Robust algorithms for the solution of the inventory problem in Life Cycle Impact Assessment Maurizio Cellura, Antonino Marvuglia – University of Palermo (Italy) Marcello Pucci –

1

S4 ENVISA Workshop 2009

Palermo, 18-20 June 2009

Robust algorithms for the solution of the inventory problem in Life Cycle

Impact Assessment

Maurizio CelluraAntonino Marvuglia

University of PalermoDep. of Energy and Environmental Researches (DREAM)

University of Palermo

Marcello Pucci

Institute for Studies on Intelligent Systems for Automation (I.S.S.I.A), National Research Council, Palermo (Italy)

2

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).

S4 ENVISA WORKSHOP 2009

A. Marvuglia, Palermo – 19/06/2009

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 other environmental exchanges (e.g. radiation).

Introduction

3

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.



Despite LCA is nowadays a universally accepted methodology, each of these phases still contains some

unresolved problems.

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.

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.

Introduction

4

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 qualityFrom: J. Reap et al., A survey of unresolved problems in life cycle assessment. Int J Life Cycle Assess (2008) 13:290–300


A. Marvuglia, Palermo – 19/06/2009Introduction

5

Allocation refers to the procedure of appropriately allocating the environmental burdens of a multi-functional process amongst its functions or products.

Kg of CO2

Kg of SO2

1

2

10

181

0.1

0

p

1

11

2

2

10

01

0.1

0

a

aa

a

p

1

11

2

2

0

181

0.1

0

b

bb

b

p

ALLOCATION FACTORS

litre of crude oil

litre of fuel

kWh of electricity

MJ of heat

Co-generation of electricity and heat

Production of electricity

Production of heat



Obviously, arbitrary allocations could lead to incorrect LCA results and less preferable decisions based on those results.

1 1 1 a b

1 1 1 a b

2 2 1 a b

Introduction

6

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.

Introduction

7



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:

A s = feconomic (or technologic)

matrix scale vector

economic functional

unit

More in general, taking into account also the environmental part of the system:

ecn

env

fAf s

B f

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.

Mathematical background

The pseudo-inverse of a matrix (with m>n) is defined

as:

m nA

1† T TA A A A

1 ecns A f envf B s

8



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 :

If the Euclidean norm is not too high (with respect to a

fixed tolerance) the solution obtained with the pseudo-inverse can

be considered acceptable.

† f A A f

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

For the normal inverse A-1 we have: 1 0 A A f f

While for the pseudo-inverse we have:

† 0 A A f f f f

f f

Mathematical background

9



One considers, for example, the system with the following inventory matrix:

1000

0

0

0

f

and let the functional unit be the following vector:

ECONOMIC MATRIX

A

ENVIRON. MATRIX

B

Production of 1000 kWh of electricity


Production of fuel

Incineration of waste

kWh of electricity 10 -500 -5

l of fuel -1 100 0

kg of organic waste 0 0 -1000

kg of chemical waste 0 0 -200

kg of CO2 1 10 1000

kg of SO2 0.1 2 30

kg of crude oil 0 -50 0

Problem description

10



In this case, using the pseudo-inverse of A, one obtains:

†

1000

0

0

0

f A A f†

200

2

0

A f

As a consequence, the discrepancy vector is: d f f

0

0

0

0

d f f

Thus the pseudo-inverse gives in this case (for this particular

choice of the functional unit vector) an exact solution to the

problem.

20d

Problem description

11



If we chose the following functional unit:

†1 1

0

0

192.31

38.46

f A A f†1

0.1923

0.0019

0.1923

A f

1

0

0

0

1000

f

Disposal of 1000 kg of chemical waste

We obtain:

Problem description

12



The discrepancy vector is in this case equal to:

1

0

0

192.31

961.54

d f f 1 2980.6d

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.

Problem description

13



range(range(AA))y = y = A s

f = As-ff

Given the system of equations

To solve this over-determined system we can use, for example,

the Ordinary Least Squares (OLS) technique, which means

solving the system

(where and )

( )m n m n AA x = f

1mf

OLSA s = f f

f is the residual error vector corresponding to a perturbation in f .

