ft
RE-ENTRY#) SYSTEMS
DEPARTMENT
Philadelphia, Pa.
FINAL REPORT
DELAWARE ESTUARY
COMPREHENSIVE STUDY
GENERAL® ELECTRIC
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FINAL REPORT
FOR
DELAWARE ESTUARY
COMPREHENSIVE STUDY
CONTRACT #PH86-64-68
U. S. Department of Health, Education, & Welfare
Public Health Service Regional Office
321 Chestnut Street
Philadelphia, Pennsylvania 19106
Tkogress k Our Most Important Pmduct
GENERAL ELECTRIC
RE-ENTRY SYSTEMS DEPARTMENT
A Department 0/ The Missile and Space Division
Autkor: R. Galtnan
Date: 7. November 1964
P. O. Box 6555 • Philadelphia,
.. 1©101
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TABLE OF CONTENTS
1.0 INTRODUCTION 1
2.0 DIGITAL SIMULATION 3
2.1 Mathematical Model 3
2.2 Programming of Model 4
2.3 Advantages and Disadvantages of Digital Simulation 11
3.0 ANALOG SIMULATION 15
4.0 HYBRID SIMULATION 19
5.0 COMPARATIVE EVALUATION OF DIGITAL, ANALOG, AND HYBRID
SIMULATIONS 23
5.1 Standards of Evaluation 23
5.2 Ability to Simulate Equations 23
5.3 Cost 26
5.4 Capacity for Growth 27
5.5 Reliability 29
6.0 CONCLUSIONS AND RECOMMENDATIONS 31
APPENDICES
A. Listing of Digital Program A-l
B. Listing of Digital Program of Hybrid Program B-l
C. Printout of Digital Program C-l
D. Digital Printout of Hybrid Program D-l
i
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LIST OF FIGURES
Figure Number Title Page
1 Analog simulation of linear 35
estuary model
2 Analog portion of hybrid program 37
3 Location of variables in analog
simulation 38
4 Portion of analog strip chart 39
recording of hybrid run
5 Average daily flow 41
ii
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NOTICE
Change the last two sentences in paragraph 2 on page 1 to read as follows:
The result was the production of a sample digital program (Appendix A)
and a sample hybrid program (Appendix B). It was not found to be necessary
to generate an analog program, as there were no conceptual problems ; only
hardware requirements.
Change the last paragraph on page 22 to read as follows:
Appendix B consists of the listing of the digital program used to supply
the data and control inputs for the hybrid program. Appendix D shows a
printout of the results of a sample run of the hybrid system.
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V
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1.0 INTRODUCTION
Pursuant to the terms of contract #PH86-64-68, between the Delaware
Estuary Comprehensive Study (DECS), a division of the U.S. Public Health
Service, and the Re-entry Systems Dept. of the General Electric Co., the
latter organization has performed a study into the problems and feasibility
of using various types of computing systems to obtain numerical solutions
to the equations comprising a mathematical model of the Delaware Estuary.
This model was supplied by the Delaware Estuary Comprehensive Study.
The particular purposes of the study were to examine digital, analog,
and hybrid computer approaches to the problem, evaluate the advantages
and disadvantages of each, and make recommendations to DECS for setting
up an operational facility. The contract specifically limited both analy-
sis of the mathematical model, and computer programming, to those activi-
ties necessary to do enough programming to produce working, although
relatively unsophisticated, digital and hybrid simulations. Without the
need to produce actual working systems, it was felt that many of the
details of the model might not properly be evaluated, and some of the
subtleties involved in the programming might be overlooked. The result
ft.
was the production of a sample digital program (Appendix -it) and a sample
£ .
hybrid program (Appendix J»r) . It was not found to be necessary to generate
an analog program, as there were no conceptual problems; only hardware
requirements.
In addition to the programming, it was also felt that some analysis
of the model was desirable, in order to estimate which types of changes
and refinements might be introduced in the future, and to consider what
allowances for such changes could be introduced into the various computa-
tional methods.
1
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The remainder of this report contains separate descriptions of digital,
analog, and hybrid approaches to the problem, along with the discussion of
the strengths and weaknesses of each. These descriptions are followed by
a comparative evaluation of all three methods in terms of the overall
goals of the Delaware Estuary Comprehensive Study. The final section is
devoted to conclusions and recommendations.
The attempt has been made to have each section of this report self
sufficient enough to read through without referring to other sections.
For this reason, there will be a certain amount of repetition, but it was
felt to be worthwhile in the interest of increased convenience and clarity.
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2.0 DIGITAL SIMULATION
(1)
2.1 Mathematical Model
If a one-dimensional estuary is divided into n segments, a mathe-
matical model representing the time variation of dissolved oxygen (D.O.)
and bacteriological oxygen demand (B.O.D.) in a body of water has been
given as
M H + [St] * [Ji]
M [c'l + [1&] * [ri • Csi+ Pl - di Li]
where denotes a square matrix of order n and J denotes a
column vector of dimension n.
The total number of equations (and the unknowns) equals 2n. The
components of matrices A and B are defined as:
- Bi,i-i - ¦*
Vi
4,1 " < I' * l+D • Qi-El-Bj
(2)
i+1 -r.
Vi J
®i,i = -0(i+i) • Qj - E,. - Ei+1 _ H
H j
Ai,i+1 = Biji+1 = Ei+1 -^i+1 . Qj
Vi
and all other terms are null.
L and C represent B.O.D. and D.O., respectively, while the remainder
of the terms in equation (1) are known quantities.
rj = f1(T) = reaeration coefficient
CSj = f2(T) = saturation value of D.O.
dj = fg(T) = decay coefficient
3
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Pj = 0 = other sources and sinks of oxygen
and
Vi = (function of Vm^ and Qj) = volume of segment i
~ Qj = f5 (function of T) = net flow ?
= eddy exchange coefficient
?Xi,^L = advection factors
Temperature T, of course, is a function of time, t.
.2 Programming of Model
Equations (1) represent 2n simultaneous first order initial value
differential equations, which can be rewritten as
(iL)i = j + Bi ±_i . + B-j^i • Li + Bi i+1 . Li+1 (3).
dt
^H§^i = rj ' Csj + Pj " dj * Li + Ai,i-1 ' ci-l + Ai,i * ci + Ai,i+1 • ci+l
i = 1, 2, 3...n
Let us now consider a simple, but similar, single differential equation
= F (x,y)
dx
Assuming that the initial condition y is known, the equations may be inte-
grated with respect to x by the use of the Runge-Kutta equation of the form
yn+i = Yn + !/6 (4)
where
Kq = hF (xn,yn) (5)
Kx = hF (xn + kh, yn + Kq)
K2 = hF (xn + %h, yn + Kx)
k3 = hF (xn + h, yn + K2)
n=0, 1, 2, 3...)
and h represents the integration interval. The actual derivation involves
considerable algebraic manipulation (see Ref. 1); let it suffice to say
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that (5) would reduce to Simpson's rule if F were independent of y.
The mathematical estuary model is simplified considerably by the
absence of most off-diagonal terms in matrices A and B. The differential
equations (3) can in fact be expressed:
(|i>. = f(t, L^!, Lt, Li+1) (6)
<3f>l - g(t, Cn, Ct, ci+1)
i = 1, 2, 3...n
where LQ, C^, L and Cn+^ are known quantities, either constants or
functions of t, and g(t) of course includes the L^.
Equations (6) are integrated by the generalized equation (4) as
follows:
(Vn+1 = (Li>n + !/6 ^ + 2Kli + 2K2t + K3i>
(ci)n+l = (ci)n + L/6 (""Oi + 2mli + 2m2i + m3i^
where
KOi
=
hf
G~n>
(^i-l)n> ^i^n' ^i+l^n3
Kli
=
hf
tn
+ %h, (L..1)n + %K0i_1, (Li)n
+ (Li+1)n + ^Ko.J
K2i
=
hf
C'n
+ hK (Li_i)n + %Kli_1, (Li)n
+ (Li+l)n + ^Kli+1]
K3i
=
hf
fcn
+ h, (Li-i)n + K2i_p (Li)n +
K2i» (Li+l>n + K2i+1]
=
kg
[j-n>
• n> (Ci>n> n]
mil
=
kg
tn
+ %h, (Ci_1)n + %moi_1, (Ci)n
+ H)i> (Ci+l)n +
m2i
=
kg
Ctn
+ %h, (Ci.1)n + Vii.p (Ci)n
+ %mli> (Ci+l)n + Wi+J
m3i
=
hg
T'n
+ h, (Ci-i)n + m2i_1, (Ci)n +
m2±> n + m2i+1*]
n = 0, 1, 2, 3...
i = 1, 2, 3...N
and h denotes the time interval of integration (7)
(1) F. B. Hildebrand, Introduction to Numerical Analysis, New York,
McGraw-Hill Book Co., 1956
5
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Equation (7), of course, does not indicate the sequence of operations
to be completed in a single step; however, it can be seen that four units
of erasable storage per variable are required: one to retain the initial
values of the step, i.e. (i = 1, 2...N), one for the current
value of & C^, one to accept the current derivative evaluation.^, mi,
and one to store linear combinations of previous derivative evaluations.
Furthermore, additional savings can be made in the computer storage of
L-£ first; a maneuver which permits sharing of locations of the intermediate
results.
After each step is complete, the new values of and Ci (i = 1, 2..N)
become the initial values of the next step. The process is repeated until
all time intervals are completed.
It is of importance to notice that the error associated with the
above solution depends upon the form of the functions (6),, and also, to
a very large degree, upon the size of the time interval h. It is there-
fore necessary for the user to be able to control the integration steps
either by pre-programmed instructions or by the use of the sense switches
on the control board of the digital computer.
To illustrate the extreme versatility of the aforementioned solution
to the estuary problem two digital computer programs have been prepared:
1) A computer program for use on the 40,000 character IBM 1620, and
2) A computer program for use on the 32K IBM 7094.
Both programs are essentially the same, and yield identical results
if all conditions are the same. The basic difference is the intentional
selection of two systems that can be considered extremes.
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The IBM 1620 is a very small, very slow digital computer. Its selection
was based on the premise that a demonstration of the feasibility of using
a small computer is in order. The IBM 7094, on the other hand, is one of
the fastest digital computers in existence today, and was selected to
demonstrate certain limitations that exist in using the smaller 1620.
The actual programming followed the general outline as follows:
(1) Initialize the program by reading into the storage variables
C5< ,
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Read in the initial conditions of B.O.D. and D.O.; i.e. and
C±, i = 1, 2, 3...N. Read in the initial values of tQ (time in
days), Q0 (discharge), and TQ (Temperature).
Read in the values of tf, Qf, and Tf at the end of the selected
time interval. The step size h is calculated internally from
h = tf-tQ.
Calculate, in the order specified (see Equation 7)
Kq., i = 1, 2, 3....N
Kii, i = 1, 2, 3 N
K2i, i = 1, 2, 3 N
K3i, i = 1, 2, 3 N
It is to be noted that (i=l) and (i = 30) are known
values or functions obtained from step 3. Similarly, the values
of KOi-1, Kjj.j, K2i1 (1 = 1) and Ko1+1, Kll+r K2.+1 (i = 30)
are considered zero. It is to be noted also, that (4N+2) storage
locations are needed for the intermediate calculations indicated
in this step.
Evaluate (L^)f (i = 1, 2, 3....N)
Calculate, in the order specified
m0i> 1 = 2> 3.•.N
m-i, i = 1, 2, 3...N
¦*¦1
m2^, i = 1, 2, 3...N
mo , i = 1, 2, 3...N
Ji
Note, however, that no additional storage is required for this
step since all intermediate calculations may be placed in the
(4N+2) storage locations previously occupied by calculations of
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step (7). As before, (i = 1) and (i = 30) are the
known values obtained from step 3. Values of mg^^, m]^
m2^ (i=l) and mo^+2.' mli+l' ant* m2i+i are considered
zero.
(10) Evaluate (Ci)f (i = 1, 2, 3...N)
(11) Print out the B.O.D. and D.O. levels of each section at the end
of the time interval h, i.e. (Li)f and (C^)f (i = 1, 2,...N)
(12) Set up new initial conditions from the end values just calculated,
thus:
(Li)c = (L±)f
i = 1, 2, 3...N
(Ci)0 = (Ci)f
to = tf
Qo = Qf
To = Tf
(13) Return to step 6, that is, to the beginning of a new cycle. Steps
6 to 13 are repeated as many times as desired, each repetition
representing a particular time interval.
It will be of interest to note that the results varied considerably
as the time interval increased or decreased. Our sample runs (see Appen-
dix 3) indicated that time intervals less than 1/5 day were satisfactory
and greater than 1/3 day were not.
One final thought remains to be discussed in connection with the
selection of the time interval. The measured quantities which are functions
of time (T = temperature, Q = discharge) are recorded as mean daily averages.
A continuous plot i>£ T or Q vs. real time in days would no doubt be disas-
trous as the temperatures tend to cycle through a 24 hour day and similar
9
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cycling perhaps may be observed for the discharge values. Nevertheless,
neither method of recording these variables, either continuously or in
steps is satisfactory for the solution of the differential equations. In-
stead, it is extremely advantageous to create a smooth continuous curve
as a function of real time. Figure 5 illustrates such approximation of
average daily discharge in ft^/sec vs. time in days. A similar plot is
obtained for temperature vs. time in days.
A Fortran II listing and sample runs are included. It is to be re-
membered, however, that this program is far from being a production type
program. It merely demonstrates the feasibility of the numerical approach
outlined herein.
Only a small sample of the problem has been run. Various time incre-
ments were chosen. It has been found that steps of 1/10 day yielded
satisfactory results. The following times, with all values between the
indicated values being linearly interpolated, were used:
Day
Discharge
Q (ft^/sec)
Temperature
T
1
3,300
33
2
4,000
33
3
4,500
33
4
4,750
33
5
5,000
33
6
4,750
33
7
4,500
33
8
4,100
33
9
4,500
33
10
4,750
34
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The values of , (j- , and E and the values of J, 1/V, and are
shown in the computer output. Similarly, the initial conditions of D.O.
and B.O.D. are as shown.
The functional relationships are given by:
Tj = .18 + 6 x 10"3 Tj
CSj = 14.563 - .37298Tj + 4.6678 x 10*3 T2j
dl = .09057 + 9.416 x 10"3Ti + 8.79 x 10"^ T.2
J J J
Vi = Vave + ^ • Q.
but = 0
The geometric boundary conditions are
Lo = L31 = 4*°
co = C31 = Cs
It should be noted that certain parameters, particularly V (volume of
section), should not be held constant, or be functions of the section alone
They are best expressed as functions of i, Q, and t. The sample problem
includes provisions for making V a function of Q in addition to i. How-
ever, at the time the problem was run, not all the input data was available
.3 Advantages and Disadvantages of Digital Simulation
The feasibility of using a digital computer has, it is hoped, been
amply demonstrated. In fact, the problem, as it exists today, can and has
been solved on a relatively small digital computer. The main advantages
of using such a small computer are:
1) Relatively inexpensive cost per hour
2) Ability to make periodic checks on the computations and make on-
the-spot corrections if trouble is encountered.
