IDENTIFICATION OF ROBUST WATER RESOURCES
PLANNING STRATEGIES
by
FRANCISCO JAVIER MARTIN CARRASCO
B.S. Civil Engineering
Polytechnic University of Madrid, Spain
(1984)
Submitted to the departments of
URBAN STUDIES AND PLANNING
and
CIVIL ENGINEERING
in
partial fulfillment of the requirements
for the degrees of
MASTER CITY PLANNING
and
MASTER OF SCIENCE IN CIVIL ENGINEERING
at the
MASSACHUSETTS
INSTITUTE OF TECHNOLOGY
June
1987
Francisco Javier Martin Carrasco
The author hereby grants to M .I .T. permission to reproduce and to
distribute copies of this thesis documents in whole or in part.
Signature of the Author
Department htJrban,8tudies and Planning and
7
Department of Civil Engineering . June,
1987
Certified by
Thesis S'gupervi'sor
Ralph A. Gakenheimer
Departrent of Urban Studies and Planning
Certified by
Theps Supervisor,
Dennis B. McLaughlin
Department of Civil Engineering
Chairman,
Phillip L. Clay
Master City Planning Committee
Certified by_
Certified by
Ole S. Madsen
Civil Engineering Master Committee
IBARI
IDENTIFICATION OF ROBUST WATER RESOURCES
PLANNING STRATEGIES
by
FRANCISCO JAVIER MARTIN CARRASCO
the departments
of Urban Studies
Submitted to
on June 25,
and Planning
a nd of Civil Engineering
1987, in parti a 1 fulfillment of the requirements for
of Master
City Planning
and Master of
the degrees
Science in Civ i I Engineering.
ABSTRACT
This thesis presents a two-objective method for
of the
objectives is
water resources planning. One
and the other is
expected net benefits,
to maximize
this thesis, robustness
to maximize
robustness. In
a measure
of the
sensitivity of the
is considered
conditions. An
to
uncertain
performance
projects
index of robustness is developed. The index is based
on the variation of net benefits
from a
project as
consequence
of
the
uncertainty in determining the
parameters of water resources systems.
by
problem
is
solved
The
two-objective
obtaining the Pareto curve, which represents
the set
method uses screening
of non-inferior projects. The
models as the optimization technique.
To prove the practical viability of the method,
a
case
study.
The
case study
applied
to
it is
select from al l possible projects to be
consists in
that represent
built in a hypothetical bas in, those
the best development altern at ive.
Thesis
supervisors:
Dr.
McLaughlin,
Dennis
Engineering
Dr.
Gakenheimer,
Ralph
Studies and Planning
professor
of
Civil
professor
of
Urban
and Civil Engineering
2
ACKNOWLEDGMENTS
This thesis
is the
final step
of two Masters
that have been possible thanks to the Fundacion Juan
March
of
Spain,
which
granted me the scholarship
that allowed me to
come
to
MIT.
I
also
want to
mention
here
some
people without whom this thesis
would have not been possible. In first place Aurelio
Menendez, who always provided the support and advice
I needed, not only for
this
thesis,
but
also for
other
academic
and
personal situations. Secondly,
professor Dennis McLaughlin, with whom I have worked
closely
during
all
the
thesis period. He made of
this thesis the most important part
of my education
at
MIT.
Besides,
he
provided
the final
economic
support. Finally, I
also
want
to
thank professor
Ralph Gakenheimer
who, although this
thesis
did not
fit in his area of main interest,
was interested
and
supportive throughout all the period.
3
I want to dedicate
this thesis to my
friend Aurelio.
4
IDENTIFICATION OF ROBUST WATER RESOURCES PLANNING STRATEGIES
TABLE OF CONTENTS
1.-
INTRODUCTION
1.1.-
1.2.
2.-
-
OVERVIEW OF A MULTIOBJECTIVE METHOD FOR
CONSIDERING ROBUSTNESS
14
ORGANIZATION OF THE THESIS
21
LITERATURE REVIEW
2.1.-
CURRENT METHODS IN WATER RESOURCES PLANNING
2.1.1.2.1.2.-
2.2.-
2.3.-
Optimization models and
Simulation techniques
25
27
27
28
ADAPTATIONS OF SYSTEM ANALYSIS TECHNIQUES TO
DECISION MAKING NEEDS
31
2.3.2.-
2.3.3.2.3.4.-
A METHOD FOR
DECISIONS
3.1.-
techniques
CURRENT USE OF PLANNING METHODS
1. Institutional resistance
to use system
analysis techniques.
2. Lack of communication between decision
makers and analysts
3. Institution's conditions to use system
analysis
4. Institutional situations LDC's
2.3.1.-
3.-
12
Multiobjective analysis
of nearly optimal
Identification
alternatives
Stochastic planning
Indices
1.Reliability index
2.Vulnerability index
3.--
Resiliency
Index
4.-
Robustness
index
INCORPORATING
ROBUSTNESS
OF THE METHOD
Screening models
3.1,1.Step 1: Derive the ideal
curve
- Decision Variables
- Uncertain Parameters
IN
PLANNING
30
30
32
36
37
43
44
44
45
47
49
50
DESCRIPTION
-
net benefits
52
5
3.1.2.-
3.1.3.-
Ideal
Step
net benefits
2:
Select
Step 3: Assess the
of the alternatives
Curve
of
net
-
-
4.-
Index
of
alternatives
performance
of Net Benefits of
an
robustness
Step 4: Compare
Final remarks
among alternatives
75
DESCRIPTION OF THE
CASE
76
STUDY
81
82
83
87
95
SCREENING MODEL
4.2.-
4.2.1.4.2.2.-
Decision variables
Objective function
4.2.3.-
Constraints
4.2.4.-
Uncertain parameters
APPLICATION OF THE METHOD TO THE CASE STUDY
4.3.1.- Step 1: Derive the ideal net benefits
curve
4.3.-
-
Ideal
net
benefits
-
Curves of
-
Expected value of net benefits for th e
alternatives
Indices of robustness for the
alternatives
Step 4: Compare among alternatives
Conclusions about robustness and
expected net benefits
4.3.4.4.3.5.-
net
113
114
115
117
121
Summary of
Characteristics
water
103
111
benefits
CONCLUSIONS
-
97
98
curve
- Optimal sets of design decision
variables
4.3.2.- Step 2: Select candidate alternatives
4.3.3.- Step 3: Assess the performance of the
alternatives
5.-
68
70
Summary
CASE STUDY
4.1.-
57
61
benefits
- Expected Value
Alternative
3.1.4.3.1.5.-
curve
candidate
the method
of the
resources
method
planning
within
the
121
123
framework
APPENDIX C
126
137
154
BIBLIOGRAPHY
169
APPENDIX A
APPENDIX B
6
INDEX OF TABLES AND FIGURES
CHAPTER
1:
Figures
FIGURE 1-1:
FIGURE 1-2:
FIGURE 1-3:
Two-objective tradeoff
curve for
alternatives
A, B, and C
16
Potential
of a basin
used for
irrigation under uncertain crop prices
17
Net benefits
uncertain
19
of an Alternative
crop prices
FIGURE 1-4:
Delta curve of an Alternative
FIGURE 1-5:
Comparison
Alternatives
of
A,
A
under
A
three delta curves, for
B, and C
20
21
CHAPTER 2:
Figures
FIGURE 2-1:
Theoretical multiobjective optimal
solution
33
FIGURE 2-2:
Typical form of utility
39
FIGURE 2-3:
Continuous and discrete
probability
functions for uncertain parameter Ck
curves
41
CHAPTER 3:
Figures
FIGURE 3-1:
Discrete probability functions for the
uncertain parameter 01 z discount rate.
54
FIGURE 3-2:
Typical ideal net benefits
curve for
a
basin under uncertain discount rate.
57
FIGURE 3-3:
Typical bar graph representation of an
hypothetical probability distribution of
values of a design decision
variable
Xu
obtained from the M1 M2 Mn optimization
runs performed in Step I
59
FIGURE 3-4:
Net benefits curve of an hypothetical
63
7
alternative
under uncertain
discount
rate
FIGURE 3-5:
Delta Curve of an hypothetical
65
alternative,
by subtracting
the net
benefits
curve (Figure
3-4) from the ideal
net benefits curve of the basin (Figure 3-2)
FIGURE 3-6:
Robustness related to the shape of the
delta
curve. Comparison between
Alternative
A,
a non-robust
alternative,
and Alternative
B, a robust
alternative.
66
FIGURE 3-7:
Typical
curves
69
FIGURE 3-8:
Derivation
form of the Pareto frontier
of
the Pareto
frontier
curve
71
CHAPTER 4:
Tables
TABLE 4-1:
Fixed and variable
possible
projects
basin
TABLE 4-2:
Parameters needed for the screening
model
91
TABLE 4-3:
Intervals
in which the continuous
probability
function
of the uncertain
parameters have been divided
99
TABLE
4-4:
costs
for
to be build
Ideal net benefits
the
in the
curve
86
102
TABLE 4-5:
Probability
distributions
of the
values of the design decision variables
104
TABLE 4-6:
Probability
reference
variables
106
TABLE 4-7:
Projects
TABLE 4-8:
Compatibility table
109
TABLE 4-9:
Reference sizes for the project
tII
TABLE 4-10:
Candidate alternatives
112
TABLE 4-11:
Expected net benefits
and index of
robustness
of each candidate
alternative
115
distribution
of the
Values of the design decision
after
custering
more likely
to be built
107
8
TABLE
4-12:
Decrease in expected net benefits and
in index of robustness
of the non-inferior
alteranti ves with respect
to the maximum
values
117
Figures
FIGURE 4-1:
Scheme of
FIGURE 4-2:
Total river
year
FIGURE 4-3:
Seasonal distribution of total
inflows for the typical year
FIGURE 4-4:
Water demands for
typical year
FIGURE 4-5:
Return of the irrigation
non-comsumed by crops
FIGURE 4-6:
Assumed probability
crop prices
distribution of (9:
84
FIGURE 4-7:
Assumed probability
electricity prices
distribution
of e2:
84
FIGURE 4-8:
Assumed probability distribution
discount rate
of
93:
85
FIGURE 4-9:
Assumed probability distribution
or decrease in
general increase
costs
construction
of
94:
88
FIGURE 4-10:
Assumed probability distributi on of e65:
general increase or decrease in i rr i ga ti on
water demands
FIGURE 4-11:
Hierarchical
relationships
FIGURE
4-12:
Pareto
the basin
77
inflows for
typical
irrigation
graph
two-objective
78
river
79
the
79
in
water
representation
among projects
of
candidate alternatives
APPENDIX
the
81
110
of
for
93
the
116
the case study
A:
Figures
FIGURE A-1:
Construction costs
FIGURE A-2:
Costruction
costs
of
of
the dams
135
the irrigations
135
9
FIGURE A-3:
Construction costs of
the power plants
136
FIGURE A-4:
Construction costs of
the
136
transfer
APPENDIX B:
Tables
TABLE B-1:
Probabilities
of
the
243
elements
of
138
the
uncertain parameter vector &
TABLE
B-2:
Optimal sizes of Dam A
139
TABLE
B-3:
Optimal sizes
of Dam B
140
TABLE B-4:
Optimal
sizes
of Dam C
141
TABLE B-5:
Optimal
sizes
of Dam D
142
TABLE B-6:
Optimal
sizes of Irrigation
143
TABLE B-7:
Optimal
sizes of Irrigation
144
145
TABLE
B-8:
Optimal sizes of Irrigation
TABLE
B-9:
Optimal sizes of hydropower
Plant A
146
TABLE
B-O:
Optimal
Plant C
147
TABLE
B-1I:
Optimal sizes of Transfer
sizes
of hydropower
148
Figures
FIGURE B-1:
Bar graph of the probability
of the sizes of Dam A
distribution
149
FIGURE B-2:
Bar graph of the probability
of the sizes of Dam B
distribution
149
FIGURE B-3:
Bar graph of the probability
distribut ion of the sizes of Dam C
150
FIGURE B-4:
Bar graph of the probability
distribution of the sizes of Dam D
150
FIGURE B-5:
Bar graph of
distribution
the probability
of the sizes of
151
Irrigation B
Bar graph of
distribution
the probability
of the sizes of
Irrigation
FIGURE B-6:
151
C
10
FIGURE B-7:
FIGURE B-8:
Bar graph of the probabil ity
distributi
on of the sizes of
152
Irrigation D
Bar graph of the probabil ity
distributi on of the sizes
of hydropower
152
Plant A
FIGURE B-9:
Bar graph of the probability
distribUti on of the sizes of hydropower
Plant
FIGURE B-10
APPENDIX
153
C
Bar graph of the probability
distributi on of the sizes of
153
Transfer
C:
Tables
TABLE C-i:
Ne t benef i ts of
A
155
TABLE C-2:
Ne t benef i ts of Al ternative B
156
TABLE
C-3:
Ne t benefi ts of Al ternative C
157
TABLE
C-4:
He t benef.zts of Al t erna t i ve
D
158
TABLE
C-5:
Ne t benef j ts of Al ternative E
159
TABLE C-6:
Ne t benef i ts of Al ternative F
160
TABLE
C-7:
Ne t benef i ts of Al ternative G
161
TABLE
C-8:
Ne t benefi ts of Al ternati
H
162
TABLE
C-9:
Ne t benef i ts of Al ternative I
163
TABLE
C-10:
Ne t benef i ts of Al terna t i ye
164
TABLE
C-11:
Ne t benef i ts of Al t erna t i ve
K
165
TABLE
C-12:
Ne t benef i ts of Al ternati ve
L
166
TABLE C- 13:
Ne t benef i ts of Al t erna ti ve M
167
TABLE C-14:
Net
168
benefits
of
Alternative
Alternative
ve
N
11
IDENTIFICATION OF ROBUST WATER
RESOURCES PLANNING STRATEGIES
CHAPTER
INTRODUCTION
1:
This thesis
particular,
water resources planning and,
deals with
with
a
strategies. Through
for
method
developing
in
robust planning
a litera ture review, I have realized that
water resources planning has a different meaning for engineers
than
for
planners.
Engineers
resources planning as a series of
optimization
models.
Engineers
under the generic name of system
planning
methods
mean
generally
consider
water
mathematical techniques and
often group these techniques
analysis. For
improvement
of
the
engineers new
mathematical
techniques, perhaps at the sacrifice of decision-making needs.
On
the
other
institutions,
social
and
hand,
planners
consider
water
objectives, decision making processes,
economical
resources
and other
issues. But planners do not usually use
analytical tools to introduce their concerns
into a practical
analysis. This
thesis should
as
narrowing
gap
the
between
be considered
theory
and
an effort for
practice,
between
12
engineers
and planners.
The
main
planning
is
objective
to
maximize
in
traditional
expected
net
water
resources
benefits.
With this
single criterion, the project or system of
projects chosen for
implementation
is
generate
benefits.
most
But
the
one
water
that
will
more
net
resources planning situations are
subject to some degree of uncertainty. The actual construction
of the
that
project begins
time
years after the plan was completed. In
interval,
planning forecasts.
some
variables
Hence, the
may
change
actual net
from
the
benefits from the
project may differ from the predicted net benefits.
As a consequence of uncertainty, there
of possible
is
a
net benefits to obtain from a project. Robustness
measure
possible net
of
the
dispersion
benefits. If
net benefits
depend heavily
other words, non robust
the value
Robust projects,
to maintain
of
that
distribution of
the distribution of net benefits
widely spread, the project
depend on
is a distribution
is considered
on the
is
non robust, because
uncertain conditions. In
projects are
those whose performance
that uncertain variables happen to take.
on the other hand,
relatively constant
are
those
which are able
net benefits under a range of
conditions.
Since
resources
of
uncertainty
planning,
a project, because
is
almost
robustness
it
is
indicates
always
present
in water
a desirable characteristic
relative guarantee
of net
benefits from the project. Another desirable characteristic of
13
a project is to produce as much net
benefits as
possible. In
most water resources systems, however, the most robust project
is not the one which produces the greatest net benefits. There
exists a
tradeoff between
projects with
robustness and net benefits:
the greatest
robustness are
those
those with lowest
expected net benefits and vice versa.
The
central
decision
making
method
does
modification
topic
method
this
for
not
represent
of
existing
decision makers'
robustness
of
thesis
water
a
is to
resources
mathematical
planning
information needs.
introduce a new
planning. The
advance,
practices
to
but the
provide
The method evaluates the
and expected net benefits of
the
projects
and ends
up with a two-objective problem.
1.1.-
OVERVIEW
OF
A
MULTIOBJECTIVE
METHOD FOR CONSIDERING
ROBUSTNESS
The
multiobjective
this thesis
is
screening model
from all
intended
is
to be
projects
an objective
problems comes from
identified
maximization
that would
the
traditional
programming.
with
of
respect
benefits
generate the
use
of
A
to
propose in
applied to screening models. A
which
that could be built
function. The
linear
method
an optimization technique
possible projects
the set of
mostly
decision-making
an
identifies
in the basin,
optimal
value of
solution to screening
optimization algorithm,
single optimal
an
unique
minus
costs.
alternative is
objective function:
The
incorporation of
14
robustness
into
screening
the
mode Is,
considered
and
decision
because
provides
expected net be nefits)
thes is
This
benefits to
that
we
projects
which
have great
candidate
are
even
or,
able
no single
robustness
both.
To
Such
alternative
an
decision
optim
and
or
expected net
great
These projects are ca lIed
alternat ives
candidate
al alternative
to exist,
exists. In fact,
it
should have the
the greatest expected net benefits.
is
problems.
hig hl y
Therefor e,
unlikely
we
need
in
to
two-objective
compare
candidate alternatives, one of w hich will be the final
Comparisons are
plotting
the
points
alternatives. Figure
Alternative A
has
1-i
c or responding
shows
three
the
lowest
choice.
the
of
is obtained
candidate
this curve.
the greatest expected net benef its,
expected
net
neither the greate st robustness
benefits. However,
makers, the
to
points
lowest robustness. Alternative C has the
but
among
made by using a two-objective tradeoff curve.
The tradeoff curve, also calle d the Pareto curve,
by
and
final project cho ice.
consider
for a single optimal alternati ve
robustness
(robustness
be
identify several alterna tive
to
better,
greatest
to
both robustness and expected net
criteria for
alte rnatives.
implies that
uncertainty
two-objective
considers
Suppose
benefits
allows
traditional
tradeoff graph.
important
be
it
a
improves
process
probab le
if
greatest
benefits.
nor the
but the
robustness,
Altern ative
B has
greatest expected net
this graph were presented t o a decision
chosen alternative would
be B ,
because
it
15
has
almost
as
great
robustness
as
C, and almost as great
expected net benefits as A.
FLISIILITY
ALTERATIV C
FLICI
~-
A-,
T
M(3 (A)
Two-objective
Alternatives A,
The Pareto
i
I
IM(C
FIGURE 1-1:
I
ALTERMI
or tradeoff
I
n
tradeoff
B, and C
curve
for
curve is the final result of the
two-objective analysis. But before obtaining the Pareto curve,
two values for every candidate alternative must be calculated.
The first value is the
alternative.
expected
Cost-benefit
net
techniques
benefits
are
widely
calculate this value. The
second value
is the
the
I
found
given
alternative.
formulation of robustness in
prior stage
to obtain
have
the literature.
the Pareto
method to measure robustness
not
curve,
I
of
the given
used
to
robustness of
any
practical
Therefore, as a
had to develop a
in water resources projects.
Robustness may be evaluated by assessing the potential of
16
the basin
to produce net benefits under uncertain
The potential
of
the
basin
is
the
obtained from an ideal project which
all hydraulic
resources that
are favorable
(for example,
more
projects
benefits
if crop
be
has. When conditions
prices are greater that
of
the
obtained
basin
from
is greater
the
irrigation
in the basin). However, if conditions are unfavorable
if crop prices drop) the potential of the basin
(for example,
decreases. Figure
benefits
can
net benefits
is always able to exploit
the basin
what was forecasted) the potential
(i.e.
expected
conditions.
of
1-2
basin
a
shows
the
assuming
potential
or
maximum net
only irrigation projects and
uncertainty in crop prices.
The upper curve shows the potential
form
of
the
distribution
of
expected
of the
net
basin in the
benefits
as a
function of crop price. The lower curve shows the uncertainty
EETED
EV NWITS
FOTENIALY
ITS f MfTMI
CM0F PFICES (:CJ
DISTIJTI(3 OF CROP PFRCES
Frob (C)
FIGURE
1-2:
Potential
of a basin used for irrigation
under uncertain crop prices
iT
in
it
because
small,
this
high. In
and
Curve,
will
crop prices
of
the potential
thesis,
basin
the
of
be that
is called
alternatives
the
is
as reference
importance
its
robustness
evaluating
for
curve
these high potentials
has
unlikely that
is
Net Benefits
Ideal
basin
that the
Therefore,
small.
values is
that crop prices take these high
the probability
see that the probability
we also
but
increases,
crop prices
basin increases as
of the
the potential
see that
curve. We
its probability distribution
represented by
prices,
crop
is
discussed below.
a measure of
Robustness of a project has been defined as
possible net benefits as
in the distribution of
the variation
net benefits
consequence of uncertainty. Variation of project
has
measured
be
to
appropriate reference
net
other reference,
benefits
net
variation of
The most
something.
is the potential of the basin to produce
any
With
benefits.
to
respect
with
is
misleading
the measure of the
measure of
a
as
robustness of projects. If we do not take the potential of the
which give
basin as reference, robust projects would be those
the
net
same
certain
may
conditions
performance
should
the most robust
yields
benefits
The net
that basin
in
fact
would
be
same net benefits
benefits
of
depends also
a given
be
ignoring that
condition,
any
be expected from the
project
exactly the
under
favorable
and
project. As
a result,
nothing, which
to
build
for
any conditions:
alternative to
on the uncertainty of
better
zero.
be built
in
the variables.
18
For example, net benefits of an
prices, as
shown in
Alternative A
Figure 1-3.
When crop
depend on crop
prices are lower
than Pi, Alternative A yields negative net benefits.
words, the
agricultural
costs
are greater
In other
than the benefits
from selling the agricultural products at price P 1 .
EIVECTE
ET EEFITS
KT RMITS & ALTEATIVE A
CmF ICM
FIGURE 1-3:
Comparison
Figure 1-3
Alternative
Net benefits
of an Alternative A under
uncertain crop prices
of
Figure
1-2
(potential of the basin) and
(net benefits of Alternative
A
is
from
realizing
curves are
is directly
Curve" for
Alternative A
the two
the shape of
two delta curves, for Alternative A
curves to be
(see Figure i-4). Delta
important because the robustness
related to
how far
the full potential of the
basin. We define the difference between
the "Delta
A) indicates
of an alternative
its delta curve. Compare
and for Alternative C
19
FOTEIIIL KT DEEFITS OFN
NET HIUlTS
EIPECTED
..
,/
f'
811I
.
.......
.
ET DEFUITS OFETERiTE I
011TI CUIE OF ETEltTITE I
CIPlCES
/icto
Delta curve of an Alternative A
FIGURE 1-4:
(Figure
1-5).
The
horizontal. This
curve
for
Alternative
crop
benefits
depend
Alternative
A
strongly
Alternative
potential
but
Alternative A
the
is
that Alternatives
basin,
for
increasingly less
A
any
the net
on
the crop
achieves
the full
other
crop
price,
attractive. We may conclude
B and C are more robust than Alternative A.
Although the assessment and
original
almost
price. However,
prices. For crop price P 2 ,
of
is
shows that the net benefits from Alternative
C are relatively independent of
from
C
contribution
of
this
criterion for decision making.
delta curves were shown
thesis,
it
of
robustness
is an
is not a sufficient
Consider two
in Figure
robust than Alternative A.
study
alternatives whose
1-5. Alternative
But Alternative A
C is more
is much closer
20
ELTA
C" rucis
It
FIGURE 1-5:
Comparison
of
Alternatives A,
three delta curves, for
B, and C
than Alternative C to reaching the potential of
any crop price. In other words, although A is
C, A always produces
more
net
benefits
the basin for
less robust than
than
C.
There is,
therefore, the need to consider robustness along with expected
net benefits as
1.2.-
in the Pareto tradeoff curve in Figure 1-1.
ORGANIZATION OF THE THESIS
Chapter 2 serves as an introduction to the
resources
The
planning methods,
first part
techniques and
of
the
called system analysis
chapter
discusses their
reviews the
is
that
in real
problems
system analysis techniques.
techniques.
existing planning
current use
situations. There are some institutional
the utility of
current water
Part of
planning
that reduce
the problem
these techniques do not completely respond to decision
21
making
information
needs.
Four
improvements to traditional
system analysis techniques are analyzed in the second
the chapter:
(1) multiobjective
analysis,
of nearly optimal alternatives,
(4) the use of
(2)
identification
(3) stochastic
planning, and
performance indices.
Chapter 3
resources
introduces the
planning.
consequence of
My
concept of
concern
robustness
with
is
a
the presence of uncertainty in water resources
in the distribution of
of uncertainty. Also
step
in water
robustness
variables and parameters. An index of robustness
based
part of
method
for
in
projects'
outcomes as consequence
this Chapter,
identifying
the
is formulated
I describe
a step-by-
robustness-net
benefits
tradeoff curve based in a screening analysis.
In Chapter 4,
in
a
the general
specific
hypothetical
application. This case is
appropriate
Possible
set
of
projects
are
and
solved,
and tables and
three
intrabasin
5
used
resources
planning
deciding
the most
with
be implemented in a basin.
irrigation
transfer. The
computer
summarizes
water resource systems,
method
limitations
are
to
3 is
outputs
areas,
hydropower
case study
are
is fully
accompanied
in
appendixes.
Chapter
the
dams,
of Chapter
water
concerned
projects
plants,
an
approach
represents
of
the
importance
and comments on
to traditional
the robustness method
the
of robustness in
improvements that
screening models. Some
and possible
improvements
also discussed.
22
LITERATURE REVIEW
CHAPTER 2:
Variables describing the system
is called a model. Models
forming what
equations,
mathematical
of
response
the
can be used to predict
through
represented
be
can
relationships
their
and
system is
of mathematical
development
the
the analysis.
for
techniques
planning
resources
water
require
to
complex enough
a
of
design
The
system
the
and to
evaluate the benefits derived from water resource project.
of formal and objective evaluation method is
Some degree
always present in
use
planning
water
some
mathematical
management
science,
many names:
analysis. This
actual
operations
planning is system
system analysis techniques
section summarizes
used in water resources planning
modeling,
engineering, etc. The
system
water resources
most standardized name in
for
techniques
analysis, or solution. There are given
research,
Therefore most
planning.
resources
and comments
on
their use
in
planning situations.
not all
Of course,
issues
reducible to mathematical form,
systems
fully
model.
In
political,
understood, or
water
resources
economical,
and
of
interest to the planner are
easy to
there
all
are
nor
water resource
identify, describe, and
are
many
institutional
other
social,
factors which can
23
only partially be
in
found
introduced
provide
designs,
optimal
information
making process. Since
is
planning
will
on
as multiobjective
such
optimal
nearly
the
is
than search
alternatives,
planning, do not end up with the optimal
stochastic
provide
of
identification
analysis,
Concern
to decision makers. New
help
planning,
methods in water resources
models.
which, rather
new ones
techniques and to devise
for
formal
adapt existing system analysis
to
literature
the
in
and
system but
for the decision-
alternatives
the decision-making oriented approach to
issue
a central
in
this
in this
be extensively considered later
these techniques
thesis,
chapter.
CURRENT METHODS IN WATER RESOURCES PLANNING
2.i.-
et
Friedman
al.
(1984]
in water
currently used
(1986]
Rogers and Fiering
Rogers
related problems.
[1979) and
is
study which
a
describe
and techniques
models
review
similar
but limited to problems which involve some optimization.
Following Rogers
groups
of
system
optimization
models
and Fiering [1986],
techniques:
analysis
and
techniques,
with search and sampling techniques,
and
techniques,
(4)
related
techniques
analysis,
and
system
analysis
discussed
(3)
(cost-benefit
techniques
The first
and
(2)
(1)
the
Analytical
Simulation combined
Probabilistic models
techniques,
Statistical
game theory).
there are five main
and
analysis,
two
only
(5) Other
input-output
are the most used
ones
that
are
here.
24
2.1.1.- Optimization models and techniques.
planning technique
variables,
define the configuration
Variables
Variables.
is
system which
the solution
provide
and constraints.
objective function,
and operation of the
value
They are formed by decision
1979].
[Rogers,
parameters,
Decision
widely spread water resources
are a
Optimization models
being
optimized. Their
to the problem. For example, when
trying to obtain the best reservoir size, the main variable
the volume
is
the reservoir.
of
Parameters. Parameters
fixed properties of
describe the
the system to be modeled. Parameters are independent and their
values do
However parameters which are
likely to
which are
or
Known,
of the
life
be frequently
demands) can
prices and
well
not
the
change during
of the model.
particular run
during one
not vary
project
(i.e.
water
varied in independent
runs creating sensitivity analysis.
function.
Objective
measure
quantitative
The
relationship
of
main objective
the
function
objective
common
most
of
decision
variables
Constraints
parameters and variables that
and
characteristics.
equations
in form of
are
the
equalities,
and
of the
the
is
a
projects.
mathematical
that
parameters
project.
relationships
among
the
system operation
constraints
are mathematical
describe
Normally
is
from the
describes the benefits minus costs
Constraints.
function
objective
The
inequalities,
integral,
and
25
differential
constraints
season is
equations.
A typical
for a reservoir:
equal
example
are the continuity
the water stored at the end
of a
to the water that was stored at the beginning
of the season, plus
all the
inflows received,
and minus all
the releases and diversions during that season.
Optimization techniques require a formal search procedure
for the set of decision variables that optimize
function while
satisfying all
the constraints. When objective
function and constraints can be expressed
equations,
set
the
of
the objective
as linear algebraic
which maximize
decision variables
objective function can be found with a technique called
programming.
Several
are available
in commercial software packages.
algorithms
the
linear
to solve linear programming
When some of the variables can only take an integer value
(zero
one),
or
the
integer programming.
way to
for
optimization problem may be solved with
The use of
constraints
introduce more
example,
programming
fixed
integer programming provides a
costs
is also available
for
in
the
facilities.
of
solution for
for special
the
decision
non-linear
cases,
variables.
problems,
but
such as quadratic
constraints remain linear, but
as,
Integer
from linear programming
objective function and constraints
functions
quadratic
linear problems,
software packages.
Non-linear programming differs
that the
into
the
in
may be non-linear
There
techniques
programming
objective
is not general
are
available
(in
which the
function takes
form).
26
Dynamic programming
linear
problems
is a method to solve linear and non-
which
have
a
sequential
problems can be divided in stages
decisions are required at
given stage
general
affects to
software
(i.e.
character.
years or seasons), and
each stage.
A decision
taken in a
the next stage. Although there
available
computational procedures
Such
for
dynamic
are relatively
is not
programming,
simple for a limited
number of stages and decisions.
2.1.2.- Simulation techniques
Simulation
performance
techniques
of
the
system
parameters. Simulation
function. In
produce
under
techniques
that case,
for each
information
different
can
on
sets
include
the
of input
an objective
simulation run, a value of
the objective function is obtained. By performing many runs, a
response
surface
formed
by
the
values
of
the
objective
function can be created. Some sampling or search procedure can
examine
the
response
surface
and
obtain
nearly
optimal
solutions.
2.2.-
CURRENT USE OF PLANNING METHODS
Assessment on
in
actual
authors.
water
the use of
system analysis planning methods
resources
planning situations differ among
Two recent surveys show very distinct results. Rogers
and Fiering
conclude that
(1986],
using
agencies and
in part results from Rogers
major consultants
[1979],
only appear to
27
use system
analysis techniques
in few cases. This pess imistic
view differs from the conclusions of
Friedman et
al.
[1984],
using results of a study performed by the U.S. Congress Off ice
of
Technology Assessment
(OTA) in 1982.
Friedman et al.
[1984)
conclude that water agencies and organizations extensively use
mathematical
problems
models
to
find
solutions
conclusions
part from the different meaning of
the
authors.
Rogers
used to solve
potential
techniques
system analysis techniques
[1986]
[1982]
only researched
surveyed techniques
any kind of water related problem. Despite their
differences, there is a
that the
in these studies result in
and Fiering
optimization techniques, while OTA
can be
institutional
currently
water resources
efficiently and effectively.
The contradictory
for
to
of
common
system analysis
improved
issues
conclusion
in
with
the
water
limit the application
both studies:
and other mathematical
understanding
resources
of
in
the
of some
agencies
techniques.
which
In what
follows, four issues are discussed.
1.
Institutional
techniques.
require
resistance
Sophisticated
great
not
easily
planners
understood
and
techniques were
mathematical
specialization.
their study conclude that
to
Rogers
complex water
by
engineers,
use
system
models
of
and Fiering
analysis
analysis
[1986]
resources models
in
are
many decision makers. Even senior
trained
before
system
used, have problem to understand
analysis
the methods.
28
Then,
those
who
departments,
supposedly
have
head
difficulties
agencies
divisions
to accept system analysis
new planning methods. In addition, the use of
for
non
specialized
people
Rogers and Fiering [1986]
confidence
2.
of
the
of
system analysis
produce wrong results. For
may
communication
Part of the
newness and
difficulties of
is also
lack of
undermine
communication
the
is
and of
the
them. But there
communication due to the use of optimization
the given
fact
economical
more the
a consequence
techniques
tools
that
"guarantee"
the best
problem. Single best solutions do not
leave any room for negotiation, and that come
in
even
decision makers and
decision makers to understand
lack of
solution for
between
complexity of
techniques as mathematical
that
as
in the new techniques.
Lack
analysts.
may
this
and
ignore
environment
part
of
in
what
the
social,
decisions
from techniques
political,
are
and
taken. When
analysts use system analysis as substitute for decision making
judgement, decision makers
imposition and
However,
are likely
to
perceive
it
as an
a threat to their authority.
if
techniques
are
used with the perspective of
providing decision making needs
for information, communication
between
analysts
makers
Miller,
1985).
under
the
and decision
Techniques are
different
decision makers need
able
is
to
improved
perform
(Meyer and
the analysis
optimization criteria and policies that
to
evaluate
their
decisions.
In this
29
sense,
system analysis can become very useful
3.
Institution's
Development
resources
of
support
system
agencies
specialized
requires
at
and
within
facilities,
institutions
do
some of
finally
planning
method
situation,
planning agencies
A
priory
developed
(DC's).
no
have
overall
in their evaluation
decide
[Friedman et al.,
systems
countries
in
already in place and
analysis
do
not
result the
exists
1984],
among
DC's
current
water
planning
in
developed
resources
is
systems,
in
resources
systems to which system analysis
effective
(Rogers, 1979).
LDC's
there
decision making process
are
lost.
in less developed countries.
than
major
water
and the possibility
should be more effective
(LDC's)
to use
was expected. To complicate
coordination
Institutional situations
While
of
conditions could not be
techniques
that
not
of sharing resources and experiences among agencies is
4.-
water
training
agencies
these four
available and system analysis
this
analysis.
least four main conditions:
introducing system analysis
system analysis,
more
system
techniques
computer
methods. Consequently, when
efficient
use
and availability of data. For Friedman et al.
agencies
strategies for
to
analysis
personnel,
people,
[1984],
conditions
and accepted.
still
only
in less
countries
systems are
made on small
large undeveloped water
Other advantage
application
in LDC's
is
is simpler and easier. Simpler
is more
that the
in the
30
sense
that
decisions
number of parties
are
more centralized, what reduce the
involved
in
the
process.
Easier because
objectives are fewer (normally national or regional economical
growth)
and
technical
more
clearly
analysis.
Institutional
similar to those on DC's,
arise
from
resources,
the
and
lack
(1979)
the
trained
of
application
reports
of
in
main
the
LDC's
are
problems could
personal,
detailed
system
simplifies
problems
although
of
what
lack or non reliability of
Rogers
cases
defined,
computational
data.
characteristics
analysis techniques
of
22
in the
developing world. Most of them were successful.
2.3.-
ADAPTATIONS OF SYSTEM
ANALYSIS
TECHNIQUES
TO DECISION
MAKING NEEDS
There are
optimal
differences between best in the real world and
in the mathematical
is unlikely
1982].