There is no exact solution if ( ) rangef A

Problem solution

14



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: TLSA A s = 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: DLSA A s = f

Problem solution

15



Classical TLS problem formulation

Frobenius normm nM

2

,1 1

m n

i jFi j

mM

OLS A S F ˆTLS Â S F

OLS approach1) Find a matrix

F' such that

2) Solve the

system

min

FrangeF AF F

TLS approach

1) Find a matrix

such that

2) Solve the

system

Â

ˆ

ˆ ˆmin

range FF AA A F F

Problem solution

0 a

f

f'

a1

2a

range(A)

a0

range(A)

a2

1a

2a

a1

^

^

range(

A)^

f^

f

16



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:

(OLS problem)

(TLS problem)1

(DLS problem)

T

T

T

E xT

T

As f As f

As f As f

s s

As f As f

s s

Ordinary Least Squares

(OLS)

Total Least Squares

(TLS)

Data Least Squares

(DLS)

(fi)

(ai) A

f

arctan(s )OLS (ai)

(fi)

f

ATLSarctan(s ) (ai)

(fi)

A

f

arctan(s )DLS

Problem solution

17



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:

1

2 1

T

TE s

A s - f A s - f

s s

0

0.5

1

OLS

TLS

D LS

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:

where is the i-th row of A and fi is the i-th element of f.

2

1

2 1

i ii

TE s

a s - f

s s

ia

1

m

iGeTLS

i

E s E s

Problem solution

18



The iterative algorithm (learning law) exploiting the gradient (steepest descent) is given by:

where: ; (t) = learning

rate

The GeTLS EXIN neuron is a linear unit with:

• n inputs (vector ai);

• n weights (vector s);

• one output (scalar );

• a training error (scalar )

21 Tit t t t t t ts s a s

1

T Ti iT

tk

t t

s a f

s s

Ti iy s a

i ia s - f

In the application showed in this work, the parameter is made variable according to a predefined scheduling (in general monotonically from 0 to 1).

Problem solution

Hence:

2

21 1

ii i i i i

T T

dE

ds t t t t

a s - f a a s - f s

s s s s

19



Substitution and

Allocation

Substitution and

Allocation

Least Squares solutions

Least Squares solutions

Production of

electricity

Production of

clay

Production of

sand

Production of

crude oil

Production of oil-derivatives

Production of natural gas

Supply of

biomass

Production of

bricks

1 MJ of electricity 1 -7.20E-03 -1.80E-02 0 -3.20E-02 0 0 -3.69E+021 MJ of heat 2.48E+00 0 0 -4.87E-02 -1.13E+00 -4.87E-02 0 -1.01E+031 kg of white clay 0 1 0 0 0 0 0 -1.37E+031 kg of red clay 0 1 0 0 0 0 0 -8.51E+021 kg of recycled inerts 0 0 0 0 0 0 0 6.90E+011 kg of sand 0 0 1 0 0 0 0 -5.00E+021 kg of gravel 0 0 1 0 0 0 0 -3.47E+021 kg of olive cake 0 0 0 0 0 0 1 -1.53E+021 kg of straw 0 0 0 0 0 0 1 -1.92E+011 MJ of crude oil 0 0 0 1 -2.34E+00 0 0 01 MJ of diesel oil 0 -4.54E-03 -3.60E-02 -5.29E-03 1 -3.61E+01 -4.06E-01 -1.41E+031 MJ of fuel oil 0 0 0 0 1 0 0 -1.11E+031 MJ of natural gas -4.27E+00 0 0 0 0 1 0 -5.52E+031 t of bricks 0 0 0 0 0 0 0 1

Rectangular technology matrix 14 8A

Illustrative case study: bricks production Illustrative case study: bricks production

Final demand vector f

00000000000001

Case study

20



The intervention matrix of the system at hand (used for the solutions obtained with the least squares techniques) is:

Production of

electricity

Poduction of

clay

Production of

sand

Production of

crude oil

Production of

oil-derivatives

Production of natural

gas

Supply of

biomass

Production of

bricksResources and raw materialsMJ of Coal 1.49E-03 5.16E-03 1.36E-02 7.46E-07 5.77E-02 7.46E-07 1.40E-03 0MJ of Lignite 2.74E-04 6.46E-03 1.63E-02 1.77E-09 2.68E-04 1.77E-09 8.32E-04 0MJ of Hydropower 0 3.82E-04 1.04E-03 9.58E-08 7.66E-03 9.58E-08 4.54E-04 0MJ of Geothermal Energy 0 2.80E-08 1.17E-07 6.08E-15 1.05E-04 6.08E-15 0 0kg of Water 0 7.80E-03 2.40E-02 1.49E-02 7.26E-03 1.49E-02 0 1.03E+03kg 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+00MJ 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+03kg 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+00Emissions to airkg 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+02kg 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+00kg 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+00kg 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+00kg 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+00Emissions to waterkg 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-01kg 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-01kg 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-03kg 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-03kg 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-05Solid wasteskg of Ash 0 1.11E-04 2.83E-04 1.48E-08 2.56E-04 1.48E-08 0 0kg of Sludge 0 2.46E-07 1.93E-06 5.30E-14 2.10E-05 5.30E-14 0 0kg of Nuclear Waste 0 2.54E-09 6.49E-09 8.82E-17 7.70E-10 8.82E-17 0 0

21 8B

Case study

21



Production of electricity and

heat

Production of red and white

clay

Production of sand and

gravel

Production of crude oil



Supply of biomass

Production of bricks

natu

ral g

as

heat

bricks

natural gas

heatelectricity

heat

electric

ity

electricity

diesel oil diesel oil

crude oil

olive cakestr

aw

sand

grav

el

fue

l oil

die

sel o

il

red

clay

wh

ite c

lay

electricity

diesel oil

diesel oil

Production of electricity and

heat

Production of red and white

clay

Production of sand and

gravel




Supply of biomass


natu

ral g

as

heat

bricks

natural gas

heatelectricity

heat

electric

ity

electricity

diesel oil diesel oil

crude oil

olive cakestr

aw

sand

grav

el

fue

l oil

die

sel o

il

red

clay

wh

ite c

lay

electricity

diesel oil

diesel oil

Flow chart of the case study on bricks production

Case study

22



Square technology matrix (physical allocation)13 13A

1

-9.52E+01

-7.14E+03

1.37E+03

8.51E+02

4.41E+02

3.47E+02

-3.51E+04

1.11E+03

-3.11E+04

-8.97E+02

1.53E+02

1.92E+01

1.00E+00

x A b

Production of

electricity

Production of

heat

Production of

white clay

Production of

red clay

Production of

sand

Production of

gravel

Production of

crude oil

Production of

fuel oil

Production of

diesel oil

Production of

natural gas

Supply of

olive cake

Supply of straw

Production of

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+021 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+031 kg of white clay 0 0 1 0 0 0 0 0 0 0 0 0 -1.37E+031 kg of red clay 0 0 0 1 0 0 0 0 0 0 0 0 -8.51E+021 kg of sand 0 0 0 0 1 0 0 0 0 0 0 0 -4.41E+021 kg of gravel 0 0 0 0 0 1 0 0 0 0 0 0 -3.47E+021 kg of olive cake 0 0 0 0 0 0 0 0 0 0 1 0 -1.53E+021 kg of straw 0 0 0 0 0 0 0 0 0 0 0 1 -1.92E+011 MJ of crude oil 0 0 0 0 0 0 1 -1.17E+00 -1.17E+00 0 0 0 0.00E+001 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+031 MJ of fuel oil 0 0 0 0 0 0 0 1 0 0 0 0 -1.11E+031 MJ of natural gas -3.41E+00 -8.54E-01 0 0 0 0 0 0 0 1 0 0 -5.52E+031 t of bricks 0 0 0 0 0 0 0 0 0 0 0 0 1.00E+00

PROCESS ALLOCATION

Prod. of electricity and heat 0.8 for electricity and 0.2 for heat

Prod. of clay (white and red) 0.5 for with clay and 0.5 for red clay

Prod. of sand and gravel 0.5 for sand and 0.5 for gravel

Production of oil-derivatives Allocation following the energy criterion.