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3) The low overall cost and the simple operating procedures permit
access to the computer at any time.
Disadvantages are:
1) Extremely slow running time causes unnecessary waste of valuable
engineering time
2) The relatively small memory capacity permits very little experi-
mentation and allows practically no room for future expansion of
the program capability.
The advantages of the large scale digital computer are:
1) Extremely fast operation reduces the actual costs tenfold or
more
2) The large storage capacity permits a large range of experimentation
and offers very few restrictions on future expansion of the program.
And the disadvantages:
1) The rental cost per month makes the acquisition of a large computer
absurd unless other uses are found for it.
2) No checks can be made while the program is running. All decisions
must be pre-programmed.
It is clear that the advantages of a large computer far outweigh the
advantages of a small one as the problem is defined today. The disadvan-
tages, however, can be overcome:
1) There are a number of high speed large scale digital computers
in the Philadelphia area. Practically all installations are will-
ing to rent time on their computers.
The advantages of renting time from one of these installations
lies in the fact that the user pays only the time actually spent
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on the machine. In addition, arrangements could be made to use
such installations' peripheral equipment which, no doubt, could
prove to be very useful if a large amount of data will have to be
handled. Too, the assistance of experienced analysts and/or
programmers could be arranged.
2) Invariably the pre-programmed decisions are far superior to the
on-the-spot decisions since more time is available for thought.
One of the major decisions that the operation would face is the
choice of the proper interval. This can be done by using a pro-
grammed step size control. Such methods have been used success-
fully. (Ref. 2)
One more possibility exists--that of an intermediate size digital
computer. By carefully weighing all the advantages and disadvantages of
the extreme systems it may be concluded that the intermediate range com-
puter is indicated only if the user wishes to have at his immediate dis-
posal a digital computer that does not have the distinct disadvantages of
a small scale computer yet is less expensive than the large scale computer.
In conclusion, it should be mentioned that the numerical solution
presented herein is not the only solution to the problem that can be ob-
tained. A number of variations to the Runge-Kutte numerical integration
scheme can be devised. It is also conceivable that a possible change from
a finite difference approach back to the original second order equation
may offer better results. One excellent example of the solution of a set
(2) Missile and Satellite Systems Program for the IBM 7090, by J. Butler
and A. Dennison, GE TIS 61SD170, 1962
13
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of second order differential equations representing heat dissipation
through a heat shield is given in Ref. 3. The general equations are of
the form:
H " H + b2 <-|f)2 + b3 > + b4
where (x, t, T).
14
(3) P. Gordon, Heat Conduction Analysis, G.E. TIS 64SD201, 1964
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3.0 ANALOG SIMULATION
One possible approach to the requirements of the Delaware Estuary
study is the use of an analog computer for the simulation of the diffusion
process. The analog computer is a continuous variable device whose basic
computing unit is the operational amplifier which can be used for such
mathematical operations as addition and integration. The computer is thus
capable of solving ordinary linear differential equations by continuous
integration. By the introduction of non-linear gear (multipliers, function
generators, etc.) the analog computer may be used for the solution of
non-linear differential equations. Since flow or diffusion equations in-
volve partial differential equations, finite difference techniques can be
used to reduce the partial differential equations to a set of coupled
ordinary differential equations which may be solved simultaneously and
continuously with respect to one variable. Usually the finite difference
techniques are supplied to the space variable, and time is chosen as the
continuous variable. This approach simplifies the boundary conditions
somewhat, and fits in with the sampling at discreet points for the problem
at hand. The analog computer is a parallel system. A component is assigned
to a particular mathematical operation, and performs only that operation.
Hence, a larger system simulation requires more equipment, rather than a
greater problem time, as would be the case with a digital computer. The
analog computer implementation of the diffusion process associated with the
dissolved oxygen analysis of the Delaware River Estuary is shown in Figure I.
The program consists essentially of two parallel systems of thirty first
order lag circuits with provision for external inputs (sources and sinks)
together with the necessary feedback loops to provide the appropriate flow
15
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balance at each of the 30 chosen stations along the estuary. Although the
mechanization is straightforward the following descriptive observations are
pertinent to the understanding of the capabilities and limitations of the
simulation.
(1) The output of each pair of integrators represents the continuous time
history of the two variables and at a particular station.
(2) Since, except for weighting coefficients, the flow balance at each
station is of the same form, the basic computing loop is serially
repeated thirty times.
(3) The boundary conditions can be satisfied, with no change in the basic
computing loop, by merely defining fictitious zeroth and thirty first
stations to provide the flow balance at the first and last station.
(4) In the implementation shown, the weighting coefficients are hand set
on fixed potentiometers; and although these coefficients may vary
from station to station (i.e. along the estuary) they cannot be varied
with time or any time varying quantity.
(5) Although the linear with constant coefficients simulation of Figure 1
is capable of providing a fairly realistic model of the estuary, it can
legitimately be criticized on a number of grounds. From the technical
point of view the proposed implementation would be somewhat sensitive
to small errors in the coefficient potentiometer settings and/or the
impedance matching of the integrator input networks. This somewhat
subtle source of error can be intuitively appreciated by noticing that
aside from forcing functions and scale factors the sum of the inputs
to each integrator is seen to be the difference of two similar quanti-
ties, each of which is itself a difference of like variables. The
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relationship ^ (ci+l " ci) " Ki}i-l (Ci " Ci-1^ which
represents the second spatial difference of the variable is of
the order of magnitude ( ^ X)2 where ^ X is the distance between
adjacent stations. If one were to contemplate more and more sampling
stations being inserted along the river (2^.X growing smaller) the
effective sum of the voltages at the input to each integrator would
rapidly fall to the millivolt level and be highly sensitive to the
aforementioned sources of "small" errors. This difficulty, which
is analagous to the loss of significant digits in a numerical compu-
tation, could be alleviated to a significant extent by first amplifying
the first differences, before computing but this would
entail introducing two additional amplifiers into the basic computing
loop. In the present problem this imposes a requirement for an
additional 120 amplifiers and appears to be a prohibitive cost for
a computational refinement.
Another serious limitation of the computing scheme of Figure 1 is
based on logistics. On commercially available analog computers, at
most 2 potentiometers are supplied per amplifier, and substantially
less than 50% of the available amplifiers are prewired to serve as
integrators. The demands of Figure 1 are for 60 integrators, a
minimum of 300 potentiometers, and an unreasonable number of function
generators, This precludes implementation of the problem upon any-
thing less than an aerospace (megabuck) computing facility.
Actually, the major drawback of the solution of this problem with
analog equipment is that it is not easily amenable to generalizations in
the direction of introducing time varying coefficients. While in theory
all that is required to introduce time varying properties into the system
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is to replace each of the fixed potentiometers with a servo-positioned
potentiometer, one is confronted in practice with the requirement of
isolating servo potentiometer outputs in order to prevent loading errors.
Again, this will increase the amplifier requirements from one per station
to at least three, but more likely to 5 amplifiers per station. Servo
multipliers are electro-mechanical tracking devices that require a
moderate amount of monitoring to assure top performance. Introducing
them into the problem not only greatly increases the cost and complexity
of the mechanization but poses a serious threat to the reliability and
repeatability of the solution, on a day to day basis.
In view of these considerations, the analog approach to this problem
does not appear practical, except for simpler simulations.
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A.0 HYBRID SIMULATION
Some aspects of the behavior of the Delaware Estuary are presented by
a mathematical model based on the diffusion equation for a one dimensional
system. By looking at a single variable, dissolved oxygen, much can be
learned about the behavior of the estuary as a function of time. A
little trickery was required to get by with less than the 15 D-A channels
which this simulation would normally require. Nine of these fifteen
channels are used to transfer the coefficients of C. There are three
equations, each with three coefficients. By assuming that the three
0 o o
coefficients of the equation can be used for the and equa-
tions, the information can be carried in three channels, thus reducing
the requirement from 15 D-A channels to 9.
An example is given of the effect this approximation has on the values
of the affected coefficients. Taking data for the first day of the
simulation (Jan. 1, 1963), and looking at the case i=10, we find that the
three sets of coefficients for the three sections, 9, 10, and 11, are
approximately
Ag,8 = 1.96 Agj9 = 3.28 ^,10 = x»49
A10,9 = 1-92 A10,10 = 3,28 A10,ll = 1*53
All,10 = 1>93 All,ll = 3,28 A11,12 = 1*53
Here, we can see that using the set A^o,k in place of both Ag ^ and A^ ^
will introduce very little error. It should be pointed out that these
elements are in a region where V changes very little from section to
section. There are occasional large changes in V, and these will intro-
duce errors. It was decided to live with these errors, as a hybrid
simulation with fewer than three sections would not provide enough
19
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information to evaluate the method with any confidence.
Figure 2 shows a schematic representation of the analog portion of the
circuit. The generic variables X, Y, and P are used, as the circuit
represents both C and L. For the L simulation,
X = L
XQ = Lq (initial conditions)
Y = J (forcing function)
P = B (variable flow and diffusion coefficients)
For the C simulation,
X = C
Xo = cG
Y = f
P = A
The simulation runs for a period corresponding to one day, at which time
the output, Xi, is sampled and sent to the digital. A new set of initial
conditions is supplied, and the simulation is run again. As there are 30
sections of the estuary to consider, and two output variables, L and C,
sixty runs will be needed to compute the behavior of the entire estuary
for one day. The equipment which is used to illustrate the technique
will operate readily in the range of 100 milliseconds per run. Thus,
a one day period will take about 6 seconds to simulate, and a full year's
simulation would take about 45 minutes.
There are four factors which can limit the speed of this hybrid simu-
lation: the time scaling of the analog, the time required to reset the
analog, the time required for data transfer, and the time required for
digital calculations. The time scaling on an analog is a function of
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the problem and the equipment. The outputs of any analog simulation are
a group of time varying voltages. The limitation is imposed by the
ability of the analog equipment to reproduce the highest frequency output
component which significantly affects the results. Most standard analog
computers will operate most comfortably in the frequency range between
0.1 and 1 cps. The upper limit is imposed by mechanical components,
such as servo multipliers. If only electronic components are used, then
the upper limit for general usage is in the range 100 to 1,000 cps, im-
posed by electronic multipliers. If only summers and integrators are used,
then the frequency response characteristics of the operational amplifiers
will limit the practical limit to the range 1,000 to 5,000 cps.
The Delaware Estuary Simulation does require the use of multipliers.
This imposes a theoretical limitation of 1,000 cps. and a practical
limitation, due to the particular equipment we are using, of 100 cps.
The time required to reset the analog is a big factor in this problem,
as it must be done sixty times for each estuary day simulated. Resetting
the analog involves stopping it, putting it into its initial condition
mode, and waiting for the feedback capacitors of the integrators to
charge.
This process, using an analog computer with electronic mode control,
takes only a few microseconds. The equipment at hand, using mechanical
relays and RC charging networks, requires about 50 milliseconds.
The time required for data transfer depends on the speed of the digital
computer and the speed of the conversion equipment. The equipment we are
using, which is faster than most, takes about 20 microseconds per D-A word
transfer, and about 40 microseconds per A-D word transfer.
21
-------
A five element simulation would require, for proper operation, 25 D-A
channels and one A-D channel. Thus, about 600 microseconds, conservative-
ly speaking, would have to be allowed for data transfer.
The time required for digital calculations is dependent on the speed
of the digital computer. The computer which was used to try our sample
problem was a GE-235. This computer is a fairly high speed machine. It
has a 6 microsecond basic cycle time with high speed arithmetic hardware.
It performs one cycle of computation in well under one millisecond.
In order to calculate possible operating speeds, it would be necessary
to know the characteristics of the type of equipment to be used. The in-
put-output operation of the digital computer will affect how many of the
above operations can be done simultaneously. In all cases, it would be
possible for the main digital program to be taking place while the analog
was working. In some cases, the data transfer routine could be performed
simultaneously with the other two. This would give us a theoretical
limitation of about one millisecond per computation, which is two orders
of magnitude better than the capability of the equipment used to study
the problem. The minimum time for a hybrid simulation using a 30 section
model would be about 60 milliseconds per day. A full year's simulation
could thus take about 20 seconds. Although mundane considerations such
as printout requirements and manual input options could result in con-
siderably longer operating times in actual practice, the minimum given
above is perfectly feasible, using presently available equipment.
3
Appendix ft. consists of the listing of the digital program used to supply
D
the data and control inputs for the hybrid program. Appendix 4 shows a
printout of the results of a sample run of the hybrid system.
-------
5.0 COMPARATIVE EVALUATION OF DIGITAL, ANALOG, AND HYBRID SIMULATIONS
5.1 Standards of Evaluation
The purpose of this portion of the report is to evaluate the advan-
tages and disadvantages of each method of simulating the equations repre-
senting the Delaware Estuary, and recommend the best one. The meaning of
best, in this case, is that method which scores highest on a rating scale
based on criteria which seem relevant to the problem. The following
criteria have been used:
Ability to simulate equations
Cost
Capability for growth
Reliability
Each of these criteria will be considered in turn.
Although numerical comparisons are desirable, it is impossible to
find performance characteristics which can be directly compared, when we
are dealing with diverse methods of simulation. We will discuss, for
instance, accuracy of results. In the case of the digital machine, the
accuracy can be increased by running the problem slower. For the analog,
the accuracy is increased by adding equipment. Therefore, the comparison
of relative accuracy will have to be based on subjective estimates as to
what constitutes acceptable speed for the digital and acceptable size for
the analog.
5.2 Ability to Simulate Equations
All three simulation methods, digital, analog, and hybrid, will work
with the same mathematical model of the estuary. These methods have been
described in sections 2, 3, and 4 of this report.
23
-------
The efficiency of the digital solution depends upon the size of the time
interval that can be used as the basis for calculation. The prototype digi-
tal simulation described in section 2 was run using time intervals of
0.0125, 0.05, 0.1, 0.2, and 0.33 days. Examination of the printouts shows
that the optimum time interval is about 0.2 days. Thus, the digital pro-
gram requires about 300 computation cycles to simulate one day of the
behavior of the estuary. The only practical limitation on the ability of
the digital computer to provide accurate solutions to the equations is the
possibility of truncation error, based on the large number of iterations
required to simulate an entire year. In the example we used, a one year
simulation requires 365 x 60 x 5, or 109,500 computation cycles. Practi-
cally speaking, this presents no problem, because the forcing functions
and coefficients are largely determined by the stored inputs; J, r, Cs, d,
, Q, and E. This constant supply of inputs would keep the pro-
blem from drifting. In addition, 109,500 cycles will not generate much
truncation error, in this case.