To
is
enter
more
improve system
variables,
complex equations. Large-scale
for research
models,
significant
because
not
increase
they
understand,
analysis
more
the
solution
first tendency
relationships, and more
models
represent
a challenge
and devise of solution techniques. But
complicate models may be
simple
mathematical
the best solution for the planning problem (Chang
et al.,
to
world. The
may
less
only
effective
because
in the real
they
in the efficiency of
be
too
complicate
large and
world than
may not represent a
the model,
for
other
but also
people to
apply, and solve.
31
System analysis techniques are supposed to
making
aim decision-
information needs. The analyst should realize that what
causes a project to be implemented is a decision, and
models,
and
techniques.
that the decision
system analysis
alternative
is
These, however, increase the chance
correct. Therefore,
to identify
projects
not the
and
analysts should use
performances and consequences of
be
able
to
clearly
present the
information, recognizing that decision makers are normally not
familiar with the technical analysis.
Four improvements to system
analysis
are
literature.
These
analysis,
identification of nearly optimal
stochastic
(2)
improvements
planning, and
properties of
are:
(4) definition of
(1)
found
in the
multiobjective
solutions,
indices to
(3)
indicate
the alternative designs.
2.3.1.- Multiobjective analysis
River basin problems are concerned with the allocation of
water
among
several
uses
and
development
Traditionally planners have used a single
maximization
efficiency
for
of
national
income,
(benefits minus costs).
river
objectives
reduction,
basin
are
to
is
environmental,
regional
etc. Solutions
forward when
development
development,
water
is
economic
also
objective:
called
economic
However, public
investment
multiobjective.
Different
recreation,
national
optimization
a single objective
alternatives.
unemployment
self-sufficiency,
problems
are
straight
considered and the rest are
32
ignored. The solution is
not that
easy when
the problem has
conflicting objectives. However, theoretically, single optimal
solution exists. The procedure
steps:
(I)
find the
to find
it consists
possibility frontier curve
projects which represent the best possible
objectives);
(2)
find the
in three
(formed by the
tradeoff among the
social indifference utility curves
(that show the social tradeoff among the objectives);
find the
curve with
tangent of
possibility
(3)
indifference
the maximum possible social
objective
the
and
frontier.
Figure 2-1
displays the method for the case of two objectives.
cum
SOCIAL lilti22
WIERI I
POSnLIY
MML t
SOLUION
OBJECTIVE I
FIGURE 2-1:
Theoretical
solution
Although theoretically simple
has
almost
determination
unsolvable
of
optimal
multiobjective
and
practical
logical,
problems.
the measure of merit for each
the process
First
the
objective. How
33
to
measure
the
benefits
according to the
markets
for
conventional
was a
number
many of
of
ecological
p
e
i
preservation?:
wild
animals?
There
these objectives, and
would be
with monetary measures of
the
problem
problem arises
possible to
merit
for
are not
then there are not
measure of benefits and costs.
market, it
convert
from
In fact,
if there
fi nd a way to end up
any
and to
objective,
in single-objective. The multiobjective
are not
because merits for different objectives
comparable.
Second
indifference
practical
analysts
when
problem
curves.
the
select
the curves,
authors do not
because
preferences
indifference curves,
determina tion
They are very difficult
applications. Many
quantifying
have
is
of
of
of
to formulate for
recommend that the
prac tical
the
social
difficulties
society. Without social
the multiobjective problem do
not
likely
single optimal solution.
The
third
problem
is
curve.
For many analysts,
step.
Almost
centered
finding
efforts
new
the
frontier
curve.
optimization,
constraints:
of the
analysis
multiobjective
and
more
The
since
effective
basic
is
f rontier
ju st
this
are
planni ng
meth ods for
problem to s olve
the objective
functi on
is
is
a
a scalar:
objective function:
and
in
studying
instead of
generation
multiobjective
in
called vector
vector
all
the
Max
Z
Z
ZR
CRi
Aij
Xi
[Zi,
Z2 '.'
Zp]
Xi
-
Bj
0
34
Xi
A vector
cannot be directly maximi zed or minimized.
a set of non-inferior
defines
1 0
the
called
solutions
be
can
obtained.
the
optimization
the
problem.
techniques
of more
yet to be
developed
up
off
trade
resources
to
study,
many
available,
for cases
worth
In their
are
method
objectives,
the
the
vector
conclude that of
not
have
practical
an efficient method
only
the
is
called subrogate
be appli ed when great computer
For Cohon
available).
four
can
[1975)
res ources problems. Also,
than four objectives
(presently,
Marks
solve
they
do
to multiobjective water
application
to
developed
techniques
This set
possibilit ies frontier curve, or
Pareto
simply frontier curve or Pareto curve. C ohon and
summarize
Only
and
best
Marks, for
me thod
cases with
is the constraint
method.
The
constraint
method
consists
in
the
following
optimization:
objective
function:
Max
Zm
m
with the new constraint:
ZI
1 LC
R
old constraints:
ZR
= Cki
and the
Aij
Xi
Then,
Zm
is maximized
other objectives.
the advantage
If
this
Xi
that sensitivity
-
Bj
1 0
1 0
subject to
problem is
XI
lower
limits Ll
in the
LP,
it has
solved using
analysis
(included in most of
LP packages) can rapidly obtain solutions for different values
35
of
LI's.
There
is
small
chance
Therefore
available.
alternatives could
no
that
social utility curves
further
be taken,
choice
are
among non-inferior
and the frontier curve is to be
presented to the decis ion makers.
Identificatio n
2.3.2.The
only
global
optim
identification
same
the object ive
problem but with
agreement
for
terms of
are significantly dif
the
in
not the
maker;
with in acceptable
function represent
involved
is
decision
alternatives
sol utions to the
fo r ne gotiation and
the
dec ision
very clos e
process. These other alternatives,
solution is
model
more possibilities
par ties
among
mathematical
d ifferent
of
ranges of
um of a
interest
of
solution
of nearly optimal solutions
making
t o the optimal
the objective function val ue, but which
ferent decisions
in terms of
the decision
variables values, are called nearly optimal solut ions.
The
solutions
is
Optimization
optimal
pr oblem
major
the
of
computer
algorithms
are
designed
nearly optimal
generate
ne arly
softwar e
to
little
reac h
optimal
available.
the single
info rmation about
solutions.
Harrington
and
interesting procedure
which has
identifying
lack
solution and they
when using
for
a LP
Gidley
[1985]
to analyze
package. The
a single optimal
suggest
a
simple
and
nearly optimal alternatives
procedure assumes a LP problem
solution:
36
objective function:
Max
and constraints:
Z
Z
= CI Xi
Xi -
Aij
Xi
and the single optimal
Bj
S 0
1 0
solution is:
=Z
Z
To generate alternative optima
the
software
package is
repeatedly applied but introducing a new constraint:
Ci
In case
1 a ZN
Xi
= i the global optimal solution ZN
that a
again. For
other nearly
any values
inferior to
nearly
optimal"
Harrington
and
nearly optimal
the
optimal
alternatives,
is obtained
a should take
i. As closer is the value to 1, "more
solution
Gidley
[1985)
solutions
is.
The
prove
exists
studies
made
by
that a great number of
within
0.5%
of
the
global
optimal objective function value.
Other
attempts
functions
in
which
variables are
varied, new
optimal
the
been
made
maximum
to
value
generate objective
of
some
decision
randomly constrained. When these constraint are
objective
solutions
The
have
functions
are
maximized,
and nearly
are obtained.
consideration
potential
not only of
producing
better
of
nearly
optimal
solutions has the
producing better decisions, but
understanding of
the nature of
also of
the decision
problem itself.
2.3.3.-
Stochastic
Planning
37
Many of
resources
system
the factors that define the
systems
is
cannot
be
known
Most
of
the
planned.
performance
with certainty when the
p1 ann ing
is
uncertain circumstances. The simplest a pproach
simple probabilistic
measure
uncertain variables,
and then
of water
(mean,
under
is to take some
median,
proceed as
made
for the
mode)
on a deterministic
problem.
There is a second approach to deal with
is
to
evaluate
the
consequences
system design. Loucks et al.
this
approach.
considers
The
maximum
(X
W)]
rev iew some
expected
net
methods under
benefits
(X
W ) are
i
given design
p
the net
is described
benefits of
probability
of
occurrence
expected net benefits
for
the system whose
by the decision variab
the uncertain parameters taR e the
of
value
given alternat ive
In
Other
of
the max-min
project
the methods
method, uses
given projects,
are
the
calculated,
a most
the
the one wit h better
which provide
s olut ion.
chosen
benefi
al.
ts
projec t
whose smallest net benefits are the greatest.
favor alternatives
thi s method
Loucks et
net
the
with greater
[1981],
pessimistic cri teri on:
smalI le st
and
when
is
Pj
essence,
pr ojects,
in dicated by
les Xi,
and
The project
Wj
chosen.
the
Wj,
average performance. It can be call ed compromise
all the
method
j
i
selects,
of uncertainty in a given
the following problem:
E [HB
where NB
(1981)
uncertainty that
It
is
from
of every
is
the one
a
a minimum of benefits
way to
every
38
year over other alternatives
more profitable
but that may produce very low net benefits
In
between
both
methods,
outcomes and other
ignoring them,
Its
is
basic
decision
point
maker's
risk premiums,
Figure
2-2).
include
analysis,
there
define
preferences,
the
in same conditions.
overreacting
is the utility
to
poor
theory.
utility function. From
indifference
situations,
and
a continuous utility curve might be drawn (see
Keeney
decision analysis
how to
to
one
on the average,
and
Raiffa
[1976],
the
traditional
book, show how to obtain utility curves and
in them other attributes than money. In utility
the new objective function value is utility,
and the
problem is to maximize it.
VTILIYT
I
UflLITT CUM
ET BIEMlTS
FIGURE 2-2:
Typical form of utility
Utility theory has many
curves
favorable points.
It represents
directly social or decision maker's preferences and tradeoffs.
39
For example, bad outcomes
ones.
Sensitivity
utility
curves,
participation.
are more heavily weighted
analysis
and
Difficulties
likely
makers,
preferences. Consensus
and then the
in
applied for different
stated
utility
assess
representing
different
have
be
conclusions
quantifying preferences to
decision
could
for
social
debate
theory
utility
perspectives
arise
rejected
and
when
curves. Different
and political groups,
of
social
values
in a utility curve might be
conclusions
than good
for
those
and
impossible,
with different
utility curves.
The
second
approach
to
deal
identify optimal alternatives under
main method of
suggested
original
this approach
in part by
Loucks
NB
Aij
Xi
where Ci,
Aij,
In stochastic
and
(1981].
programming,
We
assume the
Bj
First, the
Xi -
Bj
1 0
1 0
represent the parameters
Assume
probability distribution
There
= Ci Xi
linear programming
deterministic.
to be
al.
linear
NB,
Subject to:
has
is to
uncertain conditions. The
stochastic
et
uncertainty
deterministic formulation:
Max
no
is
with
that
some of
Ck
is
is represented
are certain steps to be taken
these
of
the model.
parameters are
uncertain,
and
its
in Figure 2-3.
to solve the
problem.
continuous probability distribution function for Ck
approximated
by a discrete
distribution
function,
as
40
}I
DISCET= FMCi0
v/i
C0TIPM FEm
MEITI
FIGURE 2-3:
p(CkI),
its correspondent
P(CR
.. ,
two
linear
kind
programming,
define operation rules of
The
model
of
as
a finite
second
dependent
step
it
is
is
on Cki,
important to
design decision
variables. Design decision
Operational
the projects
extensively discussed
discrete
(q is
of decision variables:
variables define project sizes.
is
Cj
probability associated p(Ckl)
variables and operational decision
This
.. ,
)-
In stochastic
distinguish
Ct
Continuous
and
discrete probability
functions for uncertain parameter Ck
done in graph. Each interval Ct, CR2,
number) has
PAiAwnTU
decision variables
once they
are built.
in Chapter 3.
to write the
with
i
original
= 1, 2,. ., q (q
intervals of the uncertain
consider the decision variables Xj,
with
parameter
j
optimization
is
the number
Ck).
We
= 1,2, .. , k-1,
also
to
41
with j
and the decision variables Xj,
decision variables,
be design
= R,+1, ..n ,I
decision variables. We
to be operational
can write:
NB
= C 1 Xi + C 2 X 2 + .
NB 2 = C1
NBq
Xi + C 2 X 2 +
.
= CI X, + C 2 X 2 +
depending
on
condition,
there
a
design
+ Cn Xn
2
+
..
+ Cn Xn 2
Xk-I + Ckq Xkq +
..
+ Cn Xnq
Xk-i + CR
+ CK-i
+ Ck
2
conditions:
set of operational
possible benefit
decision
XK
take
variables
uncertain
are
..
+ CR-i
the m aximum
that obtains
However,
the
+
Xk-
decision
Operational
Xk1
+ C-i
variables
because the size o f the project
can
can
different
values
for
future
decision variables
from the
projects.
take only a value,
be
not
any
adapted
to the
possible future co nditions.
The
Max
new optim ization problem is now:
[NB1
p(C gi)
+ NB 2
p(CR 2 ) +.+
NB
p(Ckql
E NBM p(Cgm)
Max
m = t, 2,..q
Subject to:
Aij XiM -
Bj
1 0
Xim ) 0
Note that the value of the superscript m
Xi"
depends on the type of decision variable:
design decision variable:
operational
m does
decision variable:
m
constraints
of
and
the
variables
problem
is
is
in
not have any value
=1, 2,..q.
The new problem continues being linear,
size
in decision variables
but the
number
of
significantly increased. The big
fact
the
main
limitation of
42
stochastic
As
with
linear programming.
a
conclusion,
uncertainty:
uncertainty,
yield
there
(1)
and
to
(2)
uncertain
are two
forecast
to
choose
benefits.
evaluating methods
main approaches to deal
consequences
et
al.
solutions,
but
(1981),
for both approaches,
that uncertainty models should not be used to
best
to
of
between alternatives which
Loucks
and models
the
eliminate
when
concludes
identify single
clearly
inferior
alternatives.
2.3.4.-
Indices
Indices
performance
are
of
criteria. All
the
quantitative
that
measures
different
alternatives
indices
discussed here
under
compare
some
given
are consequence of
uncertainty in water resources planning. Ind ices rank projects
in regarding their performance under the fol lowing criteria:
1. Probability that the project fails
2. How bad are the consequences of the f ailure
3.
Probability that the
well
i
and
2
are
reliability and vulnerability
slightly
reasonably
under different demand conditions.
Criteria
indices
p erform
project will
are
appropriate,
different,
resiliency, and
and
robustness.
respectively
indices.
although
so
For
the
3,
two
definitions
are
criterion
their
their
by
measured
proposed
In what follows,
formu Iae:
the four
ind
ices
are proposed and analyzed.
43
1.-
Reliability Index
Reliability index
has no
failure within
[1982a]
proposes
measures probability
the planning
that the
project
period. Hashimoto et al.
the fol lowing formulation:
= Prob [ Xt E S ]
Irel
where:
Irel
Xt
:
S
:
Reliability
Index of Reliability
Variable that describes the system's output at
time t (t takes the discrete values 1, 2,......)
Set of all satisfactory outputs
is
opposite
probability of failure.
as
2.-
1
to
risk,
or what
In this sense,
is
the same,
risk index
the
is defined
-Irel'
Vulnerability Index
Vulnerability
consequences of
not consider
number
of
measures
the
a failure, given that
the
length
failures,
Vulnerability only
Hashimoto
index
of
nor
time
how
magnitude
it has
until
long
the
the
of
the
occurred. It does
failure, nor the
failure
lasts.
refers to how severe the consequences are.
et al.[1982a) proposes the following formula:
Ivul
= E Sj
Ej,
E F
where:
Ivul
S
Ej
F
Index of Vulnerability
Severity of the consequences
Probability that the consequences be with severity
Sj
Set of unsatisfactory outputs
(failures)
44
3.-
Resiliency Index
Resilience
is
a
term
known
ecology, materials,
economy, and
common
definition
generalized
resiliency is associated to
situations,
and
to
in
other
structures. There
for
resiliency.
adaptation to
recovery
sciences
from
like
is not a
In general,
new (no projected)
surprises
and
adverse
situations.
In
water
resiliency.
resources
First,
for
describes how quickly the
failure. Their
of
there
system will
mathematical
is the
two
Hashimoto et al.
recovery. They propose the
If TF
are
time that
(1982a],
formulation
is
[T F]
E
is possible
1 /
(Prob
to
resilience
likely recover
from a
based on the time
following measure of
resiliency:
a system remains unsatisfactory
after a failure, the index of resiliency is
expected values,
approaches
i/TF;
considering
to define TF:
[ X
E S
I X
E F
)
where:
Xt+ i = Variable
which describes the system's output at
time t+i
Set of satisfactory outputs
Variable which describes the system's output at
time t
Set of unsatisfactory outputs (failures)
S
Xt
F
and by def inition of
I
re s
: Prob
Second, for
the
[ X
Fiering
probability that
unlike events
index of resiliency Ires:
t+i
E S
[1982a,
I X
t
E F )
b, c,
the system will
and
d]
resiliency is
operate well
occur. This concept of resiliency
enough when
is
similar to
45
of
notion
the
a
when the
partial
for these
robust to
system is
Fiering,
to
Fiering's distinction
is
and this
to
is small
systems response
the system
if
between robustness and resiliency, even
robust
According
in certain variables
changes
derivative of the
variables. But,
below.
described
robustness
is not
it may be resilient as a whole. A
certain variables
resilient system accommodates the surprise produced in several
of
its
by
variables
in
changes
relations among
derivatives:
dz/dxi
A
remaining variables.
measured as
Resiliency, therefore, should be
total
the
:
E
(dz/dxj)
(dxj/dxi)
of
the
combination
linear
all
total
derivatives dz/dxi
measures the resilience of a given system as a whole.
Other criteria and alternative measures of resiliency are
In fact he proposes
(982b].
Fiering
proposed by
different
alternative
residence
time
in
of
indices
non-failure
up to eleven
based
resiliency,
state,
and
in
combinations of
passage time between failure and non-failure states.
total derivatives
The method of the
[1982a]
used
was
resiliency of
They
that
resilient when the
the
surprises"
expected
a
system
expected
in
and Marks
systems
agricultural
concluded
"unpleasant
Allan
by
the
degradation
design
performance
in
the
[1984]
by Fiering
to measure the
developing countries.
in
planning
proposed
can be expected to be
degradation
parameters
planning
is
due to
less
than
parameters
themselves.
46
For practical
considered to
applications,
be
those
which
highly resilient systems
contain
many
are
redundances of
design. For these systems, proper operation rules minimize the
unpleasant
effects
of
surprises.
Large
systems
with many
connections have proved to be the most resilient.
4.-
Robustness Index
For Fiering
much
the
[1982a],
same.
robustness has
measure of
the
robustness and resilience mean very
However
for
Hashimoto
other meaning.
et
They consider
possibility and expenses
of
al.
[1982b],
robustness as a
adapting
a system
to future conditions different from those for which the
was
calculated.
information
It
is
the
cost
about the future. The
Prob
I
[ C(qJD)
-
of
not
having
index they propose
L(q)
1
3 L
(q)
system
perfect
is:
],
where:
Irob = Index of Robustness
q
= Future conditions
D
= A particular alternative (Project or Design)
C(qjD)
= Cost of accommodating the alternative D to
the future demand conditions q
L(q) = Minimum accommodating cost among all the
alternatives
13
= Level of robustness
Under the
al.
[1982a),
which
affect
"demand conditions"
there are
the
project
which are uncertain and
This
thesis
already introduced
grouped the
uses
term used by Hashimoto et
set of
(demand,
future conditions
costs,
prices,
...
) and
likely to vary.
a
different
in Chapter
concept
i. Robustness of
of robustness,
a project
is a
47
measure of the dispersion
project
Accordin g
as
consequence
to
this
of
of
possible
net benefits
uncertainty
criterion ,
an
index
from the
in some parameters.
of
robustness
is
develope d in Chapter 3.
48
CHAPTER 3:
A
METHOD
FOR
INCORPORATING
PLANNING DECISIONS
This chapter describes a method to
resources planning
ROBUSTNESS
ident ify robust water
strategies. The method
is a multiobjective
analysis of the water resources system. One of
is
to
maximize
expected net
robustness
and
the
is a measure of how
sensit ive
constant
rob ustness
1,
the project
In
exists.
under
is
measure of
project. The
Chapter
to maximize
analyzing robustness
the
project
parameters. A project is ca lied robust when
thesis,
is
of uncertain conditions in the bas in.
the presence
relatively
the objectives
other
benefits. The reason for
is based
project
to uncerta in
is
its perf ormance
In
is
th is
a desirable property of a
robustness,
on the distr
is
Robustne ss
conditi ons.
different
cons idere d
IN
ibution
as
of net
indicated in
benefits from
under uncertain condit ions.
mul tiobjective
an alysis,
The refore
outcome
the
several n on- inferi or develo pment
no
of
single
best
this method
is
alternatives
for
project
to define
the water
resources system. The metho d begin s by displaying all possible
projects to be bui It
method
with
is
in
applied, the
great er
identified. One
robustness
of
t he
bas in.
projects or
and
them will
be
From
these,
combinations
expected
the final
net
after
the
of projects
benefits
are
development choice.
49
3.1.-
DESCRIPTION OF THE METHOD
applying the
Previously to
the screening model for the basin.
projects
the
valu e
greatest
basin).
maximize
a
partic ular,
In
uses for the water. Projects
variables.
the optimal
most
functio n
and all
consists
expressions
of
define the
The model
contains
Both,
mathematical
are
parameters and decision variables.
etc.
maximize economic
the mathematical
If,
for
efficiency,
economic
the
criterion
the
example,
function
objective
for a typical
to be repeated during the water system
Constraints are
mathematical
the
minimize
efficiency,
expression of benefits minus
are considered
measure of
a quantitative
maximize
criterion:
unemployment,
in obtaining
constraints.
function and
objective function is
policy
that
possible
decision
their
by
constraints,
and
water system,
the
projects to build.
se t of
objective
and costs
group of
defined
are
two main el ements: objective
main
a
is
value o f every decision variable,
profit able
The
from the
to be built
solv e the screening model
To
(normally the
benefits
model
represent
poss ible projects
including all
net
screening
relati onships that
mathematical
those which provide
the objective function
to
is
function
objective
of
the basin
in
built
be
that could
possible
all
among
is an
model
The screening
selects
that
techn ique
optimization
itself, we must write
method
costs.
year, that is
is
could
to
be
Benefits
assumed
lifetime.
expressions
which show the
50
physical
conditions
institutional
etc.),
and
(water continuity,
conditions
(water
social-economical
prices and
costs,
etc.)
maximum sizes,
priorities,
etc.),
operation rules,
conditions (demands for water,
of
the
basin
or
water resources
a
screening model
system.
The
looks
mathematical
representation
of
like:
(for the typical
-
year)
objective function
Maximize:
for {XjJ
-
constraints
Gi
where Xj
(XI,
F
(Xi,
X2,
..
Xm)
,
(i equations):
X2,
. .
,
Xm)
are the decision variables.
Once the
technique
screening problem is formulated, an optimization
may
be
used
to
decision variables
Xj,
operating
These
rules.
variables
define the
(irrigation areas,
their optimal
The
models.
(i equation):
which
normally are
optimal
most
the optimum values of
values
profitable
hydropower plants,
project sizes
of
projects
etc.),
the
and
the
decision
to
be built
their sizes,
and
operation rules.
most
used
Linear
(LP),
technique.
The
screening model
screening
screening
programming
-
obtain
that
is
models
the
mathematical
is:
models
can
most
linear
screening
be solved using
available
representation
(for the typical
objective function
are
linear
optimization
of
a
linear
year)
(i equation):
51
Maximize:
for (XjI
-
constraints
(XI,
Gi
F
(XI,
X2
-
,
Xm)
F
(X) = Cj Xj
(I equations):
X2 9
. .
= Aij
Xm)
,
Xj
-
Bi 1 0
where:
Xj are
the decision variables,
parameters
3.1.1.-
and Cj,
Aij,
Bi are input
(some of them may be uncertain)
Step 1:
Derive the ideal net benefits curve
Decision Variables
Decision
variables
rules. There
screen ing
are
two
model
examples
and
0f
be
of
variables
decision
in
the
variabl es and operational
dec ision variable s define the sizes
bu ilt.
capa city
de sign
project design and operating
dec ision
Des ign
projects to
reservoir,
types
des ign
dec is I on variabl es.
of the
define
of
Area of
hydropower
irr igation, volume of
p lant
are
typical
decision variables. Op erational decision
variables def Ine the operation rules of the projects once they
are
built.
Rel eases
irrigation are
Design
from
examp les
dec is ion
of
var lables
r eservoirs and water diverted for
operational
a re
of
primar y
analysis, s ince they defi ne what p roj ects
what projec ts are not to be built. Al so,
be built, design decision variables
The
a
priori.
values of
In fact,
d ecision variables.
indicate
interest
are to
be
in our
bui It and
f or those proj e cts to
their
the design decision variables
when we search for candidate
sizes.
are not Known
alternatives
52
in Step
2, what we
decision
actually do
variables
alternative.
In
which
is
to find
define
general,
design
the
the set
candidate
decision
of design
development
variables
form a
vector X, such that:
= (Xi, X 2 ,
X
where m
,
.
XmJ
is the number of design decision variables, and where
X, may represent volume of reservoir, X 2 Irrigation
area, and
so on.
Uncertain Parameters
The reason
why predicted
and actual
project may differ is the uncertainty
the
particular
case
of
linear
parameters are a subset of the
Bi.
performance of the
in some
parameters. In
screening models, uncertain
input parameters
Cj, Aij,
and
We group this subset formed by the uncertain parameters in
a vector 0:
e
= ((
,
1
4 2,
'
n I
where n is the number of uncertain parameters
where ej
basin, and
agricultural
We need
uncertain
and so on.
the
parameter.
distributions
example:
may represent discount rate, e 2 price of
products,
to have
used
gaussian,
existing in the
The
in
distribution
probability
most
common models
engineering
lognormal,
and
Gumbel,
risk
or
of
into
a finite number of discrete
every
probability
analysis
log Pearson)
continuous distributions. For present purposes, these
divided
of
(for
are
must be
intervals with their
53
associated probabilities
If we assume that,
is
rate"
as shown
for example,
and
uncertain,
in Figure
it
is
2-3
in Chapter 2.
the parameter "discount
approached with a gaussian
distribution, Figure 3-1 shows the.division
of the continuous
distribution in discrete intervals.
Prob (1) :
VROAIllLITY
1 : DIs
functions for the
probability
uncertain parameter 94 : discount rate.
FIGURE 3-1:
In general,
T TE ()
Discrete
after
dividing
In
discrete
continuous uncertain parameter vector e1
intervals, the
is converted to:
e = Oj
where:
I = t, 2,..,n
j = 1,2, ..
where Mi
1. We
is
Mi
the number of
also obtain
intervals of
the corresponding
the uncertain parameter
discrete
probability of
each interval:
54
p(E 1 )
To
compute
this
uncertain
probability
parameters
correlation exists
some
cases
(for
are
we
have
independent;
among them.
example,
to
assume
in
other
This assumption
no
correlation
irrigation water demands and discount rate),
cases some
correlation may
be present
words, no
obvious
in
exists
between
although
in other
(for example, between
irrigation water demands and agricultural
Ideal
is
that the
prices).
net benefits curve
The
ideal
net benefits
net benefits of
ideal
net
robustness,
the
basin
under
benefits
curve
is
as
shown
curve, we
pick up
parameters
vector E,
the maximum
and,
exploit
the
full
conditions. The
reference
to
obtain one
point of
that the
We
also
for the
optimal
set
potential
calculate
this
element eiJ of the uncertain
using the screening model,
net benefits
that
uncertain
the
a particular
decision variables
defined by
is formed by the potential
in Chapter 1. To
conditions defined by ei.
design
curve
of
basin may yield under the
obtain the
conditions
of
we obtain
design
optimal
e5J.
The
set of
projects
decision variables
the basin for that situation
eij.
For the element E 1 J of the uncertain
let
X*(e
variables
NB
1
J)
represent
the
optimal
parameters
set
of
obtained from the screening model.
[X*(E9J )]
represent the
net
benefits
vector
e,
design decision
And
let
obtainable from the
55
projects
defined by the design decision variables X*(e1J). The
value NB
[X*(e 1 5J)]
eij, and,
the situation
the
ideal
net
the
ideal net benefits
curve
for:
I
j
It
is
not
=
that we
process
is
the uncertain parameter
is fully defined.
representation
correspond
project.
to
a
single
indicates,
may expect.
for
be
useful
of the
ideal
typical
zero when
t hat def ines
the
Oi3
the
of
net benefits
c urve
optimum
projects
set
to
be
of
decision
built. This
net benefits curve of
r ate is shown in Figure 3-2.
n ever takes
the discount rate equals
discount rates
element
ideal net
step s
of the idea
that point,
to the project
of
there
dis count
the system. At
The
th e maximum possible
an
form
net benefits curve
Final Iy, a lso note that, ass ociated with
th e next
in
every
is
943,
example, the final
costs
same
ideal
under uncertain
equal
a point of
n
.,Mi
variables X*(e 1ij)
of
the
important to note that the
element
A
of
curve
mathematical
uncertain parameters vector
will
When
basin for
is:
, 2,.
1,2,
curve
benefits
every
of the
[X* (8 J.- I)],
NB
does
curve.
the other elements
Therefore, the
net benefits
the potential
consequently, represents
benefits
repeated for all
vector G,
represents
costs,
the
building any project
internal
rate of
benefits from the project
and there
higher than
negative values.
the
a basin
In
this
It
is
return
are
are no net benefits. For
internal
rate of
are greater than the
return,
the
benefits
56
NET
BENEFITS
(dol lars)
,IDEAL,
I
I
I
~~
I
I
I
NET BENEFITS CUI
I
!
I
-
.
.
DISCOUNT RATE (11
Typical ideal net benefits curve for a
basin under uncertain discount rate.
FIGURE 3-2:
which yields
zero benefits and zero costs.
curve remains
internal
rate of return of the system.
net
candidate
We should
identify alternatives
benefits
or
both.
that have
decision
of
Select
Step 2:
that
design
There
have
Alternatives
variables.
Then the
ideal net
zero for discount rates higher than the
benefits
3.1.2.-
is none,
build
project to
obtained. Consequently the optimal
alternatives
that have
high robustness or,
are
Selection of
defined
by
high expected
even
better,
their design
alternatives means
selection
identifying
candidate
decision variables.
is no formal
alternatives. However,
procedure
from Step
for
i, while
obtaining the
ideal
57
net benefits
curve, we
projects
suggest the
to
procedures.
Mi M
In fact,
number of
optimal
each element Gij.
using the
use
of
enough
information
some very effective
in Step i it was
optimization
independent
Mn
2
developed
procedures
analysis
proposed
informal
necessary to perform
of
gave equal
which
runs,
sets of decision variables X*(e
From the
on the
these
3i),
1
one for
optimal
sets,
below, we can obtain candidate
alternatives.
Several
of
decision
procedures for the
variables
X*(O
1
analysis of
j)
are suggested
procedure
is to calculate the probability
values
the design decision var iables.
of
procedure
that
produces
design. C onsider
(XI,
-.
X2,
the set
, Xm,
and
in
great
of
one
into
of
variable
volume of
a reservoir). When the problem was
uncertain
Xu
Mn
2
values
optimization runs,
of
Xu"(E91
d is tr i but ion
for
Xu* (Gij)
for:
of
), one
with
j
1, 2,
have
the
we
E5j,
the
have
M,
=
optimized for the
value
M
2
is
therefore:
p(e
1 3)
.
Mi
a
discrete probability
design
X
them, the design
for eac h combination of
probability
1,2, .. , n
system's
of
Xu
was
p(e i3 ). Since we performed
the values of Xu
i
Therefore we
values
of
the
ma y, for example, represent the
element
with probability
(Eij ),
MI M
parameters
the
is a very simple
decision variables:
particu lar
(that
here. The first
This
insight
des ign
sets
distribution of
decision
Xu
the optimal
decision variable
Mn different
i
and j. The
distribution of the
Xu. The
analysis
of
58
that
likely
values
optimiza tion
Step
distribution
probability
I
for
is LP,
de sign
the
algorithm
us ed to
the vertices
for
decision variable Xu.
solve the
thre
e values.
wh Ich
the most
reduce to
If
the
screening model
in
tends to be highly
This
is
because LP
polygon formed by
a few
the
otherwise
possible values o f the design decision variable Xu-
The second
in
shown
identify
of the n-dimension
vertices
the cons traints,
infinite
to
the probabi li ty d istribution
concentr ated around two or
searches
helps
Figure
is
procedure
3-3
probabil ity distribution
fo r
a graphical
representation,
hypothetical
an
of Xu*
(eiJ).
Using
easier to visualize the most probable values
variable Xu. These values seem clustered
example
the
like
of the
graph,
it
is
of the decision
in three groups.
PROBABILITT
Prob (I)
II1I~,
DECISIIARIABLE
FIGURE
3-3:
Typical bar graph representation of an
of
probability
distribution
hypothetical
values of a design decision
variable
Xu,
obtained from the MI M2 Mn optimization
runs performed in Step I
59
The
purpose
we
expect
that
values of Xu-
likely
three more
this
of
variable,
is
procedure
do not
to
use
that
producti on
more
that
1 eft
when
other
and
l es s Pro fitable pro, jects are built
projects do not require the
relationships among
of
existence of
optimization
runs
project:
(1)
projects
that
prof itabl e
more
other projects. These
by
insp ection
resu Iting from the M,
performed
relationships among projects
are built.
only after more
identified
be
projects can
the decision variables vectors
different
H owever
built.
projects
are some
there
projects
Therefore,
are
are
are
that
is
projects
effectively are built
sti 1l
projects
of
in water
the effective
capac ity
the
table
problems
resources
of
a third
among design decision
res ourrces more
projects. Il is
a
create
that are for med by se veral
0 ther
hav e
profitable
among
combinations
char acteristic of optimizat ion
proje cts
before than
is
incompati bil ities
and
resources systems
those
With the
va riables.
the decision variables,
procedure
compatibilities
variables. A
combining the most
sugg ested.
third
The
by
possible
all
of
values
likely
profitable. These
decision
but bef ore examining
more
the
and
the likely values of every decision
identify
may
we
graphs
design
alternatives
identify
are con structed
the
of
robust
both
be
to
is to
procedure
candidate alterna tives
likely values
identify the
could
of each group, we
the averag e value
Taking
M2
in Step I. Three
suggested;
a lways
built
for
Mn
kinds
a given
when
that
60
particular project
is built
given
projects
one);
(2)
(more profitable projects that
that
particular project is built
given one);
with that
to
and
(incompatible
particular project.
they
especially
are
alternatives,
very
built when
projects
among
These tables
they
the
in
reduce
any relationship
are not difficult
more
developing
the
likely
that
with the
for small water resources
helpful
because
combinations
never
(3) projects that do not have
construct,
and
are
the
number
values
systems,
candidate
of
of
possible
the design
decision variables.
3.1.3.-
Step 3:
From Step
Assess the performance of
2, we
alternatives.
Step
alternative
this
in
have a
3
reduced but
assesses
group.
performance of
an
value
net
and
calculate them,
the curve of
Curve of
it
performance
is necessary to
net benefits of
the
use
every
two characteristics
alternative:
(2)
of
of
(1)
the
robustness
an
expected
index. To
intermediate step:
every alternative.
net benefits
For
benefits
benefits,
the
promising group
In this thesis,
define the
of
the alternatives
a
given
candidate
indicates the net
conditions. We
alternative,
benefits
assume Alternative
A
its
curve
of
net
obtained
under uncertain
defined
by the decision
variables:
XAg
with k
1, 2,..m.
61
This Alternative A,
of the
yields some net benefits
vector G,
parameters
uncertain
element Gij
the particular
under
(or,
if negative, net costs). Mathematically these net benefits are
represented by:
NB
e
[XA
j)
that reads:
XA given Eaj.
variables
To
alternative defined by the deci sion
of the
net benefits
e j]
[XA
NB
calculate
use
we
an optimiza tion
1
technique
before to
(like
LF),
cal cul a te the
an optimi zation
calculate NB
decision
[XA
element
ideal
net
E j]
is
benefits curve.
require d,
Alternativ e
Alternative
A
procure the maximum possible
net
benefits curve
benefits of
1
j.