Supply of biomass 0.8 for olive cake and 0.2 for straw

Prod. of bricks and inerts Substitution method: inerts are considered as equivalent to sand, but with an estimated correction factor of 0.85 in order to account for the difference in quality between the mass of inert materials and the mass of sand obtained from them.

Case study

23



……

Resources and raw materials

MJ of Coal 1.20E-03 2.99E-04 2.58E-03 2.58E-03 6.82E-03 6.82E-03 7.46E-07 2.89E-02 2.89E-02 7.46E-07 1.12E-03 2.81E-04 0

MJ of Lignite 2.19E-04 5.47E-05 3.23E-03 3.23E-03 8.14E-03 8.14E-03 1.77E-09 1.34E-04 1.34E-04 1.77E-09 6.66E-04 1.66E-04 0

MJ of Hydropower 0 0 1.91E-04 1.91E-04 5.19E-04 5.19E-04 9.58E-08 3.83E-03 3.83E-03 9.58E-08 3.63E-04 9.08E-05 0

MJ of Geothermal Energy 0 0 1.40E-08 1.40E-08 5.84E-08 5.84E-08 6.08E-15 5.25E-05 5.25E-05 6.08E-15 0 0 0

kg of Water 0 0 3.90E-03 3.90E-03 1.20E-02 1.20E-02 1.49E-02 3.63E-03 3.63E-03 1.49E-02 0 0 1.03E+03

kg of Ores (sand, gravel, etc.) 1.76E-05 4.39E-06 1.00E+00 1.00E+00 1.00E+00 1.00E+00 3.51E-05 2.27E-04 2.27E-04 3.51E-05 0 0 0

MJ of Crude Oil 9.71E-03 2.43E-03 9.09E-03 9.09E-03 7.14E-02 7.14E-02 1.02E+00 1.17E+00 1.17E+00 1.02E+00 3.32E-01 8.31E-02 1.27E+03

kg of other Ores (iron, copper, etc.) 1.00E-04 2.50E-05 3.68E-06 3.68E-06 1.97E-05 1.97E-05 5.60E-10 1.20E-04 1.20E-04 5.60E-10 0 0 0

Emissions to air

kg of CO2 6.66E-02 1.66E-02 1.26E-03 1.26E-03 6.65E-03 6.65E-03 4.17E-03 1.44E-02 1.44E-02 4.17E-03 3.94E+00 9.86E-01 6.02E+02

kg of CO 1.30E-04 3.26E-05 1.91E-06 1.91E-06 1.40E-05 1.40E-05 1.14E-05 1.67E-05 1.67E-05 1.14E-05 2.16E-02 5.40E-03 1.10E+00

kg of CH4 2.94E-04 7.36E-05 1.48E-06 1.48E-06 5.44E-06 5.44E-06 7.13E-05 5.10E-05 5.10E-05 7.13E-05 4.80E-03 1.20E-03 1.18E+00

kg of SO2 2.37E-05 5.92E-06 1.31E-06 1.31E-06 8.21E-06 8.21E-06 5.31E-06 1.34E-04 1.34E-04 5.31E-06 5.94E-03 1.49E-03 2.20E+00

kg of NMVOC 2.96E-05 7.40E-06 2.10E-07 2.10E-07 1.43E-06 1.43E-06 1.93E-05 2.12E-05 2.12E-05 1.93E-05 2.46E-02 6.14E-03 1.13E+00

Emissions to water

kg of COD 5.00E-07 1.25E-07 9.98E-08 9.98E-08 5.57E-07 5.57E-07 1.57E-11 3.37E-06 3.37E-06 1.57E-11 1.77E-04 4.42E-05 2.38E-01

kg of BOD 3.49E-08 8.72E-09 3.05E-09 3.05E-09 1.76E-08 1.76E-08 4.41E-13 9.47E-08 9.47E-08 4.41E-13 5.40E-06 1.35E-06 2.25E-01

kg of P 2.62E-10 6.54E-11 2.48E-11 2.48E-11 1.95E-10 1.95E-10 3.73E-18 4.92E-13 4.92E-13 3.73E-18 0 0 1.26E-03

kg of N 2.49E-08 6.22E-09 1.45E-09 1.45E-09 1.15E-08 1.15E-08 2.19E-16 5.40E-11 5.40E-11 2.19E-16 1.29E-04 3.22E-05 6.00E-03

kg of AOX 4.62E-11 1.16E-11 1.84E-12 1.84E-12 1.45E-11 1.45E-11 4.64E-18 1.00E-12 1.00E-12 4.64E-18 2.36E-07 5.90E-08 1.43E-05

