2.48E+00 0 0 0 0 -4.87E-02 -5.66E-01 -5.66E-01 -4.87E-02 0 0 -1.01E+03
1 kg of white clay 0 0 1 0 0 0 0 0 0 0 0 0 -1.37E+03
1 kg of red clay 0 0 0 1 0 0 0 0 0 0 0 0 -8.51E+02
1 kg of sand 0 0 0 0 1 0 0 0 0 0 0 0 -4.41E+02
1 kg of gravel 0 0 0 0 0 1 0 0 0 0 0 0 -3.47E+02
1 kg of olive cake 0 0 0 0 0 0 0 0 0 0 1 0 -1.53E+02
1 kg of straw 0 0 0 0 0 0 0 0 0 0 0 1 -1.92E+01
1 MJ of crude oil 0 0 0 0 0 0 1 -1.17E+00 -1.17E+00 0 0 0 0.00E+00
1 MJ of diesel oil 0 0 -2.27E-03 -2.27E-03 -1.80E-02 -1.80E-02 -5.29E-03 0 1 -3.61E+01 -3.25E-01 -8.12E-02 -1.41E+03
1 MJ of fuel oil 0 0 0 0 0 0 0 1 0 0 0 0 -1.11E+03
1 MJ of natural gas -3.41E+00 -8.54E-01 0 0 0 0 0 0 0 1 0 0 -5.52E+03
1 t of bricks 0 0 0 0 0 0 0 0 0 0 0 0 1.00E+00
 -9.52E+01 
  PROCESS ALLOCATION
 -7.14E+03 
 1.37E+03  Prod. of electricity and heat 0.8 for electricity and 0.2 for heat
 
 8.51E+02  Prod. of clay (white and red) 0.5 for with clay and 0.5 for red clay
 4.41E+02 
  Prod. of sand and gravel 0.5 for sand and 0.5 for gravel
 3.47E+02 
Production of oil-derivatives Allocation following the energy criterion.
x′ = ( A′ ) b =  -3.51E+04 
−1
  Supply of biomass 0.8 for olive cake and 0.2 for straw
 1.11E+03 
 -3.11E+04  Prod. of bricks and inerts Substitution method: inerts are considered as
 
 -8.97E+02  equivalent to sand, but with an estimated correction
  factor of 0.85 in order to account for the difference in
 1.53E+02  quality between the mass of inert materials and the
 1.92E+01  22
  mass of sand obtained from them.
 1.00E+00 

23. Case study S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
Environmental intervention matrix after physical allocation
Production Poduction Poduction Production
B′ ∈ white21×13
ℜ of
Poduction Poduction ProductionProduction
Production of Production of Production of Production of Production of Supply
of white of red Production of of Production of Production of Pr
Supply of Production
of red of of heat oil of natural gas
sand crude oil of fuel oil
of sand
electricity crude clayfuel oil clay oil
diesel olive cake
gravel straw bricks
clay clay gravel
Resources and raw materials
MJ of Coal
2.58E-03 2.58E-03 6.82E-03 1.20E-03
6.82E-03 2.99E-04
7.46E-07 2.58E-03
2.89E-02 2.58E-03
2.89E-02 6.82E-03
7.46E-07 6.82E-03
1.12E-03 7.46E-07
2.81E-04 2.89E-02
0
MJ of Lignite
3.23E-03 3.23E-03 8.14E-03 2.19E-04
8.14E-03 5.47E-05
1.77E-09 3.23E-03
1.34E-04 3.23E-03
1.34E-04 8.14E-03
1.77E-09 8.14E-03
6.66E-04 1.77E-09
1.66E-04 1.34E-04
0
MJ of Hydropower
1.91E-04 1.91E-04 5.19E-04 0 5.19E-04 0 9.58E-08 1.91E-04
3.83E-03 1.91E-04
3.83E-03 5.19E-04
9.58E-08 5.19E-04
3.63E-04 9.58E-08
9.08E-05 3.83E-03
0
MJ of Geothermal Energy
1.40E-08 1.40E-08 5.84E-08 0 5.84E-08 0 6.08E-15 1.40E-08
5.25E-05 1.40E-08
5.25E-05 5.84E-08
6.08E-15 5.84E-08
0 6.08E-15
0 5.25E-05
0
kg of3.90E-03
Water 3.90E-03 1.20E-02 0 1.20E-02 0 1.49E-02 3.90E-03
3.63E-03 3.90E-03
3.63E-03 1.20E-02
1.49E-02 1.20E-02
0 1.49E-02
0 3.63E-03
1.03E+03
kg of1.00E+00 gravel, etc.)