In
words,
A
(while
are
held
th e
when
e1 j,
design
con stant)
in
is obtained
computed
for
every
the uncertain p arameters
the
performance
take the value
indicated
by
Alternat ive A
indicates
the
net benefits
the
curve
will
op erational
net benefits.
are
A
p rocess to
because the
for Alternative A
Alternative
other
T he use of
op timization of
involves
for
variabl es
The full
when net
could be the same technique used
technique
decision variable s of
order to
that
curv e
of
that Alternati ve A
yie ld.
Fig ure
3-4
shows
of
net
uncertain
benefits
of
an
discount rate. The
hypothet ical
alternative
under
internal rate
of return of
the alternative
is for what
the net
is zero. For discount rates
higher than
the
benefits
curve
62
WT
REFITI
m
UTjW
hM=am(I
Net benefits curve of an hypothetical
FIGURE 3-4:
alte rnative
rate
internal
rate of return,
that
costs
of building
benefits
of performance
are
of
concern
in
and
robustness)
For
of
our
are
study
based
Net Benefits
of
on
of
candidate
(expected
situation
probabilities of
e 3,
e1 j. The
alternatives
value
of net
the net benefits curves.
an Alternative
a given candidate alternative, the
benefits can be calculated by
uncertain
curve takes negative
it
two measures
Expected Value
discount
from the alternative are smaller than
The
benefits
net
under uncertain
the net
values, because benefits
the
I
1
adding net
weighted
net benefits
by
expected value of
benefits for each
the
respective
are given by the net
63
benefits
curve just calculated.
before:
been calculated
the
1iJ).
[NB
[NB
(A)]
(A)]
is
E NB [XA
e
i = i, 2,.., n
j = 1, 2, .. Mi
probabilities
Therefore,
expected value of net benefi ts
E
E
p(
The
have
also
for Alternative A,
is given
by:
p(e j)
J
evidently a scalar value.
The same
process has to be
done for the rest of
candidate
alternatives.
Index
of
robustness
For a
given
candidate alternative,
to the possible variation
uncertainty.
This
potential
the
of
of
potential
benefits
curve
represented
given
We
calculated
by its net
the beginning
benefits
curve
of
that
in
as
shown
is
represented
in
Step
benefits curve,
net benefits
respect
by
the
are
be related to their shape as
The mathematical
related
to the
And the
ideal
net
other hand, for
under uncertainty are
already calculated
curve
"Delta
1.
the
the
Curve"
important because
ideal
at
net
how far the
shows
is from reaching the potential
difference
is
result from
Chapter
On the
i.
benefits
net
alternative. Delta curves
Alternative A
with
measured
Step 3. The difference between
and the
alternative
call
basin
alternative,
given
the
the
basin,
benef its which
in net
variation
robustness
of
the basin.
of
the given
robustness may
proved below.
representation
of
the
delta
curve of
is :
64
[XA
Figure
3-5
alterna
tive)
curve
s
shown
3-4
alterna
of
I
= NB [X*(E
j]
shows
tive
Figure
(dll
M44ili
I e
the
in
(net
3)]
-
delta
Figure
3-4,
e
[XA
curve
benefits
from Fi gure 3-2
NB
of
j
the
obtained
curve
of
(the assumed
hypothetical
by subtracting
the
hypothetical
ideal
net benefits
the basin).
*C
*0
IEAL E MITS CMR
e
En MUITS W to1
*N
FIGURE 3-5:
Delta Curve of an hypothetical
alternative,
by subtracting
the net
benefits
curve (Figure
3-4) from the ideal
net benefits
curve of the basin
(Figure 32)
It
3-5.
is
interesting to note two characteristics
First,
any point
Second,
the
by
these
corresponding
The
delta
ideal
the
net benefits
net
benefits
two curves
discount
curve
is
meet
rate,
in
curve
of
is not
the
exceeded
for
curve
our
is
in
alternative.
a point. Therefore,
the delta
the basis
curve
in Figure
at that
zero.
study of robustness.
65
Its measure
of differences
in
evaluate the variations
Consider
the
two
this
alternative
A
takes
benefits, which indicates
on
ei.
On
the
other
but
constant even
losses
of
for very
there are big
sensitive
e1
(its
net
potential
different values
alternatives.
The
e 1 . When ei is
to
be built.
losses
of
But
potential
performance depending
is never the best
hand, Alternative B
alternative for any value of
zero),
alternative
to
performance of
sensitive
uncertain variable
the best
net
of
be used
in Figure 3-6.
depicted
indicates
other values,
optimum can
performance
curves
under the
eil, Alternative A is
when el
the
delta
curve for Alternative
from the
delta
curve
are
benefits
of
el,
is never
like
almost
e
2
and
e9 3 . This indicates insensitive performance of Alternative B
411
FIGURE 3-6:
Cit
il
Robustness related to the shape of the
curve. Comparison between
delta
alternative,
a non robust
A,
Alternative
alternative.
B, a robust
and Alternative
66
under
the
e5. Alternative B is more robust
uncertain variable
than alternative A with respect
As
said before,
robustness
to the
shape
measure
this robustness
curves. We
the
of
its delta
were
the
robustness.
However
applied
all
to
piecewise
curve.
this
process
cases,
and
coefficient
of
variation
delta curve.
If
the
delta
is very hor izontal
of the
simpler
basin ),
if
curve
ve ry
is
of net
the
Similarly,
summary
of
has
be the
and can not be
curves
curvat ure
to
may
be
can not be
calculate
g reat
the
that form the
robustness, the
low dispersion in
with
v ariation
low
means
coefficient
<--------->
Low CV
robustness
curvature:
would
the values
has
the delta
is very
robustness,
the delta
h igh dispersion
respect
to the
of variation
low.
in
the
potential
is very high.
of
A
is:
High CV
The CV of
delta
is
of
(which
bene fits
and the
complex
how to
benefits with res pect to the potential
ti Ited
the basin),
is
(which me ans
a Iter native
of net
brief
(CV)
the coefficient
distribution
higher
of
is related
shap e of
the radius of
procedure
al ternative
and
question now is
then the radius
A
the distribut ion
The
because
calculated.
curve
alternat ive
the
ei.
parameter
an
this with
curvature,
the
linear,
of
associated with the
could measure
higher
to the uncertain
the
index
>
-----
values of
of
the
Low Robustness
High Robustness
the delta
alternative.
performance of every candidate
alternative
curve
is
After this
called the
step, the
is defined with two
67
values:
an
expected
value
of net benefits
and a robustness
index.
3.1.4.-
Step 4:
The
Compare among alternatives
performance
evaluated
in
of
Step 3.
the
candidate
Two performance measures
with every alternative:
expected net
benefits
The
objective
all
candidate alternatives and to present
an
easily
of
comparisons
candidate
and
and robustness.
collect these values for
benefits
is
an
is
and
superior but
objective
another. The
case
both
choice. Thus,
a Iternative
robustness
decision
robustness
and final
expected net
are
associated
this
information
in
that facilitate comparison and
a two-objective
benefit s
not have
is
are
was
among alternatives.
We have
net
the present step
understandable form
s e I ec t ion
alternatives
is
matter
we
superior
in
expected
alternative
say that a
another
are
expected
argument to
where
could only
to
robustness
not
problem
when
greater.
If
net benefits,
prefer one
both
only
we do
alternative to
similar when only expected
net benefits
superior but not robustness.
To easi ly visualize robustness
for all
the
projects,
benefits
are
in the horizontal axis
vert ic al.
Each
graph.
Figure
These
curves,
shows
called
net benefits
we consider a graph where expected net
alternative is
3-7
and expected
the
Pareto
and robustness
represented
t ypical
form
as
of
a
is
point
in
the
in
the
these graphs.
curves or Pareto frontier, are
68
two-objective evaluation methods.
in
common
The
vertical
robustness
Increase
to
N -
robustness by
or
like 4
axis
5).
If
CV
as we move
(where
instead
of
robustness
directly
CV,
vertical
axis.
in
is
the
representing
as
this kind of
axis,
we represent
fixed number,
N -
CV,
we represent
we
go
down
to an Alternative A,
when
curves.
formed by
robustness
from the final
B
has
than A.
evaluation
in the
al though valid,
candidate alternatives
is inferior
inferior to others. An Alternative B
which are not
order for
In
a "large"
increases
The Pareto frontier is
and
N
up
This creates an unconventional,
representation of
benef its
robustness.
represents
smaller
not
expected net
B could
Alternative
and does
both
enter
in
be omitted
the Pareto
front ier.
wS$
( - CI
UKCtD
FIGURE 3-7:
Typical form of
curves
the Pareto
K
ITS
IE (4ellan)
frontier
69
inferior alternatives,
Elimination of
direct
inspection
which fall
should
inside
the
of
of
robustness
a l so
may
requirements
Mi nimum
be
yield
expected
be
may
i Ind icates
to
to the me an value.
equal
an
twice
the
value
of
c an be
if
CV= 2,
the
the
mean.
This
CV=2
is
minimum
inde x of robustness
CV=i
robustness.
low
d ecide
cases,
very
which
smaller than
delta
in t he values of the delta
political
the delta curve is 1. In
of
dispersion
minimum
rnatives
the
acceptable v al ue. Howev er,
is
To
is h arder. An
I n some
and
than z ero are viable.
imposed.
that th e CV of
of
a
(greater o r
th e standard devi at ion
other words,
benefits
i
of view, alt
greater
politically
fo r robustness
requirements
equal
benefits
inferior and
Developme t
mi nimum require me nts
However, other
zero)
net
ar e
benefits
decision. From the analysts I poi nt
done from
alternatives
net
expected
ne t
be
those
curve
requi red.
expecte d
for
curve:
f ron tier
the
elimin a ted.
be
Pareto
can
values
is
cons idered
st andard dev iation
indicates
curve,
great
and
consequently
almost always an
unacceptable
value.
Figure 3-8
case,
with
illustrate th is
minimum required
concepts
net
benefits
for
of
an hypothetical
NBI
and minimum
requi red robustness of CV 1
3.1.5.The
this
Final
remarks
two-objective
stage.
The
outcome
analysis
is
not
of
one
the
single
water
best
system ends at
alternative,
70
LW IOMTEUS
(I
ALTTITD
~-
CY
~aLmygT
Iffia
IIU
ALWTEMI
......-
3-cl1
LM wMn KT 102M hLTUMYI1
IEECTD ET 1ITS (sliRs)
Derivation
FIGURE 3-8:
although
it
could
formed for one
There not
it exists,
there
benefits:
to
is
a
choose
necessary to
trade-off
a
give up
a final
limitation of
global
more
the
making
process,
purpose of
some net
at
the
method
is
such an
same
time.
robustness
it
But
and net
may be
and vice versa. The
could
that
limited
it
to
be
regarded as
is
an
help
number
alternatives with enough data for comparison
to general
includes
the greatest
alternative
benefits,
I think
a
the
between
best alternative
revealing
The method applies
curve
always exists
robust
the method. But
because
frontier
it should have
and expected net benefits
normally
lack of
If
the Pareto
if the non-inferior set only
alternative.
optimum alternative
robustness
be
of
a
advantage,
the decision
of
candidate
and decision.
and complex water resources
71
systems.
The
imposed
by
method does
the
optimization
screening models
A
not
of
practical
Step
have
more
limitations
technique
used
than
the
solve the
to
1.
summary of
the method
is
indicated below.
Objective:
To
identify
a
limited
a two-objective
benefits),
set
analysis
and display the
of
non-inferi or
(robustness
information
al ternatives
and
in
in
expected net
a Pareto graph.
Procedure:
STEP I.
Derive
The
the
ideal
ideal
net
net benefits
benefits
evaluating
the performance of
is derived
as
curve
Formulate the
b.)
Perform
the candidate
M
screening model
M,
Mn
one for
c. )
Obtain
the
net
d.)
Obtain
an
curve,
ideal
optimal
Select candidate
candidate
and
one
reference for
alternatives.
set
It
runs by using the
every vector
benefits
of
basin
curve:
decision
e0j
NB
[X* (G8j ))
variables:
X* (ei
)
alternatives
alternatives
of
for the
optimization
screening model,
The
serves as
follows:
a.)
STEP 2.
curve
them will
will
be
chosen
later
for
define
the
Pareto
implementation.
72
a.)
Calculate probability distributions
values
of
each design
Depict
b ar
graphs
the two
identify
of
or
a compatibility -
d.)
Generate
candidate
3.
Assess
of the
most
likely
by
design
decision variables
of
table
the candidate
alternatives
Formulate a model for every alt ernative
b.)
For each
rules
and
values of
combining the
a.)
alternative ,
optimization runs
and
incompatibility table
compatibility
the Performance
candidate
Mn
obtained
distributions,
alternatives
sat isfying the
while
2
variables
Construct
like ly values
M
Step 1
these
c.)
most
STEP
of
three
every des ign decision
the M,
decision variables,
from the optimization runs
b.)
of
in
calculate
perform
optimize
to
order
M2
Mn
operation
pos sible
maximum
the
MI
net
benefits
c.)
Obtain the net benefits
d.)
Calculate
E
[NB]
curves:
values
e)
[XA
NB
for
each
candidate
alternative:
[NB
E
= E
(A))
NB
e
[XA
j
e.)
Obtain
[[XA
f.)
the
I
Calculate
robustness
delta
=
j]
CV
of
3)]
p(E
1
1
curves
[X* (E
NB
(5)
each
)]
NB
-
values.
candidate
[XA
e j
These
measure
the
alternative
73
STEP 4.
Compare among alternatives
a. )
Plot
b.)
Eliminate
c.)
Set minimum
an
Expected
Net
Benefits
versus N -
CV
(b)
graph
and
d.)
inferior alternatives
requirements for
expected
net benefits
robustness
Present
the Pareto non-inferior
set of
alternatives
CASE STUDY
CHAPTER 4:
The
4.
Chapter
case
This
chapter
correspond to
the Rio
the
enough
to
to
not
coefficie nts
and
parameters
in A rgentina,
exposed
case study
of the
the
correspond to a real
applicability
easily
demonstrate
method described in
of the
Colorado basin
ge neral
show the
also small
of
is
does
[1979] . The size
and Lenton
enough to
study
most
However,
basin.
this
viability and usefulness
practical
Major
of
purpose
of
illustrate
is
in
large
the method, but
each
step
of
the
method.
The
models
optimization
and
alternatives
to
optimize
is
linear
simplification
function
and
expressed as
limitation of
of
1979].
linear
case
of
the
equations.
the
as
but
real
the
computer
is
is
that
the candidate
the
screening model
objective
have
to be
is
not a
simplification
Imposed
by the LP optimization
planning
optimization
run
of
(LP). Therefore, the first
This
world
a common
used to
rules
study
approximation of
by linear equations
The
programming
the method,
programming
Hence,
the
screening
used to solve the
operation
constraints
technique. Most of
linear
algorithm
situations
technique
relationships
use
[Rogers,
and constraints
practice.
the
LP package was
a personal
75
computer IBM AT. Computer time was extensive because,
the
method
does
facilities,
243 runs
not
requires
it requires
for the
many
times
runs
about 231
4.1.-
2 minutes.
of
about
(14 candidate
in 2402
total
runs).
14 minutes;
the candidate
Therefore the
to optimize
each
alternatives
computer
time was
hours.
DESCRIPTION OF THE CASE STUDY
The object of
projects
assume
of
alternative
took
optimize operating rules
took about
runs
for each, results
Each run for the screening model
run to
optimization runs:
and 243
candidate
243
sophisticated computer
independent
screening model,
operating rules of each
alternatives
very
although
are
that
to
this case
built
be
study is
in
a hypothetical
the development of
a regional
development
to decide what hydraulic
the basin
plan,
and
river
is an
we
are
basin.
We
important part
interested
in
getting as much benefit as possible from the river. But, since
there
of
are several
benefit
to
uncertain
be
obtained
concerned with obtaining a
is
insensitive
identify the
trade-off
The
four
variables
to
the
from
the basin
dams,
basin,
we
projects
three
is
amount
are also
strategy which
existing uncertainty. Our task
curve between robustness
possible
the
affect the
robust development
non-inferior set of
scheme of
that
is to
that represents the
and expected net benefits.
shown
possible
in Figure 4-1.
irrigation
There are
areas,
two
possible hydropower plants, and a possible transfer. Any of
76
MI
NY
tOPUEl PlaIT
TUNS
MN C
IniTIcE C
II1TIIA I'
MMD
IllQTIMN 0
FIGURE 4-1:
these projects
Scheme of
could be
the basin
built in the basin. Most of the data
for these projects has been taken from the
real
projects
in Chapters
in the
9 and
Rio Colorado
characteristics of
in Argentina, as described
10 of Major and Lenton
[1979].
Before formulating a screening model
we
have
to
define
the
considers
that benefits
repeated
throughout
and
the
assumed at the heads
of
main river.
enter
Inflows
typical
year.
costs for
project
lifetime.
in the model
as
the basin,
The screening model
this
the three upstream
for
typical
River
year are
inflows
tributaries
average
of
are
the
Inflows for
77
of
each season
the
typical
year.
There
are
no
tributary
inflows nor outflows in the river course. A downstream minimum
flow requirement exi sts
total
provide water
to downstream users
ecological reasons. The minimum downstream flow is
and for
X of
to
rive r upstream
the total
yearly
4-2 shows the
inflows. Figure
inflows of the river
in
40
its upstream branches.
INFL0 f3 : 2toll/&
INFLOW fi : a6 3/s
INFLOW ft : 4 3/1
FTAM FLOW
:
0 1 if1 f4 jf4
FIGURE 4-2:
river
Total
for
Inflows
3
the
typI ca I
year
The
Season
II,
I,
typical
seasonal
for
irrigation
and
zero
III,
(high flow season).
Figure 4-3
shows
ye arly
are medium
in Season
Season
and Season
of
distribution
(medium flow season);
flow season);
(low
from May to August
December
seasons. The three seasons are
to April
from January
September to
t hree
year has
III,
inflows.
in Season
as shown
The
demands
I, maximum
in Figure
from
the
for water
in Season II,
4-4.
78
PEICET Of
THETOTAL
IULV
SWSO 3
SM 2
SLSON i
Im
ASMAL DISimT
(il
uTLW
or TOTAL
Seasonal
distribution
of
total
inflows for the typical year
FIGURE 4-3:
river
0356
4150
I
sag i
EASM I
SASa3I
IflIGTIM VATER DVuADS (i3/Ha)
FIGURE 4-4:
Water
demands
typicai year
For simplicity, the
have fixed
two
Irrigation
hydropower plants
heads. Otherwise, we would obtain
in the screening model
(power is
of
and
head
for
and
flow),
to
the
are assumed to
a non-linear term
proportional
linearize
in
to
the product
it would be very time
79
consuming.
without
since
In this
case study, hydropower plants
requiring
the
dam
irrigation
the
is
construction
not
areas
do
needed
the
irrigation
case
non-consumed
losses
the
area without
water
to
to groundwater are
returns
the rest of
water
month later).
irrigation
After two
stages:
(before
in
two
seasons,
all
refers
non-consumed water.
to the return
in Season
during
I, and
irrigation
because
there
The
of
the bottom
in Season II.
connects
river with the right upstream
only
one
Water
is
pumps nor
graph
is no irrigation
transfer
direction:
from
assumed to move
some water returns
some water
(between 8
and 12
the seasonal
figure,
the top graph
water during
irrigation
is for non-consumed water
for Season
III
left upstream branch of
the
in Season
branch.
the
and
4-5 shows
No graph exits
the
later);
non-consumed water has
In this
non-consumed
inflows or
consumed by
4 months);
seasons
already returned to the river. Figure
return of
areas return
(between 4 and 8 months
returns
their
storage.
included. The water not
in the next season
of
divert water to
river. No groundwater
in the same season
However,
any water, or could also
that
crops returns to the river in three
immediately
only to
interseasonal
the
head.
construction
storing
study considers
be built
their associated dam,
create
could serve
serve for regulation and
This
to
require
associated dam. The dam
of
can
III.
The
transfer
acts
in
left branch to right branch.
by gravity, without needing elevation
operation costs.
80
PROPORTIOS
SEASU 3
SusO 2
ETM oF Tr VATEUK1 0015 U 11 SUSON I
SusM I
PR7OPorl(1
SEASON i
SE0N 3
SEASO 2
t
URN OFTHEVATER 90 CONSMED 11 SEASON
FIGURE 4-5:
4.2.-
Return of
the
Irrigation
consumed by crops
that describe
in the
to
be
particular case
mixed
of this
(10
integer
variables,
objective
below. Appendix A
summary of
model
is
mathematical
during the project
programming
and 24
(whose
function,
contains the
variables.
equations
and economic relationships existing
year
The
model
with
In the
model
34
is
is
0
or
1).
an
decision
operational variables)
value
and
that is
lifetime.
case study, the screening
project sizes
variables
formed by
identified for a typical
repeated
linear-integer
variables
is
the physical
basin. These are
assumed
the
non-
SCREEHING MODEL
The screening model
10
water
and
Decision
are discussed
constraints
screening model
formulation and
for
this screening
framework
taken from Chapter 5 of Major
and Lenton
(1979].
81
4.2.1.-
DECISION VARIABLES
There
are
screening
two
model.
variables.
types
The
Design
of
first
decision
projects of the water
variables
decision
type
"des ign"
is
are
variables
resources system.
in
the
There
this
decision
s ize of the
are
10 design
decision variables, one for each project:
VA
VB
VC
VD
AB
AC
AD
CA
CC
T
3
Volume of Reservoir A (Hm )
Volume of Reservoir B (Hm 3 )
Volume of Reservoir C (Hm 3 )
Volume of Reservoir D (Hm 3 )
Area of Irrigation B (Ha)
Area of Irrigation C (Ha)
Area of Irrigation D (Ha)
Capacity of hydropower Plant A
Capacity of hydropower Plant C
: Size of the Transfer (m 3 /s )
:
:
:
:
:
:
:
:
:
The
second
is
type
(MW)
(MW)
"o perational"
de cision variables.
Releases from reservoirs, wate r diverted to
and
water
diverted
to
the
decision variables. There are
12
corresponding
dams
times
to
three
diverted
for
seasons),
and
operation of
from the reservoirs.
operational
9
correspondi ng
(th ree
irrigation
the
water
are not any o perational
If
depends
If
the
operational
24 operation de cision variables,
3 corresponding to
the hydropower plants.
operation
are
re leases from the reservoirs
seasons),
irrigation
transfer. There
built, the
the
transfer
irrigation areas,
to
(four
the water
areas times three
diverted
to the
deci sion variable for
the
dam associated with the plant
the
plant
the
dam
directly
decision variables
on
i s given
is
the
not
f low
Is
by the releases
built,
in
the
the river.
plant
The
are:
RA, t : Releases from Reservoir A
(t
= season i,2, or 3)
82
RB, t : Releases from Reservoir B (t =
RCt : Releases from Reservoir C (t =
RD,t : Releases from Reservoir D (t =
IB,t : Water diverted to Irrigation B
ICt : Water diverted to Irrigation B
ID,t : Water diverted to Irrigation B
Tt
: Water diverted to the Transfer
In this case
design decision
study,
we
variables.
are
season 1, 2, or 3)
season 1,2,or 3)
season i,2,or 3)
(t = season 1,2,
(t = season 1,2,
(t = season 1,2,
(t = season 1,2,
only
concerned
Design decision variables
what projects
are to be built and what projects are
built. Also,
for those
variables
4.2.2.-
projects
indicate their optimal
3)
3)
3)
3)
about the
indicate
not to be
to be built, design decision
sizes.
OBJECTIVE FUNCTION
objective
The
efficiency,
or
typical
in
function
Is
other words,
of
benef its
from
the
year. Benefits
at
evaluated
come
selling
electricity
evaluated
at
mathematical
equation for benefits
= E [E 1 LS, t]
s = B,C,D
t = 1,2
+
E
s
t
minus
market
of
economic
costs for the
agricultural products
and
prices,
its
[e2
measure
a
their
B
or
or
or
or
from the produced
price.
The
general
is:
Es, t
= A,C
= 1,2,3
where:
Annual benefits ($)
Price of agricultural products ($/Ha)
Land irrigated at Irrigation s in season t (Ha)
Price of electricity ($/MWh)
Power produced at Plant s in season t (MWh)
B
P1S
Ls, t
02
Ps, t
The
c oefficients
prices)
are
assume
a
ei
(crop
prices)
and 0 2
(electricity
considered uncertain. Instead of a fixed value, we
probability
distribut ion
of their possible values
83
(Figure
4-6
and Figure
the reasons
4-7).
4.2.4. briefly
Section
explain
to consider e1 and e 2 as uncertain parameters.
Probability
dC
N%
3,1
N
3/
K
I0
30
AGIICLTUAL PlCES ($/R )
FIGURE 4-6:
Assumed probability
distribution
of
el:
crop prices
ribability
I
3'
is
rEmil"T FiCEs (w1&h
FIGURE 4-7:
Costs
projects,
for
the
are
because
typical
Assumed probability
electricity prices
only
incurred
from
no operating costs
year
result
are
from
distribution of e2:.
construction
of
considered. The
the
the
costs
amortization
of
construction costs over
discount rate (3.
C = CC
the
project
lifetime
with
a given
The formula is:
(I + 93)n E 3 ) /
[(i +
e 3 )n -
Ij
where:
C
: Annual costs (s)
CC : Total construction costs
e3 : Discount rate (. )
($)
: Projects lifetime (years)
n
lifetime
the discount
if
For example,
coefficient
03
uncertain. Figure 4-8 shows
Section 4.2.4.
sts for the typical year are 0.101
construction costs.
times the total
The
the co
50 years,
is
10x and the projects
is 9 3
rate
(discount
considered
distribution, and
its probability
briefly justi fies
is
rate)
the uncertainty in (3.
0W-0.4 \
501
/
Probability
/
/
II
a
costs
fixed
cost
variable
distribution
Assumed probability
discount rate
FIGURE 4-8:
The
12
P
DISC09ET lATE (I'J
III
of
construction of
term
any project have
independent
cost term dependent on the
on
the
size
of (3:
two terms:
project size, and
of
the
a
project. For
85
example,
the
cost of
construction of
(fixed machinery, personnel,
that
depend
volume
of
on
how
concrete,
includes four
graphs
every facility.
dams,
Figure
the
amount
that
A-2 for the
etc.)
dam
of
is
has fixed costs
and variable costs
(volume of
labor,
excavation,
etc.).
Appendix
shows
construction
irrigation areas,
costs for the
Figure A-3 for the
and Figure A-4 for the transfer.
the numerical
TABLE 4-1:
B
indicate the construction costs for
Figure A-i
hydropower plants,
summarizes
big
offices,
a dam
Table 4-1
coefficients.
Fixed
and
variable
possible
projects
to
costs
for
the
be
built
in
the
basin
FIXED (10
PROJECT
DamA
Dam B
Dam C
Dam D
Irrigation B
Irrigation C
Irrigation D
Plant A
Plant C
Transfer
/(m 3 /s)
FVA:
FVB:
FVC:
FVD:
FAB:
FAC:
FAD:
FCA:
FCC:
FT :
$)
I.50
i -50
VARIABLE
aA:
aB:
aC:
aD:
QB:
SC:
PD:
TA:
TC:
3
2
0
0
.00
.00
.25
.50
1 .00
1 .00
1 .50
1 .50
mathematical
The general
6
P :
0.0044
0. 0176
0. 0056
0. 0104
0. 000082
0. 000247
0. 000536
0. 050
0. 077
0. 140
((106 $/Hm3)
(106
(106
(106
(106
(106
(106
(106
(106
(16
$/Hm3 )
$/Hm 3 )
$/Hm 3 )
$/Ha)
$/Ha)
$/Ha)
$/MW)
$/MW)
equation for construction
costs
(CC) is:
CC
=
E4
+
s5
[ E (Vs as + FV 5
S = A, B, C, D
E (Cs
= A,C
TS
+ FCs
YVS)
YCS) +
+ E (As fs + FAs YAs) +
s = B, C, D
(T p + FT YT) I
where:
e4
: General
increase or decrease
in
construction costs
(X)
86
Vo Iume of Reservoir s (s = A,B,C,D) (Hm3)
Variabi e costs of Reservoir s ($/Hm3 )
Fixed c osts of Reservoir s ($)
Integer variabl e for Reservoir s:
Vs
as
FVS
YVs
= i
YVs = 0
As
Area of Irrigat ion s (s : BC,D) (Ha )
s ($/Ha )
:Variable costs of Irrigation
13s
FAs : Fixed costs of Irrigation s ($)
YAs : Integer variabl e for Irrigation s:
If Irrigation s is built: YAs = i
If Irrigation s is not built: YAs = 0
C5s
Capacity of hyd ropower Plant s (s = A, C)
Ts
:Variable costs of Plant s ($/MW)
FCs : Fixed costs of Plant s ($)
YCs : Integer variable for Plant s:
If Plant s is built: YCs = i
If Plant s is not built: YCs = 0
T
: Size of the Transfer (m 3 /s)
p
: Variable costs of Transfer [$/(m 3 /s)]
FT
: Fixed costs of Transfer ($)
YT
: Integer variable for Transfer:
If Transfer is built: YTs = I
If Transfer is not built: YT, = 0
If
Reservoir s
is built:
YVs
If Reservoir s is not built:
The variable
FAS,
r s IFCs,
and fixed
and
P
FT
(MW)
costs coefficients
are
found
Table
in
coefficient 94 (general increase
or decrease
costs)
Figure 4-9 shows
is
considered uncertain.
probability
distribution,
After
can be
and Section
defining benefits and
costs, the
FVs,
4-1.
QSs
The
in construction
its assumed
discusses
it.
objective function
expressed as:
Max
(decision variables]
where B
costs
4.2.4.
as,
are the
for
4.2.3.The
B - C
benefits for
a typical
a typical
year and
C are the
year.
CONSTRAINTS
screening model
of this
case study
includes seven
87
Probability
-25
FIGURE 4-9:
415 1
YAIATIO II CMSTlICilm 0STS (1)
I
Assumed probability
distribution
increase
general
construction costs
sets of constraints
(1979),
Chapter
Instead
of
or
5, this section briefly discusses the special
characteristics of
the constraints
refers to Major and Lenton
[19793
for
for this
case study, and
details.
(1) Continuity constraints. Continuity constraints
conservation of mass in the reservoirs:
in a
reservoir must
lost through
three
be
stored
evaporation
of
in
for
continuity
equations
are
necessary
requirements. The
between
to
it,
insure
that enters
all water
released, diverted,
subsurface
for every dam, one
equations
equations
-
in
and Lenton
Major
repeating
of e 4 :
decrease
There are
leakage.
per season. There
dams.
Three
or
are
also
additional
indicate the minimum downstream
explicit mathematical
notation
is:
for Dam A:
88
SA,t
SA,t+i
-
RAt
+ fit
-
eAt SA,t
where:
SA, t
RAt
fit
eAt
-
:
:
:
:
Storage in Reservoir A in season t (t = 1, 2, 3)
Releases from Reservoir A in season t
Inflows of the left branch of the river in season t
evaporation coefficient of Reservoir A in season t
continuity between Dam A and Dam B:
IBt
= RAt
-
TRt
where:
IBt
TRt
-
: Inflows in Reservoir B in season t
: Flow of the Transfer in season t
for Dam B:
SB,t +
SB,t+i
IBt
-
RBt
-
DBt
-
eBt SB,t
where:
SB,t
RBt
DBt
eBt
-
:
:
:
:
Storage in Reservoir B in season t (t = 1,2,3)
Releases from Reservoir B in season t
Water diverted to Irrigation B in season t
evaporation coefficient of Reservoir B in season t
for Dam C:
SC,t+1
SC,t + f3t + TRt -
RCt
-
DCt
-
eCt SC,t
where:
SC,t
f3t
RCt
DCt
eCt
-
:
:
:
:
:
Storage in Reservoir C in season t (t = 1,2,3)
Inflows of the right branch of the river in season t
Releases from Reservoir C in season t
Water diverted to Irrigation C in season t
evaporation coefficient of Reservoir C in season t
continuity between dams B and C and Dam D:
IDt = RBt + RCt + f2t + RIBt + RICt
where:
IDt
: Inflows
in Reservoir D
f2t
:
of
RIBt
Inflows
the
center
in season t
branch of
: Water that return to the river
Irrigation B
RICt : Water that return to the river
the
river
in
in
season t from
in
season t from
season
t
89
C
Irrigation
- for Dam D:
SD,t +
SD, t+ I
IDt
-
RDt -
DDt
-
eDt SD,t
where:
SD, t : Storage in Reservoir D in season t (t = 1,2,3)
: Releases from Reservoir D in season t
RDt
: Water diverted to Irrigation D in season t
DDt
: evaporation coefficient of Reservoir D in season t
eDt
-
for downstream requirements
RDt
+
RIDt
0.4
(40 X of
(fit + f2t
+
all
river
inflows):
f3t)
where:
Water that return to
Irrigation D
:
RIDt
The values of
= 1,2,3)
values
(t=
can
of
1, 2, 3)
(2)
reservoir
reservoir.
the
inflow parameters fit, f2t,
be calculated from Figure 4-2
the evaporation
are
indicated
Reservoir
in
are
mathematic formulation
SA, t
! VA
SB, t
SCt
SD, t
! VB
VC
! VD
coefficients
in Table
maximum
any season
There
in season t from
the river
can
three
and Figure 4-3.
eAt,
eBt,
eCt,
(t
The
and eDt
4-2.
storage.
not
and f3t
exceed
The
the
storage
volume
equations per dam. The
of
of
the
the
explicit
is:
VD are
where VA, VB, VC,
of
represent
the
volume
above)
decision variables that
the design
the reservoirs (see Section 4.2.1.
90
(3)
Water requirements for
show the
irrigation. These
relationships between
constraints
water divested for
irrigation
and
TABLE 4-2:
Project
Parameters needed for the screening model
lifetime:
n
50
years
reservoirs:
Evaporat ion from the
SEASON i
est
SEASON 2
SEASON 3
DAM A
IX
DAM B
2X
5%
3%
5X
9%
3%
7%
1o%
3%
DAM C
DAM D
Irrigation channel
losses:
Consumptive use of
the
Es
4%
6%
toy
IRRIGATION B
IRRIGATION C
IRRIGATION D
65%Xof
1%
water in
irrigation:
irrigation water
is
consumed
Coeff ic ients of hydropower plants
Head of Plant A:
Head of Plant C:
Efficiency:
Load factor:
Factor of utilization:
the
land
are
also
irrigated. Water
irrigated
losses
considered. There are
irrigation
area.
surfaces
100 m
50 m
0.65
0.86
0.68
six
in
the
equations,
Six
more
equations
in
each
period
can
irrigation project areas. The mathematical
- water requirements for
(I -
EB)
DBt =
5
"t
irrigation
two
indicate
not
channels
for each
that
the
exceed
the
formulation
is:
irrigation
LBt
91
(1 - EC)
(1 - ED)
DCt
=
DDt
=
E 5 Ft LCt
e 5 Ft LDt
where:
EB
EC
ED
(5
: water losses in Irrigation B channels
:water losses in Irrigation C channels
: water losses in Irrigation D channels
: general increase or decrease in irrigation water
demands
Ft
: irrigation water demands in season t (t
1,2)
LBt : land irrigated in Irrigation B in season t
LCt : land irrigated in Irrigation C in season t
LDt : land irrigated in Irrigation D in season t
- maximum irrigation per season:
LBt
LCt
LDt
1 AB
1 AC
S AD
where AB,
AC,
represent
the
4.2.1. above)
AD
are
surface
The values of
Table
4-2.
indicated
decrease
(4)
the
parameters
in irrigation water demands)
its
Ft
(t
(general
indicated
1,2,3)
=
in
are
increase or
is considered uncertain.
assumed probability
distribution, that
in Section 4.2.4.