Solid wastes

kg of Ash 0 0 5.6E-05 5.6E-05 1.4E-04 1.4E-04 1.5E-08 1.28E-04 1.28E-04 1.5E-08 0 0 0

kg of Sludge 0 0 1.2E-07 1.2E-07 9.6E-07 9.6E-07 5.3E-14 1.05E-05 1.05E-05 5.3E-14 0 0 0

kg of Nuclear Waste 0 0 1.3E-09 1.3E-09 3.2E-09 3.2E-09 8.8E-17 3.85E-10 3.85E-10 8.8E-17 0 0 0


Production of fuel oil

Production of

electricity

Production of heat

Poduction of white

clay

Poduction of red

clay

Production of sand

Production of

gravel

21 13BResources and raw materials






kg of Ores (sand, gravel, etc.) 1.76E-05 4.39E-06 1.00E+00 1.00E+00 1.00E+00 1.00E+00 3.51E-05 2.27E-04 2.27E-04 3.51E-05 0 0 0

MJ of Crude Oil 9.71E-03 2.43E-03 9.09E-03 9.09E-03 7.14E-02 7.14E-02 1.02E+00 1.17E+00 1.17E+00 1.02E+00 3.32E-01 8.31E-02 1.27E+03


Emissions to air

kg of CO2 6.66E-02 1.66E-02 1.26E-03 1.26E-03 6.65E-03 6.65E-03 4.17E-03 1.44E-02 1.44E-02 4.17E-03 3.94E+00 9.86E-01 6.02E+02





Emissions to water






Solid wastes




Supply of olive cake

Supply of straw




Production of diesel oil


……

Environmental intervention matrix after physical allocation

Case study

24



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 (economic allocation)13 13AProduction

of electricity

Production of

heat

Production of

white clay

Production of

red clay

Production of

sand

Production of

gravel

Production of

crude oil

Production of

fuel oil

Production of

diesel oil

Production of

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 01 MJ of heat 0 2.48E+00 0 0 0 0 -4.87E-02 -1.07E+00 -6.55E-02 -4.87E-021 kg of white clay 0 0 1 0 0 0 0 0 0 01 kg of red clay 0 0 0 1 0 0 0 0 0 01 kg of sand 0 0 0 0 1 0 0 0 0 01 kg of gravel 0 0 0 0 0 1 0 0 0 01 kg of olive cake 0 0 0 0 0 0 0 0 0 01 kg of straw 0 0 0 0 0 0 0 0 0 01 MJ of crude oil 0 0 0 0 0 0 1 -2.20E+00 -1.35E-01 01 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+011 MJ of fuel oil 0 0 0 0 0 0 0 1 0 01 MJ of natural gas -2.21E+00 -2.06E+00 0 0 0 0 0 0 0 11 t of bricks 0 0 0 0 0 0 0 0 0 0

Case study

25



Resources and raw materials






kg of Ores (sand, gravel, etc.) 1.14E-05 1.06E-05 1.05E+00 9.52E-01 1.15E+00 8.46E-01 3.51E-05 4.28E-04 2.63E-05 3.51E-05 0 0 0

MJ of Crude Oil 6.29E-03 5.85E-03 9.53E-03 8.66E-03 8.24E-02 6.04E-02 1.02E+00 2.21E+00 1.36E-01 1.02E+00 2.96E-01 1.19E-01 1.27E+03


Emissions to air

kg of CO2 4.31E-02 4.01E-02 1.32E-03 1.20E-03 7.67E-03 5.63E-03 4.17E-03 2.71E-02 1.67E-03 4.17E-03 3.51E+00 1.42E+00 6.02E+02