The solution of the problem on the digital thus presents no particular
difficulty, using a step by step approach. The size of computer required
is dependent on the type of program wanted. The sample solution was writ-
ten for an IBM 7094, which is a much larger machine than was needed, but
which was readily available. This same program was simplified enough to
fit on an IBM 1620, which is a very small general purpose machine. The
1620 program, although working, still did not, at this writing, have all
the bugs out of it. It was working well enough, however, to prove that a
machine as small as the 1620 could be used for this problem, although
not with great efficiency*
-------
The hybrid approach, described in section 4 of this report, is very
appropriate to the solution of flow problems in general. It can turn
what would be a very large, unwieldy problem on either the analog or digi-
tal machines into a problem which requires only small analog and digital
portions. In this particular case the advantage gained over the analog
approach is very great, but the advantage over the digital is not so pro-
nounced.
The hybrid solution is similar to the digital in that it solves the
equations for one section for a given time interval, and then proceeds to
the next section, working its way down the estuary. The process is then
repeated for the next time interval. The analog requirements for the
sample problem consist of three integrators, four electronic multipliers,
and three diodes. A more exact simulation, putting five sections on the
analog instead of three, would require five integrators, eight electronic
multipliers, and five diodes.
The digital requirements of the hybrid simulation are not very different
from those of the pure digital simulation. The main extra computations
which the digital simulation must perform are the integrations. These
comprise at least half of the computation time, part of which is made up
by the fact that the hybrid program has data transfer instructions, which
the digital program does not. The area in which the hybrid procedure
could provide a significant improvement over the digital is in speed of
operation. The digital program, as pointed out above, requires about
5 cycles per day of problem time. The hybrid program requires only one,
as the output variables, L, £, C, and 8 are produced continuously. The
functions and coefficients which are calculated on the digital are A, B,
25
-------
J, and f, none of which changes appreciably from day to day. The proto-
type digital program on the IBM 7094 took about 1.3 seconds to simulate
one day. This is far superior to the ten seconds of our hybrid simulation,
but much slower than the 20 millisecond capability of an advanced hybrid
system. We find, then, that a hybrid facility could supply answers at
a greater rate than a digital computer, and it might be able to do it
with a slightly smaller digital computer. However, the question remains
as to whether the advantage would be worth the expense of obtaining an
analog computer and conversion gear, and the inconvenience of operating
a facility which requires personnel familiar with three areas of computa-
tion. These aspects of the matter will be discussed in section 6.0,
Conclusions and Recommendations.
The ability of the analog simulation to provide correct answers de-
pends very much on the behavior of the input functions, and upon how many
non-linearities must be included in the mathematical model. The analog,
unlike either the digital or hybrid simulations, solves the equations for
all sections at once. It can therefore supply exceedingly rapid solutions;
a simulation of the estuary's behavior for a full year could take less
than half a second. In order to do this a great amount of equipment is
needed, as has been explained in section 3 of this report.
In summary, we find that any of the three computation methods can
provide accurate outputs.
5.3 Cost
There are many factors to consider in estimating the relative cost
of each method of simulation. The cost per run could be calculated by
26
-------
determining the equipment needed, and multiplying its rental rate by the
minimum time required to run the program. This would provide a cost per
run figure, but the problems of obtaining meaningful outputs, the use
which is to be made of the data, the preparations required to set up
series of runs, and many others, would not be taken into consideration.
In addition, an objective determination of the equipment needed cannot
be made as it depends too much upon what type of output is desired, and
to what use it will be put. Very roughly, then, we would estimate that
the cheapest facility to set up and operate would be a digital. Next
would be a hybrid, and far more than either of the others would be an
analog.
5.4 Capacity for Growth
Basically, the simulation must solve i equations of the form
ci = fi + Ai)i_1 Ci_1 + Ai}i C£ + A^1+1 Ci+1 (1)
fi ¦ rCs " dLi + Pi <2>
£± = Ji + Bi>i_1 Li_1 + B_ Lt + Bi^+1 L±+1 (3)
There are many choices which can be made within the framework of this
mathematical model. The number of sections, i, can be changed from the
present 30. Each coefficient can be considered as a constant, or as some
function of one or more parameters. To give a few examples, the coeffi-
cient depends partly on i_i> which is the flow rate from section
i-1 into section i. The model we have been using as a sample sets
equal to Q where
= a constant for section i
Q = flow into section 1, taken from a list of daily averages.
27
-------
A more exact representation might obtain Q from a set of equations
based on upper river conditions, and inputs such as rainfall, wind, and
temperature. The K^, might be a dynamic function of Q, and perhaps
other parameters.
The coefficient is also dependent on V^, the volume of section
i. This has been supplied as a fixed value for each section. Better
results could be obtained by allowing for the fact that V depends on Q.
The J1s are given as functions of i. Actually, they vary from day
to day. Therefore, instead of i, there could easily by 365i values of
J to store.
There are variations that could be explored indefinitely for every
parameter or -variable that gets into this problem. It could be made
almost as complicated as one pleases. The entire field of transient
effects hasn't been considered. All the inputs are given as daily
averages, ignoring the tidal effects and transient flow effects, except
as they average out. The errors which are allowed by this type of simpli-
fication may well turn out to be unacceptable, especially for short term
simulations.
From this partial listing, it can readily be seen that the mathemati-
cal model might turn out to require anything from a fairly small to a
large facility, depending on how it is to be used. It may be optimistic,
therefore, to assume that a computation facility can successfully be
planned on the basis of the mathematical model used for this study. It
must be assumed that whatever type of facility is used, it must be capable
of incorporating most, if not all, of the additions mentioned above, as
well as others, not mentioned, which will come readily to mind.
-------
There are three types of changes involved. An increase in the size
of the problem with no increase in complexity, an increase in the function
storage requirements, and replacement of simple functional relationships
with more complex ones. Most of these changes are readily incorporated
into a digital program, with some increase in operating time. They are
also easily incorporated into a hybrid program, unless there is an in-
crease in the number of variables to be converted, in which case care
must be taken not to exceed the number of available conversion channels.
The analog program, even of the simplified model used for this study,
is too complicated to be used without simplifications in the representa-
tion of the coefficients.
5.5 Reliability
Defining reliability as the degree to which the user can depend upon
obtaining repeatable results, the digital solution of this problem is by
far the most reliable. Next would be the hybrid, and last the analog.
It would seem that the analog approach would be more reliable than the
hybrid, as the hybrid requires an operational analog program as part of
itself. The pure analog program, however, requires very much more equip-
ment than the analog portion of a hybrid program, and the unreliability
of an analog simulation is roughly proportional to the amount of non-
linear equipment used.
29/30
-------
6.0 CONCLUSIONS AND RECOMMENDATIONS
It is recommended that no final determination as to the composition
of a computation facility should be made at this time. As has already
been pointed out in this report, the mathematical model is capable of
tremendous variations. The number of sections used, the form of the in-
put parameters and their functional dependence on other variables, choice
of effects to investigate, all will greatly affect the size and form
of the simulation.
It is recommended that at this time the DECS obtain time on an already
operational general purpose digital computer facility, and use this
facility to
1. Establish working simulations
2. Use the simulations to get the type of results desired
3. Experiment with different programs and uses.
Only actual use of a working simulation over a significant period of time
(at least several months) can establish the real requirements, expose the
weaknesses, and reveal the characteristics of an operating facility. It
is much easier to set up either a general purpose computation facility
or one to perform a single known task than it is to establish one whose
goals are known but whose specific characteristics are not. A digital
computer facility is suggested for six reasons:
1. It has been shown, in section 2 of this report, that a digital
computer can do a good job of simulating the model.
2. The problem would not tie up any reasonably large facility, as
individual runs, even simulating an entire year, should take less
than 10 minutes.
31
-------
3. There should be a number of good sized digital computer facilities
available to DECS for this work.
4. We know of no government facility which would have a hybrid system,
of the type needed, available. The hybrid facility at G.E.'s
Re-entry Systems Dept, while available, does not have electronic
mode control in its analog computers, and would therefore cost
about $75 per run. In addition, it would need more conversion
channels to investigate the job properly. An electronic mode
control computer and more conversion channels are planned for
early 1965.
5. An analog facility, as has been made abundantly clear, is just not
suited to this problem, unless it could be established that a com-
pletely time invariant set of equations can do the job. The re-
sults of the outputs obtained in this study indicate just the
opposite.
6. Even if it should turn out that a hybrid, or even an analog,
facility would best do the job, a medium or large sized digital
computer facility is flexible enough to do the studies necessary
to establish programming, operational, and hardware requirements.
It is further recommended that the computer that is used to set up an
operating simulation be a large one; no smaller than an IBM 7044 and
preferably as large as an IBM 7094. It is not suggested that these parti-
cular machines, or the machines of any particular company be used. The
two machines mentioned are just two very well known examples of medium
and large scale computers. There are three reasons for specifying a
large machine.
-------
1. Run for run, a large machine is cheaper than a small one.
2. The large machine will be capable of handling all the programs
which will suggest themselves in the course of the investi-
gations.
3. The DECS work will be a small proportion of the work load of
a large computer than of a medium sized one, and therefore
should be more convenient to use.
The second of these reasons is the most persuasive.
A tentative conclusion is that eventually the best facility for the
DECS work will be a digital computer facility; probably built around a
medium sized high speed unit. Examples of presently available computers
are:
Company
Model
ASI
6040
CCC
DDP-24
CDC
3100
DEC
PDP-7
G.E.
235
IBM
7040
Packard-Bell
440
SDS
930
Univac
418
No attempt at evaluation is made. This list undoubtedly includes unsatis-
factory machines and omits acceptable ones. These computers are all
medium sized, high speed units, and represent the general area from which
a selection might be made, provided:
-------
1. A digital computer is determined to be best for the Delaware Estuary
Simulation.
2. The projected usage is sufficiently heavy to justify establishing
and maintaining a facility.
3. The final mathematical model remains about the size of the one used
in this study.
When a set of operational programs has been obtained, along with enough
experience in their use to be familiar with their idosyncrasies, it should
be possible to run several sample programs through a number of available
machines, and use the results to help make a choice.
There are two circumstances in which it would not pay to establish
a digital facility to do the DECS simulations. These are, if the usage
will not warrant the purchase of a computer large enough to do the job
efficiently, and if the total number of runs desired is so great that
pure digital computation is too slow. In the first case, the solution
would be to obtain part-time use on an existing facility. In the second
case, the solution would be to establish a hybrid facility.
-------
PAGE NOT
AVAILABLE
DIGITALLY
-------
K>
Ol (~YX(i-i),
c^_q^7
^ ^ Inputs from Hybrid Link
Output to Hybrid Link
Figure 2. Analog Portion of Hybrid Simulation of Delaware Estuary
37
-------
Figure 3. Location of Variables Shown
for Section i of Analog Simulation - i = even integer
38
-------
PAGE NOT
AVAILABLE
DIGITALLY
-------
~ Di'sdioirge ^Seeo)oc/- ee-fj
I4O0C
/2O0O
10000
eooo
booo
4-0OO
ZOOC r
10 II
Time (da y s)
/2 75 (4<5 75<77a 19
kU
CO
Figure 5. Average Daily Flow
-------
APPENDIX A. Listing of Digital Program
11/03/64 PAGE 1
FUNCTION AK AT ( Z1, Z 2 , 13 , I I , J J , KK , MM )
COMMON KC,KC,KR,l\'S,CU0,CS0,CR0,CLi(20> , C S ( 20 ) , CK ( 20 ) , T I 999 ,2 ) t 0 ( 999
1 ,2 ) .ALPHA! 31 ) ,PHI ( 31) ,E( 31 ) , VI ( 3c) , P ( 30 ) ,FL(30,3) ,EJ (30) ,VK( 30) ,KX
2L,KYL,KXC.KYC,XLC,YLO,XCO,YCO.XL(5),YL(5),XC(5) ,YC(5 )
1 i = I I -1
J = JJ
K = KK
M = ff
V = 1 . /VI [ I ) + VK ( I ) *(J ( J > K )
IF{ 1-1 115,10,15
10 Z1 = XCO
HQ 12 L = 1iK X C
12 Z1 = Z1+XC(L)»T(J,K)«*L
c = cco
C = Cbf
R = CRC
DO 5 L = 1,KH
5 C - D
-------
1 1/C !/(><•
PAGC 1
ru^cnci:; 111"AT ( / L, i i. 3 , I I , J J, KKI
CfjKf-.a i i:r),KC,K« i, cs-, cjj.Cl i .'¦n ,cs<2 n .(.• i?it i >>r',2) ,uc)9 >
I ,2 I , ALfH-U 11 I . •'HI ( 2 1 ) , ( 11 ) . / I ( 3 i) , i'l C ) .PL I 3'i, M . 11 3 1) , VKIJO ) ,KX
2L, '
-------
1 l/<:3/64
i
CC.'TON Kr.KC,KR,\.S.CP'!,(.S',.CK;,»CI<20),CS(20),Civ(23) , T ( i)9<> , 2 ) . I ( 9C)9
li?) , ALI'H.M ? 1 I ,M'| ( 3 l),C(iLI,VI(3j),t'(3C),FL(30,i),rJ(30),VK(3a),KX
2L, KYL, KXC.KYC. XLT. YLO,
-------
c
c
c
c
A-4
ru :c i = 1,\t
1 F < M - I ) i 4 , 1 2 , 14
12 J = 1
14 CO 20 K = L , J
T( 1 ,K) = i T( 1,KI-32.)/I. S
C ( I , K I = (..(I ,K ) • a.h400.