Irrigation
water diverted for
river of
of
in Figure 4-4. The parameter e 5
discussed
are
the parameters EB, EC, ED
The values
Figure 4-10 shows
is
the
design
decision
variables that
or the irrigation areas (see Section
the non
water
return.
irrigation
with
This
the
constraint relates
return
consumed water. The mathematical
to
the main
formulation
is:
RIBt
lt [(i1
RICt
RIDt
lt E(1
Ilt [(i
-
-
EB)
EC)
ED)
DBt + EB
DCt + EC
DBtI
DCtl
DDt + ED
DDt]
92
Probability
33 1
33 1
33 '
I
~-
N
/
/
/
a~~
-501
FIGURE 4-10:
I
eve
4
VAIIATlS II IRIQTiO§ VATER DAMiDS (1)
Assumed probability
distribution of 95:
general increase or decrease in irrigation
water demands
where:
Olt : water non consumed in season 1 (1 = 1,2)
to the river in season t
water consumed by crops
The values of
4-5. The
the
value
(t
parameters Qlt can
of the
parameter 0
be
flow
with
equation for
indicates
not
each
is indicated
that
exceed
-
for
the electricity
the
representation
electricity
season
capacity
of
return
calculated from Figure
(5) Hydropower constraints. The first
relates
that
1, 2, 3)
set
production.
each
plant.
production
the
plants.
in Table
of
constraints
There
The
4-2.
is
one
second set
in each season can
The mathematical
is:
electricity production:
PAt
PCt
= (2.73 i0-6) RAt HA SA Rt
= (2.73 10-6) RCt HC sC kt
93
where:
power produced at Plant A
power produced at Plant C
PA, t
PC, t
HA
HC
head
season t
season t
(t
1,2,3)
Plant A
head of Plant C
efficiency of Plant A
efficiency of Plant C
number of seconds in season t
sA
sC
Rt
-
of
in
in
maximum electricity
(
PAt
PBt
ht
ht
If
per season:
u CA
If U CC
where:
ht
: number of
If
u
: load factor
: factor of utilization
hours
in
season
t
= 1, 2, 3)
(t
and CA and CC are the design decision var iables
the capacity of
the
hydropower
plants
(see
above)
The values
are
indicated
of the
in Table
HC,
parameters HA,
transferred
transfer.
TRt
in
sA, sC,
If,
and u
4-2.
(6) Transfer size . These constraints
water
that represent
Section 4.2.1.
each
season
In a general mathematical
to
limit
the
the amount of
capacity of the
form:
1 T
where:
TRt
: Flow of the Transfer
in season t
and T is the design decision variable that
of the Transfer (see Section 4.2.1. above)
(7)
Conditionality
constraints
indicate that
corresponding
dam
has
and
to
also
maximum
build
to
be
(t
represent the size
sizes.
an
= 1,2,3)
Conditionality
irrigation
built.
area the
The maximum size
94
if a project
constraints ensure that
are
included.
ten
are
There
project. The mathematical
built
is
of these
the fixed costs
equations,
one for each
is:
representation
irrigation-dam conditionality:
-
AB - AMAXB YVB
0
AC - AMAXC YVC 1 0
AD - AMAXD YVD
0
-
maximum size constraints:
VA - VMAXA YVA 1 0
0
VB - VMAXB YVB
VC - VMAXC YVC 5 0
VD - VMAXD YVD 1 0
AB - AMAXB YAB 5 0
-
AMAXC YAC
AD -
AMAXD YAD
AC
0
0
CA - CMAXA YCA
CC - CMAXC YCC 5 0
T - TMAX YT ! 0
1 0
The
coefficients
AMAXC, AMAXD,
for the
VMAXA,
VMAXB,
VMAXC,
VMAXD,
AMAXBI
and TMAX represent maximum sizes
CMAXA, CMAXC,
projects. These values
can be obtained from Figures B-
i through Figure B-4.
4.2.4.In
UNCERTAIN PARAMETERS
this
study
case
uncertain parameters:
(3) discount
water demands.
constraints.
rate,
we
consider the existence of
(1) crop prices,
(2)
electricity prices,
(5)
irrigation
the
irrigation
four uncertain parameters
affect the
(4) construction costs,
and
Irrigation water demands affect
The
other
objective function. This section briefly discusses
for the uncertainty
five
the reasons
in these parameters.
95
Crop
from the
prices
and
projects. Market
agricultural
products
crop productions,
prices
4-6
electricity prices
are
and
shows
the
common for
depending on climatological
conditions,
international
estimated
prices
are
normally tend to rise.
subsidized when
incentive
used
for
electricity cost
are
very
uncertain and affected
Electricity
an
fluctuations
affect the benefits
In
exports. Crop
for complex factors.
distribution
more
as
imports and
stable
rural
of
than
areas,
crop
crop
agricultural
an uncertain
prices
and
may be
irrigation
development.
variable
prices.
electricity
energy for pumping and
Figure
This
as
makes
difficult to estimate
(Figure 4-7).
The meaning and
stressed
in
other parts
discount rate
is
construction,
years.
the
money
this thesis. We
interest rate
that
has
variable. Figure
objective function.
those
of
to
benefits
in
have
We have
costs,
all
are reduced.
real
of
a
assumed
Projects
costs
could assume
that
money borrowed for
returned yearly during 50
also
projects
situations.
construction
been
is
strongly affected
distribution
study.
construction. Therefore,
construction
factor
of the
has
4-8 shows the assumed
discount rates for the case
Construction costs
the
of
of the discount rate
Therefore, the objective function
for this
of
importance
if
that
there
are more
costs
Most of
a
direct
are
the
effect
on the
only costs are
is an
increase
in
expensive and net
a
very uncertain
the elements that define
project
(labor
costs,
row
96
materials,
fuel,
etc)
are
subject to
uncertainty. The
probability distribution for construction
figure
water
demands
are
uncertainty. First, unexpected water
irrigation
greater
channels
amount
requirements.
and
of
water
Second,
differ from
the
be
with
planted
factors
create
demanded for
forecasts.
great
irrigation,
an
This
will
the
in
of
soil
great
in
happen
require
a
irrigation
requirements
and
are
not
plants may
Third, different crops may
water
requirements.
uncertainty
as
to
could
satisfy
in the
can be seen
here that the
important aspect
People of
that area
of
want,
benefit as
they can.
effect
uncertainty.
of
"guarantee"
that
they can
projects.
projects
built
shown
All
these
amount of water
in Figure
4-10.
APPLICATION OF THE METHOD TO THE CASE STUDY
We have supposed
bad,
to
conditio ns
different
a
losses
irrigation
initial
subject
installat ions.
perfectly defined. Physical
is
is
4-9.
Irrigation
4.3.-
costs
assumed
Our
job
projects to
what
course,
They
if
to
as
want
be built,
a rural
obtain as
area.
much net
robust
projects
that
uncertain conditions happen to
analysts
Figure
projects
the river
equally concerned with the
expect satisfactory
indicated in
and
of
development of
But they are
even
still
the
development of
is
4-1,
should
we have
performance from the
to decide, from all
what
be
projects
the
should be
not be built. Also for the
to decide
their sizes.
These
97
have
projects
to provide
acceptable net
benefits
and also the
robustness that people want.
The method described
their robustness
evaluates
in
the
the basin,
expected net
4.3.1.-
Step
e =
in deciding how
to obtain more
follows describes
is solved using the method. The description
i: Derive the
the vector
how
is
indicated in Chapter 4.
net benefits
ideal
There are five uncertain
Therefore,
compared with
give up
the method. What
the step by step process
in
involved
the parties
benefits. The subsequent decision making process
study
the case
and
benefits. After
net
consists
people willing to
beyond the scope of
based
among
process
The comparison
others.
suited to solve this
each candidate alternative may be
much robustness are
is
expected
and
a decision making process
in
that,
is
a few candidate alternatives
method identifies
The
case.
in Chapter 4
parameters
in
curve
this case
study.
e is formed by five elements:
e 1 , 0 2, 0 3 , E 4 , E51
where:
agricultural products prices ($/Ha)
electricity prices ($/MWh)
discount rate (X.)
E3
)4 = construction costs (Z increase)
e 5 = irrigation water demands (X increase)
e
=
E2
Figures 4-6
of
the
method,
through 4-10
uncertain
we
divide each
have
showed
parameters.
to
divide
probability curve
the probability distribution
To
them
in
use
these
in discrete
three
curves
in the
intervals. We
intervals.
Table
4-3
98
summarizes the values
for
of
the
intervals and their probabilities
every uncertain parameter.
TABLE 4-3:
Intervals
in
which
the
continuous
probability distribution
of the uncertain
parameters have been divided
VARIABLE
ist
INTERVAL
Value Prob,
Discount Rate
Constr. Costs
Irrig. Demands
82
-252
-50Z
301
502
331
fox
even
even
50Z
351
332
122
+25Z
+502
Agric, Prices
Electr. Prices
30
10
301
251
40
20
401
401
50
30
2nd
INTERVAL
Value Prob.
Now, for example,
rate,
e33
has
= 12%.
three values
The
same
parameters. The
243
elements.
To
associated:
be
e
is
clarify
e,
2
3
said
UNITS
202
152
332
1
301
351
$/a
$/Mh
2 Increase
I Increase
parameter
03
1
G 3 , discount
=6,
8
for
the
3
2
meaning
one of
,9 4 1,
3
5
].
its
=
ox,
and
other uncertain
therefore formed by:
the
consider
[e 1 1 ,e2 2 ,0
element
uncertain
can
vector
parameters vector
the
the
3rd
INTERVAL
Value Prob.
of
3-3-3-3-3
the
=
uncertain
elements, for example
This
element
indicates
uncertain conditions defined by (according to Table 4-3):
G)
02
)3
E4
E5
From
table
element
the
= agricultural products prices = 30 $/Ha
= electricity prices = 20 $/MWh
= discount rate = 10%
= construction costs = 25X decrease
= irrigation water demands = 50% increase
4-3
we
[9 1 1 ,e 2 2,0
uncertain
reasonable
3
can
also
2, 4 1 , e 5 3 ] . We
parameters,
assumption.
The
what
obtain the
assume
in
independence between
this
probability
probability of the
of
case seems
[011,(2
2
a very
32941,
0 5 3 ) is:
99
p
[e 1 1 ,e2 2 e3 2 E
= 0.50
0.50
3
1
4
,0
5
0.33
]
0.30
After doing this for each
p(e
0.40
0.010
the
shows
B
Appendix
p(E
)
=
2
3
) p(e 4
1
3
) p(e
5
)
1.000 %
243
of the
elements
ha ve defined all the uncertain
of occurrence.
probability
their
with
situations
2
the
of
one
parameters vector, we
uncertain
2
p(e 1 i)
probabili ty
Table B-1
each
of
occurrence
of
in
uncertain element.
Ideal
net
benefits curve
We have
defined 243 possible
correspondent
possibilities
benefits
curve
indicates
obtained
from
the
future
of
occurrence.
basin
for
each
we
curve
screening model.
[E1 1 , E 2 2, E 3 2
1
4
For
water demands
are multiplied
build.
possible
the
net
and
situation,
The
are
ideal
by
t hat
possi ble
the
ideal net
of
to
screening
products
1.50.
solve
the
model
for
prices
discount rate
by
can be
0.75,
and
are 30
is
10%,
irrigation
The computer
gives us
that
cot Ild
be obtained for this
the
optimal
si zes
of
process
has
al so
situations.
screening model
The
I he
multiplied
benefits
same
future
in
net
the 2 43
package
LP
prices are 20 $/MWh,
cos ts
maximum
point
agricultural
construction
the
the
example,
, 953],
electricity
$/Ha,
utilize
one of
e ir
th
ideal
The
the maxi mum net benefits
future situations. To calculate one
benefits
situations with
net
have
to
In total,
to be
benefits
De
the
projects
done for the
243
to
other 242
optimization runs of
performed.
curve
can
not be graphically
100
represented,
Table
because
4-4 shows
curve for the
E4 1 ,0 5 3 ],
of
1.03
Optimal
is
defined
study. We
ideal
that
indicates
curve
net benefits
[
for
space.
1,
22,E32,
net benefits
million $.
sets
In
of design decision variables
this
study
case
there
variables. Then, the vector X
= {XI,
X
of the
can see
net benefits
ideal
a 5-dimensions
in
the numerical values
case
the
it
X2 ,
X3 ,
X4 , X 5 ,
are
design
ten
decision
is formed by:
X 6,
X7,
X 8,
X9,
x
10
1
where:
= Volume of Reservoir A (Hm 3 )
3
X2 = Volume of Reservoir B (Hm )
3
X3 =Volume of Reservoir C (Hm )
3
X4 =Volume of Reservoir D (Hm )
Area of Irrigation B (Ha)
X5
X6 =Area of Irrigation C (Ha)
Area of Irrigation D (Ha)
X=
Capacity of Plant A (MWh)
X8
X9 = Capacity of Plant C (MW)
X0 =Size of the Transfer (m 3 /s)
X,
From the
ideal
net benefits
to obtain
runs performed
243 optimization
243
curve, we also obtained
the
set s of
optimal
generically represented them by:
decision variables. We
X* [(]
To
the
illustrate
particular value X*[e
optimal
set
run
optimization
particular for
we
obtained
of
meaning
1
1
of
X* [E],
,e 2 2 ,E 3 2 ,E 4 1,E
decision
performed
for
[ei 1 , (
obtained
Xj
2
2
,(
the
represents the
]. This
variables
decision variable
the optimal value:
3
5
consider
lets
2
3
from
e4 1 ,E
3
5
].
the
In
(volume of Reservoir A),
Xj*[G1t,922,932,E
i E531
101
TABLE 4-4:
Ideal net benefits
curve
WATER REQUIREMENTS = -50 1 : WATER
REQUIREMENTS = even
DISCOUNT RATE:
(million
$)
i WATER REQUIREMENTS = +50 %
|EI.Pr.=10 EI.Pr.=20 EI.Pr.=30!E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25 1 !Agr.Pr.=50
3.67
5.19
6.71
4.01
5.51
7.01
4.49
5.99
7.50
1.50
2.20
2.91
1.80
2.52
3.26
2.26
3.00
3.75
0.80
1.27
1.73
1.14
1.57
2.03
1.63
2.01
2.51
CONSTR.:Agr.Pr.=30!
COSTS= !Agr.Pr.=40
even |Agr.Pr.=50
3.30
4.79
6.31
3.60
5.09
6.61
4.02
5.51
7.00
1.20
1.90
2.60
1.50
2.20
2.90
1.80
2.53
3.27
0.50
0.97
1.43
0.98
1.27
1.73
1.47
1.69
2.03
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
425 1 |Agr.Pr.=50
3.00
4.40
5.92
3.30
4.70
6.21
3.60
5.03 1
6.52 .
0.90
1.60
2.30
1.20
1.90
2.60
1.64 |
2.20
2.90
0.34
0.67
1.13
0.82
1.04
1.43
1.30
1.53
1.75
------- :---------- ----------------------------- :-----------------------------|-----------------------------
REQUIREMENTS = even
WATER
REQUIREMENTS = -50 1 | WATER
DISCOUNT RATE:
: WATER REQUIREMENTS = +50 1
10 1
E1.Pr.=10 EI.Pr.=20 EI.Pr.=30|E.Pr.=10 EI.Pr.=20 EI.Pr.=30|E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.|Agr.Pr.=30
COSTS= Agr.Pr.=40
-25 X Agr.Pr.=50
3.39
4.91
6.43
3.69
5.20
6.72
4.16
5.64:
7.15 |
1.29
1.99
2.69
1.59
2.29
2.99
1.92 |
2.67
3.41 |
0.59
1.05
1.52
1.03
1.35
1.82
1.51
1.74
2.17
CONSTR.!Agr.Pr.=30
COSTS= !Agr.Pr.=40:
even Agr.Pr.=50!
3.02
4.42
5.94
3.32
4.72
6.23
3.62 :
5.06 :
6.54 |
0.92
1.62
2.32
1.22
1.92
2.62
1.64 1
2.21 :
2.91
0.35
0.68
1.15
0.83
1.05
1.45
1.26
1.53
1.76
------- :---------- !-----------------------------|-----------------------------|-----------------------------
CONSTR.|Agr.Pr.=30:
COSTS= !Agr.Pr.=40
+25 1 |Aqr.Pr.=50
2.65
4.05
5.45
2.95
4.35
5.75
3.24
4.65 !
6.05:
WATER
REQUIREMENTS = -50
DISCOUNT RATE:
12 1
0.55
1.25
1.95
0.96
1.55
2.25
1.44 |
1.84 1
2.54:
0.20
0.42
0.78
0.63
0.85
1.08
1.11
1.33
1.55
z : WATER
REQUIREMENTS = even I WATER REQUIREMENTS = +50 1
IE1.Pr.=10 EI.Pr.=20 E1.Pr.=30!E1.Pr.=10 EI.Pr.=20 E1.Pr.=30|E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.:Agr.Pr.=30:
COSTS= !Agr.Pr.=40!
-25 1 Agr.Pr.=50
3.18
4.63
6.15
3.48
4.93
6.45
3.82 1
5.31 '
6.81 |
1.08
1.78
2.48
1.38
2.08
2.78
1.73
2.38
3.08 '
0.43
0.85
1.31
0.92
1.14
1.61
1.40
1.62
1.91
CONSTR.|Agr.Pr.=30|
COSTS= |Agr.Pr.=40!
even 1Agr.Pr.=50!
2.74
4.14
5.57
3.04
4.44
5.86
3.34
4.74
6.16 1
0.64
1.34
2.04
1.01
1.64
2.34
1.49
1.94 |
2.64 I
0.24
0.46
0.87
0.68
0.90
1.17
1.16
1.38
1.60
CONSTR.IAgr.Pr.=30|
COSTS= |Aqr.Pr.=40I
2.30
3.70
2.60
4.00
2.90 |
4.30 !
0.39
0.90
0.77
1.20
1.26 |
1.59
0.08
0.28
0.47
0.66
0.95
1.14
+25 % |Agr.Pr.=50I
5.10
5.40
5.70 |
1.60
1.90
2.20
0.50
0.88
1.37
102
0 Hm 3 .
Besides
this
optimal value
E 5 3 ), there are other 242 optimal
to
the
other
242
possible
Appendix B shows the
The
same
variables.
B-3
243
[0 1 1 ,
for
values of
values of
2
2
,e3 2 641
X1 , corresponding
future situations.
optimal
Table B-2
in
X 1.
could be said for the other nine design decision
There
are 243
through B-il
show the
variables X 2
of X 1
optimal values
optimal
for each one.
values
through X 1 0 . These tables
for
Tables
design decision
are the basis
for next
step.
4.3.2.-
Step 2:
We have
Select candidate alternatives
243 values
for
each
design
When w e consider one decision vari ab le,
get
th e probability
looR in g at
Hm 3
table B-2,
an d 0 Hm
3
of
t o calculate
X1 .
And the
dec is i on
distri
the
of
the
these
h av e the
With thi s,
be
Xj,
value S.
can
we
For
we
probabil ity
have
all
probability d istribution
same can
variables.
butions
B-1 we
to occur.
0f
for example
Xi,
o nl y two different values:
there are
. In table
optima I value
need
distribution
decision variable.
said
Table
4- 5
values
of
for the
shows
the
ten
of
84
e ach
the data we
the val ues
of
other nine des ign
the
probabil ity
desig n
decis ion
variables.
The
probability distributions
also be
plotted. Figures B-I
representation of
provide
a
in Table
4-5 can
through B-10 show the bar graphs
the probability
quick means
indicated
distributions. These graphs
for identifying the most
likely values
103
TABLE 4-5:
Probability distribution
of
the
of the design decision variables
VOLUME DAN A
0
81
I
88.501
0
0.001
11.501
9
100.001
PROB. N
0
8
100.00!
VOLUME DAN B
99.281
0.121
100.00'
AREA IRRIGATION B
FROD. I
---------
0
81
3/t
FROe.
MV
100.001
Jo3
TRANSFER SIZE
CAPACITY PLANT A
PROB.
Hm3
values
Ha
-
12.411
87.591
FROB.
0
16550
16150
27880
100.001
9
-----------
0.911
83.111
11.501
3.881
100.001
VOLUM
DAM
C
3
AREA
IRRIGATION C
FROB.
O
105
I
to. 581
59.421
---100.001
Ha
FROB.
0
18410
19660
10.581
48.161
10.661
---------
CAPACITY PLANT A
0
6
to
100.001
------ ----
-
VOLUE DAN D
0
21
122
123
100.001
-
-----
61.141
38.111
0.T21
---------
--
-
-
AREA IRRIGATION D
PROB. I
1a3
-----
PROB.
mw
68.45
10.661
3.501
11.0
------
00.00
Ha
PROB. I
0
1550
1650
0
206TO
20110
20818
68.15:
1.431
9oo
o
8.851
0.38
14.441
2.651:
3.501
104
of the
B-7,
design decision variables.
Consider for example Figure
for the design decision variable
D. The
values of
(2) around
is formed
X 7 are
1600 Ha, and
cluster
is
formed
the second
1550,
the
of
1650,
of
20670,
The weighted
instead
of
by the
the
is
1640
considering
cluster.
cluster obtaining
The
three
20710 Ha
clustering
more
likely
shows
the reference values of
built.
Table
likely to be built,
Analyzing
4-7
B-2
to
this
cluster,
values,
we
only
done for t he third
value.
identify
the t
(with their respe
variable.
decision
zero means
su mmarizes
each value).
t he refere n ce value
as
decision
their s izes,
Tables
in
close
values
clustering procedure. A siz e of
not
Then,
1640
the
zero. For
included
serves
probabilities) for each des ign
The
the three values
very
refere nce
20840 Ha.
obtained
the reference
proced ure
third
can be
be analogously
as
The
is obviously
Ha.
t hree
can
It
and
probability of
consider the representative value
for
Ha.
representative value
(weighted
average
(1) 0 Ha,
second cluster
1720
20770,
first cluster
cluster, the
cluster
groups:
0 Ha. The
and
by obtaining the weighted average of
in the
irrigation
(3) around 20700 Ha. The first cluster
by values
reference value of
area of
clustered in three
exclusively by the value
is formed by values of
=
X7
ive
ct
afte r the
the proje ct is
the projects that
are m ore
and their probability.
through
B-I,
we
can
infer
s ome
re Iationships among projects. Consider, for example, Dam A.
is easy to see, by comparing
or
Tabl e 4-6
variables
that
NO
Table
B-2
with Table
B-3,
It
that
105
TABLE 4-6:
Probability
distribution
of
the
reference values
of
the
design decision
variables after clustering
VOLUME DAM A
Hs3
CAPACITY PLANT A
PROD.
I
mV
--------- -----------
84
0
0.001
100.001
9
100.001
100.001
------------
PROD.I
3/1
----
0
99.281
8
0.721
100.00'
----------------------------------------------------------------------------------------
YOLUME DAM
B
AREA
IRRIGATION I
PROD. I
033
---------
0
84
Ha
-
PROD. I
-----------------------
1?.41
0
87.591
16510
27880
VOLUME DAM C
------------
:
---------------------a--
88.501
11.50
0
PROD. 1
TRANSFER SIZE
0.9
95.211
3.881
AREA IRRIGATION C
3
PROB
0
105
40.581
59.421
-------
PROD. I
Ha
0
40.581
8680
59.121
-------
VOLUME DAN D
"3
PROD. I
-
a
0
6
61.141
38.141
0!2
- - - -
.0
100.001
---------------------------------------
AREA IRRIGATION D
Ha
PROD.
P
a
21
123
MV PROa.I
----------------------------------------------------------------------------------------
100.0010
---------
CAPACITY PLANT A
10.661
20.60
2010
100.001
:0o.00
160
10.66
20.60
a
a
a
ata
enii10g6
a
TABLE
PROBABILITY
100.00
95.21
%
87.59
x
built.
is
Also
we
project with
Table
X
%
%
%
%
%
X
Dam
can
Dam A:
f or Dam
B-2
when Dam A
Dam D,
%
(size different that
when
Therefore,
X
built
is
B
that
they
are
mutually
facil
Study
We
Irrigation C
7)
that
built.
built.
between
see
are
ities
now the
We
times
Dam
A
and
to
MW
Ha
Hm 3
Hm 3
1640 Ha
3
21 Hm
27880 Ha
MW
10
8 m 3 /s
zero),
A
Dam B
Dam B
is
is
is never
never built.
an
incompat ibl e
e x c Ius i ve.
the t ables ,
Compa ring
we see
that
are always bu i lt:
and P I ant
A.
profi t able
This m eans
than
Da m A,
a l ready built be fore
be
relationship
betwe en
Dam A and
from the ir respective Table s (B-2 an d B-
Dam
concl ude
c an
D,
more
ha ve
some times Dam A is
Othe r
6
20710
123
84
oth er facilities
Irr igat ion
MW
Ha
Hm 3
Ha
3
105 Hm
Dam
say
that these f our facil.ities
building Dam A.
built,
then
I our
built,
the se
is
A with the rest of
Irri gation B,
because
9
16570
84
18680
Plant A
Irrigation B
Dam B
Irrigation C
Dam C
Plant C
Irrigation D
Dam D
Dam A
Irrigation D
Dam D
Irrigation B
Plant C
Transfer
59.42 x
59.42 X
when Dam A
SIZE
PROJECT
%
38.14
20.60
20.60
11.50
10.66
10.66
3.88
0.72
0.72
to be built
more likely
Projects
4-7:
bu ilt
and
is
built
A
that
Irrigation
there
C:
Irrigation
C
is
and
Irrigation C is
is
no
no
one
unique
also
not
relation
is more or less
107
than
profitable
other.
the
called
the Compati bilities table.
table
shows
Irrigat
at
D,
and
that are above
that have to be
Dam D)
Irri gation D can
and
Dam
D
projects
(for example,
built together or not
built without building
not be
not
can
it
be built
without
also
building
D.
for de velopment
create candidate alternatives
To
plants,
strategies.
But,
combinations
among
(Table 4-8)
number of
when
we
candidate
have the
we
have
decided
alternative,
we
in this
projects
must decide
reference sizes for the
Figure
is very reduc ed.
what
their
projects.
that the
the compatibility
graph
or the hierarchical
possible combinations
planning
con straint
projects have to respect
14 possible combinations are allowed
Once
the
impose
the
can be made
result in d ifferent
will
that
of
irrigation
(dams,
Many combinat ions
and transfer).
projects
these
projects
possible
the
combine
we
basin,
table
(Figure
form
"hierarchical"
are
There
built.
all:
Irr igation
with
also
ion D
areas,
information
same
The
a project, those projects
to build
be
have to
Dam
incompatible
Top project s are more profitable than bottom ones. When
deci de
built
in
project, this
the
projects,
related projects.
graphicall y represented
be
4-11).
we
the non
given
For any
profitable
more
the
and
project s,
can
we obtain Table 4-8,
other des ign decision variables,
the
all
for
comparison process made with Dam A
Repeating the same
It
the
4-11),
In fact,
only
case study.
are
part
of
a
sizes. We already
is reasonable
to
108
------
table
Compatibilities
TABLE 4-8:
facilities that )AY or
Facilities that
Additional facilities that
:
KATNOT be built
are NEVER built
are ALWAYS built:
......................................
Initial
facility:
PLANT A
TIT
IRRIG B
DAN A
DAN
a
DAM C 111I C PLANT C
DAN D IRRIG D
DAN A
PLANT A
TRI
1 1IG B
DANE
DADC IRIGC PANT C
DAM D IRRIG D
-------------
----------
------------
----------
---------------
TRI DAN A
DAM B
PLANT A
DAN C
IRRIG B
IRRIG C
PLANT C
DAN D 11IG D
-------------
----------
------------
------------------------------
TIRFDAN A
DAN B
DAM C
PLANT A
D
DAM
IRIG B
IIIG C PLANT C
IRRIG D
----------------
------------
------------------------------
TEI DAN A
DAM B
PLANT A
I1IG B
DAN C IRIG C PLANT C
DAN D I116 D
----------------------
--------------------------------------------
IIG B
1IG C
.
.
.
TRI DAN A
PLANT A
DAM I
PANT C
DAN C
DAMD I1IG D
-------------------------
----------
------------------------------
TRI DAN A
DAN B
DAN C IRRIG C PLANT C
PLANT A
IRIG B
IRRIG D
DAN D
-
-
-
-
-
-
--
-
---
-----
- -
-
- -
- -
-
-
- -
-
- -
- -
-
- -
- -
-
-
-
-
-
-
-
-
-
-
-
-
-
DAN A
-
-
TRF
DAN E IRIG B
PLANT A
DAN C IRIG C PLANT C
DAN D 11IG D
PLANT A
DAM A
DAM
I IUIGB
PLANT C
TRY
DAN C IRIs C
DAN D 1IG D
-------------------------------------------------------------
PLANT A
TEF
PLANT C
DAM A
DAN IRIG
DAN C IHIG C
D IMIG D
DAM
t019
FIGURE 4-11:
Hierarchical
representation
relationships among projects
think that the sizes
alternative
project,
reference
for
a
most
likely
sizes
has
with similar
project.
a
main
In
included
of
in the candidate
the projects. For each
reference
probability. In some cases
si ngle
project
the projects
are the ref erence sizes
the
greatest
of
size
two
or
is
reference
case
size
study,
that
probabilities of
of
the
main
of
summarize s the reference size for
Now
projects,
we
have
three
most
identified
14
and for each project we
reference
identi f ied
however,
stands
have
probability
with
even
Secondary reference sizes
the
the one
probabilities can be
this
of
clea rly.
at least
size.
each
Table
half
4-9
every project.
possible
have
combinations
identified
of
its
110
Reference sizes
TABLE 4-9:
for
the projects
SIZE
PROJECT
Dam A
105
123
16570
18680
20710
A
Plant
9
6
Plant C
Transfer
reference
size.
alternatives,
alternatives
of
are
4.3.3.-
this
have
described
satisfy
the
identified
in
Compatibility
Assess the performance of
Step 3:
former step
step we have
of net
we
to asses
conditions
have
the
are
possible future situations. To
under
alternative
independently
evaluate
under each one of
Lets
%
20.60
100.00
38.14
0.72
X
X
%
%
X
X
%
4-10.
These
by combinations
table,
and each
14 candidate
performance
alternatives.
of
every
one of
in the basin.
benefits
Uncertain
candidate
95.21
59.42
X
the alternatives
them under the uncertain conditions existing
Curves
%
14 candidate
Table
are formed
11.50
87.59
59.42
20.60
its reference size.
From the
In
are
we
candidate because
projects that
project has
8
Therefore
that
Hm 3
Hm 3
Hm 3
Hm 3
Ha
Ha
Ha
MW
MW
m 3 /s
84
84
Dam B
Dam C
Dam D
Irrigation B
C
Irrigation
D
Irrigation
PROBABILITY
study
the
the
the
approximated
assess
by defining 243
performance
the
of a
uncertain conditions we have to
performance
of
the
alternative
243 possible future situations.
performance of Alternative A. We consider
111
TABLE 4-10:
ALT ERNOI1VE A:
Plant A:
Candidate alternatives
9 MW
ALTERNATIVE 1:
Plant A:
9 MW
Irrigation B: 16,570 Ha
ALTERNATIVE F:
Plant 1:
irrigation B:
Dan F:
ALTERNAI VE C:
Plant A:
Plant C:
9 MW
Dam A:
Plant C:
84 Hmr3
6 MW
16,5/0 Ha
84 Hm31
9 MW
6 MW
ALTERNiiVE D:
9 1W
Irrigation B: 16,570 Ha
Dam A:
84 Hm
ALTERNATIVE J:
9
Plant A:
Irrigation B: 16,570
Dam A:
84
Irrigation C: 18,6
Dam C:
105
MW
Ha
Hm3
Ha
HM3
Plant A:
ALTERNATIVE E:
91MW
Plant A:
Irrigation P: 16,570 Ha
Dam B:
Plant C:
84 Hm
6! MW
ALTERiJAIVE F:
Plant A:
9 MW
w
Irrigation B: 16,570 Ha
Dan F:
84 Hm3
Irrigation C: 18 ,680 Ha
De C:
105
ALTERNAT IVE G:
Plant A:
9
irrigatian B: 16, 57')
DaR :
84
Irrigation D: 20,710
Dam D:
123
MW
Ha
Hm
3
Ha
ALTERNATIVE H:
Plant A:
MW
w
Irrigation P: 16,570
Dam B:
84
irrigation C: 18,680 Ha
DamC:
105
Irrigation D: 20,710 Ha
Hm3
Dan D:
123
ALIERNATIVE K*:
9 MW
Plant A:
Irrigation B: 16,570 Ha
84 HW3
Datq A:
Irrigation D: 20,710 Ha
123 H 3
Dam D:
ALTERNATIVE L:
9 MW
Plant A:
Irrigation B: 16,570 Ha
84 H 3
Dam A:
Irrigation C: 18,680 Ha
Dan C:
105 Hm3
Irrigation D: 20,710 Ha
123 HW3
Darn0:
ALTERNATIVE M:
Plant A:
9 MW
Irrigation B: 16,570 Ha
Dam B:
84 HI3
Plant C:
6 f1W
Irrigation D: 20,710 Ha
123 Hm3
Dam D:
ALTERNATIVE N:
Plant A:
9 MW
Irrigation B: 16,570 Ha
84 Hm3
Dam A:
6 MW
Plant C:
Irrigation D: 20,710 Ha
Dan, D:
123 HW3
i 12
the
future
calculate
benefits
[0 1 1 ,022,032 E)41
we obtain
the optimal
yield maximum net
Alternative A would
the
evaluate
2
use
We
,E3 2 ,4IE
and
the LP
3
5
].
To
of
9E4
1
and
policies that
A.
Alternative
benefits
package,
irrigation
are not
We
but only about the
operating rules,
net
[e1 1 t62 2 ,3
conditions
].
for
benefits
maximum
of
3
5
releases
about the
really concern
To
that
2
[ 1 1 ,e 2 2 ,E 3 2 , e 4 1 ,e 5 3 ] does
situation defined by
conditions
value
2
optimize operating rules for Alternative A under the
we
come,
[E 1 1 ,0
by
maximum net
the
if the
produce
defined
situation
Alternative
A
under
9(5 3 1.
of Alternative A for all
performance
the
possible future situations, we must run the LP program a total
of
243
future situation E9i.
maximum net
benefits
A.
for Alternative
Alternatives
B
same
M.
through
the
Alternatives
optimization runs
the
C-2
Tables
under
was
repeated for
to
be
to
C-14 show their
uncertain situations.
of net benefits of
to C-14 are called curves
C-i
it
has
process
benefits
These Tables
In total,
indicates the
Appendix C
in
resulting from the 243 optimization runs
The
respective maximum net
candidate
C-1
Table
for each possible
rules
optimize operating
to
times,
A to N.
necessary
to
perform
to calculate the curves
14'243
of net
=
3402
benefits for
14 candidate alternatives.
Expected value
The
of
expected
net
benefits
value
of
for
net
the
alternatives
benefits
for
a candidate
113
are the probabilities
The weights
given
is
also required
is
to
net benefits
of
first stage
of
the
to
calculate both we need again
conditions
uncertain
compute
The
expected
va lue
shown
alternative
in
is
of
use
Chapter
equal
4,
to
f ar
how
potential
probabilities
the
B-1.
the
the
last
The
of
the
net
the
of
the
to
Index
values
(note
To
of the
is to
stage
of
the
stan dard
is just the
value
ex pected
alway s has
delta
ideal
the delta curve
of
values
var iation
the
by
The
delta c urv e of
in dica t es
coefficient of variation
divid ed
deviation
As
Ta bI e
in
the coef ficient of
delta curve
is
between the curve
reaching the
t he
deviation of
standard
it
4-4).
(Table
cal Ied
the
But
to c alculate the expected value and
is
second stage
for
al ternative
alternative and the
is
fro m
is
alternative
basin. The
curve
curve
delta
The
alternative.
candidate
the
candid a te
The differen ce
benefits curve.
candidate
calculate the difference
the
benefits
net benefits.
net benefits
ideal
the
benefits.
in Table 4-11.
curve of
its
calculated from
also
net
net
expected
robustness for a given
index of
The
of
curve
only the
alternatives
robustness for the
of
is
of net
the uncertain conditions
g iven
every candidate alternative are
Indices
of
curve
benefits
the
of
The values
Table B-1.
in
of
its
from
net
values
the
of
weighted average
of
value
expected
The
benefits.
calcul ated
directly
is
alternative
that
the
be greater than ze ro).
of
coefficient
Robustness
of
an
of variation of the
114
values
of
of
its
robustness
delta
of
curve. Table
the
TABLE 4-11:
candidate
4-11
alternatives.