Emissions to water






Solid wastes




Production of

electricity

Production of

heat

Poduction of

white clay

Poduction of

red clay

Production of

sand

Production of

gravel


21 13BResources and raw materials






kg of Ores (sand, gravel, etc.) 1.14E-05 1.06E-05 1.05E+00 9.52E-01 1.15E+00 8.46E-01 3.51E-05 4.28E-04 2.63E-05 3.51E-05 0 0 0

MJ of Crude Oil 6.29E-03 5.85E-03 9.53E-03 8.66E-03 8.24E-02 6.04E-02 1.02E+00 2.21E+00 1.36E-01 1.02E+00 2.96E-01 1.19E-01 1.27E+03


Emissions to air

kg of CO2 4.31E-02 4.01E-02 1.32E-03 1.20E-03 7.67E-03 5.63E-03 4.17E-03 2.71E-02 1.67E-03 4.17E-03 3.51E+00 1.42E+00 6.02E+02





Emissions to water






Solid wastes




Supply of straw






21 13B

…………

Environmental intervention matrix after economical allocation

Case study

26



The GeTLS algorithm was applied with a linear scheduling of the parameter, and for different values of:• number of iterations (t=5,…,500)• learning rate (=1.0E-04,…,1)

The chosen solution is the one with the minimal residual, that is the Euclidean norm:

For every combination of n and the solution vector sGeTLS was

computed, along with:• the corresponding final supply vector• the discrepancy vector

GeTLSf A s

d f f

2

21

n

ii

d

d

By using the Least Squares techniques the system was solved in its rectangular form, without using any allocation or substitution.

Case study

27

0 100 200 300 400 500

0,0

0,1

0,2

0,3

0,4

1,0000

1,0001

1,0002

1,0003

1,0004

1,0005

; m

in(|d|)

Iterations

min(|d|)



0 50 100 150 200 250 300 350 400 450 5001

1.0001

1.0002

1.0003

1.0004

1.0005

Iterations

min

(|d|)

0,341

Case study

28



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11

1.005

1.01

1.015

alfa

|d|

10 Iterations

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11.0003

1.0004

1.0004

1.0005

1.0005

1.0006

1.0006

1.0007

1.0007

alfa

|d|

5 Iterations

0 0.1 0.2 0.3 0.4 0.5 0.6 0.71

1.005

1.01

1.015

1.02

1.025

1.03

1.035

alfa

|d|

100 Iterations

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.451

1.0005

1.001

1.0015

alfa

|d|

500 Iterations

Case study

29



Obtained scale vectors

Case study

ProcessPhysical allocation

Economic allocation

OLS DLS TLS GeTLS

Production of electricity -9.52E+01 1.47E+00

Production of heat -7.14E+03 -5.73E+03

Production of white clay 1.37E+03 1.37E+03

Production of red clay 8.51E+02 8.51E+02

Production of sand 4.41E+02 4.41E+02

Production of gravel 3.47E+02 3.47E+02

Production of crude oil -3.51E+04 -2.80E+04 4.77E-04 8.23E-04 7.74E-04 5.32E-04

Production of fuel oil 1.11E+03 1.11E+03

Production of diesel oil -3.11E+04 -2.25E+05

Production of natural gas -8.97E+02 -6.26E+03 7.71E-06 -1.80E-05 -1.70E-05 8.60E-06

Supply of olive cake 1.53E+02 1.53E+02

Supply of straw 1.92E+01 1.92E+01

Production of bricks 1.00E+00 1.00E+00 2.17E-07 2.85E-07 2.68E-07 2.43E-07

1.00E-04 1.30E-04 1.23E-04 1.12E-04

2.05E-04 4.51E-05 4.23E-05 2.28E-04

1.12E-04 1.21E-04 1.13E-04 1.24E-04

5.20E-04 4.50E-04 4.23E-04 5.80E-04

LCI technique

7.63E-05 1.76E-04 1.65E-04 8.51E-05

ProcessPercentage Difference between Economic and

Physical allocation

Production of electricity 101.54

Production of heat 19.75

Production of white clay 0.00

Production of red clay 0.00

Production of sand 0.00

Production of gravel 0.00

Production of crude oil 20.23

Production of fuel oil 0.00

Production of diesel oil -623.47

Production of natural gas -597.88

Supply of olive cake 0.00

Supply of straw 0.00

Production of bricks 0.00

30



The eco-profiles g of the functional unit can be obtained by the product g = B·s, where B is the intervention matrix, containing the environmental interventions of the unit processes.