20 WOT lC.SC?.TIf1( I , K ) , IJI I ,K) ,T( I ,K)
*EAC IN FU\CT I UNS GF SECTIONS ( ALPHA , Ph I , E ) AfjD (J,i/J,K>
N S i = N S ~ I
GO TC (25,55), KKCN
25 WOT 10,803,TITLE
CO 30 1=1,NS1
RIT 2,1,J,ALPHA!1),PHI(I ) ,F[I )
30 WOT 1C.3G4, J, ALPHA! [ I , PHI ( I ) ,E ( I )
WOT 10,843,TITLE
CQ 22 1=1,NS
KIT 2,1,J,E J I I),V I( I I,VKI I ) ,PI I)
32 WOT 1C , 330, J,E J( I ) , VI ( I ) ,VK I I >„ P < I )
REAC I r\ POLYNOMIALS UF R, CS, C
ERASE CR,CS,CD
WOT 1?,3C5,TITLF
R I T 2i 1,KR,CRC,ICR! I ) ,1=1,KRI
KIT 2,1,KC,CSG,(CSC I),1=1,KC)
HIT 2,l.KD.CrcO,(CDtI),1 = 1 ,KU)
1 = 0
WOT 10,807,1,CR0,CS0,CDC
DO 40 1=1,9
40 WOT 1C ,BC7,I,CR1 1),CS( I ),CD(I )
READ IN POLYNOMIALS OF LC, L31, CO, C31
ERASE XL.YL.XC.YC
WOT 10,931
RIT 2,1,KXL,XLO,I XL(I ) , 1 = 1,KXL)
RIT 2,1,KYL,YLO,(YL(l),1=1,KYL)
HIT 2,1,KXC,XCO,IXCiI ) , 1 = 1,KXC )
RIT 2,1,KYC,YC0,(YC(I) , 1 = 1,KYC )
I = 0
WOT 10,S3?,I,XL0,YL0,XCC,YC0
CO 45 1=1,5
45 WOT 10,632,I,XL( I),YL( I),XCI I ) ,YC
-------
h2 = .?«ll
I.SUTh = H It.
LU 25T J = 2,I\,T
N = J-l
t F ( J-2)7C,10,70
60 r.c t5 1 = 1,\s
uoi 1*1> = cm
FL1 1,1 ) = FL1I)
65 ZC( 1 + 1 ) = EL(I )
WuT 10,8 21,TIKI1,1),(UCII+1),I=1,NS>
WOT 10,R22,(ZOII + 1 I,1=1,NS)
CU TO 80
7C Ln 75 1=2,NS1
FL(I-1.1 ) = FL([-1,3)
U0( I ) = I'A I I I
75 201 I ) = ZM I )
,201I + 1) ,I,M,1)
CU 9C I = 2,N S1
9c z111) = zrii)*hz»;po(i)
CO <55 1 = 2,NSl
95 iPlll) = BrAT IZ1I 1-1),Z1(1 ) , 21 I I*1) , I ,N>2)
CO 1?C 1=2,NS1
10C 2211) = L'j I 1 )+H?»ZPl( I )
CO 1C5 I =2 ,ivlSl
105 ZP21H = UVATl Z?l 1-1) ,221 I ) ,Z2 (1*1) , I ,\l,2)
DO 11C 1=2,NS1
11C 23(1) = 201 I)+H»ZP2(I )
CO 115 1=2,MSI
115 ZP3II) = HI" ATI Z 3 ( 1-1 ) , Z3( I ) ,Z3( 1 *1 ) , I , J,1 I
CO 12C i=r,,\si
Z = AVATtUK 1-1 ) ,U1( 1 ) ,U1 ( 1 + 1 ) , I ,N,2,?)
CO 15C 1=2,!MS1
130 U2(I) = UOI1)+H?«UPllI)
CO 185 I=2,i\Sl
135 UP2IM = AVAKU2I 1-1 ) ,U2( I ) ,U2 ( I *1 ) , I ,N,2,2)
CO 19C I = 2 • N S1
190 U3( I) = U01 I ) + H • U P 2 II )
CO 195 I= 2,Mb 1
1 J5 L'P.MI) = AI'AriU1 ( I-I ) ,'.'31 I ) ,U? ( l« 1 ) , I , J, 1 , 2 )
CO 2 CC 1=2,MSI
2Cr IJ4I1) = L<".| I ) +HSI>TH« ( UPC I I ) +2. • IUP1 ( I ) *UP2 ( 1 ) ) »UP3I 1 ) I
IF (N-(\/KHH)»
-------
11/C3/64
PAGE 4
1*0T lC>32X,4IIT:M,y//)
3C2 FUtC'AT IF4 P . 2« lPElb.f tCPF12.3/)
903 FG«.VAT I1H1,?4X,12A6///31X, 17HSF„TICA ALPHA,11 *,jHPHI,13X,1HE/
1/)
e04 FORMAT (I3S.2X,1P3E15.5/1
005 FLSf'AT ( LH1,24X,12A6///49X,23HPC.LYNUMIAL CCE FF I C I CIMT b//46X , ltlk , 16X
1 ,41-Cl S ) , 13X , 1HU//)
307 FORMAT ( J2X,?HC( i II .:H) ,l»3E17.c/l
8C8 FORMAT ( 1HI ,?4X, 12A6///S1X, lelllMTI AL CC\P I T Ii.NS / /42 X , 1 7HS ECT I ON
1 D.O.,1CX.6HU.G.D.//)
e09 FORMAT I146,lPE17.5,Ei5.5/>
82C FQRfAT ( 1H1 , 24X, 12 A6///44X , 52HF I.mAL ANSWEKS - 0.0. A'-JD R.0.0. V S
1 SECTICX VS TINE//)
821 FORMAT I3HC TIME =,Ft..1.91! D.O. , 1 PfcE 1 b . 6 / I 2 3X , 6E 16. 6 ) )
822 FORMAT ( LUC , 16 X , fchR . 0 . C . . IP6E 11. 6/ 1 2 ? X , f,C I 6. C.) )
930 FORMAT I 135,?X,1W4E15.5/1
831 FORMAT ( 111'' / //46 X , 2IIL 0 .1 5X , 3HL • 1,1 4 X , 2HC0 . ! b X , 3HC 3 1 / / )
832 FORMAT (32X,2HC1,11,2H) .1P4E17.S/)
843 FORM,AT ( 1H1.24X, 12A6///31X , 7HSEC T I CN, 7X , 1HJ .ITX, 3H1/V, 1 3X, 1HK, 15X ,
HHP// )
E.NC(l.l,0,0,0,l,l,l,C.1.0.0.0,C.G)
A-6
-------
APPENDIX B. Listing of Digital Portion of Hybrid Program
DELAWARE ESTUARY COMPRFHFNS 1VE STUDY PARF 1
HYBRID riFMONSTRAT I ON PRORLEM
DIMENSION TKOTtO]
DIMENSION TTtlSS]
DIMENSION V "r 32 I .e [ 32 ] , ALPHI32 J . AJ 1321 . PHI 1321 , FLI32] .CI 3? J ,
1 R11 3?1 .'R3 T32 1 .RETA Ml ,HDLI32i .F t321 »uncf321
DIMENSION THD[366 1 . QHDI3661
COMMON DUMt49]'.'SCIN[15 1.SCOUT 1101 ,DINll5l.SPARE,Y3»Y2.Yl,P3,P?.P
11.X3.X9,X1'.'nOUT.NIN
2 . SP ACF f 231
READ I. Sf-TNtil
READ 1 . f S(?OUT I 11 . T=1.10 J
PPINTtOOl'.SCINM 1 . [ SCOUT "t I 1 . 1=1.101
1001 F0RMATr2*1.3HINPUT SCALF ¦ F6.3.2X15H0UTPUT SCALFS= 10F6.2]
SCINM 1 = SCINM 3 /4095 .
DO 25 I = 1.10
25 scout r ii = 40957/ scoutiii
PRINT 10 0?.'SC1N(11 . t SCOUT f I 1 . 1=1.1 01
1002 FORMAT[1X20WNORMALIZEn FACTORS- F6.3, 2X10F6.21
MN = 1
NOUTs 9
Rfad i.rTT'ril . I = l. l n 31
DO 2 T s1V183
LFFT « TT fI1/10 0 0.
FT = i FFT
THD12*1-11 =FT
TWO t 2*I1 » TT r I 1 -FT«1 000 .
2 CONTINIIF
PRINT33.[TWDCI1.T=1.3661
RFAD 3. [OHDI I 1.1=1.3661
PR I NT33 .iOHntI 1.1=1.3661
3 FORMAT M 2F6 • 0 1
DO 15 I = 1.366
THDtll = r T H D T I i - 32. l*r"5./9. 1
OHD til = OHD I I 1 *86400 .
15 CONTINUE
PRINT 1011
1011 F0RMATt?X32HDAli.Y TFMPFRATUPES t nFG , CFNT . 1 = 1
PRINT*.t THD tI 1.1=1 .3661
PRINT 101?
1012 FORMAT f 2 V 30HDAILY FLOW RATE trU.FT./OAYl = 1
PRINT33 ,' tOHDI I l'.'I =1.3661
33 FORMAT! 1X 6F1?,0/1X6F12 . 0 1
PRINT 1013
1013 FORMAT f 2X 41HINTTIAL CONDITlONStC AND i' IN MG./LITER 1 1
DO 35 I = 1.32
READ 4. ctll . FLill
PRINT 44, C t T 1 7 ELtll
4 FORMATr?F12.21
35 CONTINUE
KN » A190
44 FORMAt r 2X 3HC= F10.2. 5X3HL" F10.21
PRINT 1014
1014 FORMAT[?X 18H STATION CONSTANTS 1
READ 1252'.' r A J [ T S.Vt II . 1=1.321
12S2 FORMATt36X2Fl2l0 1
B-l
-------
r. nPLiUARF ESTUARY COMPHfHPNSIVE STUPY °4GF 2
DO 1240 » = 2.32
vri-ii * vttj
1260 CONTINIIF
READ 12537 TALPHh1,PHI I IJ.EI 11.I«1.321
1253 FORMATf3Fl2.0]
DO 5 I = 1 '• 3?
E 11 1 ¦ FtM*100n0D0.
v 111 ¦ viii/inonoooooo.
PRINT 1111.I .
1 alpwhi. PH» t 1 1 ,c (M , » J [ I 1 . V [ I 1
"5 CONTINIIF
1111 FORMAT! 2* 7HST»T10N 13. ?X7H.AIPHA« r5.2,2X4HPHI»F5.?.?X2MF«F1
i5.8,2X?HJsF5.2.2*4Hl/V«F15.81
1 FORMAT F 5F1?.2 J
R F A D1. ( B F T A [ I J I s 1.H1
PRINT ?011. tBBTAtll .1=1.8]
?011 FORMAT f ?X5HRETA» 5F12. 8/I(IX3F1 ? . 81
D020JK = li365
T s TwnfjKl
Q ¦ QHnt'iK]
Bill] s (A1 PH(11*0*CM 1 1 * V 111
D ¦ BFTAili*BETA>21*T *RETA[3]*T*•?
b? « i r -alphI2i*phi rn i*o-etii-Fr2i i*vrn-n
B3111 ¦ f f? r ? 1 -PHT [?1*01*V111
Pl'BlII1
P2"B2
P3«B3I11
YloAJM 1
Y2«AJ I?1
Y3>AJ I 31
Xl'ELI11
X2»FLI?1
X3«EL(31
I >2
CALL STRTr
30 CALL tnniT
CALL uni.Dt
HDLI!-11=niN[l]
IF!I-3214S.40.4ft
45 CONTINUF
CALL "SFTlr
V
ldz
V
RCS
V
RPL
V
RRU
NRD1
V
Lnz
V
ST A
1
V
STA
2
VTNS1
LHA
7938.2
V
STA
IKOT.1
V
I NX
2.1
V
I NX
1*2
V
RXH
18.1
V
RRU
**2
V
RRU
tn<;i
B-2
-------
r
DFLaWARF ESTUARY COMPftFHFNSIVE STUnY
PAGE
3
V
LOA
795?
V
STA
I K T T
PRINT 3011,t
3011
FORMAT [ 2X16UL COMPUT AT I ON I a
PRINT30027X1, P1.P2.P3.Y1,
3002
FORMATr?X3HXl*Ei1.4,
1 3HP2aBil.*."lX3HP3aE11.4.1X3HY1 = E11
1 lX4MniNa Ell.4]
PRiNTSOOl'." 1 IKOTII II . I 1=1.91 . fKIT
3001
FORMAT f 1109]
wNRni
LOA
ONE
V
STA
K
VMP2
LDA
K
V
Ano
ONE
V
STA
K
V
SUB
MIL
V
BM I
V
RRU
MR2
Ml
niMi ]
1X3HP1»P11.4.1X
CALL STRIA
Bitn ¦ iAt'PHtn»o*Eii n*vm
B2 St l-ALPHf 1*1 ]*PMI "r I n*0-E( n -Ef !tl i i*w n i -r
BSfllerPI1*1l-PHT tI*1]*01*w tII
Pi«Birii
P2»B2
PS«B3f 11
I S 1*1
Yl»AJfT-11
Y2»AJ[f1
YSsAjf T*11
XlaEL(1-11
X2sFLlM
XSsELI 1*11
GO TO 30
40 CALL MOl nA
CSs RPTA[A1+BFTA171«T*BETA[S]*T**?
CI11 = CS
C r 32) S CS
R s RBTA t"!l*T*BFTA [4]
DO 50 t » ?•31
Fill ¦ »*cs - D* FLII1
El r 11 ¦ Hni, 111
¦50 CONTINUE
FflJ ¦ P*rS;D*EL[ll
F 1321 » R*CS - D* EL f 3?1
PRINT 51..IK, IFLll I. Ja?.31J
Al= Rttll
A3 »ir-ALPH(21 *PH I[1J]*Q*F[11-E r ?11 *V[11-R
A3 • R3fll
P1«A1
P2 = A2
P3 = A3
Y1 a F [ 1 1
Y2aFtiM
Y3 ar f 3 ]
XlaCH I
B-3
-------
r DELAWARE FSTlJiRY CnMPRFHFNS1VF STUnY
X2 = C t'1
X3=cni
CALLSTRTC
I *2
40 call innAT
CALL WOLDA
Hocti-n ¦ oInti']
IPII -39] A?.70770
65 CONTINUE
CALL SCTtr
w Lnz
V RCS
V RPL
V RRU NRn?
v lpz
V STA 1
V STA 2
VTNS? L1A 7936.?
V STA IKOT.'l
V T NX 2.1
V !NX 1.2
V RVH 18.1
V RRU **2
V RRU TNS?
V LDA 795?
V STA IKIT
PRINT 3021¦I
3021 FORMATr2X16HC rOMPUTATI ON.J a 14]
PR INT3 0 027x1. P1.P2.P3.Y1. niNMI
PR INT30 0l7 tIKOT[I II.!1=1.91.T KIT
vNPn?