Ores (sand, 1.00E+00 1.00E+00 1.76E-05
1.00E+00 4.39E-06
3.51E-05 1.00E+00
2.27E-04 1.00E+00
2.27E-04 1.00E+00
3.51E-05 1.00E+00
0 3.51E-05
0 2.27E-04
0
MJ of Crude Oil
9.09E-03 9.09E-03 7.14E-02 9.71E-03
7.14E-02 2.43E-03
1.02E+00 9.09E-03
1.17E+00 9.09E-03
1.17E+00 7.14E-02
1.02E+00 7.14E-02
3.32E-01 1.02E+00
8.31E-02 1.17E+00
1.27E+03
kg of3.68E-06 (iron, copper, etc.)
other Ores
Emissions to air
kg of1.26E-03
CO2
3.68E-06
1.26E-03
1.97E-05 1.00E-04
6.65E-03 6.66E-02
…
1.97E-05 2.50E-05
5.60E-10 3.68E-06
6.65E-03 1.66E-02
1.20E-04
4.17E-03 1.26E-03
1.44E-02
3.68E-06
1.20E-04
1.26E-03
1.44E-02
1.97E-05
5.60E-10
6.65E-03
4.17E-03
1.97E-05
0
6.65E-03
3.94E+00
5.60E-10
0
4.17E-03
9.86E-01
0
…
1.20E-04
1.44E-02
6.02E+02
kg of1.91E-06
CO 1.91E-06 1.40E-05 1.30E-04
1.40E-05 3.26E-05
1.14E-05 1.91E-06
1.67E-05 1.91E-06
1.67E-05 1.40E-05
1.14E-05 1.40E-05
2.16E-02 1.14E-05
5.40E-03 1.67E-05
1.10E+00
kg of1.48E-06
CH4 1.48E-06 5.44E-06 2.94E-04
5.44E-06 7.36E-05
7.13E-05 1.48E-06
5.10E-05 1.48E-06
5.10E-05 5.44E-06
7.13E-05 5.44E-06
4.80E-03 7.13E-05
1.20E-03 5.10E-05
1.18E+00
kg of1.31E-06
SO2 1.31E-06 8.21E-06 2.37E-05
8.21E-06 5.92E-06
5.31E-06 1.31E-06
1.34E-04 1.31E-06
1.34E-04 8.21E-06
5.31E-06 8.21E-06
5.94E-03 5.31E-06
1.49E-03 1.34E-04
2.20E+00
kg of2.10E-07
NMVOC 2.10E-07 1.43E-06 2.96E-05
1.43E-06 7.40E-06
1.93E-05 2.10E-07
2.12E-05 2.10E-07
2.12E-05 1.43E-06
1.93E-05 1.43E-06
2.46E-02 1.93E-05
6.14E-03 2.12E-05
1.13E+00
Emissions to water
kg of9.98E-08
COD 9.98E-08 5.57E-07 5.00E-07
5.57E-07 1.25E-07
1.57E-11 9.98E-08
3.37E-06 9.98E-08
3.37E-06 5.57E-07
1.57E-11 5.57E-07
1.77E-04 1.57E-11
4.42E-05 3.37E-06
2.38E-01
kg of3.05E-09
BOD 3.05E-09 1.76E-08 3.49E-08
1.76E-08 8.72E-09
4.41E-13 3.05E-09
9.47E-08 3.05E-09
9.47E-08 1.76E-08
4.41E-13 1.76E-08
5.40E-06 4.41E-13
1.35E-06 9.47E-08
2.25E-01
kg of2.48E-11
P 2.48E-11 1.95E-10 2.62E-10
1.95E-10 6.54E-11
3.73E-18 2.48E-11
4.92E-13 2.48E-11
4.92E-13 1.95E-10
3.73E-18 1.95E-10
0 3.73E-18
0 4.92E-13
1.26E-03
kg of1.45E-09
N 1.45E-09 1.15E-08 2.49E-08
1.15E-08 6.22E-09