EXPECTED NET
4.3.4.-
BENEFITS
NB Curve
2.818
0.000
2.000
0.769
1.231
Alternative
Alternative
Alternative
Alternative
B
C
D
E
1.698
0.710
1.657
1.913
0.858
0.846
0.825
1.057
1.142
1.154
1.175
0.943
Alternative F
2.718
1.343
0.657
Alternative G
Alternative H
Alternative I
Alternative J
Alternative K
Alternative L
Alternative M
Alternative N
2.331
1.722
1.871
2.693
2.292
1.706
2.545
2.507
0.455
0.374
1.008
1.079
0.437
0.381
0.972
0.872
1.545
1.626
0.992
0.921
1.563
1.619
1.028
1.128
Step 4:
Compare
information
where
horizontal
indicated
- CV,
among
alternatives
expected value
in
where
study.
in
and
value
of
robustness
a "large"
4-12 shows
Each
number.
the Pareto
alternative
net
benefits
is
net
is
benefits
in
is
the
and the
to organize
The best method is
Chapter 4, robustness
N is
of
alternative, we have
a clear form
expected
axis
2. Figure
case
CV
(N=2)
0.496
index of robustness for every
graph
N -
INDEX OF
ROBUSTNESS
Alternative A
Once we have the
this
indices
Expected net benefits and index of
robustness of each candidate alternati ve
ALTERNATIVE
Ideal
summarizes the
to use a
in
the
vertical.
As
should be represented by N
In
this
case,
two-objective
represented
as
N
is
equal
to
graph for our
a point
in
this
graph.
The non-inferior set
is formed by Alternatives H,
K, G,
115
ALYT1YE I
1.2
I
hLTT
ALTIMYJI
E
1.8
ALTED&YIVE
1.4
1.2
TETIVE J
1
ALTETETIET
A
LTIEATIVE I
LTENTIVI F
0.8
ILEYWAIVE D
8
0.4
0.2
I
I
1
2
PECTD IET
FIGURE 4-12:
N,
M,
J,
and
Alternative
provided
by
n E
H,
3
EFNUITS
(9llin$J
Fareto
two-objective
graph
for
the
candidate alternatives of the case study
F.
The
and
the
Alternative
maximum
robustness
maximum
F.
expected
With
respect
values, we can calculate the percent
and expected
net benefits of
of
is
provided by
net
benefits
is
to these extreme
losses
in
robustness
the other alternatives
(Table 4-
12).
Alternatives K and G seem to be
These
alternatives
(only about
15X
in the "compromise"
have relatively high expected net benefits
less expected net
are relatively robust
benefits than
(only about 20X
less
Any other of the non-inferior alternatives
has
significant
decrease
benefits. The final
outside of
zone.
choice
the scope
of
F),
and they
robustness
than H).
(H, M,
N,
J,
or F)
in either robustness or expected net
of
this
an
alternative
thesis,
to
because
implement is
it
depends on
116
TABLE
and in
Decrease in expected net benefits
of the non-Inferior
index of robustness
to the maximum
with respect
alternatives
va ILies
4-12:
DECREASE IN
EXPECTED NET
BENEFITS
NON-INFERIOR
ALTERNATIVE
process.
conclusion
as
However,
that
Alternatives
K and G.
may obtain some conclusions
benefits
we
shows
study the
the
most
expected
(those
Therefore, Alternative
(Plant A,
which
we
expected net
net benefits.
to
likely
be
the
greater
contains
Irrigation B with its
Table
The
built.
projects
of more than 50%
have
F, which
of
value
most
the
with probability
are the alternatives
projects
in Figure 4-12,
and their
project
likely
alternatives which contain
built
being chosen are
and robustness.
First,
4-7
the
study,
about the relationship between the
alternatives
the
of
characteristics
case
benefits and robustness
the Pareto curve
analysis of
From the
the decision-making
the
of
expected net
Conclusions about
4.3.5.-
in
possibility of
have more
alternatives
16.87%
21.75%
133. 50%
160. 29%
188.92X
259. 58X
involved
parties
the
among
IN
OF ROBUSTNESS
--------
36.64%
15.67%/
14.26%
7.77%Y
6.36%.
0.93%
Alternative H
Alternative K
Alternative G
Alternative N
Alternative M
Alternative J
Alternative F-
agreement
DECREASE
INDEX
to be
in Table 4-7)
net benefits.
the most profitable
associated Dam B,
and
117
C
Irrigation
with
the
formed
g reatest
less
by
For
benefits.
projects
t han
bui It
expected
Plant
C
than
the
the
E w ith
C,
the
in
condition S
formed by more
by
seven
c as e,
of
some
expected
projects
than
net
benefits
comparing Alternative
should expec t that those
projects. are the most
m itigate
the
p rojects.
In fact,
alternative
of
effect
(Alternat ives H
in an
discussed at
exception.
possibilit ies
mor e
the most
for this
reason
We
are
that
the
robust. But,
and
of
varying
unfavorable
two alternatives
and L,
cons istent relationship
included
built
Irrigatio n D. However
interrelated
have
project s
is no
the
robustness.
projects) are
there
be
alternati ve s containing
greater
Irrigation
may be
to
to
D,
they
rule s
operation
interesting
An
of
study
becaus e
smaller
I with Altern ative K. The
alternatives formed by more
robust,
has
G
o r Alternative
chapter,
Se cond, we
same
pr obability of
F.
likely
ry happens, a s can be seen by
of
the
I rr igation C by
lower
altern atives containing
G,
by
substitut in g Plant C by
T he r ef ore,
have
to
formed
Al ternative
is less
4 -7 ).
Table
expected net
A Iterna tive
than
on D
circumstances
end
a
happ en s when
Irrigati
D
Alternative
s pe c ial
the
Irrigation
rule
smaller
subs tituting
than
s ame
contra
is
D has
ar e supposed
Plant
G
Irrigation
(s ee
C
bu t
h ave
Since
benefits
Other alternatives
benefits.
Alternati ve
F,
is the alternative
C),
Dam
proj ects
profitable
this
exception to
net
Alternative
ne t
Irrigation
expected
example,
Irrigation D.
being
associated
its
with
eac h one formed
except in
this
between the number
its
robustness. For
118
hiie rarchical
From the
Dam A and Dam B
are
the
s ame
in
present
is
with Dam
D,
I,
N,
Table
This
4-7).
smaller
have
explains
that
expected
net
include
upst ream
the
of
conf igurat ion
the
Dam B .
of
for
This
for
used
regulati on
poss ibi I it y
conf ers
grea t e r
There
is
composition of
that
include
a
of
of
and
A
Plant
Irrigation
B,
can
to the
robustness
consistent
the alternatives
Irrigation D
on
sever al
alternat
relat
of
uses
G
(see
al ternatives
look i nrg at
by
Dam A is
addition
of
of
also be used
t he Transfer.
of
the water
ives.
ionship
and robu stness:
(with its
and
used for reg ulati on
in
regulati
Dam B
4-1.).
A,
for
F
of alternat ives which
Dam
Dam A for
using
second
be
However,
regulation
of
than
Figure
(see
can only
Dam B
Irrigation B.
water of
be in g
system
the
Al ternatives
M,
Dam B may be explained
of
instead
A
Dam
benefits)
contai ning Dam A
alternatives
benefits
be
a Iternative
than
built
greater robustness
containing Dam B. The
of
E,
B,
to be
likely
less
is
respectively. Dam A
an
case
the
is
This
that
exp ected net
less
Alternatives
over
K
and
J,
having
with Dam B.
than
A
found
We
alternative.
that
not
c an
and
proj ect s
see
we
4-11),
(Figure
incompatibl e
( although
robust
more
gra p h
Dam B by
replace
which
alternatives
first
The
robustness.
and
alternatives
to the
refers
relationship
Dam A.
the
of
composition
the
between
consistent relationships
however, two
are,
There
projects.
six
by
each formed
and M,
N
than Alternatives
robust
more
is
project,
one
only by
formed
A,
Alternative
example,
between the
alternatives
associated Dam D)
are more
119
robust
than
example,
sa me
the
Alternative B
Alternative
D
etc.
project
built
the
an
to
be
increase
alternative
using
the
extra
downstream
river
and
the
this
C uriously,
that
water
because
flow
is
is
no t
D),
exist ence of
clear
conc lusion
relationship
always
and
of
characteristi cs
net
of
very
explanation
excess
otherwise
will
this
be
use
lost
section of
Then,
alternatives
in
over the
from Irr igations
efficient
D
last chance of
in
benefits
b etween
of
B and C
although
the water
with
Irrigation
than without
Irrigation
provides
wit h
the
the
possibility of
individual
as a whole.
that, while there
configuration
net benefits,
respect
of
is
the
expected
seem to be more
syster
th e
in
likely
any condition.
Irrigation D)
the
a
is
a
G,
than
Irrigation
tributary.
this section
their
exits
D
present
ext ra water
relat ionship
alternatives
that
middle
Irrigat ion D under
The
including
F
not
poss ible
the water
the reas on why some
small er expected
using
is
make
D have
the
when
is
D
For
D.
Alternative
Alte rnative
The
return flows
the
does
4-7),
I rrigation
than
K,
Irrigation
(that
th e
of
robust
Irrigation
(see Table
ro bustness
is
without
Alternative
Extra wat er
Irrigation D
(and
in
less
requirem ents)
downstream.
the
is
tha n
Alternative H,
of
alternative
to
their
of
is a
the
no such clear
robustness. The
projects
important than the
(as
Dam A
and
configuration
120
CONCLUSION
CHAPTER 5:
the
summarizes
in
characteristics
of
The
first part
thesis.
this
The
the method within
genera I framework of water resources planning.
the
S umar! A
the method
of
This
reason for,
The
uncertain
p rocess
to maximize
is
water
resources
the
is
existence of
planning water systems.
of
th e concept
problems,
determi nistic
have
the
in
for water
to maximize robustness.
is
robustness
considering
parameters
ional
tradit
obj ective
other
the
method
obj ectives
the
(th e
benefits
and
objective)
of
One
planning.
net
expected
two-objec tive
thesis presented a
resources
In
the
parts.
described
planning method
second par t discusses
two
in
divided
is
chapter
This
robustness does
of
not
any meaning.
The
thesis.
of
consideration
Robustness
is
of
performance
correspond
maintain
a
measure
a
a proj e ct
performance
relia ble
Ch
apter
2,
of
to robust
performance,
I
t he
orig ina I
is
sensit iv ity
to uncer ta in conditions .
that the un certain parameters
In
robustness
projects,
t hat
independently
in
of
this
the
Inse ns itive
are
able to
of the value
happen to take.
reviewed
some
indices
to
measure
121
robust
ness
exposed
and
d)
do
in
to use
robustness
may
be
benefits
a
correspond
this
thesis.
in the
index.
the
of
consequence
b]
Fiering
to
index
value of
net
obtain
the
is
and
the
of
the
of
possible
net
depending
the
hand,
on
the
robust
project
a
project
but we
of
parameters.
indicates
input
of
the
as
can not
project,
the
is
benefits
If
net
this
non-robust
is a wide range of
dispersion
indicates
of
it
c,
robustness
distribution
uncertain
variation on
small
from a
possible
b,
robustness
uncertainty, we
benefits
there
indices
I developed a new
robustness
disperse,
project:
of
quantify
analysis,
of
The
[1982a,
concept
to
of the
very
performance
performance
and
two-objective
a function
distribution
systems.
Therefore,
This
to
performance of
other
water
uncertainty. Because of
able
distribution
of
distribution
single
as
[1982a,
not
it
to
related
predict
resiliency
by Hashimoto
indicated
and
and
project
parameters.
of
the
because
the
values
performance,
On
similar even under different
is
conditions.
The
planning
in
this
process
to
obtain
alternatives
thesis
requires
to
the
is
solve
use
technique.
The
method
facilities
that
could
the
part
the
of
indices
of
screening
begins
be
by
built
the
method
is
a set of
non-inferior
robustness
general
two-objective
performing the four steps described
of
the
of
method
problem.
models
as
displaying
of
the
proposed
The
method
optimization
all
basin.
possible
in
the
Then,
after
in
Chapter 3,
the result
development
strategies
122
for
the
basin,
to
prove
to
a n, hypotheti cal
in
th e
of
interesting
inf
the
the
net
benefits
the
bas in
Cu rve.
wo
lI
d
al tern
at i ve.
benefits
of
projec ts
of
Character istics
planning
as
also
mo St
was
applied
that,
method
a whole
method
shows
in
the
produces
and
on the
evaluates
yield
can
for
never
goal
the
to
the
candidate
the
ideal
profitable
incompatible
after
on
the
net
exceeded for
information
least
insight
maximum
each uncertain
be
of
provides
summary,
In
the
possible
and
met hod
the
time
section
any
within
planning
on
net
the
projects
combinations
method
the
potential
the
of
water
the
and
amount
expenses
discusses
the
of
in
is
method
Different
decision makers.
characterized by
This
the
of
purpose
information to
by
it
4,
is
the
resource
framework
The
and
Chapter
projects.
and
basin
the
The
a good
can have
In
is
the
as
This
cI ose
detected.
are
we
applied,
method
Also co mpa tible and
identified.
are
as
The
proj ects.
individual
the
benefits
The met hod
curve.
of
ideally
any
to be
method,
e xample,
net
is
the
basin
the
These
alternatives
of
curve,
future conditio n.
candidate
graph.
t o p roduce net benefits, represented by
of
benefits
on
For
basin
th e
form
Pareto
ormation
potential
ideal
ity
main ad vantages
proj ects.
individual
Pareto
study.
case
fin ding
of
a
iabil
practical
the
One
process
presented
relevant
to
provide
planning methods
are
information provided
generating
improvements
the
to
information.
the
planning
123
process
that
about,
the
method
described
in
particularly
multiobjective
There
maximizes
is
However,
deal
solution
et
But
the
since
limited
very
some methods
still
linear
(stochastic
of
practical
subject of
are
basin,
of
the
an
and
there
hydrological
in theory,
that may,
optimal
unique
described by
Loucks
these methods
utility of
require
they
if we
situation of
the
obtain
programming,
objective
our estimation of
in
the randomness
and
uncertainty
1981).
al.,
conditions
are
There
itself.
with
uncertainty
economical
the
procedure,
resources systems
is
is also uncertainty due to
process
and
optimization
the real
represents
water
great uncertainty. There
and
wrong with this
nothing
the
physical
bring
uncertainty
solution that
that the model
the
with
resources
were certain
basin.
will
thesis
water
Traditional
only provide one
measure.
dealing
this
analysis.
Uncertainty.
models
in
extraordinary
is
computer
facilities.
The method
to
developed
with the
deal
uncertainty
this
task by performing
Any
available
problems
method
fact
the case
Another
to
study
thesis
issue.
is
optimization runs.
independent
solve
specially suited
The method accomplishes
deterministic
(screening models,
not requires
does
XT personal
many
algorithm
be used
can
this
in
optimization
in particular).
sophisticated computer equipment;
in Chapter 4 was fully solved using
an
The
in
IBM
computer.
advantage
of
the
method
is
that
there
is
no
124
the
limitatoion to
want
the
analysis,
number,
time
is
not
method
The
in
the
when
future
only
it
quantifies
is
the
ability and
between the
basin
the
in
computer
uncertainty
of
our
by
cond itions
in
a
measure
of
thesis
This
calls
projects
to
in
the area
of
n
is
of
sensiti vity
the
predict
surely
not
projects.
the
because
about uncertainty,
can
the
of
uncertainty
of
effect
the
under
planning
for
serves
we
present,
performance
robustness
stud y
relations hi ps
concerned
are
projects. We
uncertainty
actual
not
also
but
uncertainty,
th e
a nd mo re
.
form
mathematical
but
it se If,
exist ing
the
modeling
variables
t o the
the
. Of course, the
reased,
inc
we
in
parameters
applied
De
limi Lation
represent
to
knowledge
is
runs
the method
by
given
still
If
consider.
can
we
uncertain
ot her
necessary. The
is
uncertainty
can
method
optimization
of
of
or
more
include
to
level
uncertainty.
The
in
uncertainty
projects.
this
projects
outcomes
of
the
However,
the
project
example,
pollution)
practical
to
problem of
measure
criterion
use
them.
may
that
can
have
be
considering
Methods
alone
are
is
the
t he most
merits
correct
ed
quantifi
ot her
could
c riterio
evalua te
to
criterion
evaluation
To
Economic
analysis.
Multiobjective
utilized
planning.
resources
water'
re search
the
advances
further
me thod
when
of
the
all
the
in m onetary terms.
non-economic
effects
(for
interest decision maRers.
non-economic
found
in
effects
the
The
is how
literature
to
125
quantify
ecological,
these
methods
recreational
are
general,
other
and
and
are
some
effects.
Some
of
for specific
just
cases.
The
this
thesis
said before,
this
of
method
As
techniques.
net benefits
as
part
alone.
under
the
countries,
issue of
may be critical.
In
subsistence
net
some
produces
that
produce
nothing.
The method
is
to
be
Pareto
projects
given
curve
is
Pareto
or
in
other
words,
in
curve
to
is
easily
the
i s the
dec is ion
that
the
visuali2
resources
projec t
result
makers.
First,
when
regions.
that
to
than
provide
of
The
a
but may also
the
finding
method,
advantage
about
a
set
of
and
it
of
t he
the alternati ve
and under stood.
is much
decision
always
implement
non-inferior
the
the decision-making process
curve.
projects
to
problem by
information
ed
developing
underdevelo ped
preferable
two-objective
Pareto
The
reasons why
the
is
be
performance
implemented
reliable
solves the
or,
curve
projects.
benefits
to be
to
element
may produce muc h more net benefits,
project
Pareto
a
of
and
consider
criteria
project
of water
people
the
implemen t
to
cases,
those
the
areas
rural
robustness
to
robustness
for decision making
process
projects may be
These
are
important
in
In
conditions.
of
an
reliability
indicates
future
a mean
is
Robustness
it
the multiobjective
studies
net benefits
in the decision making
considered
because
thesis
of
a two-objective problem. We
of
that neither robustness nor
used
part
is
There
are
two
improved with
making
involves a
12 6
collective
the
probaba I it y
solu
o
s
sI
this
object ive wate r
the
of
proc eS
decision
guaranteed
of
agreement
is
not
the
the
one
of
thi s
not
exist.
the
tradeoffs
and
the
only
makers
to be
t he
case
planning
resource
best
nego t iat ion.
roum for
negotiati on,
objectives,
the
method
the
This
in
concept
compromise
s in g Ie
traditional
that
present
alternative
does
to
that
is
provide
much
two- objective
methods
like
best alternative
does
of
is
in
increases
not
process
A dec is ion-mak ing
among
in
curve
finding
techn iques
solution.
the
of
optima I
one
However,
thesis,
Pareto
thi
necessary to define
s
case
robustness
and
net benefits.
Secondly,
between
analysts
decision-making
and
group
it
in
the
objectively compared
case
of
non-infer'ior
being
inferior,
improves
strongly supports
the
Pareto
with other
it
The
candidate
should
be
communication
a
if
example,
a particular project,
of
performance
graph.
the
For
makers.
decision
the analyst may evaluate
locate
also
the
project and
can
project
projects,
disregarded
in
and
favor
now be
in
the
of
a
project,
127
APPENDIX A
128
FORMULATION OF THE SCREENING MODEL
MAX
BENEFITS
-
DISCOST
Alter parameters
DISCOST - 0.0817 COST =C
COST - u.75
CONSTRUC = 0
IRRIGAT ).f
IRRAREA =(
IRRAREA = 0
IRRBEN - 0. (3
ELECBEN - 0.1 FOWER = -)
Benefits
and costs
BENEFITS - IRRBEN - ELECBEN = 0
CONSTRUC - DAMCOST - IRRCOST - ELECCOST - TRFCOST = 0
DAMCOST - 1.5 YDAMA 1.5 YDAMB - 3r YDAMC - 2 YDAMD - 0.44
VOLDAMA - 1.76 VOLDAMB - 0.56 VOLDAMC - 1.04 VOLDAMD = (
IRRCOST - 0.25 YIRRB - 0.5 YIRRC - YIRRD - 0.082 AREAB - 0.247
AREAD = C)
.56
AREAC
ELECCOST - YFLANTA 1.5 YPLANTC - 5 CAFACA - 7.7 CAPACC = 0
TRFCOST - 1.5 YTRF - 0.14 TRFSIZE = 0
Continuity
9.513 SA2 - 9.418 SA1
SA3
+
+
RELA3
RELB1
+
RELB2 +
SA1
-
9.228
9.418
SEC
SB3
SB1
SC2
SC
C1
SD
=
-
9.037 SB
9.418 SB3
- 9.037 SCi1
- 8.657 SC2
- 9.228 SC3
- 8.847 SD1
.3:2
9.51
SD-
+
+
+
Reservoir
3B1
-
8.562 SD 2
68
9.228
SD
+
+
+
+
+
=
=
-
VOLDAMEL-
=
0I
=~
0
=
0
=
=
=
(
-
S3
C1
-
OLDAMhB
VOLDAMB
-
~VOLDAMC
IRRB1 IRRB2 RELB. - RELA3 +
RELC1 + IRRC1 RELC2 + IRRC2 RELC: - TRF3 =
RELDI + IRRDI -
+ RELD2
+
+
+ RELD3 -
IRRD2 RELB3
-
RELA1 + TRF1 = 0
RELA 2 + TRF 2 =
TRF3 = 0
TRF1 = 6.6
TRF2 = 3.4
10D
IRRTB1
RELBI - RELC1 RELB2 -
RELC2
RELC3
IRRTB3
-
-
IRRTB2 -
-
-
IRRTC3 =
)
=
U
VOLDOMC
(9
VOLDA4MC
0
SD1 - VULDAMD '=
SD2 - VOLDAMD 1=
=
SD7 - ViLDAMD
Irrigaton loses and
-
5.28
2.72
8
volumesE
VLDAMA
VOLDAMA
OVOLDA
5812
SC2
8C3
=A
=
IF:RTD1
IRRTD2 >= 2.72
7 j>=
IFRTD3.
-
A2-
=
9.323 SB1
-
FELL
RELD2
FE L D.
RELA1
RELA2
+
SA3
IRRTC2 =.
9.513 SDI1
+
S2
9.513
9.513
9.513
9.513
9.513
9.513
9.513
9.513
9.51
IRRTC1
water
requirements
for
irrIga.tion
129
129
IRRB1 - LANDB11
2.125 IRRB2 - LANDB2 =)
2. 08 IRRC1 - LANDC1 =(
2. 8 IRRC2 - L-ANDC2 =0
1.912
RFRD1 - LANDD1 =
1.9c92 IRRDZ - LANDD2 =
IRRIGAT - LANDB1
LANDB2 - LANDC1 - LANDC2
LAND1 - AREAB=
LANDB2 - AREA1B
= 0)
LANDC1 - AREAC=
LANDC2 - AREAC
0
LANDD1 - AREAD
= (
LANDD2 - AREAD <=0
Irrigation
return
flow
IRRTB1 162 IRRB1 ).135 IRRB2 = C)
IRRT2 -IRRB1
- 0.03)
0.026 IRRB2 = 0
IRRTB3 ). 184 IRRB1 - 0.214 IRRB2 = 0
IRRTC1 - D.167 IRRC1 - D.140 IRRC2 = 0
2. 12
IRRTC
-
C.C31
IRRC1 -
LANDD1
-
-
LANDD2
C)
=
C.027 IRRC2 = 0
= 0
IRRTC3 - C. 191 IRRC1 - 0.222 IRRC2
IRRTD1 -1
. 178 IRRD1 - 0.149 IRRD2
IRRTD2 - 0.033 IRRD1 - 0.029 IRRD2
IRRTD3 - 0. 203 IRRD1 - 0.237 IRRD2
Hydropower constraints
5.37'61 POWERA1 - RELA1 := 0
5.361 FOWERA2 - RELA2
= C
5.361 F0WERA3 - RELA3 <= C
1(.722 FOWERC1 - RELC1 <= 0
10.'722 FOWERC2 - RELC2 <= C
10. 722 FWERC3 - RELC3 <= C)
0. 059 POWERA1 - CAFACA
0
0. 059 FOWERA2 - CAFACA := C)
C. 059 F0WERA3 - CAFACA (= 0
0.059 FOWERCI - CAFACC
0
.j059 POWERC2 - CAPACC
=0
C). 059 FOWERC3 - CAFACC <= C
POWER - FOWERA1 - FOWERA2 - FOWERA3
= 0j
1 r-ansfers
size
TRFi - TRFSIZE := 0
TRF2 - TRFBIZE <= U
TRFSIZE <= 0
Condi ionalitv
and maxiLLfn
ize
AREAB - SQ YDAMB
AREAC - 75 YDAMC
Y YDAMD <
AREAD 'LDAMA
YDAMY= (9
VOLDAMB - 7 YDAMB
=
VOLDAMC - 10 YDAMC
.=
VOLDAMD () YDAMD <=
REA
- 50 YIFB <:.= 0
AREAC - 75 YIRRC i= (9
= C)
= 0
= 0
-
FOWERC1
-
POWERC2
-
FOWERC3
130
AFEAD
CAPACA
-
9) YI RRD
YF'LANT
CAPCC-
3YP'LANTC
TRFEZE
-
END
INTEGER
INTEGER
INTEGER
INTEGER
INTEGER
INTEGER
INTEGER
INTEGER
INTEGER
INTEGER
YDAMA
YDAMB
YDAMC
YDAMD
YIRRP
YIRRC
'YIFRD
YFLAN'TA
YFLANTC
YTRF
30
(
=
(9
.=0
YTRF
131
SUMMARY OF VARIABLES AND PARAMETERS OF THE SCREENING MODEL
Design decision variables
Volume of Reservoir A (Hm 3 )
Volume of Reservoir B (Hm 3 )
Volume of Reservoir C (Hm 3 )
Volume of Reservoir D (Hm 3 )
Area of Irrigation B (Ha)
Area of Irrigation C (Ha)
Area of Irrigation
D (Ha)
Capacity of hydropower Plant A
Capacity of hydropower Plant C
Size of the Transfer (m3 /s)
VA
VB
VC
VD
AB
AC
AD
CA
CC
T
Operational
RA, t
RB, t
RC, t
RD, t
B, t
IC, t
D, t
Tt
(MW)
(MW)
decision variables
Releases from Reservoir A (t =
Releases from Reservoir B (t =
Releases from Reservoir C (t =
Releases from Reservoir D (t =
Water diverted to Irrigation B
Water diverted to Irrigation C
Water diverted to Irrigation D
Water diverted to the Transfer
season
season
season
season
(t
s
(t
s
(t
s
(t = s
1, 2, or
3)
3)
3)
3)
1, 2,
1, 2,
1, 2,
1, 2,
1, 2, or
1, 2, or
1, 2, or
eason
eason
eason
eason
or
3
or 3)
or 3)
or 3)
Other variables and parameters
B
C
CC
e
E2
E)3
E)4
E)5
Lst
LBt
LCt
LDt
Ps, t
PA, t
PC, t
n
as
Ts
Annual benefits ($)
Annual costs ($)
Total construction costs ($)
Price of agricultural products ($/Ha)
Price of electricity ($/MWh)
Discount rate (X)
General increase or decrease in constr. costs (%.
general increase or decrease in irrigation water
demands
Land irrigated at Irrigation s in season t (Ha)
land irrigated in Irrigation B in season t
land irrigated in Irrigation C in season t
land irrigated in Irrigation D in season t
Power produced at Plant s in season t (MWh)
power produced at Plant A in season t (t = 1,2,3)
power produced at Plant C in season t
Projects lifet ime (years)
Variable costs of Reservoir s ($/Hm3 )
Variable costs of Irrigation s ($/Ha)
Variable costs of Plant s ($/MW)
132
Transfer [$/( M3 /s
R eservoir s ($)
I rrigation s ($)
P lant s ($)
T ransfer ($)
Integer variable for Reservoir s:
If Reservoir s is built: YV, = i
p
:Variable
FVS
FAs
FCS
FT
:
:
:
:
YVS
:
Fixed
Fixed
Fixed
Fixed
costs o f
costs
costs
costs
costs
)]
of
of
of
of
EB
If Reservoir s is not built: YV S= 0
: Integer variable for Irrigation s:
If Irrigation s is built:
YAs = I
If Irrigation s is not built:
YAs =0
: Integer variable for Plant s:
If Plant s is bu ilt:
YCS = 1
If Plant s is no t built: YC, = 0
: Integer variable for Transfer:
If Transfer is b Ui It: YTS = 1
If Transfer is n ot bu i It: YTS = 0
: Storage in Reser vo ir A in season t (t = i 2, 3
Storage in Reservoir B in season t (t = 1 2, 3
Storage in Reservoir C in season t (t = 1 2, 3
Storage in Reservoir D in season t (t = 1 2, 3
Re l ease s from Reservo ir A in season t
Re I ease s from Reservo i r B in season t
Release s from Reservo ir C in season t
Release s from Reservo ir D in season t
Inflows in Reservoir B in season t
Inflows in Reservoir D in season t
Water d iverted to Irr ig at ion B in season t
Water diverted to I rrigati on C in season t
Water diverted
to I rrigati
on D in season t
evaporation coeffic ient of Reservoir A in season
evaporation coeffic ient
of Reservoir B in season
evaporation coeffic ient of Reservoir C in season
evaporation coeffic ient
of
Reservoir D in season
: Inflows of the left branch of the river in
season t
: Inflows of the center branch of the river in
season t
: Inflows of the right branch of the river in
season t
: Water that return to the river in season t from
Irrigation B
: Water that return to the river in season t from
Irrigation C
: Water that return to the river in season t from
Irrigation D
:water losses in Irrigation B channels
EC
:water
YAS
YCS
YT
SA t
SB, t
SC, t
S D, t.
RAt
RBt
RCt
RDt
IBt
IDt
DBt
DCt
DDt
eAt
eBt
eCt
eDt
fit
f2t
f3t
RIBt
RICt
RIDt
ED
t
Qlt
return to
losses
in
Irrigation
C
t
t
t
t
channels
: water losses in Irrigation D channels
:irrigation
water demands in season t (t = 1, 2)
: water non consumed in season I (1 = 1, 2) that
the river
in season t (t = 1, 2, 3)
133
HA
HC
water consumed by
head of Plant A
head of Plant C
SA
SC
kt
ht
efficiency of Plant A
efficiency
of Plant C
number of seconds in season t
number of hours in season t (t
if
u
load factor
factor of utilization
TRt
Flow of
crops
the Transfer
in
season
1,2,3)
t
(t
=
1,2,3)
134
(1FUR
Fixed
(14
20
$)
larlile
(1.6 $A3)
iA 1.5
3 1.5
S.0044
6.6if6
~C
3.S
II
2.
*.w5%
6.61W4
w I
w I
C
low
15W
EEil
FIGURE A-i:
Construction costs of
SIZE (O)
the dams
COST
(1364$)
Tariable
Fired
(186
j(1g6$/83)
IflIGTIM 0
3,
IUIQTI 31 0.25
IBIGTIO C 0.50
Ifl6QTIm D i.0
8.00002
O.000247
l.0536
20
IuI&ATIcm C
10
250M
SO0
(Ba)
InIGTIOU SIZE
FIGURE A-2:
Costruction costs of the irrigations
135
Fired
(l'$
tilable
(16g/fgU
PAIaT A .
PLAWi
C .58
M.g50
8.11
(186 s)
26
MIT C
to
FnAl' A
IN
mno CWCTf
Construction costs
FIGURE A-3:
of
the power plants
(16 u
le
Filed
Tism 150
Turble
.14
36
1
TnAmTE
SilE (*3/i)
FIGURE A-4:
Construct i on
costs
of
the transfer
136
APPENDIX B
137
TABLE
B-1:
Frobabilities of
uncertain
WATER
REQUIREMENTS = -50
DISCOUNT RATE:
8
%
the
243
elements
parameter vector
|
MATER REQUIREMENTS = even
|
of
the
&
WATER
REQUIREMENTS = +50 %
========================================================
E1.Pr.=10 EI.Pr.=20 EI.Pr.=30|E1.Pr.=10 E1.Pr.=20 El.Pr.=30:El.Pr.=10 EI.Pr.=20 E1.Pr.=30
CONSTR. Agr.Pr.=30:
COSTS= Agr.Pr.=40:
-25 1 :Agr.Pr.=50
0.3751
0.5001
0.3751
0.6001
0.8001
0.600%
0.5251:
0.7001
0.5251:
0.3751
0.5001
0.3751
0.6001
0.8001
0.600%
0.5251:
0.7001:
0.5251:
0.3751
0.5001
0.3751
0.600%
0.8001
0.600%
0.5251
0.7001
0.5251
CONSTR.Agr.Pr.=30:
COSTS= Agr.Pr.=40:
even :Aqr.Pr.=50:
0.2621
0.3501
0.2621
0.4201
0.5601
0.420%
0.367%:
0.4901:
0.3671:
0.262!
0.350%
0.2621
0.420%
0.560%
0.4201
0.3671:
0.4901:
0.3671:
0.2621
0.3501
0.2621
0.4201
0.560%
0.4201
0.3671
0.4901
0.3671
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
+25 1 :Agr.Pr.=50:
0.1121
0.150%
0.1121
0.1801
0.2401
0.1801
0.1571:
0.210%
0.1571:
0.1121
0.1501
0.112!
0.180%
0.240%
0.1801
0.157%|
0.2101:
0.157%:
0.1121
0.1502
0.1121
0.1801
0.2401
0.180%
0.1571
0.2101
0.1571
WATER
REQUIREMENTS = -50 1%
DISCOUNT RATE:
=
10
=
=
=
:WATER REQUIREMENTS = +50 %
WATER REQUIREMENTS = even
=
=
=
=
=
=
=
=
:El.Pr.=10 E1.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.220 E1.Pr.=30:E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25 % :Agr.Pr.=50:
0.625%
0.8331
0.6251
1.0001
1.3331
1.0001
0.8751:
1.1671:
0.8751|
0.6251
0.8331
0.6251
1.000%
1.3331
1.000%
0.875%|
1.1671:
0.875%:
0.6251
0.8331
0.6251
1.0001
1.333%
1.000%
0.8751
1.167%
0.875%
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
even :Agr.Pr.=50:
0.4371
0.5831
0.4371
0.7001
0.933%
0.7001
0.612%:
0.8171:
0.6122:
0.4371
0.583%
0.4371
0.700%
0.9331
0.7001
0.612%:
0.8171:
0.6121:
0.4371
0.583
0.4371
0.700!
0.933%
0.7001
0.6121
0.8171
0.6121
0.3001-. 0.2621:
0.4001 0.350%:
0.300% 0.2621:
0.1871
0.250%
0.1871
0.3001
0.4001
0.3001
0.2621
0.3501
0.2621
|----- --
------------
-
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
+25 % :Agr.Pr.=50:
0.187%
0.2501
0.1671
0.300%
0.4001
0.3001
----------------------------------
0.262%:
0.3501:
0.2621:
-
------------------------------------
WATER
REQUIREMENTS = -50 1
DISCOUNT RATE:
0.1871
0.250%
0.1871
- ---------------------------
:WATER
REQUIREMENTS = even : WATER
REQUIREMENTS = +50 %
:EI.Pr.=10 El.Pr.=20 E1.Pr.=30:EI.Pr.=10 El.Pr.=20 EI.Pr.=30 E.Pr.=10 E1.Pr.=20 EI.Pr.=30
CONSTR. :Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25 % :Agr.Pr.=50:
CONSTR. :Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
even
:Agr.Pr.=50:
0.2501
0.3331
0.2501
0.400% 0.3501:
0.5331 0.4671:
0.400% 0.3501:
0.2501
0.3331
0.250%
0.4001
0.5331
0.400%
0.350%:
0.4671:
0.350%:
0.2501
0.3331
0.2501
0.4001
0.533%
0.4001
0.350%
0.4671
0.3501
0.1751
0.2331
0.1751
0.280% 0.245%:
0.373% 0.3271:
0.280% 0.2451:
0.1751
0.2331
0.1751
0.2801
0.373%
0.2801
0.2451:
0.3271:
0.245%:
0.1751
0.2331
0.1751
0.2801
0.3731
0.2801
0.2451
0.3271
0.2451
0.1051
CONSTR. :Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
0.0751
0.120%
0.1051
0.0751
0.1201
0.1051
0.0751
0.120%
0.100%
0.1601
0.1401:
0.100%
0.160%
0.140%:
0.100%
0.160%
0.1401
+25 % :Agr.Pr.=50:
0.0751
0.120%
0.1051
0.075%
0.120%
0.1051
0.0751
0.120%
0.1051
138
TABLE
Optimal sizes
B-2:
of Dam A
REQUIREMENTS = even
REQUIREMENTS 2 -50 1 : MATER
WATER
DISCOUNT RATE:
WATER REQUIREMENTS = +50 1
E1.Pr.=10 E1.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 El.Pr.=30 E1.Pr.=10 EI.Pr.=20 E1.Pr.=30
0.00
0.00
0.00 -- 0.00
0.00
0.84
CONSTR. Agr.Pr.:300.00
0.00
0.00
0.00
0.00
0.84
COSTS= Aqr.Pr.=40
0.00
0.00
0.00
0.00
0.00
0.00
-25C IAgr.Pr.=50
|--------------------------:------------------------------------------0.00
0.00
0.00
0.00
0.00
0.00
25NSTRAgr.Pr.:30:
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR.:Agr.Pr.=30
0.00
0.00
0.00
0.84
0.00
0.84
COSTS= :Agr.Pr.=40:
|Agr.Pr.=50|
0.84
0.84
0.84
CNSTR. Agr.Pr.=30
COSTS= :Agr.Pr.=40:
+251 !Agr.Pr.=50
0.00
0.00
0.84
0.00
0.00
0.84
0.00
0.00
0.84
even
------
--------
|
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
------
-------------------
----------------
-------------
REQUIREMENTS = even
REQUIREMENTS = -50 1 | WATER
WATER
DISCOUNT RATE:
10 1
-
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
------------------0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
: WATER REQUIREMENTS = +50 1
|E1.Pr.=10 El.Pr.=20 EI.Pr.=30:E.Pr.=10 EI.Pr.=20 EL.Pr.=30:E1.Pr.=10 EI.Pr.=20 El.Pr.=30
CONSTR. Agr.Pr.:30COSTS= Agr.Pr.=40:
-25C :Agr.Pr.=50|
0.00
0.84
0.84
0.00
0.84
0.84
0.00 -- 0.00
0.00
0.84
0.00
0.00
0.00
0.00
0.00
0.00 |
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
COSTR.Ar.Pr.=30!