LHA
ONE
V
STA
K
VMRl
LIJA
K
V
ADD
ONE
V
STA
K
V
SUB
MIL
V
RMI
V
RRU
MR1
CALL STRTa
Pl»RlJ 11
P2 ¦ ri -ALPH [f *1 ] * PHIfI 1!*0-EfI)-<=[t *11l*v[I] -R
P3=R3f 11
I = I *1
YlsFM-11
Y2=FIt1
Y3=F[T*1)
Xl=Ctf-1 I
X2=CII1
xs=rr t*i i
GO TO 60
70 CALL HOLnA
DO BO I a 2.31
RO Cin « wnrru
PPINT 81. [C[M.1 = 2.311
81 FORMAT (//¦!X2Hr= / [5XAF15.81 »
B-4
paGF
-------
r
HFLiWARF FSTIJ4RV CnMPRFHFNS T VF STUnY
?1 FOPMATf 5X3HT1AY I4,2X3H,L=/t5X6P15.RH
?0 CONT1NI1F
call wm.n»
CALL BXIT
VK
vonf
nec
1
VM! L
nsc
9000
END
»*RF 5
B-5/B-6
-------
APPENDIX C. Printout of Digital Program
?NTRV PI. I NT S TC SU3R0UT I NE S KTOUFSTEP FROF LIBRARY,
< FPT ) ( TSMfl) ( SL I ) UTN)
LCADINC MAP
UNUSED 5TCRACE- 13461 THRU
KAV.E
JKIGIN
ENTKV
NAf'F
URIG I N
CfJTRY
N AMU
ORIGIN
ENTRY
NAME
OK IGIN
CNTRY
NAME
ORIGIN
• f Air,1*
OC 144
JO l 5 5
AMAT
07712
C7717
BMAT
10352
10357
C-XPl 2
10666
10671
ANYER
IC744
SPkER
1C 7 4 4
11'.4 3
OVFLrf
i n 7 4 4
11436
UNFLW
10744
11441
B*CDE
10744
11423
EPMDE
10744
( SVN)
10744
1C 7 ^ 3
(SIX)
10744
lw752
(FPT)
1 0744
10754
FRRI1R
10744
11116
ERUP
10744
ERSIN
1 C 74 4
11447
{TSHM)
11476
11511
( TSH )
11476
11506
SETPu
11562
11766
SETLN
11562
PJCES
11362
1*016
L INES
11562
11755
SPGHDR
11562
120C0
WPCHDK
11562
11746
(STHD )
11562
ISTHMJ
11562
11577
( STK)
11562
11571
(SLO)
12144
12144
( SLI )
12162
12162
(WTO
12200
UER)
122CG
12212
( KDC )
12270
12321
(RER)
12270
12277
(RTN)
12J36
14165
(FIL)
12336
I IOH)
12336
12340
(EXEN)
14340
14342
(EXE)
14340
14342
< TEF )
15240
15333
(RCH)
15240
( ETT )
15240
15331
(RUN)
15240
15330
(REW)
15240
15327
(WEF )
15240
15326
(8SR)
15240
(URS )
1 524C
15324
( RDb )
15240
15323
( (OS)
15240
15243
1 TRC )
15240
15335
( TCO)
15240
( IGU)
1 = 272
15375
E X1 T
15434
15435
( TES)
15460
15460
11445
11414
11455
11761
11574
12243
14154
15332
15 325
15334
LCGICAL
MAChlNE
TOTAL
TOTAL
NOISE
RECORDS
TOTAL REDUNDANCIES
POSITIONING
TAPE
TA^C
WR ITES
READS
WRITING
HEAPING
WRITING
READING
ERRORS
l
A
0
60
0
0
0
0
o
£
2
392
428
0
0
0
0
0
3
6
i
87
108
0
0
0
0
0
4
A
4
?96
322
0
0
0
0
0
5
A
2
0
257
0
0
0
0
0
6
A
3
62C
3
0
0
0
0
0
7
d
4
lib
117
0
0
1
0
0
18
B
9
0
63
0
0
0
0
0
EXECUTION
C-l
-------
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rELAWArtF. EblliAKY STUL'Y -
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2
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6
7
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10
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16
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ALPHA
9.7UGC0E-C1
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8.24000E-01
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nCLSW.pir rsrn/V.Y btULiY
TEST CALf.ULAT I CNr> FCH YEAk 1963
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rrLAVAS= r S, 11, Y j T l.L Y - Tr SI C A L C U L A 11 LNS FOR YEA!< 1963 - 3i LAYS
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C-14
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"ELAWAKF r S TIJ A ,< Y STl.DY - TEST C .\LCI1L A I I fJNS FUK Y f AR 1963 - 31 i.V'vS
INITIAL CHNDITIUNS
SE.CT I ON
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0 • 0 • 0 a
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C-16
ELVriArfE 17 b r IJ A K Y STlilJY
rcsr r.„LCI)LM lOMS FOi< YEAR 1963 - 31 DAY b
MNAL ANSdKS
o.u. and 3.L-.D. vs section vs time
1 . 340001E
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1 .330000!:
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PELAWArtE ESTIMKY STUDY -
TtST CALCULATIONS FCJR
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3.376BI3:
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9.781294E
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1.069274E
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4.45'j2B7E
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4.BC0187E
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B.O.D.
TIME = 3.C C.O.
B.0.0.
TIME = 4.0 IJ.C.
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LTI-AwAkT r S TI! A .< Y iTUlY - TL51 CALCiJLAT HiT. F-OH YEA* 1963 - 6 LAYS
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k*«5 JIams
1 .34'. "'OCF
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1.61C215E 01
9.321912E PP
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1.3H319E r.l
1.1C7996E 01
9.343402E 00
9.G33C35E CC
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4.25')41tE PC
4.y95427E PC
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7.623252E
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9. 3C2663E
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9.65 7888 E
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1.155514?
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1.414990E 02
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1.947195E
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C-29
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TIME = 5.0
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1.13713'JE 02
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3.807329E 00
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£.9C699GE 00
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TIME
TIME
TIME
T IME
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C-34
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2.0 C.L'I.
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APPENDIX D. Digital Printout of Hybrid Program
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4400 .
3860 .
3420 .
2950
2170 .
2050 .
2230 .
2320 .
2630 .
3180
2920 .
2350.
2380 .
2600.
2690 .
2820
2570 .
2350 .
2850 .
3550.
4590 .
4700
4d20 .
401 0 .
3450 .
3420 .
3210.
3120
331 0.
9570 .
3620 .
3950.
3650.
4920
4420 .
3720 .
3000 .
2700 .
2760 .
2940
3320 .
3490 .
2620. •
2420 .
2P 8 0 .
2760
3 340 .
21°0 .
2530 .
2760 .
?»20 .
2970
3120.
3880 .
2760 .
2650,
31 >0.
1990
1*70 .
1 920 .
2770 .
2450 .
3730 .
2240
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2650 .
2940 .
2700 .
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2400
3140 .
3040 .
2340 .
3330 .
279 0 .
2760
-------
'51 0 .
'340 .
7 3 4 0.
2590 .
7760 .
7760 .
'790 .
78«0 .
7260 .
19O0 .
71*0.
7290 .
7260.
7700 .
"*75Q .
3390 .
371 0 .
7840 .
'500 .
7330 .
750(1 .
2280 .
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1930 .
3«O0 .
'450 .
'330 .
2180 .
1 970 .
1890.
1 750 .
1 950 .
7260 .
221 0 .
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7260 .
1 930 .
1 670 .
1 640 .
1730 .
t 6 "» 0 .
1820.
1 nOO .
1 73 0 .
1370 .
1300 .
7 o (J 0 .
1910.
1 930 .
1 860 .
1 6°0 .
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7 9 >• 0 .
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9200 .
79*0 .
5860 .
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4500 .
4160 .
3580 .
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7780 .
3060.
3180.
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4270.
4750.
4240.
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6550 .
8050 .
8400 .
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5980 .
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17100.
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0.
PA T LY
TEMPERATURES rDFG
i-ent .
=
1 .
1.
1 .
1 .
l.
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1 .
1 .
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1.
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7 .
8 .
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9.
10
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10 .
11 .
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11 .
11.
12
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18
18.
19.
19.
20.
20.
20.
19.
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19
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19.
19.
20.
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70 .
21.
21 .
21.
72.
21.
21
27.
27.
23.
23.
23.
23.
23.
23.
23.
23.
23.
23
24".
24.
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24.
24.
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25.
25.
25.
25.
25.
25
25.
26.
26.
26.
26.
76.
26.
26.
26.
26.
26.
26
26.
26.
26.
26.
27.
78.
28.
28.
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27
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27 .
27.
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daily
FLOW
RiTF
rru.FT
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=
2851 20 u o o.
388800000.
69i?ooooo.
608756000.
67208ouoo.
613*40000.
7?576iooo.
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4^117^(10.
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794880000.
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97131 ?HOC0.
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77^4400 0 0.
73fl7?0000 .
asi^snono.
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712«D0000 .
84153*000.
574*60000.
686*30000 .
548*40000.
611712000 .
475200000 ¦
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346464000 .
412Q92 0 0 0 ¦
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177120090.
203040 000.
20304(1000 .
346444000 .
222048060 .
321404000.
301536000.
189516000•
243»32000.
165888000 .
228960000 .
176556000 .
20217(5000 .
24OR320BO.
233280000 .
201312000.
211680000 .
168480000.
139968000.
149472000.
160704000.
794880000.
309-512000 .
274752000.
350784000.
626400000 .
915840000.
5978B8000 .
466560000.
483«40000 .
8931 6(1100 .
667440000.
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686*80000.
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5PJ5R4000.
652320000.
1054080000.
578P80000.
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362B81000.
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274752000
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165024000
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0'
D-4
INTTT4L
PONDITIONSfC aNTI
L IN MR
Cs
0.
L =
4.00
Cs
13.40
L =
5.30
Ce
13.3(1
L-
5.30
Cs
13.2(1
La
5.30
Cs
13.00
L =
5.60
Cs
12.80
L-
5.70
e»
12.80
L =
5.40
Cs
12.50
L =
6.00
c*
12.20
i. =
6.90
Cs'
11.90
L ¦
7.60
Cs
11.60
Ls
B. 30
Cs
11.20
L =
9.40
Cs
10.20
1 =
12.20
Cs
9.60
L®
11.00
e«
8 . 90
L =
11.40
Cs
8.00
L =
11.90
Cs
8. 4(1
L =
12.40
-------
8.40
L»
12.30
Cs
8.60
L»
10.60
Cs
8.90
L =
9.10
Cs
9.40
L s
9.60
Cs
9.80
L 0
9.90
Cs
10.10
L 5
10.20
Cs
10.30
La
9.90
Cs
10.50
L s
9.50
Cs
10.80
L s
9.10
Cs
11.00
l_ s
8.70
Gs
11.30
L *
8.30
Cs
11.50
l_ s
7.80
Cs
11.90
L»
7.20
Cs
12.40
L 8
6.40
Cs
0.
L*
4.00
STATION CONSTANTS
1
2
3
4
5
6
7
B
9
10
STATION
STATION
STATION
STATION
STATION
STATION
STATION
STATION
STATION
STATION
ST At I ON 11
STATION 12
STATION 13
STATION 14
STATION 15
STATION 16
STATION 17
STATION 18
STATION 19
STATION SO
STATION 21
STATION 22
STATION 23
STATION 24
STATION 25
STATION 26
STATION 27
STATION 28
STATION 29
STATION 30
STATION 31
STATION 32
BETA"
AL»HA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
alpha
AL'PHA
ALPHA
ALPHA
ALPHA
alpha
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
ALPHA
alpha
alpha
0,09057000
0.97
PHIs
0.03
Es
0.53000000?
08
J=
0.
1/V«
0.41S20000F-08
0.90
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0.12
En
0.12900000?
09
J=
0.78
1 / V c
0.27470000F-08
0.82
PHIs
0.21
Es
0.18500050=
09
Js
0.57
1/V"
0.71740000F-08
0.77
PHIs
0.26
•Es
0.21700000c
09
J=
0.23
1/Vs
0.1BR00000F-08
0.64
PHIs
0.29
Es
0.247000005
09
J=
0.26
1/Vs
0•1 5970 0 0 OF-08
0 .64
PHIs
0.35
Es
0.29400000F
09
J=
0.07
1/V"
0.13730000F-08
0.6S
PHis
0 .47
Es
0.47700000F
09
J=
0.16
1/Vs
0.?i980000F-08
0 .55
PHIs
0.55
Es
0.85700000?
09
J:
0.10
1/Vs
0.19M000OF-08
0.56
PHIs
0.56
Es
0 .884000001=
09
Js
0.17
1/Vs
0.18760000F-08
0.56
PHIs
0.56
Es
0.95700000?
09
Js
0.14
1/Vs
0.17180000F-08
0.56
PHis
0.56
Es
0.10520000?
10
J:
3.90
1/Vs
0.15B70000F-08
0.56
PHIs
0.56
E>
0 .112300001?
10
J =
0.19
1/V =
0.15270000F-08
0.56
PHIs
0.56
Es
0.11400000S
10
Js
0.03
1/Vs
0.1441000008
0.56
PHIs
0.56
Es
0.12560000c
10
Js
2.04
1/Vs
0.12420000F-08
0.63
PHIs
0,63
Es
0.10170000?
10
Js
3.80
1/Vs
0.53800000E-09
0.68
PHIs
0.68
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0.84600000c
09
Js
2.45
1/Vs
0.49300000F-09
0 .69
PHIs
0.69
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0.90600000?
09
Js
0.31
1/Vs
0.45800000F-09
0.69
PHis
0.69
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0.9790 0 0 0 OF
09
Js
0.55
1/Vs
0.41700000F-09
0.70
PHIs
0.70
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0.10910000?
10
J =
0.27
1/Vs
0.37100000F-09
0.70
PHIs
0.70
Es
0.12330000?
10
Js
0.59
1/Vs
0.34100000F-09
0.7?
PHis
0.72
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0.17320000?
10
Js
0.01
1/V"
0.6610 0 0 0 OF-09
0.7S
PHIs
0.73
Es
0.26230000?
10
J=
0.97
1/V"
0.63500000F-09
0.74
PHis
0.74
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0.28130000?
10
Js
0.61
1/Vs
0.58900000F-09
0.74
PHIs
0.74
Es
0.30530000?
10
Js
0.02
1/V =
0¦55B000O0F"09
0.74
PHis
0.74
Es
0 .31380000c
10
Js
0.01
1/Vs
0.53800000F-09
0.74
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0.74
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0.32540000?
10
Js
0.01
1/Vs
0.5?0 0 0 0 0 0F-09
0.74
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0.74
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0.33920000?
io
Js
0.01
1/Vs
0.48700000E-09
0.74
PHIs
0.74
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0.37040000?
10
Js
0.10
1/Vs
0.44500000F-09
0 .75
PHIs
0.75
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0 .27060000?
10
J=
0.06
1/Vs
0.40B0000OF-09
0.79
PHIs
0.75
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0.21980000?
10
Js
0.03
1/Vs
0.35600000E-09
0.75
PHIs
0.75
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0.21980000?