Physical allocation

Economic allocation

OLS DLS TLS GeTLS

Resources and raw materialsMJ of Coal 6.83E+02 8.56E+02 1.73E-04 1.86E-04 1.79E-04 1.81E-05

MJ of Lignite -9.93E+00 -9.28E+00 4.91E-05 5.28E-05 5.08E-05 5.95E-06

MJ of Hydropower 9.08E+01 1.14E+02 2.05E-05 2.20E-05 2.12E-05 2.15E-06

MJ of Geothermal Energy 1.26E+00 1.57E+00 2.40E-07 2.58E-07 2.49E-07 2.40E-08

kg of Water -4.50E+02 -4.03E+02 2.45E-03 2.63E-03 2.54E-03 2.67E-04

kg of Ores (sand, gravel, etc.) -3.04E+03 -3.00E+03 1.19E-02 1.28E-02 1.23E-02 1.41E-03

MJ of Crude Oil 6.17E+04 7.05E+04 1.40E-02 1.50E-02 1.45E-02 1.47E-03

kg of other Ores (iron, copper, etc.) 3.19E+00 3.76E+00 7.88E-07 8.48E-07 8.16E-07 7.46E-08

Emissions to airkg of CO2 4.59E+02 5.27E+02 2.38E-03 2.56E-03 2.46E-03 7.15E-04

kg of CO 2.97E+00 3.37E+00 7.30E-06 7.86E-06 7.56E-06 3.31E-06

kg of CH4 -2.81E+00 -2.70E+00 4.73E-06 5.10E-06 4.90E-06 1.05E-06

kg of SO2 -4.17E-01 -1.11E+00 6.83E-06 7.36E-06 7.08E-06 1.43E-06

kg of NMVOC 3.39E+00 3.63E+00 7.84E-06 8.44E-06 8.12E-06 3.72E-06

Emissions to waterkg of COD 1.81E-01 1.64E-01 5.84E-07 6.29E-07 6.06E-07 8.42E-08

kg of BOD 2.24E-01 2.23E-01 5.03E-07 5.42E-07 5.21E-07 5.54E-08

kg of P 1.26E-03 1.26E-03 2.80E-09 3.01E-09 2.90E-09 3.05E-10

kg of N 2.44E-02 2.64E-02 4.00E-08 4.31E-08 4.15E-08 1.94E-08

kg of AOX 4.80E-05 5.15E-05 8.06E-11 8.68E-11 8.35E-11 3.64E-11

Solid wasteskg of Ash -2.83E+00 -3.60E+00 1.41E-06 1.52E-06 1.46E-06 1.58E-07

kg of Sludge -2.50E-01 -3.14E-01 5.10E-08 5.49E-08 5.29E-08 5.17E-09

kg of Nuclear Waste -3.75E-06 -6.15E-06 2.07E-11 2.23E-11 2.14E-11 2.46E-12

Case study

Percentage Difference between Economic and

Physical allocation

Resources and raw materialsMJ of Coal 25.33

MJ of Lignite 6.55

MJ of Hydropower 25.55

MJ of Geothermal Energy 24.60

kg of Water 10.44

kg of Ores (sand, gravel, etc.) 1.32

MJ of Crude Oil 14.26

kg of other Ores (iron, copper, etc.) 17.87

Emissions to airkg of CO2 14.81

kg of CO 13.47

kg of CH4 3.91

kg of SO2 -166.19

kg of NMVOC 7.08

Emissions to waterkg of COD -9.39

kg of BOD -0.45

kg of P 0.00

kg of N 8.20

kg of AOX 7.29

Solid wasteskg of Ash -27.21

kg of Sludge -25.60

kg of Nuclear Waste -64.00

31


A. Marvuglia, Palermo – 19/06/2009Case study

32


A. Marvuglia, Palermo – 19/06/2009Case study

33


A. Marvuglia, Palermo – 19/06/2009Normalization

In physical problems influenced by variables of different nature, these variables are often expressed by numbers whose range of variation can differ even by several orders of magnitude.