COSTS= :Agr.Pr.=40:
even1Aqr.Pr.=50:
0.00
0.00
0.84
0.00
0.00
0.84
0.00
0.00:!
0.84
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-- ----CONSTR.:Aqr.Pr.=30:
+25n :Agr.Pr.=50:
----------------- ------0.00
0.00 !
0.00
0.00
:----------------------- :Agr.Pr.=40:
0.00
0.00
COSTS=
DISCOUNT RATE:
:
:
:
------------
|
---------
0.00
0.00
0.00
0.00 |:
--------------------------0.00
0.00
0.00
0.00::
0.00
0.00
0.00: ----------------------------
MATER REQUIREMENTS = -50
I
|
0.00
REQUIREMENTS = even
WATER
2------------------------------------------------0.000
0.00
0.00
0.00
0.00
CONSTR.EAgr.Pr.=30|
0.0
----0.0
0.00------------0.00-- 0.00
0
COSTS:- :Ag
0.00
0.00
0.00
0.00
0.00
COSTS= Agr.Pr.=40|
0.00
0.00
0.00
0.00
0.84
:Agr.Pr.=50:
-25
: WATER
REQUIREMENTS = +50 1
0.00
.0
0.00----------------0.00
0.00 :
0.00
0.00
0.00
0.00
0.00---0.000.00
0.00
0.00
0.00
0.00
0.00
0.00:1
0.00
0.00
0.00
0.00
0.00
0.00
:
0.00
0.00
0.00
0.00
0.00
0.00
0.00:
0.00
0.00
0.00
0.00
0.00:
0.00
0.00
0.00
:
0.00
0.00
0.00
0.00
0.00
0.00 :
0.00:
0.00
0.00
0.00
0.00
0.00 |
0.00:
0.00
0.00
0.00
0.00
0.00
0.00
Agr.Pr.=30:
CONSTR
0.84
0.84
0.00
CONSTR=
Agr.Pr.=430:
0.00
0.00
0.00
even
Agr.Pr.:50
0.84
0.00
CONSTR.:Aqr.Pr.=30:
0.00
COSTS= :Agr.Pr.=40
+251 :Agr.Pr.=50
0.00
0.00
:
139
TABLE B-3:
Optimal sizes
of Dam B
MATER
REQUIREMENTS x -50 1 WATER REQUIREMNTS = even : WATER REQUIREMENTS = +50 1
DISCOUNT RATE:
EI.Pr.=10 El.Pr.=20 EI.Pr.=30 E1.Pr.=10 E1.Pr.=20 El.Pr.=30;E.Pr.=10 EI.Pr.=20 EI.Pr.30
CONSTR. Agr.Pr.:30
COSTS= Agr.Pr.=40:
-25C :Agr.Pr.=50:
0.00
0.00
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
-0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
CONSTR. :Agr.Pr.=30:
COSTS :Agr.Pr.=40:
even !Agr.Pr.=50:
0.84
0.00
0.00
0.84
0.00
0.00
0.84
0.84
0.00
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
CONSTR.:Agr.Pr.=30:
COSTS= :Aqr.Pr.=40:
4251 :Agr.Pr.=50:
0.84
0.84
0.00
0.84
0.84
0.00
0.84
0.84
0.00
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
WATER
REQUIREMENTS = -50 1
DISCOUNT RATE:
10 I
:WATER
REQUIREMENTS = even ; WATER REQUIREMENTS = +50 Z
-=========================================================================================
:E1.Pr.=10 EI.Pr.=20 El.Pr.=30:E.Pr.=10 El.Pr.z20 E1.Pr.=30|E.Pr.=10 EI.Pr.=20 E1.Pr.=30
C-SR-grP.3
CONSTR.:Agr.Pr.=30:
COSTS :Agr.Pr.=40:
-251 :Aqr.Pr.=50:
C-.
0
0.84-0.-4--4------------------------------------------
0.84
0.00
0.00
0.84
0.00
0.00
0.84 :
0.00 :
0.84 1
84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84 |
0.84
0.84
.
0.84 |.4
CONSTR Agr.Pr.=30|
COSTS :Agr.Pr.=40:
riven :Agr.Pr.:5o:
0.84
0.84
0.00
0.84
0.84
0.00
0.84
0.84 |
0.00 :
0.84
0.84
0.84
CONSTR.:Agr.Pr.=30:
COSTS: :Agr.Pr.=40:
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84---------0.84 -0.84
-
-----
----25-1
.-.
:---------------i-----|--------------
Ar.Pr.-5- :
0.84
0.84
0.84
WATER
REQUIREMENTS = -50 1
DISCOUNT RATE:
12 1
0.84
0.84
0.84
0.84
0.84
-------
484
0.84
0.84
0.84
0.84
:
0.84
0.84
0.84
0.84
0.84
0.84
.4
8
0.84
0.84
0.84
0.84
0.84
0.84
0.00
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
---------------------0.84
0.84
0.84
: WATER
REQUIREMENTS = even : WATER REQUIREMENTS = +50 1
:EI.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:El.Pr.=10 EI.Pr.=20 EI.Pr.=30
--
------- :
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-5 1 :Agr.Pr.=50:
-
------
0.84-0.8--------------|-
0.84
0.00
0.00
0.84
0.84
0.00
0.84
0.84
0.00
------------------------
--------- 0------------------
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
-----------------
0.84
0.84
0.84
0.84
0.84
0.84
----------------
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
;Agr.Pr.=50:
0.00
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
+25 1 :Agr.Pr.=50:
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.84
0.00
0.84
0.84
0.00
0.84
0.84
0.00
0.84
0.84
CONSTR. :Agr.Pr.=30:
COSTS= |Agr.Pr.=40:
even
140
TABLE B-4:
DISCOUNT RATE:
Optimal
sizes of Dam C
WATER REQUIREMENTS = -50 1 : WATER REQUIREMENTS =evn
MATER REQUIREMNTS x +50 2
:EI.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30 E.Pr.=10 EL.Pr.=20 EI.Pr.=30
C-SP-gr
R. 3
1
0.00
0.00
05--.05--.0--.05--.00
---.
--.
0
COSTR Agr.Pr.=30:
1.05
0.00
0.00|
1.05
1.05
0.00
1.05
0.00
0.00
COSTS: Agr.Pr.=40
1.05
0.00
0.00:
1.05
0.00
0.00
1.05
1.05
0.00
25 :Aqr.Pr.=50:
1.05
0.00
0.00|
1.05
0.00
0.00
1.05
1.05
0.00
CA- -STR
1.05
.
1.05
05-1.05--.00--.00--.00 -.
--.
00
CONSTR.:Agr.Pr.=30:
1.05
1.05
0.00
1.05
1.05
0.00
1.05
0.00
0.00
COSTS: :Agr.Pr.=50:
1.05
1.05
0.00
1.05
1.05
0.00
1.05
1.05
0.00
CONSTR.:Aqr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25 1 :Agr.Pr.=50:
--
10U=============
- - - -
--
----------
1.05
1.05
1.05
-
-
-
-
1.05
0.00
0.00
-
-
-
-
1.05
1.05
1.05
-
-
-
-
-
1.05
1.05
1.05
- - -
-
0.00
1.05
1.05
-
-
-
0.00
1.05
1.05
-
-
-
-
-
-
0.00
0.00
1.05
-
-
-
-
0.00
0.00
0.00
-
-
-
-
-
-
----------------------------------------------------------
DISCOUNT RATE:
10
1.05
1.05
1.05
ATER REQUIREMENTS
: ATER REQUIREMENTS
-50
: MATER REQUIREMENTS
even
1-
--
500
-
CEI.Pr.10 EI.Pr.20 EI.Pr.30:E1.r.10 EI.Pr.20 EI.Pr.30:E1.Pr.10 EI.Pr.20 E 0.Pr.0
--
-- --- --- - -- -
CONSTR.:Agr.Pr.=50:
1.05
COSTS:
--
-
-
--
1.05
0.00
1.05
1.05
0.00
1.05
1.05
0.00
Cg-r---.
-R A = ----- .0
COSTR.:Agr.Pr.=30:
1.05
1.05
.
1.05
COSTS:
1.05
0.00:
:Ar.Pr.=40:
-25% 1Agr.Pr.:50:
:Ar.Pr.=40:
1.05
-
---
-
-
--
-
---------RATE:
-
---
:
1.05
1.05
0.00
:
1.05
1.05
0.00
-5 15
1.05
1.05
:----------------------
-
--
:
-
-
-
-----
-
-
-
-
-
-
1.05
0.00
0.00
1.05
1.05
0.00
1.05
1.05
0.00
---------------------00
-0 .
1.05
0.00
0.00
0.00
0.00
1.05
1.05:
i.0s
0.00
0.00
1.05
1.05
1.05
0.00
0.00
1.05
1.05
0.00 El
1.05
1.05
0.00
0.00
1.05
1.05
---0.00
0.00
1.05
-------------------------------
--------W ATER
-
0.00
!Agr.Pr.=50:
1.05
1.05
0.00
1.05
12I
============================
=======
CONSTR.
1.05 EAgr.PPr..30:
1.05
1.05 E.1.05
COSTSR :Agr.Pr.=40:
1.05
1.05
1.05
1.05
+25C :Agr.Pr.=50:
1.05
1.05
1.05
1.05
DISCOUNT
--
1.05
even
-----
-
1.05
0.00
0.00
0.00
----------
------------------------------------------------
REQUIREMENTS
=
-50
MATER
21
REQUIREMENTS
=
even
:
MATER
REQUIREMENTS
=
+50
1
12 1
:E2.Pr.1A E1.Pr.=20 EI.Pr.=30E.Pr.=10 El.Pr.=20 E0.Pr.=30E.Pr.10 EI.Pr.:20 EI.Pr.=30
---------------------------------------
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25e :Agr.Pr.=50:
1.05
1.05
1.05
----------------------------
1.05
1.05
1.05
0.00
0.00 |
0.00
1.05
1.05
1.05
1.05
1.05
1.05
0.00
1.05
0.00
0.00
1.05
1.05
------------------.
0.00
1.05
1.05
.
.
0.00
0.00
1.05
.
.
CONSIR. :Agr.Pr.=3o:
1.05
1.05
1.05
:
1.05
0.00
0.00
0.00
0.00
COSTS=
:Aqr.Pr.=4o:
1.05
1.05
1.05:
1.05
1.05
1.05:
0.00
0.00
0.00
even
:Agr.Pr.:so:
1.05
1.05
1.05:
1.05
1.05
1.05:
1.05
1.05
0.00
CONSTR.
:Agr.Pr.=30:
1.05
1.05
1.05
0.00
0.00
COSTS= :Agr.Pr.=40:
+25 2 :Agr.Pr.=50
0.00
0.00
0.00
1.05
1.0
0.00
1.05
1.05
1.05
1.05 :
1.05
1.05
1.05
1.05
0.00
1.05
0.00
0.00
0.00
0.00
0.00
0.00
0.00
141
TABLE B-5:
Optimal
sizes
of Dam D
WATER REQUIREMENTS = -50 1 : MATER REQUIREMENTS x even
DISCOUNT RATE:
81
WWATER REQUIREMENTS z +50 1
;EI.Pr.=10 EI.Pr.=20 EI.Pr.=30 E1.Pr.=10 EI.Pr.=20 E1.Pr.=30!E.Pr.=10 EI.Pr.=20 E1.Pr.=30
-------------- :----------------------------- :-----------------------------|:-- ------------------- -------
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40!
-25 1 :Agr.Pr.=50
0.21
0.21
0.21
1.23
1.22
1.22
1.23
1.22
1.22
0.00
0.00
0.21
0.00
1.23
1.23
1.23
1.23
1.23
------- |:---------- :----------------------------- -----------------------------
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40!
even 1Agr.Pr.=50
0.00
0.21
0.21
0.00
0.21
0.21
1.23
1.23
1.23
0.00
0.00
0.00
0.00
0.00
0.00
0.00
1.23
1.23
------- |:---------- :----------------------------- ------ ------------------- ---
CONSTR.|Agr.Pr.=30
COSTS= 1Agr.Pr.=401
+25 1 1Agr.Pr.=50
0.00
0.00
0.21
0.00
0.00
0.21
0.00
1.23
1.23
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
WATER REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even
DISCOUNT RATE:
10
0.00
0.00
0.00
0.00
0.00
1.23
0.00
0.00
1.23
--------------------------0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
1.23
------------------------ ---0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
WATER REQUIREMENTS = +50 1
:EI.Pr.=10 E1.Pr.=20 E1.Pr.=301E1.Pr.=10 EI.Pr.=20 El.Pr.=30 E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.1Agr.Pr.=30t
CDSTS= lAgr.Pr.=40:
-25 1 1Agr.Pr.=50
0.00
0.21
0.21
0.00
0.21
0.21
1.23 1
1.23 1
1.22 1
0.00
0.00
0.00
0.00
0.00
0.00
1.23
1.23
1.23
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
1.23
CONSTR.:Agr.Pr.=30
COSTS= 1Aqr.Pr.=40!
even :Agr.Pr.=50:
0.00
0.00
0.21
0.00
0.00
0.21
0.00 |
1.23
1.23 1
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00 0.00 |
0.00 1
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
C
-NSTR.
-Agr.Pr.=300.00
COSTS= Agr.Pr.=401
0.00
+25 Z lAgr.Pr.=50:
0.00
WATER REQUIREMENTS = -50 2 | WATER REQUIREMENTS = even
DISCOUNT RATE:
12 1
WATER REQUIREMENTS = +50 Z
:EI.Pr.=10 EI.Pr.=20 EI.Pr.=30 E.Pr.=10 EI.Pr.=20 EI.Pr.=30 E1.Pr.=10 El.Pr.=20 E1.Pr.=30
CONSTR.
-Agr.Pr.=300.00
COSTS: Agr.Pr.=40
0.21
0.21
-25 C Agr.Pr.=50
0.00
0.21
0.21
1.23 -- 0.00
1.23
0.00
1.23 1
0.00
0.00
0.00
0.00
0.00
0.00
1.23
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR. Agr.Pr.=30:
COSTS: 1Agr.Pr.=40~
even Agr.Pr.=50
0.00
0.00
0.21
0.00
0.00
0.21
0.00
0.00
0.21
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR. Agr.Pr.=30
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
COSTS: |Agr.Pr.=40
+25 |Agr.Pr.=50
0.00
0.00
0.00
0.00
0.00 ,
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
14 2
TABLE
B-6:
Optimal sizes
of Irrigation B
I
WATER REQUIREMENTS = -50
DISCOUNT RATE:
1 : WATER REQUIREMENTS = even
|
WATER REQUIREMENTS = +50 1
:EI.Pr.=10 E1.Pr.=20 EI.Pr.=30|E1.Pr.=10 E1.Pr.=20 E1.Pr.=30:E1.Pr.=10 E1.Pr.=20 E1.Pr.=30
16.75
16.75
27.88
16.55
27.88
27.88
16.55
27.88
27.88
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
CONSTR. Agr.Pr.=30 -- 16.55
16.75
COSTS= Agr.Pr.=40|
16.75
even :Agr.Pr.=50:
16.55
16.75
16.75
16.55
16.55
16.75
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.75
16.55
16.55
16.75
16.55
16.55
16.75
16.55
16.55
16.55
16.55
16.55
16.55
16.55 :
16.55 1
16.55 1
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
CONSTR. Agr.Pr.=30;
COSTS= Agr.Pr.=40
-25 Z Agr.Pr.=50
CONSTR. Agr.Pr.=30:
COSTS= Agr.Pr.=40:
+25 1 :Agr.Pr.=50|
WATER REQUIREMENTS = -50 1 IWATER REQUIREMENTS = even
DISCOUNT RATE:
! WATER REQUIREMENTS = +50 1
====================================
10 1
|E1.Pr.=10 EI.Pr.=20 E1.Pr.=30:E1.Pr.=10 E1.Pr.=20 E1.Pr.=30:E1.Pr.=10 EI.Pr.=20 E1.Pr.=30
CONSTR|Ar.Pr.=30:
2OSTS= :Agr.Pr.=40:
-25 i|Agr.Pr.=50
16.55
16.75
16.75
16.55
16.75
16.75
16.55 |
16.75
27.88
16.55
16.55
16.55
16.55
16.55
16.55
16.55 -- 16.55
16.55 1
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
O0MSTR.:Agr.Pr.=30
O0STS= |Agr.Pr.=:40
even |Agr.Pr.=50
16.55
16.55
16.75
16.55
16.55
16.75
16.55 1
16.55
16.75
16.55
16.55
16.55
16.55
16.55
16.55
16.55 1
16.55 |
16.55 |
16.55
16.55
16.55
16.55
16.55
16.55
0.00
16.55
16.55
-NSTR.Agr.Pr.=30 -- 16.55
16.55
16.55
+25 Z :Agr.Pr.=50:
:-----:--------------12---
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55 |
16.55 |
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
OSTS= |Agr.Pr.=40
ISCOUNT RATE:
-----------------------------------------
WATER REQUIREMENTS
-50
I WATER REQUIREMENTS = even
WATER REQUIREMENTS = +501
IE1.Pr.=10 E1.Pr.=20 E1.Pr.=30 E1.Pr.=10 EI.Pr.=20 EI.Pr.=30|E1.Pr.=10 E1.Pr.=20 EI.Pr.=30
ONSTR.1Agr.Pr.=30
COSTS= :Aqr.Pr.=40
-25 Z Agr.Pr.=50
16.55
16.75
16.75
16.55
16.55
16.75
16.55
16.55
16.75
16.55
16.55
16.55
16.55
16.55
16.55
16.55 :
16.55 |
16.55 |
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
ONSTR. IAgr.Pr.=30
Agr.Pr.:40W
16.55
16.55
16.55
16.55
16.55 1
16.55
16.55
16.55
16.55
16.55
16.55 1
16.55
16.55
16.55
16.55
16.55
16.55
16.55
Agr.Pr.=50
16.75
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55 |
16.55 |
16.55 |
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
16.55
10.00
16.55
16.55
0.00
16.55
16.55
0.00
16.55
16.55
COSTS=
even
CONSTR. :Aqr.Pr.=30:
COSTS= :Agr.Pr.=40:
:Agr.Pr.=50:
+25
143
TABLE
Optimal
B-7:
sizes of
Irrigation C
T
WATER
REQUIREMENTS z -50 1 1 WATER
REQUIREMENTS = even
DISCOUNT RATE:
8
WATER REQUIREMENTS = 450 Z
===
EI.Pr.=10 EI.Pr.=20 EI.Pr.=30 E1.Pr.=10 EI.Pr.=20 EI.Pr.=30 E.Pr.=10 EI.Pr.=20 EI.Pr.=30
--
-- --- -- -
CONSTR. Aqr.Pr.=30
COSTS= Agr.Pr.=40
-25 1 Agr.Pr.=50
-
-
--
-
19.66
19.66
19.66
-
--
-
-
0.00
0.00
0.00
-
- - : - - - - -- -- - - -:
-
-
-
-
-
-
-
0.00
0.00
0.00
-
-
-
-
-
-
-----
--
18.47
18.47
19.66
-
-
-
-
-
-
-
-
-
-
-
18.47
0.00
0.00
-
-
-
-
-
-
.--
0.00
0.00
0.00
-
-
-
-
--
--
-
--
--
18.47
18.47
18.47
-
-
-
-
-
--
--
0.00
18.47
18.47
- -
---
--
--
-
0.00
0.00
0.00
---
--
--
CONSTR.:Agr.Pr.=30:
COSTS= |Aqr.Pr.=40
even tAqr.Pr.=50
18.47
19.66
19.66
18.47
19.66
19.66
0.00
0.00
0.00
18.47
18.47
18.47
18.47
18.47
18.47
0.00
0.00
0.00
18.47
18.47
18.47
0.00
18.47
18.47
0.00
0.00
0.00
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40
+25 1 !Agr.Pr.=50
18.47
18.47
19.66
18.47
18.47
19.66
18.47
0.00
0.00
18.47
18.47
18.47
18.47
18.47
18.47
0.00
18.47
18.47
0.00
18.47
18.47
0.00
0.00
18.47
0.00
0.00
0.00
------------------------------- ----------------
--
-
- ---
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-------
-
-
-
-
-
-
WATER REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even
DISCOUNT RATE:
10 1
-
--
-
-
-
--------
--
--
--
--
-
| WATER
REQUIREMENTS = +50 1
!EI.Pr.=10 EI.Pr.=20 EI.Pr.=30|E.Pr.=10 EI.Pr.=20 El.Pr.=30 E.Pr.=10 EI.Pr.=20 E1.Pr.=30
--
-
--
-
:I--
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- - -
-
-
-
-
-
- ---
-
-
--
-
--
--
--
----
--
-
CONSTR.:Aqr.Pr.=30:
18.47
18.47
0.00
COSTS= Agr.Pr.=40
19.66
19.66
0.00
-25 1 :Agr.Pr.=50
19.66
19.66
0.00
---- :-----:-----------------------CONSTR.2Agr.Pr.=30
18.47
18.47
18.47
COSTS= Agr.Pr.=40
18.47
18.47
0.00
even |Agr.Pr.=50
19.66
19.66
0.00
18.47
18.47
18.47
18.47
18.47
18.47
0.00
0.00
0.00
18.47
18.47
18.47
0.00
18.47
18.47
0.00
0.00
0.00
18.47
18.47
18.47
18.47
18.47
18.47
0.00
18.47
18.47
0.00
18.47
18.47
0.00
0.00
18.47
0.00
0.00
0.00
CONSTR.:Agr.Pr.=30
COSTS= :Agr.Pr.=40:
+25 1 :Agr.Pr.=50
18.47
18.47
18.47
0.00
18.47
18.47
0.00
18.47
18.47
0.00
0.00
18.47
0.00
0.00
18.47
0.00
0.00
0.00
DISCOUNT RATE:
12 1
18.47
18.47
18.47
18.47
18.47
18.47
18.47
18.47
18.47
WATER REQUIREMENTS = -50 Z WATER REQUIREMENTS = even
| WATER REQUIREMENTS = +50 1
E1.Pr.=10 EI.Pr.=20 EI.Pr.=30 E1.Pr.=10 E1.Pr.=20 El.Pr.=30 E1.Pr.=10 El.Pr.=20 EI.Pr.=30
CONSTR. Agr.Pr.=30
18.47
18.47
0.00
18.47
18.47
0.00
0.00
0.00
0.00
COSTS= Aqr.Pr.=40
19.66
19.66
0.00
18.47
18.47
18.47
18.47
18.47
0.00
-25 1 Agr.Pr.=50
19.66
19.66
0.00
18.47
18.47
0.00
18.47
18.47
18.47
---------------------------------------------------- ----------------------------CONSTR. Aqr.Pr.=30
18.47
18.47
18.47
18.47
0.00
0.00
0.00
0.00
COSTS= Agr.Pr.=40
18.47
18.47
18.47
18.47
18.47
18.47
0.00
0.00
0.00
Agr.Pr.=50
19.66
19.66
19.66
18.47
18.47
18.47
18.47
18.47
0.00
CONSTR. Agr.Pr.=30
18.47
18.47
18.47
0.00
0.00
0.00
0.00
0.00
0.00
COSTS= Agr.Pr.=40:
+25 1 Agr.Pr.=50
18.47
18.47
18.47
18.47
18.47
18.47
18.47
18.47
18.47
18.47
0.00
18.47
0.00
0.00
0.00
0.00
0.00
0.00
even
0.00
144
TABLE
B-8:
Optimal
sizes
WATER REQUIREMENTS = -50
DISCOUNT RATE:
of
Irrigation D
WATER
REQUIREMENTS = even
WATER REQUIREMENTS = +50 Z
W
El.Pr.=10 EI.Pr.=20 EI.Pr.=30 E1.Pr.=10 EI.Pr.=20 EI.Pr.=30 E.Pr.=10 E1.Pr.=20 E1.Pr.=30
CONSTR. Aqr.Pr.=30
1.65
20.67
20.67
0.00
0.00
20.67
0.00
0.00
0.00
COSTS= |Agr.Pr.=40
-25 Z :Agr.Pr.=50
1.65
1.72
20.84
20.84
20.84
20.84
0.00
1.55
20.67
20.67
20.67
20.67
0.00
0.00
0.00
0.00
20.67
20.67
CONSTR.:Agr.Pr.=30
COSTS= |Aqr.Pr.=40
even :Aqr.Pr.=50
0.00
1.65
1.65
0.00
1.65
1.65
20.67
20.67
20.77
0.00
0.00
0.00
0.00
0.00
Q.00
0.00
20.67
20.67
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
20.67
CONSTR. Agr.Pr.=30COSTS= Agr.Pr.=40
+25 Z :Agr.Pr.=50
0.00
0.00
1.65
0.00
0.00
1.65
0.00
20.61
20.77
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
DISCOUNT RATE:
10
:
WATER REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even
WATER
W
REQUIREMENTS = +50 Z
:-================================================
:EI.Pr.=10 EI.Pr.=20 E1.Pr.=30 E1.Pr.=10 El.Pr.=20 EI.Pr.=30|E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.JAgr.Pr.=30
COSTS= tAgr.Pr.=40:
-25 Z Agr.Pr.=50
0.00
1.65
1.65
0.00
1.65
1.65
20.67
20.77
20.84
CONSTR. Agr.Pr.=30- -COSTS= Agr.Pr.=40
even :Agr.Pr.=50:
0.00
0.00
1.65
0.00
0.00
1.65
0.00
20.67
20.77
-ONSTR. Agr.Pr.=30
COSTS :Agr.Pr.=40:
+25 C :Agr.Pr.=501
0.00
0.00
0.00
0.00
0.00
0.00
:
0.00:
0.00
0.00
20.67 :
20.67 t
20.67 |
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
WATER REQUIREMENTS = -50 tI WATER REQUIREMENTS = even
DISCOUNT RATE:
-
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
20.67
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-0.00 0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
:WATER REQUIREMENTS = +50 Z
==
12 I
:El.Pr.=10 EI.Pr.=20 El.Pr.=30 E1.Pr.=10 El.Pr.=20 EI.Pr.=30:El.Pr.=10 E1.Pr.=20 EI.Pr.=30
-------------- ------------------------------ ----------------------------- -----------------------------
CONSTR.|Agr.Pr.=30:
COSTS= |Agr.Pr.=40
-25 Z Agr.Pr.=50
0.00
1.65
1.65
0.00
1.55
1.65
20.67
20.67
20.77
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
20.67
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR. Agr.Pr.=30 -COSTS= Agr.Pr.=40:
even :Agr.Pr.=50:
0.00
0.00
1.65
0.00
0.00
1.55
0.00
0.00
1.55
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR. :Agr.Pr.=30:
COSTS= Agr.Pr.=40;
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
+25 1 Agr.Pr.=50
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
145
TABLE
Optimal sizes
B-9:
DISCOUNT RATE:
WATER REQUIREMENTS = -50 1
of hydropower Plant
WATER
W
REQUIREMENTS = even
A
| WATER REQUIREMENTS = +50 Z
E1.Pr.=10 EL.Pr.=20 El.Pr.=30:EI.Pr.=10 El.Pr.=20 EI.Pr.=30:EI.Pr.=10 EL.Pr.=20 EI.Pr.=30
CONSTR.
-Ar.Pr.=-300.09
0.09
0.09 |
0.09
0.09
0.09
0.09
0.09
0.09
COSTS: Agr.Pr.=40
-25C Agr.Pr.=50
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40
even Aqr.Pr.=50
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
CONSTR. Agr.Pr.=30
COSTS= Aqr.Pr.=40
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
+25 I
Agr.Pr.=50
:WATER REQUIREMENTS = -50 Z | WATER REQUIREMENTS = even
DISCOUNT RATE:
101
WATER REQUIREMENTS = +50 1
::=========================================================================================...
CEI.Pr.=0 E1.Pr.=20 E0.Pr.=30E0.Pr.=10 E0.Pr.=20 E0.Pr.=301.Pr.=10 E0.Pr.=20 El.Pr.=30
CO2STR. Agr.Pr.=30
COSTS=: Agr.Pr.=40-25
I
:Agr.Pr.:50 W
0.09
0.09
0.09
0.09
0.09
0.09
0.09 |
0.09:
0.09
0.09
0.09
0.09
CONSTR. :Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
even 1 '.Agr.Pr.=50:
Agr.Pr.= 50
+25
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09 :
0.09
0.09 I
----
-----
----------------0
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
0.09
0.09
DISCOUNT RATE:
0.09
0.09
0.09
0.09
0.09
0.09 :
0.09
0.09 :
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09 |
0.09
0.09
0.09
0.09
0.09
0.09 |:I
0.09
0.09
0.09
0.09
0.09
0.09
----------------
0.09
0.09
WATER REQUIREMENTS = -50
0.09
0.09
0.09
0.09
WATER REQUIREMENTS
------------------
0.09 :
0.09
even
0.09
0.09
0.09
0.09
0.09
0.09
WATER REQUIREMENTS = +50
-
:EI.Pr.=10 EI.Pr.=20 El.Pr.=30 E1.Pr.=l0 E1.Pr.=20 EI.Pr.=30!E1.Pr.=10 EI.Pr.=20 E1.Pr.=30
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40:
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
IAgr.Pr.=50:
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
CONSTR.Agr.Pr.=30:
COSTS :Agr.Pr.=40:
even :Agr.Pr.=50:
0.09
0.09
0.09
0.09
0.09
0.09
0.09:
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
-- 0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.40
+25
:Agr.Pr.=50:
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09 :
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
0.09
-25
146
TABLE B-O:
Optimal sizes of hydropower Plant C
I
WATER REQUIREMENTS = -50 Z
DISCOUNT RATE:
8-------
E-----------------------------:
' WATER REQUIREMENTS = +50
1
----------------------------- ----------------------------0.00
0.00
0.00
0.00
0.06
0.06
0.06
0.06
0.06
0.00
0.00
0.00
0.06
0.00
0.00
0.06
0.06
0.06
0.06
0.06
0.06
0.00
0.00
0.00
0.00
0.06
0.06
0.00
0.00
0.06
0.00
0.00
0.00
0.06
0.00
0.00
0.06
0.06
0.06
0.00
0.06
0.06
0.00
0.00
0.00
0.00
0.00
0.00
0.06
0.00
0.00
0.00
0.00
0.06
0.06
0.06
0.06
0.00
0.00
0.06
CONSTR. :Agr.Pr.=30:
COSTS= Agr.Pr.=40
-25 1 :Aqr.Pr.=50:
0.00
0.00
0.06
0.06
0.06
0.06
0.00
0.06
0.06
CONSTR. :Agr.Pr.=30:
COSTS= !Agr.Pr.=40:
even !Agr.Pr.=50:
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR. :Agr.Pr.=30:
COSTS= Agr.Pr.:40+25 1 Agr.Pr.=50!
0.00
0.00
0.00
0.00
0.00
0.00
DISCOUNT RATE:
WATER REQUIREMENTS = even
M
t WATER REQUIREMENTS = -50 Z : WATER REQUIREMENTS = even
: WATER
REQUIREMENTS = +50 1
:EI.Pr.=10 EI.Pr.=20 E.Pr.=30;E.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
------ :---------------i----------------------CONSTR.:Agr.Pr.=30:
0.00
0.00
0.06
0.00
0.00
0.06
COSTS= Aqr.Pr.=40:
0.00
0.00
0.00
0.00
-25 1 :Agr.Pr.=50:
0.00
0.06
!----------------------------0.00
0.00
0.00
0.06
0.06
0.06
0.00
0.00
0.00
0.06
0.00
0.00
0.06
0.06
0.06
0.00
0.00
0.00
0.00
0.06
0.06
0.00
0.00
0.00
0.00
0.00
0.00
0.06
0.00
0.00
0.00
0.06
0.10
Agr.Pr.=50:
0.00
0.00
0.00
0.00
0.00
0.06
0.00
0.06
0.06
rCONSTR. :Agr.Pr.=30
COSTS= Agr.Pr.=40:
+25 1 :Aqr.Pr.=50!
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.06
0.00
0.00
0.06
0.00
0.00
0.00
0.00
0.06
0.06
0.06
0.06
0.00
0.00
0.06
: WATER
REQUIREMENTS
= +50 1
CONSTR. :Agr.Pr.=30
COSTS= :Agr.Pr.=40:
even
WATER REQUIREMENTS = -50 1
DISCOUNT RATE:
WATER REQUIREMENTS = even
:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 El.Pr.=20 EI.Pr.=30:E.Pr.=10 EI.Pr.=20 EI.Pr.=30
0.06
0.06
0.00
0.00
0.06
0.00
0.00
0.06
0.06
0.00
0.06
0.00
0.00
0.06
0.00
0.00
0.00
0.00
0.00
0.06
0.06
0.00
0.06
0.06
0.06
0.06
0.00
0.00
0.06
0.06
0.00 '
0.00
0.00
0.00
0.06
0.06
0.06
0.10
0.06
0.06
CONSTR. :Agr.Pr.=30:
0.00
0.00
0.06
0.00
0.00
0.06
COSTS= :Agr.Pr.=40:
0.00
0.00
0.06
0.00
0.00
0.00
-25 Z :Agr.Pr.=50:
0.00
0.00
0.06
0.00
0.00
CONSTR. Aqr.Pr.=30
COSTS= !Agr.Pr.=40:
even :Aqr.Pr.=50:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR. Agr.Pr.=30
COSTS: Agr.Pr.=40:
+25 1 'Aqr.Pr.=50:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00 '
0.00
0.00
0.00
1 47
T ABL E
B-1 1 :
DISCOUNT RATE:
Optimal sizes of Transfer
WATER REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even
| WATER REQUIREMENTS = +50 1
E1.Pr.=10 El.Pr.=20 EI.Pr.=30|E.Pr.=10 El.Pr.=20 El.Pr.=30:E.Pr.=I0 EI.Pr.=20 EI.Pr.:30
CONSTR.