10
Js
0.03
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0.
0.
PHIs
0.
Es
0.
Js
0.
1/Vs
0.
0001055
01
2001377
0.00941600 0
14.56299999 -0.37298000
L COMPUTATION,!! 2
XI" 0.4000F 01 Pis 0.1362E 01 P2s-0.1871F
0001463 0002075 0002075
L COMPUTATION, Is 3
Xls 0.5380F "01 Pie OT1O58E
0002075 0002075 0002075
L COMPUTATION.!¦ 4
XI" 0.5300E 01 Pis 0¦91?9E
0002075 0002075 0002172
L COMPUTATION. I« 5
Xli 0.5300E 01 PI: 0.81?9E 00 P2s-0
0002075 0002172 0002217 0000514
L COMPUTATION. I ¦ 6
X1s 0.5600F 01 Pis 0.686&E
00S2172 0002217 0002171
L COMPUTATION.Is 7
00087900 O.lflOOOOOO
0,00466780
0.00600000
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01 P2s-0,1508F 01 P3s 0.3469E-00 Yls 0.7800F
0000661 2001152 0000216 0000477 0000351
00 P2«•0
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2001031
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3000735
P3s 0.3010E-00 Yls 0.5700F
0000173 0000351 0000136
P3s 0.306BF-00 Yls 0.2300E"
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1118F 01 P3s 0.3106E-00 Yl= 0.?60at;-
2000712 0000177 0000152 0000034
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00 niN=
0000136
00 niN=
00001 5?
00 niN:
0000034
no niN=
0000101
0.997BF 01
0403773
0.4313E 01
0401563
0.5201? 01
0402051
0 • 49'87F 01
0401775
0.5280? 01
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0002217 0002121 000231*
L r0MPuTiT10N.Ii 8
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0002121 0002314 000260*
L COMPUTATION,Is 9
Xls 0.6000F 01 Pl= 0.?0l*E
0002314 0002604 00030?*
L COMPUTATION, I« 10
*1= 0.6'0OF 01 Pis 0.1958E
0002604 0003024 0003243
L COMPUTATION,!« 11
Vis 0.7600F 01 Pis 0.191BE
0003024 0003243 000360*
L COMPUTATION, I« 12
Xls 0.8300F 01 Pis 0 .19?3E
0003243 0003604 0004701
L COMPUTATION, Is 13
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0003604 0004701 000431*
L COMPUTATION, Is 14
XI¦ 0 .1220F 02 Pis 011875E
0004701 0004314 0084436
L COMPUTATION, Is 15
XI* 0 ¦ HOOF 02 Pis 0 ,176OE
0004314 0004436 000460*
L COMPUTATION,Is 16
Xl« 0.1140F 02 Pis 0.6446E
0004436 000*604 0004752
L COMPUTATION,Is 17
Xls 0.1190F 02 Pis 0".51?7E
0004604 0004752 00047?6
L COMPUTATION,Is 18
Xls 0.1240F 02 Pl= 0.5051E
0004752 0004726 0004172
L COMPUTATION, 1= 19
XI* 0.1Z30F 02 Pi's 0r49fl'E"
0004726 0 P 04172 0003507
L COMPUTATION,Is ?0
Xls 0.1060F 02 Pis 0¦479'E"
0004172 0003507 000365'
L COMPUTATION,Is 21
Xls 0•910 OF 01 Pis 0 i 4W90E*
0003507 0003655 0003753
L COMPUTATION,Is 32
Xl« 0,960 OF 01 Pl= 0.1281E
0003655 0003753 0004050
L COMPUTATION,1= 23
Xls 0¦99flOP 01 Pis 0.179»E
0003753 0004050 0003753
L COMPUTATION,Is ?4
Xls 0.10?0F 02 Pt = 0.17B1E
0004050 0003753 0003631
L COMPUTATION, 1= ?5
Xls 0.9900F 01 Pis O.lBplE
0003753 0 0 0 3631 0003507
L COMPUTATION, 1= 96
Xls 0 .950OF 01 Pis 0.1B02E
0003631 0003507 000336$
L COMPUTATION, I> ?7
xi« o.oinoc ni dis 0.i8o2e
0003507 00033*5 0003243
L rOMPUTATION,Is ?8
*ls 0.S700F 01 P1= 0.175SE
0003365 0003243 0003075
L COMP'lT »T ' OM, I = ?9
00 P2s.0.12?lF 01
0000402 2000764
01 P2s.0.30R0F 01
0001116 2002356
01 P2s«0¦3553P 01
0001470 2002657
01 P2s»0•3550F 01
0001441 2002656
01 P2sa0.3548P 01
0001421 2002655
01 P2s«0.3550P 01
0001423 2002656
01 P2»°0.3552P 01
0001442 2002657
01 P2» = 0 .3549F 01
0001377 2002656
01 P2se0¦2944p 01
0001320 2002266
00 P2»«0.1105F 01
0000407 2000705
00 P2s»0.9612F 00
0000321 2000612
00 P2s=0•9601P 00
0000316 2000612
00 P2s«0.9605F 00
0000311 2000612
00 P2s.0.9583F 00
0000304 2000611
00 P2sio.ll09F 01
0000310 2000706
01 P2sc0.297BF 01
0001014 2002304
01 P2s.o.3549F 01
0001340 2002656
01 P2s.o.3551P 01
0001331 2002657
01 P2s = l).3551F 01
0001351 2002656
01 P2s=0.3535F 01
00013*1 2002650
01 P2S.0.3552F 01
0001341 2002657
01 P?sin,3553F 01
0001316 2002657
P3s 0.4S26F-00 Vis 0.7000F-01 PtN = 0.490AF 01
0000271 000003* 0000101 0000050 0401755
P3S 0.1536F 01 Yls 0.1600F-00 PIN = 0.474JP fll
000116* 0000101 0000050 0000105 0401713
P3 s 0.1*37F 01 Yls 0 . 1 000E-00 nIN= 0.5074P 01
0001114 0000050 0 B 0 0105 0000071 0402017
P3s 0.1496E 01 Yls 0 .1700F-00 PINs 0.5739F 01
0001144 0000105 0000071 0003075 0402227
P3s 0.1533E 01 Yls 0.1400F-00 IlINs 0.7233F 01
0001 3 43 0000071 000.1075 0000115 0402711
P3S 0.1527P 01 Yls 0.3900E 01 PIN = 0.R816E 01
0001161 0003075 0000115 0000014 0403415
P3S 0.1495E 01 Yls 0.1900F-00 OINs 0.9411E 01
0001144 0000115 000001* 0001503 0403607
P3s 0.1578E 01 Yls 0.3000F-01 DIN: 0.9714E 01
0001206 0000014 0001503 0003024 0403705
P3s 0.1038E 01 Yls 0.20*OF 01 PIN= 0.U70E 02
0000651 0001503 00030 2* 0 0 01753 04 04533
P3s 0.350Hf=-00 Yls 0.3800F 01 PINs 0.1218F 02
0000217 0003024 0001753 0000176 0404675
P3s 0.3497E-00 Yls 0.?450E 01 PIN = 0.1*28? 02
0000217 0001753 0000176 00003*1 0405553
P3s O.3576E-0O Yls 0.3100F-00 PINs 0.1141E n2
0000222 0000176 00003*1 0000156 0404440
P3s 0.3711P-00 Yls 0.5500F 00 PINs 0.1103E 02
0000227 0000341 0000156 0000361 0404323
P31 0.3829F-00 Yls 0.2700F-00 PINs 0.9841E 01
0000234 0000156 0000361 0000004 0403737
P3s 0.5206E 00 Yls 0.5900F 00 PINs 0.R864F 01
0000325 0000361 0000004 0000615 0403427
P3s 0.1595E 01 Yls 0.1000F-01 PINs 0.B454E 01
0001215 0000004 0000615 0000371 0403303
P3s 0,1652F 01 Yls S.9700F 00 IlINs 0.RB35E 01
0001244 0000615 0000371 0000010 0403421
P3s 0.1674F 01 Yls 8.6100F 00 PINs 0.9001F 01
0001255 0000371 0000010 0000004 0403463
P3s 0.1633F 01 Yls 0.2000F-01 niNs 0.8557E fll
0001234 0000010 0000004 0000004 0403330
P3s 0.1637P 01 Yls 0.1000E-01 PINs 0.H112E 01
0001236 <1 0 000 04 0000004 0000004 0403175
P3s 0.1654P 01 Yls 0.1000F-01 PINs 0.7751P 01
0001245 0000004 0000004 0000050 0403063
P3S 0.1700= 01 Yls 0.1000F-01 PINs 0.7409F 01
0001270 0 0 Oo 0 04 0000050 0000030 0402755
-------
*1= O.R300F fll Pis 0 .1 7 4 3 E 01 P23.I1
800324'' 0003075 0002702 0001331
L COMPi IT A T I ON, I s SO
Xls 0.780 OF 01 P1= 0.1191E 01 P2 = .0
0093075 0002702 0002436 0000747
L COMPUTATION, Is 31
Xls 0.7200F 01 P1= 0.85B4E 00 P2 = = 0
000270? 0002436 0001463 0000537
DAY 1 • L °
0.43125763E 01 0.520l465?F 01
0.50744811E 01
0.11697192F 02
0.573B7057E 01
07l2l75824R 02
0.886446B8E Q1 0.84542124F 01
0.740903&4E 01
2
pp = -n
0000661
000040?
0.77509157E 01
C COMPUTATION,Is
Xl» 0.14S6F 02 Pis 071362E 01 P2s = 0
0005573 0005267 0005243 0001055
C COMPUTATION, Is 3
Xls O.134 0F 02 P1s 0.1058E 01
0005267 0005243 00T!52l6
C COMPUTATION. Is 4
Xls 0.1330F 02 Pis 01912'E 00 P2s = n
0005243 0005216 000514$ 0000565
C COMPUTATION.Is 5
*1= 0.1320F 02 P1s 0.81?9E 00 P2s-0
0005216 0005145 000507* 0000514
C COMPuTaTION,Is 6
*1" 0.130OF 02 Pis 0.686BE 00 P2s = o
0005145 0005074 000507*
C COMPUTATION,Is 7
Xl«
-------
xi= i.»«noc ni pi
0013?67 0003340
r. rcpnTij ion, r =
*1= n.nonoc ni pi
nona^an op 03436
C C0mP"T *Tion,i =
*1= n.«»00f ni oi
0003436 0r03604
C COMPUTATION,1=
*1= o.'Jinoc ni pi
0003*0* 0003726
C rOMP'lTiTIOM, 1 =
xt = n.osnop ni pi
00037?*, 0 P 0 * 0 23
C COMPUTATION, r =
*1= 0 .1Oloc 02 01
00P4P23 0 0 04074
C COMPUTATION, I =
xi* p.i030F 02 pi
o o o 4 n 74 0O04145
c roMPnT*Tton,i =
X3 = 0.1 050P 02 B1
0004145 0O04243
e rO"PilT*TION,Js
*1= n.i oaof 02 pi
00n4?4"* 0014314
C COMPmTaTION,I=
XI* P.llOOF P2 PI
00(14314 000443 1
C COMPUTATION,i =
XI = n.1130P 02 P1
8004411 0004462
c roMPnTation, i*
XI = 0.1150F n2 PI
000446? 0 0 04604
c computation, i =
Xl = n.llUOF 02 PI
0004604 0 0 0 4752
s 0.4909E-
0003436
?0
= 0.47936'
000360*
31
r 0 , 4S23F 01
0;8932B*49F 01
0Tl0?12454E
2
0 .1 6 0 * E
0002050
3
071207E 01 P2»»P
0001 77* 0000756
4
0 .10?1E 01 P2 = in
.12840049F 02 0.13269R4lg 02 0.12195360F 02 0.H 746032E 02
.11199023C 02 0.1PB37607E 02 0.10300366P 02 0.9689S657E Ol
¦80146520F 01 0.79267I99E 01 0.R0P302R1F 01 0.83174603E Ol
•8932S**9f 01 0.93040293E 01 0.95189255F 01 0.9733R217E Ol
.10417582P 02 0.3P876679E 02 0.11228327F 02 O.H970696E 02
01 P2==t
0001220
0002070
5
0.9008E
P 0 01 75 *
6
0.7*flBE
0001712
7
9.68?0E 00 P2r.o
0002016 0000*27
8
0.15?5E 01 PZ=-0
^OBSF 01 P3« 0.35P1E-00 Yl» 0.
2001527 0000222 OOPODOO 0000*77
l)IN« 0.1197F 02
0000351 040*623
0000642
00 P?=»0,
0000560
00 P2==P.
0 0 0 046 ?