For this reason, when any approach is used to solve a multiple-input problem, the solution could be polarized by those variables expressed by extreme values. In order to prevent this, normalization is often used.

1

1 0 0max , 1, ,

0

0

10 0max , 1, ,

i

in

a i m

a i m

H

Normalization Matrix

nA A H

Normalized Economic

Matrix

max( )n

ff

f

Normalized functional

unit

If A is invertible, the de-normalized solution sden is obtained as

follows: 1n n n

s A f maxden n n s H s f

34


A. Marvuglia, Palermo – 19/06/2009Condition number

For a generic square and invertible matrix A the condition number with respect to the Euclidean norm is defined as:

where is the Euclidean norm of the matrix, that is defined as , where x is a generic conformable vector for the matrix A.

1k A A A

A1

max

x

A A x

For square matrices the norm can be computed as the square root

of the largest eigenvalue max of the product , that is: TA A

maxTA A A

The condition number provides an upper bound to the magnification factor that measures how a relative change in f propagates as a relative change in s.In fact the relation holds: 1

s fA A

s f

Matrices with an high condition number are called badly conditioned or ill-conditioned, and they can provide unstable solutions to the system.

In our cases: 1.33 004 k EA 197.78normk A

35


A. Marvuglia, Palermo – 19/06/2009Normalization

The traditional solutions (i.e. those obtained with the physical allocation and the economic allocation) found after de-normalization are the same as those obtained without using any normalization procedure.

Process OLS DLS TLS GeTLS


Production of heat

Production of white clay

Production of red clay

Production of sand

Production of gravel

Production of crude oil 2.67E-03 2.45E-03 2.33E-03 5.32E-04



Production of natural gas -1.84E-05 -1.67E-05 -1.59E-05 8.60E-06


Supply of straw

Production of bricks 1.07E-06 9.86E-07 9.39E-07 2.43E-07

7.93E-05 7.55E-05 1.12E-04

LCI technique

5.83E-04

2.57E-03

1.13E-03

8.50E-05

4.70E-04

5.35E-04

1.04E-03 9.94E-04 2.28E-04

5.09E-04 8.51E-05

4.32E-04 4.11E-04 1.24E-04

2.37E-03 2.25E-03 5.80E-04

36



CONCLUSION

The application of the least squares techniques can be useful in presence of multi-functional processes and in all those cases in which the technology matrix is rectangular but the pseudo-inverse method does not provide an exact solution.

The substantial advantage in using these techniques lies into:1. the possibility to circumvent the drawback of the traditional solution of the inventory

problem, which needs to use some computational tricks to transform the rectangular technology matrix into a square and invertible matrix.

The only subjective choice in the GeTLS solutions are the learning rate and the number

of iterations, but there is a guide criterion (the norm of the discrepancy vector |d| )2. The visualization of the error surfaces, which allows the identification of the most

critical components of the solution.

The solution of the inventory problem in LCA is a complicated task, especially in presence of multi-functional processes or open-loop recycling, that are characterized by a rectangular, and thus non-invertible technology matrix.

37



POSSIBLE IMPROVEMENTS

Application of a weighted version of the algorithm, in which higher weights are assigned to the most important components of the solution.

…..

Implementation of a learning rule with a dynamically variable learning rate () as a function of the local gradient of the error surface.

38

DREAMDipartimento di Ricerche Energetiche ed Ambientali

Prof. M. CelluraDr. A. Marvuglia

Leiden – 26/03/2009

Thank you for your attention

Robust algorithms for the solution of the inventory problem in Life Cycle Impact Assessment Maurizio Cellura, Antonino Marvuglia – University of Palermo (Italy) Marcello Pucci –

Technology

life cycle assessment

life activities

products life result

int j life cycle

introductions4 envisa

b s b fenv

inventory data

environmental exchanges