:Ar.Pr.=300.00
0.00
COSTS :Agr.Pr.=40:
0.00
0.00
-25 I |Agr.Pr.=50
0.00
0.00
0.00
0.00
0.00
CONSTR2
Agr.Pr.=30
COSTS= Agr.Pr.=40
even Agr.Pr.=50
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR.AAgr.Pr.=30
COSTS= Agr.Pr.=40
+25 1 :Agr.Pr.=50
0.00
0.00
0.00
0.00
0.00
0.00
-
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00:
0.00:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-
:
:
:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
------- |----------- ----------------------------- ------------------------------------------------------------
DISCOUNT RATE:
10
| WATER REQUIREMENTS = -50 %| WATER
REQUIREMENTS = even
WATER REQUIREMENTS = +50 1
W
|=
:El.Pr.=10 El.Pr.=20 EI.Pr.=30 E.Pr.=I0 El.Pr.=20 EI.Pr.=30:E.Pr.=10 El.Pr.=20 EI.Pr.=30
CONSTR. Agr.Pr.:30
COSTS :Agr.Pr.=40:
-25C Agr.Pr.=50
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00 -- 0.00
0.00
0.00
0.00:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
CONSTR Agr.Pr.=30
COSTS=: Agr.Pr.=40even Agr.Pr.=50
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
6.00
0.00
0.00
CONSTR. Agr.Pr.=30
OSTS= Agr.Pr.=40
+25 1 Agr.Pr.=50
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00 :
0.00 |
0.00 :
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
DISCOUNT
WATER REQUIREMENTS =-50 1I WATER REQUIREMENTS = even iWATER REQUIREMENTS = +50 1
RATE:
:E1.Pr.=l0 EI.Pr.=20 EI.Pr.=30 E1.Pr.=10 E1.Pr.=20 EI.Pr.=30!E1.Pr.=10 El.Pr.=20 E1.Pr.=30
ONSTR. !Agr.Pr.=30
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
OSTS= :Aqr.Pr.=40
0.00
0.00
0.00
0.00
0.00
0.00:
0.00
0.00
0.00
0.00
0.00
0.00
:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00:
0.00
0.00:
0.00
0.00
0.00
0.00
0.00
0.00
0.00 :
0.00 :
0.00:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
:
:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
8.00
0.00
0.00
0.00
0.00
:
0.00
.00
0.00
0.00
0.00
0.00
0.00
-2
:Agr.Pr.=50:
ONSTR.:Agr.Pr.=30:
OSTS= Agr.Pr.=40
even
Agr.Pr.=50
CONSTR. :Agr.Pr.=:30
COSTS= :Agr.Pr.=40:
+25
Agr.Pr.=50
:
148
70%~
-..
enx
5D%
D
401
301
20%
1Xx
20
40
no
100
d
\CKLUME
I
I
I
I
I
I
I
I
I
I
I
I
0
120
~1
140
m3
Bar graph of the probability
distribution of the sizes of Dam A
FIGURE B-1:
tont
-J
E0t -
1C
I
5
20
;
I
II
on
I
II
S0
I
I
II
1np
II
iI
i2n
I
I
t
14n
'--U ME Hm 3
FIGURE B-2:
Bar graph of the probability
distribution
of the sizes
of Dam B
149
a901
-T
5ax
-
sax IL
20%
a%
0
20
40o
I
Do
so
I
I
101
I
I
120
I
v
140
'.OL1ME Hm
FIGURE B-3:
100%
Bar graph of
distribution
the probability
of the sizes of Dam C
-
all-
IT
jdlt
20t
rat
-a
I
Ii
21
40
I
I
I
on
in0
1211
14')
'.COi3ME rAm3
FIGURE B-4:
Bar graph of the probabili ty
distribution of the sizes of Dam D
150
00%
4-
E-
-
50X -
40%2a% -
413% -
20%3m -
1
,4,
0
FYGURE F-5:
1001
II
"""
a
12
1.
(Thumamd}
AREA Na
20
24
25
Bar graph of the probability distribution
of the sizes of Irrigation B
-
01-
T
4
r
r
e
-y
___________
---
-T
12
-r11
*(Thounmdv'e
FTGUwR
F-6:
lID
liii
--
---a1)
24
28
ME Ho
Bar graph of the probability distribution
of the sizes of Irrigation C
151
aos
50%
E-
40%
20x
10%
Wx I ."
0
+4
12 .
1
(Thaummdu}
20
24
25
MU!L.4Na
FYGUWE B-T:
Bar graph of the probability distribution
of the sizes of Irrigation D
1nOT-
sa( 803 -
Jolt
+4
FIGURE U-6:
10I
12
14
Bar graph of the probability distribution
of the sizes of hydropower Plant A
152
5D'
CL
LL
4B.1
70%
0%
am.
10%
0
a
4
a
10
12
14
[CZP.ZCW MAI
FYGURE V-9:
Bar graph of the probability distribution
of the sizes of hydropower Plant
C
E
I.J
a
10% -
i
2
4
I
i
I
I
8
a
I
- T
1?
-I
i
14
':!ZE m.3/s
FIGURE B-O:
Bar graph of the probability
distribution of the sizes of Transfer
153
APPENDIX
C
t54
TABLE C-i:
Net benefits
of Alternative
WATER REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even
DISCOUNT RATE:
A
: WATER REQUIREMENTS = +50 1
:E1.Pr.=10 El.Pr.=20 El.Pr.=30|El.Pr.=10 El.Pr.=20 El.Pr.=30:El.Pr.=10 El.Pr.=20 EI.Pr.=30
0.21
0.21
0.21
0.51
0.81 :
0.51
0.51
0.81
0.81
0.21
0.21
0.21
0.51
0.51
0.51
0.81
0.81
0.81
-- 0.21
0.21
0.21
0.51
0.51
0.51
0.81
0.81
0.81
0.18
CONSTR. :Agr.Pr.=30:
COSTS= |Agr.Pr.40 -- 0.18
0.18
even |Agr.Pr.=50:
0.48
0.48
0.48
0.78
0.78
0.78
0.18
0.18
0.18
0.48
0.48
0.48
0.18
0.78 1
0.78 -- 0.18
0.18
0.78 1
0.48
0.48
0.48
0.78
0.78
0.78
0.15
0.15
0.15
0.45
0.45
0.45
0.75
0.75
0.75
0.15
0.15
0.15
0.45
0.45
0.45
0.75
0.75
0.75
0.15
0.15
0.15
0.45
0.45
0.45
0.75
0.75
0.75
CONSTR.-Ar.Pr.=30
COSTS= Agr.Pr.=40
-25C :Agr.Pr.=50
CONSTR.|Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
+25 1 |Agr.Pr.=50!
1
REQUIREMENTS = even
WATER
REQUIREMENTS = -50 1 : WATER
DISCOUNT RATE:
10 1
WATER REQUIREMENTS = +50 1
W
:El.Pr.=10 El.Pr.=20 EI.Pr.=30|E1.Pr.=10 E1.Pr.=20 E1.Pr.=30:El.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.Agr.Pr.=30:
COSTS= Agr.Pr.=40!
-25 C |Agr.Pr.=50:
0.19
0.19
0.19
0.49
0.49
0.49
0.79
0.79
0.79
0.19
0.19
0.19
0.49
0.49
0.49
0.79
0.79
0.79
0.19
0.19
0.19
0.49
0.49
0.49
0.79
0.79
0.79
CONSTR.|Agr.Pr.=30:
0.15
0.45
0.75
0.15
0.45
0.75
0.15
0.45
0.75
COSTS: -Agr.Pr.=40--- 0.15
0.15
even Agr.Pr.=50:
0.45
0.45
0.75
0.75
0.15
0.15
0.45
0.45
0.75
0.75
0.15
0.15
0.45
0.45
0.75
0.75
0.12
0.12
0.12
0.41
0.41
0.41
0.71 !
0.71
0.71
0.12
0.12
0.12
0.41
0.41
0.41
0.71
0.71
0.71
0.12
0.12
0.12
0.41
0.41
0.41
0.71
0.71
0.71
CONSTR.:Agr.Pr.=30;
COSTS= :Agr.Pr.=40|
+25 1 !Agr.Pr.=50:
REQUIREMENTS = even
WATER REQUIREMENTS = -50 1 : WATER
DISCOUNT RATE:
--
: WATER REQUIREMENTS = +50
1
EI.Pr.=10 EI.Pr.=20 E1.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:EI.Pr.=10 E1.Pr.=20 EI.Pr.=30
CONSTR. Agr.Pr.=301
COSTS= Agr.Pr.=40|
-25 X Agr.Pr.=50|
0.!7
0.17
0.17
0.47
0.47
0.47
0.76
0.76
0.76
0.17
0.17
0.17
0.47
0.47
0.47
0.76
0.76
0.76
:
ONSTR. Agr.Pr.=30|
COSTS: Agr.Pr.=40
0.12
0.12
0.42
0.42
0.72
0.72
0.12
0.12
0.42
0.42
0.72
0.72
|0.12
Agr.Pr.=50:
0.12
0.42
0.72
0.12
0.42
0.72 |
even
-
0.08
0.02
0.08
0.38
0.38
0.38
0.68
0.68
0.68
0.08
0.08
0.08
0.47
0.47
0.47
0.76
0.76
0.76
0.12
0.42
0.42
0.72
0.72
0.12
0.42
0.72
-----------------------------
-------- --------------------------
CONSTR. Agr.Pr.=30t
COSTS= Ar.Pr.=40:
+25 1 Agr.Pr.=50:
0.17
0.17
0.17
0.38
0.38
0.38
0.68
0.68
0.68
0.08
0.08
0.08
0.38
0.38
0.30
0.68
0.68
0.68
155
TABLE
C-2:
DISCOUNT RATE:
Net benefits
of
Alternative
WATER REQUIREMENTS = -50 1 WATER
W
REQUIREMENTS = even
B
WATER REQUIREMENTS = 450 1
EI.Pr.=10 EI.Pr.=20 El.Pr.=30:El.Pr.=10 El.Pr.=20 El.Pr.=30:EI.Pr.=10 El.Pr.=20 El.Pr.=30
I
---------------------------------------------------------------CONSTR. Agr.Pr.=30
1.^1
2.21
2.51
0.92
1.22
1.52
0.59
0.89
1.19
COSTS= :Agr.Pr.=40:
2.57
2.87
3.17
1.25
1.55
1.85
0.81
1.11
1.41
-25 1 lAgr.Pr.=50;
3.24
3.53
3.83
1.58
1.88
2.18
1.03
1.33
1.63
CONSTR.:Agr.Pr.=30:
COSTS= |Agr.Pr.=40
even |Agr.Pr.=50
1.79
2.45
3.11
2.09
2.75
3.41
2.39
3.05
3.71 !
0.80
1.13
1.46
1.10
1.43
1.76
1.39
1.72
2.06
0.47
0.69
0.91
0.76
0.99
1.21
1.06
1.28
1.50
CONSTR. Agr.Pr.=30
COSTS= |Agr.Pr.=40
1.67
2.33
1.96
2.63
2.26
2.92
0.67
1.00
0.97
1.30
1.27
1.60
0.34
0.56
0.64
0.86
0.94
1.16
2.99
3.29
3.59
1.34
1.63
1.93
0.78
1.08
1.38
+251 Agr.Pr.=50!
--------------------------------------------------------------------------------------DISCOUNT RATE:
WATER REQUIREMENTS = -50 1 | WATER REQUIREMENTS = even : WATER REQUIREMENTS = +50 1
10 1
E1.Pr.=10 El.Pr.=20 EI.Pr.=30|E1.Pr.=10 EI.Pr.=20 El.Pr.=30:E.Pr.=10 EI.Pr.=20 E1.Pr.=30
CONSTR.2Agr.Pr.=30!
COSTS= Agr.Pr.=40:
-25 1 Agr.Pr.=50:
1.83
2.49
3.15
2.12
2.79
3.45
2.42
3.08
3.75
0.83
1.16
1.49
1.13
1.46
1.79
1.43
1.76
2.09
0.50
0.72
0.94
0.80
1.02
1.24
1.10
1.32
1.54
CONSTR.2Agr.Pr.=30|
COSTS= |Agr.Pr.=40:
even |Agr.Pr.=50!
1.67
2.33
3.00
1.97
2.63
3.29
2.27
2.93
3.59
0.68
1.01
1.34
0.98
1.31
1.64
1.28
1.61
1.94
0.35
0.57
0.79
0.65
0.87
1.09
0.95
1.17
1.39
CONSTR.2Agr.Pr.=30|
COSTS= :Agr.Pr.=40:
1.52
2.18
1.82
2.48
2.12
2.78
0.53
0.86
0.83
1.16
1.13
1.46
0.20
0.42
0.50
0.72
0.79
1.01
+25 1 Agr.Pr.=50|
2.84
3.14
3.44
1.19
1.49
1.79
0.64
0.94
1.24
: -------------------- ----------------------------- ----------------------------DISCOUNT RATE:
WATER REQUIREMENTS = -50
i
WATER REQJIREMENTS = even
WATER REQUIREMENTS = t50 1
12 1
El.Pr.=10 El.Pr.=20 El.Pr.=30 E .Pr.=10 El.Pr.=20 E1.Pr.=30;EI.Fr.=10 EI.Pr.=210 Ei.Pr.z 30
CONSTR. Agr.Pr.=301
COSTS= Agr.Pr.=40
-25 1 Agr.Pr.=50
1.74
2.40
3.06
2.04
2.70
3.36
2.34
3.00
3.66
0.75
1.08
1.41
1.05
1.38
1.71
1.34
1.67
2.0
0.42
0.64
0.86
0.71
0.94
1.16
1.01
1.23
1.45
CONSTR.Agr.Pr.=30
COSTS= lAgr.Pr.=40
even
Agr.Pr.=50:
1.56
2.22
2.88
1.86
2.52
3.18
2.16
2.82
3.48
0.57
0.90
1.23
0.86
1.20
1.53
1.16
1.49
1.82
0.24
0.46
0.68
0.53
0.75
0.97
0.83
1.05
1.27
CONSTR. :Agr.Pr.=30:
COSTS= Agr.Pr.=40
1.38
2.04
1.68
2.34
1.97
2.64
0.38
0.72
0.68
1.01
0.8
1.31
0.05
0.27
0.35
0.57
0.65
0.87
2.70
3.00
3.30
1.05
1.34
1.64
0.50
0.79
1.09
+25 1
Agr.Pr.=50
156
C-3:
TABLE
DISCOUNT RATE:
Net
benefits
of Alternative
WATER
REQUIREMENTS = -50 1
C
WATER REQUIREMENTS x even ' WATER REQUIREMENTS z +50 Z
:El.Pr.=10 E1.Pr.=20 El.Pr.=30:E1.Pr.=10 E1.Pr.=20 EI.Pr.=30:E1.Pr.x10 EI.Pr.=20 EI.Pr.z30
CONSTR. Agr.Pr.=30:
COSTS= |Agr.Pr.=40!
-25 1 :Agr.Pr.=50:
1.91
2.57
3.24
2.21
2.87
3.53
2.51
3.17
3.83
CONSTR. Agr.Pr.=30
COSTS= lAgr.Pr.=40:
even |Agr.Pr.=50:
1.79
2.45
3.11
2.09
2.75
3.41
2.39 I
3.05 1
3.71 1
CONSTR.|Aqr.Pr.=30:
COSTS= |Agr.Pr.=40:
+25 1 |Agr.Pr.=50:
1.67
2.33
2.99
1.96
2.63
3.29
2.26 1
2.92 I
3.59 1
:---|------- :-----------------------
----
1.22
1.55
1.88
1.52
1.85
2.18
0.80
1.13
1.46
1.10
1.43
1.76
0.67
1.00
1.34
0.97
1.30
1.63
0.59
4.81
1.03
0.89
1.11
1.33
1.19
1.41
1.63
1.39
1.72
2.06
0.47
0.69
0.91
0.76
0.99
1.21
1.06
1.28
1.50
1.27
1.60
1.93 ;
0.34
0.56
0.78
0.64
0.86
1.08
0.94
1.16
1.38
11
WATER
REQUIREMENTS = -50 Z : WATER REQUIREMENTS = even
Z
=
=
=
=
=
------------------
---------------------
- ---------------------- ---- 01
DISCOUNT RATE:
10
0.92
1.25
1.58
=
=
=
=
=
1 WATER REQUIREMENTS = +50 Z
=
=
=
=
=
=
=
IEI.Pr.z10 EI.Pr.=20 EI.Pr.z30!E.Pr.z10 EI.Pr.=20 EI.Pr.=301E1.Pr.=10 E.Pr.=20 E.Pr.=30
-------------------------
------:a-----------------------------------
CONSTR.Agr.Pr.=30
COSTS= ;Agr.Pr.=40
-25 1 |Agr.Pr.=50
----
------
1.83
2.49
3.15
2.12
2.79
3.45
2.42 |
3.08
3.75 |
0.83
1.16
1.49
2.27 1
2.93
3.59
0.68
1.01
1.34
0.98
1.31
1.64
0.53
0.86
1.19
0.83
1.16
1.49
!---------------------a
CONSTR.1Agr.Pr.=30
COSTS= !Agr.Pr.=40
even IAgr.Pr.=50:
---- ------
1.67
2.33
3.00
:----------------
CONSTR.!Aqr.Pr.=30
COSTS= :Agr.Pr.=40:
+25 1 :Agr.Pr.=50:
1.52
2.18
2.84
1.13
1.46
1.79
------
1.97
2.63
3.29
-
2.12
2.78
3.44
12 1
0.50
0.72
0.94
0.80
1.02
1.24
1.10
1.32
1.54
0.35
0.57
0.79
0.65
0.87
1.09
0.95
1.17
1.39
S
1.28
1.61
1.94
---
: -----
1.13
1.46 |
1.79 1
--------
0.20
0.42
0.64
--------
0.50
0.72
0.94
0.79
1.01
1.24
----
--------------------
---------------------------------
DISCOUNT RATE:
---
------------------
1.82
2.48
3.14
1.43
1.76
2.09
WATER
REQUIREMENTS = -50 1 WATER REQUIREMENTS = even
WATER REQUIREMENTS = +50 1
=========================================================================================
E.Pr.=10 EI.Pr.=20 EI.Pr.:30 E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.:Agr.Pr.=30:
COSTS= .Agr.Pr.=40
-25 Z |Agr.Pr.=50
1.74
2.40
3.06
2.04
2.70
3.36
2.34
3.00
3.66
0.75
1.08
1.41
1.05
1.38
1.71
1.34
1.67
2.01
0.42
0.64
0.86
0.71
0.94
1.16
1.01
1.23
1.45
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40
1.56
2.22
1.86
2.52
2.16
2.82
0.57
0.90
0.86
1.20
1.16
1.49
0.24
0.46
0.53
0.75
0.83
1.05
2.88
3.18
3.48
1.23
1.53
1.82
0.68
0.97
1.27
even
Agr.Pr.=50
CONSTR. :Aqr.Pr.=301
1.38
1.68
1.97
0.38
0.68
0.98
0.05
0.35
0.65
COSTS= :Agr.Pr.=40;
2.04
2.34
2.64
0.72
1.01
1.31
0.27
0.57
0.87
+25 1 Agr.Pr.=50
2.70
3.00
3.30
1.05
1.34
1.64
0.50
0.79
1.09
157
WATER
REQUIREMENTS a -50 1
DISCGUNT RATE:
D
of Alternative
Net benefits
TABLE C-4:,
WATER REQUIREMENTS x even
WATER REQUIREMENTS = +50 1
W
:EI.Pr.=10 El.Pr.=20 EI.Pr.=30!E1.Pr.=10 EI.Pr.=20 El.Pr.=30:E1.Pr.=10 El.Pr.220 E1.Pr.=30
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40
-25 Z Agr.Pr.=50
--
1.89
2.55
3.21
1-
--
-
-
COKSTR.JAgr.Pr.=30
COSTS= Agr.Pr.=40
even Agr.Pr.=50
-----; -
-
-
-
-:-
-
-
-
-
DISCOUNT RATE:
10
-
-
-
-
-
-
-
-
-
2.48
3.14
3.80
---
-
2.05
2.71
3.38
-
-
-
1.63
2.29
2.95
--
-
---
1.76
2.42
3.08
COMSTR. :Agr.Pr.=30:
COSTS= Agr.Pr.=40
+25 1 Agr.Pr.=50
-- - - - - -
-
2.18
2.85
3.51
-
--
--
-
-
-
-
-
-
-
---
-
-
---
-
-
-
---
-
-
- ------
-
1.48
1.81
2.15
---
-
1.06
1.39
1.72
-
-
-
0.63
0.96
1.29
-
-
1.19
1.52
1.85
0.76
1.09
1.43
2.21
2.88
3.54
-
--
2.35
3.01
3.67
1.92
2.58
3.25
-
-
0.89
1.23
1.56
- - -
-
-
-
-
- --
==
=
==
===
=
=
-
-
-
--
:
-
-
-
--
-
----
-
-----
-
-
-
1.15
1.37
1.59
--
--
0.73
0.95
1.17
0.30
0.52
0.74
-- - - -- -
0.86
1.08
1.30
0.43
0.65
0.87
:
:
-
WATER
REQUIREMENTS = -50 I : WATER
REQUIREMENTS = even
1
-
1.22
1.55
1.88
-------
----
:
- -
1.35
1.68
2.01
-
0.93
1.26
1.59
-
0.56
0.78
1.00
-.
.
.
0.59
0.82
1.04
-
-
-
-
-
---
-.--
0.89
1.11
1.33
-
-
-
1.02
1.24
1.46
--
-
--
: WATER
REQUIREMENTS = +50 1
===
=
=
=
IE.Pr.zlO EI.Pr.=20 EI.Pr.z30 E1.Pr.=10 EI.Pr.=20 E.Pr.=30!E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
------
----
!---------------
CONSTR.JAqr.Pr.=30
COSTS= ;Aer.Pr.=40:
-25 1 Agr.Pr.=50
1.80
2.46
3.12
2.09
2.75
3.42
-
--------------
2.38
3.05
3.71
---------------
0.80
1.13
1.46
1.10
1.43
1.76
1.39
1.72
2.05
0.47
0.69
0.91
0.76
0.98
1.21
1.06
1.28
1.50
------:-----:-------------------------------------------.-------
CONSTR.!Agr.Pr.=30:
COSTS= :Aqr.Pr.=40:
even Agr.Pr.=50:
1.63
2.30
2.96
1.93
2.59
3.25
2.22
2.88
3.55
0.64
0.97
1.30
0.93
1.26
1.60
1.23
1.56
1.89
0.31
0.53
0.75
0.60
0.82
1.04
0.9"
1.12
1.34
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
+25 1 :Agr.Pr.=50:
1.47
2.13
2.80
1.76
2.43
3.09
2.06
2.72
3.38
0.48
0.81
1.14
0.77
1.10
1.43
1.06
1.40
1.73
0.15
0.37
0.59
0.44
0.66
0.88
0.73
0.95
1.18
WATER REQUIREMENTS = -50 1 WATER REQUIREMENTS = even
DISCOUNT RATE:
12 1
: WATER REQUIREMENTS = +50 Z
:EI.Pr.=10 EI.Pr.=20 EI.Pr.=30 E1.Pr.=10 EI.Pr.=20 EI.Pr.=30 E.Pr.=10 El.Pr.=20 EI.Pr.=30
CONSTR. Agr.Pr.=30
COSTS= !Agr.Pr.=40
-25 1 Agr.Pr.=50
1.70
2.37
3.03
CONSTR. :Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
even tAgr.Pr.=50:
CONSTR. Agr.Pr.=30
1.32
COSTS= :Agr.Pr.=40:
1.98
+25 1 Agr.Pr.=50
2.64
2.94
2.00
2.66
3.32
2.29
2.95
3.62
1.51
1.80
2.17
2.84
2.47
3.13
0.71
1.04
1.37
1.00
1.34
1.67
1.30
1.63
1.96
0.38
0.60
0.82
0.67
0.89
1.11
0.97
1.19
1.41
2.10
0.52
0.81
1.10
0.19
0.48
0.77
2.76
3.42
0.85
1.18
1.14
1.47
1.44
1.77
0.41
0.63
0.70
0.92
0.99
1.22
1.61
1.91
0.32
0.62
0.91
-0.01
0.29
0.58
2.27
2.57
0.66
0.95
1.24
0.21
0.51
0.80
3.23
0.99
1.28
1.57
0.43
0.73
1.02
158
WATER
REQUIREMENTS = -50 2
DISCOUNT RATE:
E
of Alternative
Net benefits
C-5:
TABLE
ATER REQUIREMENTS=
even
WATER REQUIREMENTS z +50 1
W
:El.Pr.=10 EI.Pr.=20 EI.Pr.=30 El.Pr.=10 El.Pr.=20 EI.Pr.=30 E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR. Agr.Pr.=30
COSTS= :Agr.Pr.=40
-25 2 Agr.Pr.=50
1.98
2.64
3.30
2.46
3.13
3.79
2.95
3.61
4.27
0.99
1.32
1.65
1.47
1.80
2.13
1.96
2.29
2.62
0.66
0.88
1.10
1.14
1.36
1.58
1.63
1.85
2.07
CONSTR.!Agr.Pr.:30!
COSTS= Agr.Pr.=40
even :Agr.Pr.=50:
1.82
2.48
3.14
2.30
2.96
3.62
2.79
3.45
4.11
0.82
1.15
1.49
1.31
1.64
1.97
1.79
2.12
2.46
0.49
0.71
0.93
0.98
1.20
1.42
1.46
1.68
1.90
-- - - !-
-
-
-
-
---
-
CONSTR.JAgr.Pr.=30:
COSTS= lAgr.Pr.=40
+25 1 Agr.Pr.=50
---
---
1.65
2.31
2.98
--
--
2.14
2.80
3.46
---------------
---
-----
2.62
3.28
3.95
--
-
0.66
0.99
1.32
---
-
-
1.15
1.48
1.81
- -----------
---
1.63
1.96
2.29
---------------------------------------
------
0.81
1.04
1.26
-------
1.30
1.52
1.74
--------
WATER
REQUIREMENTS = -50 1 i WATER REQUIREMENTS = even
DISCOMNT RATE:
-
0.33
0.55
0.77
WATER REQUIREMENTS = +50 1
W
:EJ.Pr.=10 El.Pr.=20 El.Pr.=30:El.Pr.=10 E1.Pr.=20 EI.Pr.=30:E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.2Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25 %|Agr.Pr.=50
1.86
2.53
3.19
2.35
3.01
3.67
------ :--------
CONSTR.IAqr.Pr.z301
COSTS= :Agr.Pr.=40
even :Agr.Pr.=50:
----
-----
-- - - : --
-
2.83 1
3.50 1
4.16
--
----
1.66
2.32
2.98
-
DISCOUNT RATE:
12
-
-
1.46
2.12
2.78
-
-
-
-
-
-
-
-
0.87
1.20
1.53
1.36
1.69
2.02
1.84
2.17
2.50 1
1.15
1.49
1.82
1.64
1.97 1
2.30 |
--------------------
2.15
2.81
3.47
2.63 1
3.29
3.95 :
1.94
2.61
3.27
2.43
3.09
3.75
-
-
-
-
-
-
0.47
0.80
1.13
-
-
-
-
-
-
1.03
1.25
1.47
1.51
1.73
1.95
0.34
0.56
0.78
0.82
1.04
1.26
1.31
1.53
1.75
0.14
0.36
0.58
0.62
0.84
1.06
1.11
1.33
1.55
-
-
-------------
-
------
0.95
1.28
1.61
--
---------
0.54
0.76
0.98
1-------
0.67
1.00
1.33
------------------------------------
--------------
CONSTR.:Agr.Pr.=30!
COSTS= IAgr.Pr.=40
+25 2 !Agr.Pr.=50
-
------------------------------
----------------
-
1.44
1.77
2.10
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
WATER
REQUIREMENTS = -50 2 WATER
REQUIREMENTS = even : WATER REQUIREMENTS = +50 Z
=
:E1.Pr.=10 EI.Pr.=20 E1.Pr.=30!El.Pr.=10 EI.Pr.=20 EI.Pr.=30 E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.tAgr.Pr.=30
COSTS= |Agr.Pr.=40
-25 X |Agr.Pr.=50:
1.75
2.41
3.07
2.23
2.90
3.56
2.72
3.38 1
4.04
0.76
1.09
1.42
1.24
1.57
1.90
1.73
2.06
2.39
0.43
0.65
0.87
0.91
1.13
1.35
1.40
1.62
1.84
CONSTR.:Agr.Pr.=30:
COSTS= |Agr.Pr.=40,
even Agr.Pr.=50
1.51
2.17
2.83
1.99
2.66
3.32
2.48
3.14
3.80
0.52
0.85
1.18
1.00
1.33
1.66
1.49
1.82
2.15
0.19
0.41
0.63
0.67
0.89
1.16
1.38
1.11
1.60
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40
+25 1 Agr.Pr.=50:
1.27
1.93
2.59
1.75
2.42
3.08
2.24
2.90
3.56
0.28
0.61
0.94
0.76
1.09
1.42
1.25
1.58
1.91
-0.05
0.17
0.39
0.43
0.65
0.87
0.92
1.14
1.36
159
TABLE
Net benefits
C-6:
DISCOUNT RATE:
WATER
REQUIREMENTS = -50 1
of Alternative
WATER
REQUIREMENTS -
F
even
WATER REQUIREMENTS = +50 1
W
El.Pr.=10 EI.Pr.=20 EI.Pr.=30 E1.Pr.=10 E1.Pr.=20 El.Pr.=30:El.Pr.=10 EI.Pr.:20 EI.Pr.:30
CONSTR. Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25 1 :Agr.Pr.=50:
-----:-----
-----
3.89
5.30
6.70
4.19
5.59
6.99
1.50
2.20
2.90
1.79
2.49
3.19
2.09
2.79
3.49
-------------------------------
CONSTR.:Agr.Pr.z30!
COSTS= |Agr.Pr.=40!
even :Agr.Pr.=50:
---- :
3.60
5.00
6.40
3.30
4.70
6.10
3.59
4.99
6.39
0.80
1.26
1.73
:
-- :-------
3.89
5.29 1
6.69 1
1.19
1.89
2.60
-
1.09
1.56
2.03
--------
---
1.49
2.19
2.89
1.79 1
2.49 1
3.19
0.49
0.96
1.43
0.79
1.26
1.73
1.19
1.89
2.59
1.49 1
2.19 I
2.89:
0.19
0.66
1.13
0.49
0.96
1.43
:-------------------------------------------------
CONSTR.:Agr.Pr.=30|
COSTS= lAgr.Pr.=40:
+25 1 :Agr.Pr.=50|
2.99
4.39
5.79
3.59 |
4.99 1
6.39 '
3.29
4.69
6.09
----------- --------------
0.89
1.59
2.29
1.39
1.86
2.33
1.09
1.56
2.03
-
--
|------------------
-
0.79
1.26
1.72
----------------
---------------|----------------------------------------------------------------------- --
WATER
REQUIREMENTS = -50 1
DISCOUNT RATE:
WATER
W
REQUIREMENTS = even
WATER REQUIREMENTS = +50 1
W
IE1.Pr.=10 E1.Pr.=20 EI.Pr.z301E1.Pr.=10 El.Pr.=20 E1.Pr.=301El.Pr.z10 E1.Pr.=20 EI.Pr.=30
------
: -----------------------------
CONSTR.JAgr.Pr.=30:
COSTS= |Agr.Pr.=40:
-25 1 :Agr.Pr.=50:
----i -- --
3.38
4.78
6.18
-- :----
3.68
5.08
6.48
----
---------------------:
3.98
5.38
6.78
1.28
1.98
2.68
1.58
2.28
2.98
1.88
2.58 1
3.28 1
----------------------
-----
0.88
1.35
1.81
1.18
1.65
2.11
0.21
0.68
1.14
0.51
0.98
1.44
0.81
1.27
1.74
-0.16
0.30
0.77
0.14
0.60
1.07
0.43
0.90
1.37
----------------
COSTR.!Agr.Pr.z30:
COSTS= !Agr.Pr.=40:
even :Agr.Pr.=50:
3.01
4.41
5.81
3.31
4.71
6.11
3.61
5.01 1
6.41 1
0.91
1.61
2.31
1.21
1.91
2.61
1.51 1
2.21
2.91 1
CONSTR.1Agr.Pr.=30:
COSTS= :Agr.Pr.=401
+25 1 1Agr.Pr.=501
2.64
4.04
5.44
2.94
4.34
5.74
3.24 1
4.64
6.04
0.54
1.24
1.94
0.84
1.54
2.24
1.13
1.84
2.54
DISCOUNT RATE:
12 1
0.58
1.05
1.52
WATER REQUIREMENTS = -50 Z : WATER
REQUIREMENTS = even
: WATER
REQUIREMENTS = +50 1
1E.Pr.=10 EI.Pr.=20 EI.Pr.=30:El.Pr.=10 EI.Pr.=20 EI.Pr.=30!EI.Pr.=10 EI.Pr.=20 E1.Pr.=30
-------------- |:---------------------------
----------------------------- -----------------------------
CONSTR. Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25 1 |Agr.Pr.=50:
3.17
4.57
5.97
3.47
4.87
6.27
3.77
5.17
6.57
1.07
1.77
2.47
1.37
2.07
2.77
1.67
2.37
3.07
0.37
0.84
1.31
0.67
1.14
1.60
0.97
1.44
1.90
ONSTR. Agr.Pr.=30
COSTS= :Agr.Pr.=40!
even |Agr.Pr.=50!
2.73
4.13
5.53
3.03
4.43
5.83
3.33
4.73
6.13
0.63
1.33
2.03
0.93
1.63
2.33
1.23
1.93
2.63
-0.07
0.40
0.86
0.23
0.70
1.16
0.53
0.99
1.46
CONSTR.|Agr.Pr.=30:
2.29
2.59
2.89
0.19
0.49
0.78
-0.51
-0.21
0.08
COSTS=
Agr.Pr.=40:
3.69
3.99
4.29
0.89
1.19
1.49
-0.05
0.25
0.55
+25 1 Agr.Pr.=50:
5.09
5.39
5.69
1.59
1.89
2.19
0.42
0.72
1.02
160
TABLE
C-7:
of Alternative
WATER
REQUIREMENTS = -50 1 1 WATER
REQUIREMENTS = even
DISCOUNT RATE:
8
Net benefits
I
G
. WATER REQUIREMENTS = +50 1
|========================================================--=-
E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 El.Pr.=30 E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.!Agr.Pr.=30
COSTS= :Agr.Pr.=40:
-25 1 :Agr.Pr.z50:
3.45
4.94
6.43
------- |----------
3.75
5.24
6.73
4.05
5.54 :
7.03
-------------------------------
CONSTR.JAgr.Pr.=30
COSTS= !Agr.Pr.=40
even lAgr.Pr.=50-
3.01
4.50
5.99
3.31
4.80
6.29
~~~~
2.58
4.07
5.55
2.88
4.36
5.85
1.52
2.26
3.01
1.82
2.56
3.30
0.47
0.97
1.47
-------------------------
3.61 |
5.10
6.59 1
--------------------:
CONSTR.|Agr.Pr.=30:
'COSTS= |Agr.Pr.=40
+25 1 |Agr.Pr.=50
1.22
1.96
2.71
0.78
1.53
2.27
1.08
1.82
2.57
0.34
1.09
1.83
0.04
0.53
1.03
--
0.64
1.39
2.13
1.07
1.57
2.06
0.34
0.83
1.33
0.63
1.13
1.63
-0.10
0.39
0.89
0.20
0.69
1.19
----------------------------
1.38
2.12
2.87
----------
3.17 |
4.66
6.15 1
0.77
1.27
1.77
0.94
1.69.:
2.43
-----------------
-0.40
0.10
0.59
DISCOUNT RATE:
WATER REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even : WATER REQUIREMENTS = +50 1
10%-------------1
EI.Pr.=10 EI.Pr.=20 EI.Pr.=301E1.Pr.=10 EI.Pr.=20 EI.Pr.=301E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
------
:
--------------
-----------------------
--------------
CONSTR.Agr.Pr.=30:
COSTS= :Agr.Pr.=40!