,16?5F 01 P3= 0.3126F-00 Ylr 0.7800F 00 DIN= 0.3951F fll
2001232 0000200 0000*77 0000351 0000136 0401451
, 1385F 01 P3* 0 .2670F-00 Yl» 0 . 5700F 00 MN= 0 .4654E 01
2001070 0000155 0000351 00001S6 000015? 0401671
, 120RF 01 P3= 0.2733F-00 Yl« 0 .230 0F-00 HIN= 0.4830F 01
2 0 0 0 757 0000157 0000136 0000152 0000034 0401735
.1152F 01 P3« 0.2769F-00 Yl« 0.?600!=-00 n!N* 0 . 5001 F 01
2000730 0 0 0 0161 0000152 0000034 0000101 0402000
, 1243F 01 P3= 0.414RC-00 Yl* 0.7000F-01 DIN: 0.4400F fll
2000776 0000251 no00034 0000101 0000050 0401605
.30O0F 01 p.3» n.146PP PI rl* 0. 1 600F-00 niNr 0.4063F 01
-------
0001715 0 0 0 2 016 0 0 022?6
L r0«onT»T10N,l= 9
Xl = 0.5 074F ni oi= 0.20R1E
0002016 0002226 0002710
L COMPIjT AT I ON, Is 10
Xls 0.5739c 01 Pl= 0.2021E
0002226 0 0 02710 n003«l<»
L COMPUTATION, Is 11
*1= 0.7233F 01 Pis 0 .19 7 7 E
0002710 0003414 0003606
L COMPUTATION,Is 12
Xls n.B8l6F 01 P1s 071977E
0003414 000360ft 000370*
L COMPUTATION,I: 13
Xls 0.O4HE 01 Pl= 0.2013E
0003606 0003704 00045*2
L COMPUTATION,I= 14
*1= 0.9714F 01 P1s 0.192*E
0003704 0004532 0004674
L COMPUTATION,Is 15
*1= 0 .117 0 F 02 Pis 0.180ZE
0 0 04532 0004674 0005552
L COMPUTATION,Is 16
*1= 0.121BF 02 Pis 0.6652E
0004674 0005552 0004437
L COMPUTATION,Is 17
Xls 0.14J8F 02 Pis 0.53?9E
0005552 0004437 0004322
L COMPUTaTION,I= 18
*1= 0.1141F 02 P1s 0 ,5 2 4 2 E
0004437 0 0 04322 0003736
L COMPUTATION,Is 19
*1= 0.U03F 02 P1s 0 i 5Os*E
0004322 0003736 0003*26
L COMPUTATION,Is 20
Xls 0.9841E 01 Pis 0.4952E1
0003736 0003426 0003302
L COMPUTaTION.Is 91
Xl= 0.R864C 01 P1s 0.5035E
0003426 0003302 0003420
L COMPUTATION,Is 22
Xls 0 • R454F 01 Pis 0.130'E
000330? 0003420 00034ft2
L COMPUTATION,Is ?3
Xl= 0.1835F 01 Pis o;i827E
0003420 0003462 0003327
L COMPUTATION,Is 24
Xls 0.9001F 01 P1s 0.1807E
0 0 03462 0003327 000317"
L COMPUTATION,Is 25
Xls 0,8557F 01 P1s O.I846E
0 0 03327 0003174 0003 062
L COMPUTATION,1= 26
Xl= n.91l2F 01 P1s 0.18?6E
0003174 0003062 0002754
L COMPUTATION,Is 27
Xls 0.7751F 01 P1s 0.18?5E
000306? 0002754 000263®
L COMPUTATION.Is ?8
*ls 0.7409F 01 Pis 0.1776E
0002754 000263ft 0002602
L COMPUTATION,Is ?9
Xls 0,7 0 ?8F ni P1= 0.17ft3f
0002636 0n02602 0002413
L COMPUTATION,Is 30
Xls 0.6391C 01 P1= 0.1210E
00011*0 2002362 0001126 0000101 0000050 0000105 0401500
01 P2 = .0.3554C 01 P3 s 0.1 "<7 0C 01 Y1 s 0 . 1 000F-00 OINs 0.4313F 01
0001524 2002660 0 0 010ft0 0 000050 0000105 0000071 0401563
01 P2==0.3550C 01 03s 0.1432C 01 Yls 0.1700E-00 PIN= 0.490RF 01
0001 473 2002656 0001 112 0000105 0000071 0003075 0401755
01 P2s = 0 ¦ 3548F 01 P3 s 0 .1475F 01 Yls 0.1400F-00 IIINs 0.6696E 01
0001451 2002655 0001133 0000071 0003075 0000115 0402533
01 P2==0.3551F 01 P3s 0.1472F 01 Yls 0.3900F 01 HINs 0.88161= 01
0001451 2002656 0001132 0003075 0000115 0000014 0403415
01 P2s«0 ¦ 3552F 01 P3» 0.1443F 01 Yls 0 .1900F-00 DINs 0.9128F 01
0001470 2002657 0001116 0000115 0000014 0001503 0403515
01 P2 = = 0.3549C 01 P3s 0 .1529F 01 Yl* 0 .3000F-01 DIN = 0 .9060E 01
0001423 2002656 0001161 0000014 0001503 0003024 0403477
01 P2 s = 0.2949F 01 P3s 0.9905F 00 Yls 0.20*01= 01 DINs O.lllic 02
0001342 2002270 0000625 0001503 0003024 0001753 0404 343
00 B2 = = 0.1107F 01 P3s 0 .3287F-00 Yls 0.3800F 01 PINs 0.1291F 02
0000420 2000706 0000206 0003024 0001753 0000176 0405123
00 P?s-0¦9615F 00 o3= 0.3291F-00 Yls 0.2450E 01 nINs 0.13R7E 02
0000332 2000612 0000206 0001753 0000176 0000341 0405427
00 P2==0.9602F 00 P3s 0.3384F-00 Yls 0.3100E-00 PINs 0.1155F 02
0000326 2000612 0000212 0000176 00003<1 0000156 0404474
00 P2S=0,9607F 00 P3s 0.3533E-00 Yls 0.5500F 00 DINs 0.1012F 02
0000320 2000612 0000220 0000341 0000156 0000361 0404031
00 P2S-0.95B3F 00 P3s 0 .3670E-00 Yls 0 .2700F-00 lilNs 0.9138E 01
000031? 2000611 0000226 0000156 0000361 0000004 0403517
00 P2ss0•1109F 01 P3s 0.5058C 00 Yls 0.5900F 00 OINs 0.R415F 01
0000316 2000707 0000317 0000361 0000004 0000615 0403273
01 P2S-0.2978F 01 P3= 0.1566C 01 Yl= 0 .1000F-01 fl!N= 0.7678F 01
0001030 2002304 0001201 0000004 0000615 0000371 04030*4
01 P2S-0 .3549F 01 P3S 0.1*24F 01 Yls 0 .9700F 00 I)IN= 0.7B68E 01
0001354 2002656 0001230 0000615 0000371 0000010 0403H3
01 P2 s o0 ,3551F 01 P3s 0.164RE 01 Yls 0.6100E 00 nINs 0.7985F 01
0001344 2002657 0001242 0000371 0000010 0000004 0403143
01 P2ss0¦3551F 01 P3= 0.160RC 01 Yl» 0.2000F-01 DIN= 0.7429E 01
0001364 2002656 0001222 0000010 0000004 0000004 0402761
01 P2S-0.3535F 01 P3s 0.1613C 01 Yls 0.1000F-01 nINs 0.6950E 01
0001353 2002650 0001224 0000004 0000004 0000004 0402617
01 P2S-0.3552F 01 P3s 0.1631= 01 Yls 0.1000F-01 HINs 0.6569F 01
0001353 2002657 0001233 000000* 0000004 0000050 0402501
01 P2s-0 .3553= 01 P3s 0.1678? 01 Yls 0 . 1000F-01 niN= 0 .6266F 01
0001327 2002657 0001257 0000004 0000050 0000030 0402403
01 P2 = »0 .2949F 01 P3s 0.1089= 01 Yls O.'lOOOF-OO nINs 0.6012E 01
0001321 2002270 On00675 OOP0050 0000030 0000014 0402317
01 P?s-0.2097F 01 P3s 0.7910c 00 Yl= 0.6000C-01 niN= 0.5924F 01
D-9
-------
O.lOjlF 01 P2==0
0005234 000064?
5
0.90 0 8E 00 P2 = -0,
(i o o *7o o o o o n s s o
oo^o? ono?4ij 00020*2
L CG»P'lT»TION, 1= .11
x]= n.*3io= ni o'= o.R7«8e on o?::p
no n?«in on= 11 0.46544566F 01 P
0.4.5125763 = 01 0.49oB4?'K}F 01 0
«.1U11111F 02 0. 12o08425F 0? 0
0.*4151404= 01 0.76776557F- 0) £1
0 .65689866F 01 0.626617S3F 01 0
r rOMPHTiTIONI, 1= 2
*1= 0.1436F 02 Pl= 0.1604E 01 P 2 = = 0
OJD n5 573 0005232 0005H4 0001 220
C rOMPHT»TION,1s 3
*1= 0.1326F n2 P1= 0.1207E 01 P2 = = 0
0005?3? 0P05 114 000510* 0000756
r rOMPI!T«T1QM, 1 = 4
*1= 0.12B8F 02 P1 :
0005114 0005104
C nOMPilTiTTQM, 1 =
Xl= 0.1ZR4= (12 Pi:
00051114 0005234
C nO«PiiTiT 7 ON, I = 6
*1= 0.1397F 02 P1s 0.74ft8E 00 P2=-n,
0 0 05?34 0^04700 1)004544 000046?
C rOHPilTtTlOW, 1= 7
*1= n.i2?0= 02 P1: D.6??0E 00 P2=-0,
0004700 0P04544 0004467 0000427
c roMBuT»TiOM,is a
Xl= 0.1175F 02 Pis 0.15?5E 01 P2 = -0,
0004544 0 0 04467 00044*6 00011*0
C rOMPUTATION,I= 9
*1= 0.1153= 02 Pl = 0.20«1E 01 P?=iO,
0004467 0^04466 0004364 0001524
C rOMPi.lTiTION, 1= 10
*1= 0.1152= 02 Pis 0.20216 01 P2=-0,
000446ft 0004364 0004252 0001473
t rOMPHTiTION,1= 11
*1= 0.1120c 02 P1= 0.1977E 01 P2=»0,
0004364 0004252 0004D7* 0001451
C fTOHP|jT»T!OKI, 1= 12
*1= 0.1084F 02 P1= 0I1977E 01 D2s-0,
000425? 0 P 04 0 7 4 0003677 0001451
C riOMPIITiTION, 1= 13
*1= 0.1030= 02 Pts 0.2013E 01 P2=-0,
0004074 0 0 0367 7 0003450 0001470
C rO"PUT»TIOM,!= 14
yi= 0.9690F 01 P1 = 0.19?4E 01 P2=-0.
0 0036 77 0 0 034 50 0 0 03232 00 01423
C r0»PI)T4TI0W, 1= 15
*1= n.9952P 01 Pis 0.1«o2E 01 p?==0.
0 00 345(1 OP 03232 0 003150 0 00134?
Of 0 1 7^7 ?0(11533 000(1503 n D f 0 0 3 0 0000 01 4 00 00014 04 0 2275
C nO"PUT»TlOM, 1 =
*1= 0.«259F 01 oi:
0003?3? 0003150
r rowPiiT»TiOM,i =
16
0.6652E 00 P2=.0,
0003126 0000450
1 7
*1= n.a0l5= 01 Pis 0.53?9E 00 P2==0,
0003150 0003126 000316* 0000312
C. nO"PiJT»T'OW, Is 18
Xl= 0.7977= 01 »is 0.5242E 00 D2=»0,
0 0 031 2* 00031*6 (1003246 0001326
D-10
C rO"P>lTiTTOM, 1 =
*1= 0.aOH3 = 01 P1
0 0 0 3166 000324ft
C. rQMpiiTjTIOM.Is
19
o.50r4e on °? = = o
0003360 0000,320
20
.1661= 01 P3= 0.6909c 00 Yl= 0.3000F-01 DINs 0 .5504F (11
?oni?5i o n o 04 "*2 nni'0014 oooooi4 ooooooo 0402147
,48302808= 01 0.5C 01??1OE 01 O.44OO40fl4F 01 0.4O634921 E Ol
,66959707c 01 0.8«156?P8F 01 0.91282Q51F 01 0.9059s2
-------
0003?46 0 0 0 33*0 0003444
C COMPUTATION,Is 31
*1= 0.R679C 01 Pis 0.5035E
0003360 00 03444 0 0 03444
C COMPUTATION,I= 22
Xl= n.q933F 01 B1= 0 .1 3 0 9 E
0003444 0003444 0003560
C COMPUTATION,Is 93
Xl = 0•8933F 01 P1s 0.18?7e
0003444 0003560 0 0 0363*
C COMPUTATION,1= ?4
Xls 0.9304F 01 Pis 0.1807E
0003560 0003634 0003710
C COMPUTATION,Is 25
*1= 0.9519F 01 Pis 0.1846E
0003634 0003710 000376*
C COMPllT AT I ON, I s 36
Xls 0.9734F 01 Pis 0".18?6E
0003710 0003764 0004052
C COMPUTATION,Is ?7
Xls 0.9949F 01 Pis 0.18?5E
0003764 0004052 000412*
C COMPUTATION,1= 28
Xls 0.10?1F 02 Pis 0.1776E
0004052 0004124 11004262
C COMPUTATION,Is 29
Xls 0.1042F 02 Pis 0.1763E
0004124 0004262 0004372
C COMPUTATION,1= SO
Xls 0.5088E 02 Pis 0.l2l0E
0 0 04262 0004372 00046?2
C COMPUTATION,Is 31
Xls 0.1123F 02 Pi's 0.8748E
000437? 0004622 0005573
0000312 2000655 0000236 r>00l262 000l3.il 0001351 0403203
00 P2=-0.1196c 01 P3= 0.505RC 00 Yls 0.1781C 01 PINs 0.H513F 01
0000316 2000752 0000317 0001331 0001351 000133? 0403317
01 P2s = 0 ¦ 3065F 01 P3s 0.156SC 01 Yls 0.1820F 01 niNs 0.8630F 01
0001030 2002350 0001201 0001351 0001332 0001323 04033*7
01 P2s.0 .3636l= 01 P3» 0.1624F 01 Yl» 0 .1 783F 01 (lINs 0.84B4E 01
0001354 2002722 0001230 0001332 0001323 0001345 0403311
01 P2s = 0 .3638F 01 P3» 0.1648F 01 Yls 0.1767F 01 PINs 0.87571= 01
0001344 2002722 0001242 0001323 0001345 0001366 0403401
01 P2s = 0 .3638F 01 P3a 0.1609E 01' Yls 0.1810F 01 niNs 0.8991F 01
0001364 2002722 00012?2 0001345 0001366 0001404 0403461
01 P2s = o. 3622F 01 P3« 0.1613c 01 Yl» 0.1853F 01 niN= 0.9255F 01
0001353 2002714 0001224 0001366 000J404 0001*2? 0403547
01 P2 = = 0 .3639F 01 P3s 0.1631F 01 Yls 0.1888F 01 PINs 0.9441P nl
0001353 2002723 0001233 0001404 0001422 0001441 0403615
01 P2s«0.3640F 01 P3s 0.1678F 01 Yls 0.1920F 01 PINs 0.9656F 01
0001327 2002723 0001257 0001422 0001441 0001446 0403671
01 P2s = 0 ¦ 3037F 01 P3» 0.1089F 01 Yl« 0.1957)= 01 PINs 0.9890F 01
0001321 2002334 0000675 00014.41 0001446 0001475 0403751
01 P2s = 0.21S4F 01 P3s 0.7910F 00 Yls 0.1970F 01 niNs 0.1032F 02