2.19E-16 1.45E-09
5.40E-11 1.45E-09
5.40E-11 1.15E-08
2.19E-16 1.15E-08
1.29E-04 2.19E-16
3.22E-05 5.40E-11
6.00E-03
kg of1.84E-12
AOX 1.84E-12 1.45E-11 4.62E-11
1.45E-11 1.16E-11
4.64E-18 1.84E-12
1.00E-12 1.84E-12
1.00E-12 1.45E-11
4.64E-18 1.45E-11
2.36E-07 4.64E-18
5.90E-08 1.00E-12
1.43E-05
Solid wastes
kg of 5.6E-05
Ash 5.6E-05 1.4E-04 0 1.4E-04 0 1.5E-08 5.6E-05
1.28E-04 5.6E-05
1.28E-04 1.4E-04
1.5E-08 1.4E-04
0 1.5E-08
0 1.28E-04
0
kg of 1.2E-07
Sludge 0 9.6E-07 0 5.3E-14 1.2E-07
1.05E-05 1.2E-07
1.05E-05 9.6E-07 9.6E-07 5.3E-14 1.05E-05
1.2E-07 9.6E-07 5.3E-14 0 0 23 0
kg of 1.3E-09 Waste
Nuclear 1.3E-09 3.2E-09 0 3.2E-09 0 8.8E-17 1.3E-09
3.85E-10 1.3E-09
3.85E-10 3.2E-09
8.8E-17 3.2E-09
0 8.8E-17
0 3.85E-10
0

24. Case study S4 ENVISA WORKSHOP 2009
A. Marvuglia, Palermo – 19/06/2009
We also used a different kind of allocation (economic instead of mass allocation)
PRODUCT PRICE PRODUCT PRICE
electricity (€/MJ) 0.034 gravel (€/kg) 0.011
heat (€/MJ) 0.013 fuel oil (€/kg) 0.4
white clay (€/kg) 2.2 diesel oil (€/kg) 0.025
red clay (€/kg) 2 olive cake (€/kg) 0.16
sand (€/kg) 0.015 straw (€/kg) 0.065
Square technology matrix A′′ ∈ℜ13×13 (economic allocation)
Production Production Production Production Production Production Production Production Production Production
of of of of of of of of of of
electricity heat white clay red clay sand gravel crude oil fuel oil diesel oil natural gas
1 MJ of electricity 1 0 -3.77E-03 -3.43E-03 -1.04E-02 -7.62E-03 0 -3.01E-02 -1.85E-03 0
1 MJ of heat 0 2.48E+00 0 0 0 0 -4.87E-02 -1.07E+00 -6.55E-02 -4.87E-02
1 kg of white clay 0 0 1 0 0 0 0 0 0 0
1 kg of red clay 0 0 0 1 0 0 0 0 0 0
1 kg of sand 0 0 0 0 1 0 0 0 0 0
1 kg of gravel 0 0 0 0 0 1 0 0 0 0
1 kg of olive cake 0 0 0 0 0 0 0 0 0 0
1 kg of straw 0 0 0 0 0 0 0 0 0 0
1 MJ of crude oil 0 0 0 0 0 0 1 -2.20E+00 -1.35E-01 0
1 MJ of diesel oil 0 0 -2.38E-03 -2.16E-03 -2.08E-02 -1.52E-02 -5.29E-03 0 1 -3.61E+01
1 MJ of fuel oil 0 0 0 0 0 0 0 1 0 24 0
1 MJ of natural gas -2.21E+00 -2.06E+00 0 0 0 0 0 0 0 1
1 t of bricks 0 0 0 0 0 0 0 0 0 0