-25 1 IAgr.Pr.=50:
3.14
4.63
6.12
3.44
4.93
6.42
3.74
5.23
6.72
0.91
1.65
2.40
1.21
1.95
2.70
1.51 1
2.25 1
2.99
0.16
0.66
1.16
0.46
0.96
1.46
0.76
1.26
1.75
CONSTR.Agr.Pr.=30:
COSTS= IAgr.Pr.=40
even :Agr.Pr.=50:
2.60
4.09
5.58
2.90
4.39
5.88
3.20
4.69 |
6.18 1
0.37
1.11
1.86
0.67
1.41
2.16
0.96 1
1.71 1
2.45 |
-0.38
0.12
0.62
-0.08
0.42
0.91
0.22
0.72
1.21
---------
--- - ------------------------
CONSTR.'Agr.Pr.=30;
COSTS= lAgr.Pr.=401
+25 1 lAgr.Pr.=50
2.06
3.55
5.04
2.36
3.85
5.34
---------------------------- ------------------ ------- ----
2.66 1
4.15 1
5.63
--------|---------- ----- ------------------------
-0.17
0.57
1.32
0.13
0.87
1.61
0.42 1
1.17 1
1.91 |
-0.62
-0.12
0.37
-0.32
0.18
0.67
i-------------------------- ---|-----------------------------
WATER REQUIREMENTS = -50 1: WATER REQUIREMENTS = even
DISCOUNT RATE:
-0.92
-0.42
0.08
| WATER REQUIREMENTS = +50 1
1E1.Pr.=10 EI.Pr.=20 El.Pr.=30 E1.Pr.=10 E1.Pr.=20 EI.Pr.=30|E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR. Agr.Pr.=30
COSTS= lAgr.Pr.=40
-25 1 lAgr.Pr.=50
2.84
4.33
5.81
3.14
4.62
b.11
3.43
4.92
6.41
0.60
1.35
2.09
0.90
1.65
2.39
1.20
1.94
2.69
-0.14
0.36
0.85
0.16
0.65
1.15
0.46
0.95
1.45
CONSTR. :Agr.Pr.=30:
COSTS= Agr.Pr.=40
even :Agr.Pr.=50:
2.19
3.68
5.17
2.49
3.98
5.47
2.79
4.28
5.77
-0.04
0.71
1.45
0.26
1.00
1.75
0.56
1.30
2.05
-0.78
-0.29
0.21
-0.48
0.01
0.51
-0.19
0.31
0.81
ONSTR. Agr.Pr.=30
1.55
1.85
2.15
-0.68
-0.38
-0.08
-1.43
-1.13
-0.83
COSTS= |Agr.Pr.=40!
+25 1 lAgr.Pr.:50'
3.04
4.53
3.34
4.83
3.64
5.13
0.06
0.81
0.36
1.11
0.66
1.40
-0.93
-0.43
-0.63
-0.13
-0.33
0.16
161
Net benefits
C-8:
TABLE
WATER REQUIREMENTS = -50 1
DISCOUNT RATE:
H
of Alternative
WATER
W
REQUIREMENTS = even
|
WATER REQUIREMENTS = +50 1
.EI.Pr.=10 E1.Pr.=20 EI.Pr.=30.E1.Pr.=10 E1.Pr.=20 El.Pr.=30:E1.Pr.=10 El.Pr.=20 El.Pr.=30
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40|
-25 1 iAgr.Pr.=50:
3.14
4.70
6.26
3.44
5.00
6.56
3.73
5.30
6.86
0.79
1.58
2.36
1.09
1.87
2.65
1.39 1
2.17
2.95
0.01
0.53
1.06
0.31
0.83
1.35
0.18
0.96
1.74
0.48
1.26
2.04
0.78
1.56 1
2.34
-0.60
-0.08
0.44
-0.30
0.22
0.74
-0.00
0.52
1.04
-0.14
0.64
.1.42
0.16
0.94 1
1.72 1
-1.22
-0.70
-0.18
-0.92
-0.40
0.12
-0.62
-0.10
0.42
0.61
1.13
1.65
---- !-----:---------------------------------------------------
CONSTR.!Agr.Pr.=30t
COSTS= :Agr.Pr.=40;
even :Agr.Pr.=50:
----! -----
--
---
2.82
4.38
5.94
3.12
4.68:
6.241
-
---
:-----------------------------------------------------
CONSTR.:Agr.Pr.=30
COSTS= 1Agr.Pr.=40:
425 1 :Agr.Pr.=50!
----
2.52
4.08
5.64
--
1.91
3.47
5.03
-
2.21
3.77
5.33
-
-
-
-
2.50
4.07 1
5.63 |
-
-
-
-
- --
-0.44
0.35
1.13
-
- --
- -- -
- - - ---- --
-
-
-
-
-
WATER REQUIREMENTS = -50 1 ; WATER
W
REQUIREMENTS = even
DISCOUNT RATE:
-
-
-
-
-
-
-
-
-
-
-
-
-
-
WATER REQUIREMENTS = +50 1
W
.El.Pr.=10 EI.Pr.=20 EI.Pr.=30!,El.Pr.=10 EI.Pr.=20 El.Pr.=30:El.Pr.=10 EI.Pr.=20 EI.Pr.=30
: ---------------
-:-----------------------------
------------
CONSTR..Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
-25 1 :Agr.Pr.=50:
2.70
4.26
5.82
3.00
4.56
6.12
3.30 1
4.86
6.42 I
0.36
1.14
1.92
0.66
1.44
2.22
0.96 |
1.74 I
2.52
-0.42
0.10
0.62
-0.12
0.40
0.92
0.18
0.70
1.22
CONSTR.!Agr.Pr.=301
COSTS= :Agr.Pr.=40:
even IAgr.Pr.=501
1.94
3.50
5.06
2.24
3.80
5.36
2.54 1
4.10
5.66
-0.40
0.38
1.16
-0.10
0.68
1.46
0.20
0.98
1.76 1
-1.18
-0.66
-0.14
-0.88
-0.36
0.16
-0.59
-0.06
0.46
CONSTR.Agr.Pr.=30:
COSTS= IAgr.Pr.=40:
+25 I1Agr.Pr.=50:
1.18
2.74
4.30
1.48
3.04
4.60
1.78
3.34
4.90
-1.16
-0.38
0.40
-0.86
-0.08
0.70
-0.57
0.22 1
1.00 :
-1.94
-1.42
-0.90
-1.64
-1.12
-0.60
-1.35
-0.83
-0.30
DISCOUNT RATE:
WATER
W
REQUIREMENTS = -50 1 I WATER
REQUIREMENTS = even I WATER REQUIREMENTS = +50 1
El.Pr.=10 E1.Pr.=20 EI.Pr.=30:E1.Pr.=10 El.Pr.=20 El.Pr.=30:E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.!Agr.Pr.=30
COSTS= 'Agr.Pr.=401
2.27
3.83
2.57
4.13
2.87 1
4.43 |
-0.07
0.71
0.23
1.01
0.53
1.31
-25 1 Agr.Pr.=50
5.39
5.69
5.99 |
1.49
1.79
CONSTR.:AgrPr.=301
COSTS= Agr.Pr.=40
1.37
2.93
1.67
3.23
1.97
3.53
-0.97
-0.19
:Agr.Pr.=50:
4.49
4.79
5.09
CONSTR.|Agr.Pr.=30
0.46
0.76
COSTS= :Agr.Pr.=40
2.03
2.32
+25 X :Agr.Pr.=50
3.59
3.89
even
-0.85
-0.33
-0.55
-0.03
-0.25
0.27
2.09
0.19
0.49
0.79
-0.68
0.11
-0.38
0.40
-1.75
-1.23
-1.46
-0.94
-1.16
-0.64
0.59
0.89
1.18
-0.71
-0.42
-0.12
1.06
-1.88
-1.58
-1.28
-2.66
-2.36
-2.06
2.62
-1.10
-0.80
-0.50
-2.14
-1.84
-1.54
4.18
-0.32
-0.02
0.28
-1.62
-1.32
-1.02
|
162
TABLE C-9:
Net benefits
of Alternative
I
WATER
REQUIREMENTS = -50 Z : WATER REQUIREMENTS = even
DISCOUNT RATE:
| WATER REQUIREMENTS = +50 1
E1.Pr.=10 El.Pr.=20 E1.Pr.=30 E1.Pr.=10 EI.Pr.=20 E1.Pr.=30:E1.Pr.210 EI.Pr.=20 EI.Pr.=30
--
-
-
-
-:-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
:-
-
-
--
-
-
-
-
--
-
-
-
CONSTR.!Agr.Pr.=30
COSTS= Agr.Pr.=40
-25 1 Agr.Pr.=50
1.95
2.62
3.28
2.44
3.10
3.76
2.92
3.58
4.24
0.96
1.29
1.62
1.44
1.77
2.10
1.92 :
2.25 1
2.58
0.63
0.85
1.07
1.11
1.33
1.55
1.59
1.81
2.03
CONSTR.:Agr.Pr.=301
COSTS= :Agr.Pr.=401
even Agr.Pr.=50
1.78
2.45
3.11
2.26
2.93
3.59
2.74
3.41
4.07
0.79
1.12
1.45
1.27
1.60
1.93
1.75 :
2.08
2.41 !
0.46
0.68
0.90
0.94
1.16
1.38
1.42
1.64
1.86
CONSTR.2Agr.Pr.=30
COSTS= |Agr.Pr.=40
+25 1 |Agr.Pr.z50
1.61
2.27
2.94
2.09
2.76
3.42
2.57
3.24
3.90
0.62
0.95
1.28
1.10
1.43
1.76
1.58
1.91
2.24
0.29
0.51
0.73
0.77
0.99
1.21
1.25
1.47
1.69
------- :----------
-----------------------------
------------------------------------
----------
WATER
REQUIREMENTS = -50 1 ' WATER
REQUIREMENTS = even
DISC00LT RATE:
10 1
-
' WATER
REQUIREMENTS = +50 1
.EI.Pr.=10 El.Pr.=20 El.Pr.=30!E1.Pr.=10 EI.Pr.=20 EI.Pr.=30!E.Pr.=10 EI.Pr.=20 EI.Pr.=30
-------------- :------------ ----------------- |--------------------------- --------------------------- --
CONSTR.JAgr.Pr.=30
COSTS= |Agr.Pr.=40
-25 1 IAgr.Pr.=50
------- |:----------
CONSTR.IAgr.Pr.=301
COSTS= |Agr.Pr.=40
even |Agr.Pr.=50
----
-----
CONSTR.:Agr.Pr.=30:
COSTS= |Agr.Pr.=40
+25 1 :Agr.Pr.=50
----
1.83
2.50
3.16
2.31
2.98
3.64
2.79
3.46
4.12
0.84
1.17
1.50
----------------------------
1.62
2.28
2.95
2.58 1
3.25
3.91
1.89
2.55
3.22
2.37 1
3.03
3.70
0.63
0.96
1.29
0.42
0.75
1.08
0.51
0.73
0.95
1.59
1.92
2.25 1
0.90
1.23
1.56
1.38:
1.71 !
2.04
-
0.30
0.52
0.74
---
--
0.99
1.21
1.43
-----------
1.11
1.44
1.77
:---------------------------------------
DISCOUNT RATE:
12 1
1.80 1
2.13
2.46
----------------
--------------------
---------------
1.41
2.07
2.74
--------
2.10
2.77
3.43
1.32
1.65
1.98
1.47
1.69
1.91
-
0.78
1.00
1.22
1.26
1.48
1.70
0.56
0.79
1.01
1.05
1.27
1.49
---------------
0.08
0.31
0.53
--------------
WATER
REQUIREMENTS = -50 1 WATER
W
REQUIREMENTS = even | WATER REQUIREMENTS = +50 Z
====================================================================
EI.Pr.=10 E1.Pr.=20 EI.Pr.=301E1.Pr.=10 EI.Pr.=20 El.Pr.=30;E1.Pr.=10 E1.Pr.=20 EI.Pr.=30
CONSTR.:Agr.Pr.=30:
COSTS= Agr.Pr.=40
-25 2 Agr.Pr.=50
1.71
2.38
3.04
2.19
2.86
3.52
2.68
3.34
4.00
0.72
1.05
1.38
1.20
1.53
1.86
1.68
2.01
2.34
0.39
0.61
0.83
0.87
1.09
1.31
1.35
1.57
1.79
CONSTR. Agr.Pr.=301
COSTS= !Agr.Pr.=401
1.46
2.13
1.94
2.61
2.42
3.09
0.47
0.80
0.95
1.28
1.43
1.76
0.14
0.36
0.62
0.84
1.10
1.32
Agr.Pr.=50
2.79
3.27
3.75
1.13
1.61
2.09
0.58
1.06
1.54
CONSTR. Agr.Pr.=30
1.21
1.69
2.17
0.22
0.70
1.18
-0.12
0.37
0.85
COSTS= :Agr.Pr.=401
1.87
2.35
2.83
0.55
1.03
1.51
0.11
0.59
1.07
2.54
3.02
3.50
0.88
1.36
1.84
0.33
0.81
1.29
even
+25 I
Agr.Pr.=50,
163
TABLE C-10:
Net
J
of Alternative
benefits
WATER REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even
DISCOUNT RATE:
8 1
: WATER REQUIREMENTS a +50 1
======================================================
EI.Pr.=10 EI.Pr.=20 E1.Pr.=30:EI.Pr.=10 E1.Pr.=20 EI.Pr.=30!E.Pr.=10 EI.Pr.=20 EZ.Pr.=30
CONSTR.:Agr.Pr.=30:
COSTS= Agr.Pr.=40
-25 1 Agr.Pr.=50:
-- - - i
-
-
-
-
3.59
5.00
6.41
-i--
-
-
-
3.89
5.29
6.70
-
-
-
-
-
4.18
5.59
7.00
-
-
-
-
-
1.48
2.18
2.89
--
-
-
-
1.77
2.48
3.18
-
-
-
-
-
2.07
2.77
3.48
-
-
-
-
0.78
1.24
1.71
-
-
-
-
-
-
1.07
1.54
2.01
-- -
---------
1.36
1.83
2.30
-
-
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40
even :Agr.Pr.=50:
3.28
4.69
6.10
3.58
4.98
6.39
3.87
5.28
6.69
1.17
1.87
2.58
1.46
2.17
2.87
1.76
2.46
3.17
0.47
0.94
1.40
0.76
1.23
1.70
1.05
1.52
1.99
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40
+25 1 Agr.Pr.=50:
2.97
4.38
5.79
3.27
4.68
6.08
3.56
4.97
6.38
0.86
1.57
2.27
1.16
1.86
2.56
1.45
2.15
2.86
0.16
0.63
1.10
0.45
0.92
1.39
0.74
1.21
1.68
DISCOUNT RATE:
10 1
WATER
REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even
: WATER REQUIREMENTS = +50 1
=====
E1.Pr.=10 E1.Pr.=20 EI.Pr.=30'El.Pr.=10 EI.Pr.=20 El.Pr.=30|EI.Pr.=10 E1.Pr.=20 EI.Pr.=30
CDNSTR.:Agr.Pr.=30:
COSTS= |Agr.Pr.=40:
-25 1 :Aqr.Pr.=50
---- :
-----
3.37
4.78
6.19
----
3.67
5.07
6.48
3.96
5.37
6.78
1.26
1.96
2.67
1.55
2.26
2.96
1.85
2.55
3.26
0.85
1.32
1.79
1.14
1.61
2.08
0.17
0.64
1.11
0.47
0.94
1.41
0.76
1.23
1.70
-0.21
0.26
0.73
0.09
0.56
1.02
0.38
0.85
1.32
--------------------------------------
---------------
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40|
even |Agr.Pr.=50:
2.99
4.40
5.81
3.28
4.69
6.10
3.58
4.99
6.40
0.88
1.58
2.29
1.17
1.88
2.58
1.47
2.17
2.87 1
CONSTR.:Agr.Pr.=30:
COSTS= !Agr.Pr.=40:
425 1 :Agr.Pr.=50:
2.61
4.02
5.43
2.90
4.31
5.72
3.20
4.60
6.01
0.50
1.20
1.90
0.79
1.49
2.20
1.08
1.79
2.49
DISCOUNT RATE:
0.56
1.03
1.49
WATER
REQUIREMENTS = -50 1 : WATER
REQUIREMENTS = even
WATER REQUIREMENTS = +50 1
El.Pr.=10 El.Pr.=20 EI.Pr.=30 El.Pr.=10 EI.Pr.=20 EI.Pr.=30 El.Pr.=10 El.Pr.=20 EI.Pr.=30
-- ------------------------------------------------------------------------------------------------
CONSTR. Agr.Pr.=30
COSTS= |Agr.Pr.=40
-25 1 Agr.Pr.=50
3.16
4.57
5.97
CONSTR. Agr.Pr.=30
2.70
COSTS=
Agr.Pr.=40:
4.11
even
Agr.Pr.=50
5.52
CONSTR. Agr.Pr.=30
COSTS=
Agr.Pr.=40:
+25 1 Agr.Pr.=50
3.45
4.86
6.27
3.74
5.15
6.56
1.04
1.75
2.45
1.34
2.04
2.75
1.63
2.34
3.04
0.34
0.81
1.28
0.63
1.10
1.57
0.93
1.40
1.87
3.00
3.29
0.59
0.88
1.18
-0.11
0.18
0.47
4.41
4.70
1.29
1.59
1.88
0.36
0.65
0.94
5.81
6.11
2.00
2.29
2.59
0.83
1.12
1.41
2.25
2.54
2.84
0.14
0.02
4.25
0.84
-0.10
0.20
0.49
5.07
5.36
5.65
1.55
0.72
1.43
2.13
-0.27
3.95
0.43
1.13
1.84
-0.57
3.66
0.37
0.67
0.96
1 64
TABLE
DISCOUNT RATE:
8
Net
C-il:
benefits
of Alternative
K
WATER REQUIREMENTS = -50 1 !WATER REQUIREMENTS = even
I
| WATER REQUIREMENTS = +50 1
============================================================
EI.Pr.=10 EI.Pr.=20 E1.Pr.=30 E1.Pr.=10 EI.Pr.=20 E1.Pr.=30|E.Pr.=10 El.Pr.=20 El.Pr.-30
CONSTR. :Aqr.Pr.=30
COSTS= Agr.Pr.=40
-251 Agr.Pr.=50
3.43
4.92
6.41
3.73
5.22
6.71
4.02
5.51
7.00
1.19
1.94
2.69
1.49
2.23
2.98
1.78
2.53
3.27
0.45
0.95
1.44
0.74
1.24
1.74
1.04
1.53
2.03
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
even Agr.Pr.=50
2.99
4.48
5.97
3.28
4.77
6.26
3.57
5.07
6.56
0.75
1.49
2.24
1.04
1.79
2.53
1.34
2.08
2.83:
0.00
0.50
1.00
0.30
0.79
1.29
0.59
1.09
1.59
0.30
1.05
1.79
0.60
1.34
2.09
0.89
1.64:
2.31 :
-0.15
0.35
0.85
0.15
0.64
1.14
---- ;
-----
!---------------------------------------
CONSTR.IAgr.Pr.=30
COSTS= :Agr.Pr.=40:
+25 2 :Agr.Pr.=50:
11----
2.54
4.03
5.52
1-
2.83
4.33
5.82
--------
3.13
4.62
6.11
---
I
----------
----------------
------------------------------
DISCOUNT RATE:
------------------
--------
MATER REQUIREMENTS = -50 1
-0.44
0.05
0.55
----------------------------
: WATER REQUIREMENTS = +50 1
WATER REQUIREMENTS = even
!E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 E1.Pr.=20 EI.Pr.=30:E.Pr.=10 EI.Pr.=20 E1.Pr.=30
NSTR.:Aqr.Pr.=30:
COSTS= :Agr.Pr.z40:
-25 1 :Agr.Pr.=50:
3.12
4.61
6.10
3.41
4.90
6.39
3.70
5.19
6.69
2.56
4.06
5.55
2.86
4.35
5.84
3.15
4.64
6.14
TR.:Agr.Pr.=30:
COSTS= Agr.Pr.=40;
even Agr.Pr.=50
-----
2.60
4.09
5.58
1.47
2.21
2.96
0.13
0.63
1.13
0.43
0.92
1.42
0.33
1.07
1.82
0.62
1.37
2.11
0.92
1.66
2.41
-0.42
0.08
0.58
-0.12
0.37
0.87
0.17
0.67
1.16
2.01
3.51
5.00
------:----
2.31
3.80
5.29
-0.22
0.52
1.27
0.07
0.82
1.56
0.37
1.11
1.86
-0.67
-0.18
0.32
-0.38
0.12
0.61
-
---------------------------
---------------
0.72
1.22
1.72
---------------
:----------------------------
CONSTR.:Agr.Pr.=30
COSTS= :Agr.Pr.=40
+251 :Agr.Pr.=50:
DISCOUNT RATE:
12 1
1.17
1.92
2.66
---- a-------------------------------------
---- :------
----:
0.88
1.62
2.37
----------------------
-0.97
-0.47
0.03
------
-
WATER REQUIREMENTS = -50 1 WATER REQUIREMENTS = even : WATER REQUIREMENTS = +50 1
-================================-===============
:EI.Pr.=10 E1.Pr.=20 EI.Pr.=30 E.Pr.=10 EI.Pr.=20 E1.Pr.=30 E1.Pr.=10 E1.Pr.=20 E1.Pr.=30
CONSTR.:Agr.Pr.=30
COSTS= :Agr.Pr.=40
-25 1 :Agr.Pr.=50:
2.80
4.30
5.79
3.10
4.59
6.08
3.39
4.88
6.38
0.57
1.31
2.06
0.86
1.61
2.35
1.16
1.90
2.65
-0.18
0.32
0.82
0.12
0.61
1.11
0.41
0.91
1.40
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40
even :Agr.Pr.=50:
2.15
3.64
5.13
2.44
3.94
5.43
2.74
4.23
5.72
-0.09
0.66
1.40
0.21
0.95
1.70
0.50
1.25
1.99
-0.83
-0.33
0.16
-0.54
-0.04
0.46
-0.24
0.25
0.75
CONSTR. Agr.Pr.=30
COSTS= :Agr.Pr.=40
+25 1 Aqr.Pr.=50
1.50
2.99
4.48
1.79
3.28
4.77
2.08
3.58
5.07
-0.74
0.00
0.75
-0.45
0.30
1.04
-0.15
0.59
1.34 |
-1.49
-0.99
-0.49
-1.19
-0.70
-0.20
-0.90
-0.40
0.10
165
TABLE C-i2:
DISCOUNT RATE:
81
Net benefits
of Alternative
WATER REQUIREKENTS = -50 1 | WATER REQUIREMENTS = even
L
: WATER
REQUIREMENTS a +50 1
==================================2
|EI.Pr.=10 E1.Pr.=20 EI.Pr.=30|E1.Pr.=10E1.Pr.=20 EI.Pr.=30|E.Pr.=10 EI.Pr.=20 EI.Pr.=3U
CONSTR.:Agr.Pr.=3o:
COSTS= !Aqr.Pr.=40|
-25 1 :Agr.Pr.=50:
3.14
4.72
6.29
3.44
5.01
6.58
3.73
5.30
6.88|
0.78
1.57
2.36
1.08
1.86
2.65
1.37
2.16
2.95:
-0.00
0.52
1.05
0.29
0.82
1.34
0.59
1.11
1.63
CONSTR.Agr.Pr.=30;
COSTS= |Agr.Pr.=40:
even |Agr.Pr.=50|
2.52
4.09
5.67
2.81
4.39
5.96
3.11
4.68 :
6.25 |
0.16
0.95
1.73
0.45
1.24
2.03
0.75
1.54
2.32
-0.63
-0.10
0.42
-0.33
0.19
0.72
-0.04
0.49
1.01
-0.46
0.32
1.11
-0.17
0.62
1.40
0.13
0.91
1.70
---- :
-----
:---------------:-----------------------------------------
CONSTR.!Agr.Pr.=30:
COSTS= :Agr.Pr.=40:
+25 Z :Agr.Pr.=50:
1.90
3.47
5.04
2.19
3.76
5.34
2.49 1
4.06 1
5.63 |
---
-1.25
-0.72
-0.20
-0.95
-0.43
0.09
-0.66
-0.14
0.39
-----------------------------------------------------------------------------------DISCOUNT RATE:
WATER REQUIREMENTS = -50 1 : WATER
REQUIREMENTS = even : WATER REQUIREMENTS = +50 1
10
=
E1.Pr.=10 E.Pr.=20 EI.Pr,=30|E1.Pr.=10 E1.Pr.=20 E1.Pr.=30!E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
-----------------------------------------------------------------------------------------------
CONSTR.Agr.Pr.=30:
COSTS= Agr.Pr.=40|
-25 Z Agr.Pr.=50!
2.70
4.28
5.85
3.00
4.57
6.14
3.29 |
4.86 1
6.44 |
CONSTR.Agr.Pr.=30:
COSTS= Agr.Pr.=40|
even 1Agr.Pr.=50;
1.93
3.50
5.08
2.23
3.80
5.37
2.52
4.09
5.67
0.34
1.13
1.92
0.64
1.42
2.21
0.93 1
1.72 |
2.50 1
-0.44
0.08
0.60
-0.15
0.37
0.90
0.14
0.67
1.19
-0.43
0.36
1.15
-0.13
0.65
1.44
0.16
0.95 |
1.73
-1.21
-0.69
-0.17
-0.92
-0.40
0.13
-0.63
-0.10
0.42
------- :----------|:----------------------------- ----------------------------- -----------------------------
CONSTR.Agr.Pr.=30:
COSTS= 1Agr.Pr.=40:
+25 1 lAgr.Pr.=50:
DISCOUNT RATE:
12 1
1.16
2.73
4.31
1.46
3.03
4.60
1.75
3.32
4.90
-1.20
-0.41
0.37
-0.90
-0.12
0.67
-0.61
0.18
0.96
-1.98
-1.46
-0.94
-1.69
-1.17
-0.64
-1.40
-0.87
-0.35
WATER
REQUIREMENTS = -50 1
WATER
REQUIREMENTS = even : WATER
REQUIREMENTS = +50 1
=========================
:EI.Pr.=10 El.Pr.=20 E.Pr.=30 E1.Pr.=10 EI.Pr.=20 E1.Pr.=30|E1.Pr.=10 El.Pr.=20 EL.Pr.=30
CONSTR. 1Agr.Pr.=30:
COSTS= Agr.Pr.=40
2.27
3.84
2.56
4.13
2.86
4.43
-0.09
0.69
0.20
0.99
0.50
1.28
-0.88
-0.35
-0.58
-0.06
-0.29
0.23
-25 1 lAqr.Pr.=50
5.41
5.71
6.00
1.48
1.77
2.07
0.17
0.46
0.76
CONSTR. :Agr.Pr.=30:
-1.01
-0.71
-0.42
-1.79
-1.50
-0.22
0.07
0.37
-1.27
-0.98
-0.68
Agr.Pr.=40:
1.35
2.93
1.65
COSTS=
3.22
1.94
3.51
even
Agr.Pr.=50
4.50
4.79
5.09
0.57
0.86
1.15
-0.75
-0.45
-0.16
CONSTR. :Agr.Pr.=:30
COSTS= Agr.Pr.=40
+25
:Aqr.Pr.=50:
0.44
2.01
3.58
0.73
2.30
3.88
1.02
2.60
4.17
-1.92
-1.14
-0.35
-1.63
-0.84
-0.06
-1.34
-0.55
0.24
-2.71
-2.19
-1.66
-2.42
-1.89
-1.37
-2.12
-1.60
-1.07
X
-1.21
1 66
TABLE
Net benefits
C-13:
WATER REQUIREMENTS = -50 1
Z
DISCOUNT RATE:
of Alternative
WATER REQUIREMENTS = even
M
: WATER REQUIREMENTS = +50 1
EI.Pr.=10 E1.Pr.=20 EI.Pr.=30:EI.Pr.=10 EI.Pr.=20 E1.Pr.=30|E.Pr.=10 EI.Pr.=20 EI.Pr.230
CONSTR.2Agr.Pr.=30:
COSTS= :Agr.Pr.=40
-25 1 :Agr.Pr.=50:
3.52
5.01
6.50
4.00
5.49
6.98
4.49
5.98
7.47
1.29
2.03
2.77
1.77
2.51
3.26
2.26
3.00
3.74
0.54
1.04
1.53
1.03
1.52
2.02
1.51
2.01
2.50
CONSTR.IAgr.Pr.=30
COSTS= Agr.Pr.=40
even :Aqr.Pr.=50:
3.04
4.53
6.02
3.53
5.01
6.50
4.01
5.50
6.99
0.81
1.55
2.30
1.29
2.04
2.78
1.78
2.52
3.27
0.06
0.56
1.06
0.55
1.04
1.54
1.03
1.53
2.03
CONSTR.:Agr.Pr.=30
COSTS= Agr.Pr.=40:
+25 1 :Agr.Pr.=50:
2.56
4.05
5.54
3.05
4.54
6.03
3.53
5.02
6.51
0.33
1.07
1.82
0.82
1.56
2.30
1.30
2.04
2.79
-0.41
0.08
0.58
0.07
0.57
1.06
0.56
1.05
1.55
WATER REQUIREMENTS = -50 1 | WATER REQUIREMENTS = even : WATER REQUIREMENTS = +50 Z
DISCOUNT RATE:
:EI.Pr.=10 E1.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:E.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.Agr.Pr.=30
COSTS= |Agr.Pr.=40|
-25 1 !Agr.Pr.=50|
----
-----
3.18
4.67
6.16
3.67
5.15
6.64
4.15 1
5.64
7.13
0.95
1.69
2.44
1.43
2.18
2.92
1.92
2.66
3.41
:
:-------------------------------------------------------- ----
CONSTR.Aqr.Pr.=30:
COSTS- Agr.Pr.=40t
even :Agr.Pr.=50:
2.59
4.08
5.57
3.07
4.56
6.05
3.56
5.05
6.54
0.36
1.10
1.85
0.84
1.59
2.33
1.33
2.07
2.82
CONSTRAgr.Pr.=30:
COSTS= :Agr.Pr.=40|
+25 1 :Agr.Pr.=50!
2.00
3.49
4.98
2.48
3.97
5.46
2.97
4.46
5.95
-0.23
0.51
1.26
0.25
1.00
1.74
12
0.69
1.18
1.68
1.17
1.67
2.17
---------------------
:
:
-0.39
0.11
0.60
0.10
0.59
1.09
0.58
1.08
1.57
0.74 1
1.48
2.22
-0.9
-0.48
0.01
-0.49
0.00
0.50
-0.01
0.49
0.98
-----------
-----------------------------------
DISCOUNT RATE:
0.20
0.70
1.20
WATER
REQUIREMENTS = -50 1 : WATER REQUIREMENTS = even
: WATER REQUIREMENTS = +50 2
=====
E1.Pr.=10 EI.Pr.=20 EI.Pr.=30|E1.Pr.=10 E1.Pr.=20 EI.Pr.=30|E1.Pr.=10 EI.Pr.=20 El.Pr.=30
CONSTR.:Agr.Pr.=30'
COSTS= |Agr.Pr.=40
-25 1 'Agr.Pr.=50
2.85
4.34
5.82
3.33
4.82
6.31
CONSTR. Agr.Pr.=30
2.15
COSTS= Agr.Pr.=4O0
even :Agr.Pr.=50:
3.63
5.12
CONSTR. :Agr.Fr.=30
COSTS= Agr.Pr.=40
+25 1 Agr.Pr.=50
1.44
2.93
4.42
3.82
5.31
6.79
0.61
1.36
2.10
1.10
1.84
2.59
1.58
2.33
3.07 '
-0.13
0.37
0.86
0.35
0.85
1.35
0.84
1.34
1.83
2.63
3.12
-0.09
0.40
0.82
-0.83
-0.35
0.14
4.12
5.61
4.60
6.09
0.66
1.40
1.14
1.89
1.63
2.37
-0.34
0.16
0.15
0.65
0.63
1.13
1.93
3.42
4.91
2.41
3.90
5.39
-0.79
-0.04
0.70
-0.0
0.44
1.18
0.18
0.93
1.67
-1.53
-1.04
-0.54
-1.05
-0.55
-0.06
-0.56
-0.07
0.43
167
TABLE
DISCOUNT RATE:
C-14:
Net
benefits
of Alternative
: WATER REQUIREMENTS = -50 1 i WATER REQUIREMENTS = even
N
: WATER REQUIREMENTS = +50 1
|E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30:E1.Pr.=10 EI.Pr.=20 EI.Pr.=30
CONSTR.:Agr.Pr.=30:
COSTS= :Agr.Pr.=40|
-25 1 :Agr.Pr.=50:
3.50
4.99
6.48
3.98
5.47
6.96
4.46
5.95
7.44
1.26
2.01
2.75
1.74
2.49
3.23
2.22
2.97
3.71
0.52
1.01
1.51
1.00
1.49
1.99
1.48
1.97
2.47
COMTR.|Agr.Pr.=30:
COSTS= :Agr.Pr.=40|
even |Agr.Pr.=50:
3.01
4.50
5.99
3.49
4.98
6.48
3.97
5.46
6.96
0.78
1.52
2.27
1.26
2.00
2.75
1.74 :
2.48 :
3.23
0.03
0.53
1.02
0.51
1.01
1.50
0.99
1.49
1.99
CONSTR. Agr.Pr.=30!
COSTS= :Agr.Pr.=40:
+25 2 :Agr.Pr.=50:
2.53
4.02
5.51
3.01
4.50
5.99
3.49 |
4.98:
6.47 :
0.29
1.04
1.78
0.77
1.52
2.26
1.25 :
2.00
2.74 :
-0.46
0.04
0.54
0.02
0.52
1.02
0.51
1.00
1.50
---------
I
----------------
DISCOUNT RATE:
10
WATER
REQUIREMENTS = -50 1 1 WATER
REQUIREMENTS = even : WATER REQUIREMENTS = +50 1
=
!EI.Pr.=10 EI.Pr.=20 EI.Pr.=30!E1.Pr.=10 E1.Pr.=20 EI.Pr.=30 E.Pr.=10 E1.Pr.=20 EI.Pr.=30
-----------------------------------------
CONSTR.:Agr.Pr.=30!
COSTS= :Agr.Pr.=401
-25 1 :Agr.Pr.=50:
3.15
4.64
6.14
3.63
5.13
6.62
4.11 1
5.61 1
7.10
0.92
1.66
2.41
1.40
2.14
2.89
1.88
2.62
3.37
0.17
0.67
1.17
0.65
1.15
1.65
1.13
1.63
2.13
CONSTR.!Agr.Pr.=30:
COSTS= :Agr.Pr.=40
even :Agr.Pr.=50:
2.55
4.04
5.54
3.03
4.53
6.02
3.51 :
5.01 :
6.50 |
0.32
1.06
1.81
0.80
1.54
2.29
1.28
2.02
2.77
-0.43
0.07
0.57
0.05
0.55
1.05
0.53
1.03
1.53
----
|:--------- : ---------------------
CONSTR.!Agr.Pr.=30:
COSTS= 1Agr.Pr.=40|
+25 1 |Agr.Pr.=50
2.43
3.93
5.42
----------------------------- -------------------------
2.91 1
4.41 |
5.90|
-0.28
0.46
1.21
0.20
0.94
1.69
0.68
1.42
2.17
WATER
REQUIREMENTS = -50 1 | WATER
REQUIREMENTS = even
DISCOUNT RATE:
12
1.95
3.44
4.94
I
-1.03
-0.53
-0.04
-0.55
-0.05
0.45
-0.07
0.43
0.93
: WATER REQUIREMENTS = +50 1
=========================================================.=============================
IE1.Pr.=10 E1.Pr.=20 EI.Pr.=30:E.Pr.=10 EI.Pr.=20 E1.Pr.=30:E1.Pr.=10 EI.Pr.=20 El.Pr.=30
CONSTR. Aqr.Pr.=30:
COSTS= Agr.Pr.=40:
-25 1 Agr.Pr.=50
2.81
4.31
5.80
3.30
4.79
6.28
3.78
5.27
6.76
0.58
1.32
2.07
1.06
1.80
2.55
1.54
2.28
3.03
-0.17
0.33
0.83
0.31
0.81
1.31
0.79
1.29
1.79
CONSTR. Agr.Pr.=30
COSTS= Agr.Pr.=40
2.10
3.59
2.58
4.07
3.06
4.55
-0.13
0.61
0.35
1.09
0.83
1.57 1
-0.88
-0.38
-0.40
0.10
0.08
0.58
Agr.Pr.=50:
5.08
5.56
6.05
1.36
1.84
2.32
0.11
0.59
1.07
CONSTR. Agr.Pr.=30:
COSTS= Agr.Pr.=40:
+25 ' Aqr.Pr.=50
1.39
2.88
4.37
1.87
3.36
4.85
2.35
3.84
5.33
-0.85
-0.10
0.64
-0.37
0.38
1.12
0.11
0.86
1.60
-1.59
-1.10
-0.60
-1.11
-0.62
-0.12
-0.63
-0.14
0.36
even
168
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