0000757 2001577 0000503 0001446 0001475 0001547 0404101
00 P2sa0 • 1748F 01 P3» 0.6902E 00 Vl« 0.2026E 01 I11N= 0.1083E 02
0000546 2001314 0000432 0001*75 0001547 0001630 0404251
C =
0.l3?991*5E 02 0.12732601E 02
0•10O25519E 02 0.1099B779E 02
0.85323565F 01 0T773137V7E 01
0,8512P205E 01 0.863003&6F 01
.12693529F 02 0.13377289E 02
•10710623F 02 0.10251526E 02
.76532356F 01 0.77313797E 01
0.12009768F 02 0.11101343E 02
0.97338217F 01 0.92454212E 01
0.79169719F 01 0.81416361E 01
0.94407814F 01 0.96556776E 01
L COMPUTATION,!* 2
Xls 0 , 40 0 OF 01 Pis 0.1777E 01 P2si0
0001*63 0001450 0001670
L COMPUTATION,Is 3
Xls 0.3951F 01 Pis 0.1313E 01 P2s»0
0001450 0001670 000173* 0001031
L COMPUTATION,Is 4
Xls 0.4654F 01 Pis 0.1099E 01 P2ss0
0001670 0001734 0001777 0000701
L COMPUTATION,Is 5
Xls 0.4830c 01 Pis 0T9636E 00 P2s.O
0001734 0001777 000160* 0000612
L COMPUTATION,Is 6
Xls 0.5001F 01 Pis 0.7931E 00 P2s.O
0001777 0001604 0001477 0000504
L COMPUTATION.Is 7
Xls 0.44(|0F 01 Pis 0.7187E 00 P2s-I)
0001604 0001477 0001562 0000446
L COMPUTATION,Is 8
Xls 0.4063c PI P1: 0.15B4E 01 P2=.0
0001477 0P01562 00017*4
L COMPUTATION,Is 9
Xls (1.4313F 01 P1 s 0 . ?1 ?8E
000156? 0001754 0002532
.84835165F 01 0.87570207E 01 0.89914530C 01 0.9?551892E 01
.98901099F 01 0 .10319902E 02 0.10B27839C 02 0 .H7H844E 02
¦ 224 3F 01 P3s 0.3370F-00 Yls 0. niN= 0.1171E 02
0001327 2001627 0000212 0000000 0000477 0000351 0404535
0001 21 0
01 P2==0,
0 0 015*7
, 1708F 01 P3s 0.2882F-00 Yl= 0.7800F 00 PINs 0.3814F 01
2001274 0000166 0000*77 0000351 0000136 0401415
.1438c 01 P3= 0.2428F-00 Yls 0.5700F 00 HINs 0.4264F 01
2001116 0000143 0000351 0000136 0000152 0401551
• 1240F 01 P3= 0.2495F-00 Yls 0.2300F-00 I1IN= 0.4498F 01
2000774 0000146 0000136 0000152 0000034 0401631
. 1176F 01 P3 s 0.2^2(^-00 Yls 0.2600F-00 niNs 0 .4830F 01
2000742 0000147 0000152 0000034 0000101 0401735
•1259C 01 P3s 0.3P7RC-00 Yls 0.7000C-01 ntNs 0.4039F 01
2001004 0000236 00 0 0 034 0000101 0000050 040147 3
,3098c 01 P3s 0.140OC 01 Yls 0.1600F-00 niNs 0.3531F 01
2002365 0001101 0000101 0000050 0000105 0401323
3554C 01 P3s 0.13??c 01 Yls 0.1000F-00 PIN:
? 0 0 266 0 0001 015 10 0 0050 0000105 0000071
0.3678F 01
0401361
D-ll
-------
D-12
L COMPUTATION , 1= 10
*3= n.a'Jnst- m pu o.20*7e
0001754 0002532 n o o i 4
L COMPUTATION, 1= 11
*1= Q.66Q6C ni PI = O.3O18F
00(1253? 0P014J4 00(1351 4
L COMPUTATION, Is 12
*1= n.«8l6F m Pi= 0.?0i5E
0003414 0003514 0003476
L COMPUTATION,1= 13
*1= n.2s = 0.2953c 01 P3s 0.9S65E 00 Yls 0.2040F 01 IlINs 0.108JE 02
0001356 2002272 0000607 0001503 0003024 0001753 0404251
00 P2 = = 0.110BF 01 P3s 0.3129C-00 Vis O.3B0OF 01 IlINs 0.1270F 02
0000426 2000706 0000200 0003024 0001753 0000176 0405051
00 P2s = 0.961 7C 00 P3» 0.3144C-00 Vis 0.2450F 01 D I N= 0.1558F 02
0000340 2 0 0 0 612 0000200 00 01 753 0000176 0000341 04061 65
00 P2==0.9603F 00 P3s 0.3?46 = -00 Vis 0.3100F-00 IlINs 0.1140F 02
0000334 2000*12 0000204 0000176 0000341 0000156 0404437
00 P2s=0,9609c 00 P3s 0.3406C-00 Vis 0.5500F 00 OfNs 0.9617E 01
0000325 2000612 0000213 0000341 0000156 0000361 0403661
00 P2=-0.95A3C 00 P3s O^SBC-OO Yl= 0 .2700F«00 n T Ns 0.8513F 01
000 0317 2000611 0000221 0000156 0000361 0000004 0403317
00 P2s = 0,1109C 01 P3= 0 .4952F-00 Vl« 0 .5900E 00 n'lNs 0.7917F 01
0000322 2000707 0000312 0000361 0000004 0000615 0403125
01 «>2 = .0 .2979F 01 P3s 0.1545C 01 Yl = 0 .1 000E-01 OIN= 0.701SE 01
0001040 2002304 0001170 0000004 0000615 0000371 0402635
01 P2S»0,3549F 01 P3s 0.1604F 01 Yl= 0.9700F 00 nINs 0.7067F 01
0001364 2002*56 0001290 0000615 0000371 0000010 0402*47
01 P2s = 0 .3551F 01 P3s 0.1*29? 01 Vis 0.6100E 00 IlINs 0.7116E 01
0001353 2002657 0001232 0000371 0000010 0000004 0402661
01 O2s„0.35*1C 01 °3S 0,159OF 01 Vis 0.2000F-01 IlINs 0.650tC 01
0001373 2002656 0001213 0000010 0000004 0000004 0402463
01 P2 = »0.3535c 01 P3s 0.1596F 01 Vis 0.1000F-01 TlINs 0.5963F 01
0 0 01362 2002*50 0001215 0000004 0000004 0000004 0402305
01 P2=s0,3552F 01 P3s 0.1614F 01 Yls 0.1000F-01 nIN= 0.556RF 01
00013*2 2002*57 0001255 0000004 0000004 0000050 0402164
01 P2s-0.3553C 01 P3s 0.1663F 01 Yls 0.1000F-01 niNs 0.5319F 01
0001335 2002657 0001250 0000004 0000090 0000030 0402101
01 P?s-0 .2949F 01 P3* 0 . 1074F 01 Vis O.lOOOF-OO PINs 0.5094C nl
0001327 2002270 00006*7 0000050 0000030 0000014 0402023
01 P2=.0.2097c 01 P3s 0.777«F 00 Vis 0.6000F-01 MNs 0.5034C 01
00007*4 2001533 0000476 0000030 0000014 0000014 040?0?l
00 P2s-0.1661C 01 P3s 0.67P7C 00 Yls 0 . 30 0 0<=-01 niN= 0.4759F 01
U 0 0 055 ? 9001951 0000495 0 0 C 0 014 000005 4 OOOOOOO 0401715
-------
D A Y 3 ¦ L =
n.^at44078F ni
0.4?637363F 01 0
0
0.36776557F 01 0.42637363F 01
0.10827839E 0? 0.12703297F 02 0
0.79169719F 01 0.70183150F 01 0
fl.55677656F ni 0.53l86813E 01 0
C COMPUTATION,!= 2
*1= (1.1 436P 02 Pis 0 i 1777E 01 P? = = 0
0005577 0005242 00050*6 0001327
C COMPUTATION.Is 3
Vie 0.1330F 02 P1 s 0ll3l3E 01 P2=o0
000524? 0005056 0005046 0001031
C COMPUTATION,Is 4
*1= 0.1273F 02 Pis 0.1099E 01 P2 = = 0
0 0 05 05 A 0005046 (1005262 0000701
C COMPUTATION,Is 5
Xls 0.1269F 02 P1s 0 . 9 6 3 6 E 00 P2s.O
0005046 0005262 0004632 0000612
C COMPUTATION,Is 6
XI" 0.1338F 02 Pl= 0.7931E 00 P2s-o
000526? 0004632 0004340 0000504
C COMPUTATION,Is 7
Xls 0.1201F 02 Pis 0•71fi7E 00 P2s-0
0004632 0004340 000427* 0000446
C COMPUTATION,Is 8
Xl = 0•111 OF 02 Pis 0.158*E OS P2s-o
0004340 0004274 0004313 0001210
C COMPUTET ION,Is 9
Xl« 0.1093E 02 Pis 0.2128E 01 P2s-0
0004274 0004313 00042?0 0001547
C COMPUTATION,1= 10
Xls O.llOOF 02 Pis 0•2067b 01 P2s-0
0 0 04313 0004220 0004062 0001516
C COMPUTATION,Is 11
Xls 0.1071F 02 Pis 0I2018E 01 P2s-o
0004220 0004062 00037l0 0001472
C COMPUTATION,1= 12
Xl= 0.1025F 02 Pl= 0.2015E 01 P2=-0
000406? 0003710 0003544 0001471
C COMPUTATION,Is 13
Xls 0.9734F 01 P1s 0.2050E 01 P2=-0
0003710 0003544 00033?2 0001507
C COMPUTATION,Is 14
Xls 0.9245F 01 Pl= 0;i95'E 01 P2s-0
0003544 0003322 00030*6 000144?
C COMPUTATION,Is 15
Xls 0.B532F 01 P1s 0.1833E 01 P2s.O
000332? 0003056 0003036 0001356
C COMPUTATION,Is 16
Xls 0.7731F 01 Pis 0.6800E 00 P2s,0
0003056 0003036 0003056 0000426
C COMPUTATION,Is 17
Xls 0.7653F ni Bl= 0l547*E 00 P2s-0
0003036 0003056 00031?* 0000340
C COMPUTATION,Is 18
Xls 0.7731F 01 P1s 0.537BE 00 B2s.O
0003056 0003124 0003202 0000334
C COMPUTATION,Is 19
Xls 0.7917F 01 P1s 0.520 9E 00 P2s-0
0003124 000320? 0003316
C COMPUTATION,Is ?0
Xls 0.R142F 01 o1= 0 i 5065E 00 P2s-0
000320? 0003316 0003346 0000317
C COMPUTATION,Is ?i
Xls 0.S513F 01 oi= 0.5139E 00 P?s»n
000331* 01033*6 000331 0 QQQM2?
.44981685F 01 0.4»30?fi0BE 01 0.403907?OF 01 0.35311355E 01
.6153R462F 01 0.R57142R6F 01 0.R9914530F 01 0.86495726E 01
•15575092F 02 0.11404151E 02 0.O6166056F 01 0.8512B205E 01
,70671551F 01 0.71159951E 01 0.65006105c 01 0.59633700E 01
.50O40171C 01 0.50M2491E 01 0.47521367F 01 0.40683761E 01
,2330F 01 P3= 0 .3370F-00 Yl» 0.2248F 01 n T Ns 0.4068F 01
2001673 0000212 0001630 0001632 0001576 0401501
•1796F 01 P3S 0.2882E-00 Yls 0.?253F 01 DINs 0.1334E 02
20013 40 0000166 0 0 01632 0 0 01576 0 0 01567 0 4 05253
1526F 01 P3" 0.2428E-00 Yls 0.?185F fll IlTNs
2001161 0000143 0001576 0001567 0001561
0.1276E 02
0405065
.1327F 01 P3= 0.2495F-00 Yls 0.2168F 01 niNs 0.1259F 02
2001040 0000146 0001567 0001561 0001610 0405021
¦1263F 01 P3= 0.2528F-00 Yls 0.2152F 01 DINs 0.1351F 02
2001006 0000147 0001561 0001610 0001626 0405317
a1346F 01 P3" 0.3R7RF-00 Yl» 0.?209E 01 niNs 0.1197E 02
2001050 0000236 0001610 0001626 0001614 0404623
.31R6F 01 P3s 0.1409E 01 Yls 0.2242F 01 niNs 0.1076E 02
2002431 0001101 0001626 0001614 0001564 0404?33
•3641F 01 P3 = 0.1322F 01 Yls 0.2218F fll OINs 0.1058F 02
2002724 00010 35 0001614 0001564 0001456 0404167
«3637F 01 P3= 0.1387F 01 Yls 0.2161E 01 PINs 0.1 059E 02
2002722 0001067 0001564 0001456 0001333 0404171
.3635F 01 P3" 0.1433F 01 Yl» 0.1989F fll DINs 0.1037F 02
2002721 0001112 0001456 0001333 0001316 040*113
• 3638F 01 P3= 0 .1434F 01 Yls 0.1785F 01 DINs 0.9871F 01
2002722 0001113 0001333 0001316 0001321 0403745
, 3639F 01 P3» 0.1405F 01 Yl« 0.1755F 111 OINs 0.9363F (11
2002723 0001077 0001316 0001321 8001200 0403575
•3636F 01 P3s 0.1493F 01 Yls 0.1762? 01 OINs 0.R952F 01
2002721 0001143 0001321 0 0 01 2 0 0 000107? 0403451
. 3040F 01 P3= 0.9565F 00 Yls 0.1565E 111 DINs 0.R239F 01
2002335 00006117 0001200 0001 072 0001024 0403227
. 1195F 01 P3 s 0,3129F-00 Yl» 0.1392E 01 I1IN= 0.7350E 01
2000752 0000200 0001072 000102* 0001157 0402741
•1049F 01 P3= 0.3144F-00 Yls 0.1300E 01 PINs A.7311F fll
2000656 0000200 0001024 0001157 0001247 0402731
¦ 1048F 01 P3s 0.3746F-00 Yls 0.1523E 01 PIN= 0 .7546F (11
2000655 000 0204 0001157 0001 247 (1001316 0403011
.1048F 01 P3» 0.3406C-00 Yls 0.1659F 01 niNs 0.7849F nl
0000325 2000656 0000213 0001247 0001316 0001752 0403107
1046F 01 P3= 0.355RC-00 Yls 0.1 754E 01 (}INs 0.R073F nl
2000655 0 0 0 0 2?1 0001316 0001352 0001407 0403165
1196F 01 03s 0.4O5PF-00 Yls 0.1824F 01 IlINs 0.R405F (11
200075? 000031 2 00 0135? 0001 407 00 01*00 0a03?71
D-13
-------
c |-0*PI|T s T ' 'JN:, I = 52
*1= O.BOTOC 01 D1= 0.133ft
0003344 0f03310 0 0 034 0 0
r rOxP'lT*TION,!s 53
*1= n.«4n4E ni D1= 0.1B47E
000331n 0 0 0 3400 00034ft0
C rOMPIlTiTlOM, 1= 34
*1" 1.P757F 111 P1 = 0 .1 8?6E
0 0034 00 0 0 03460 00 03546
C COMPUTATION, Is 55
Xl= 0.B991C 01 Pl = 0.186«E
0003460 0003546 0003614
C COMPUTATION,Is 56
*3= 0.9255F 01 "1: 0.1843E
0003S46 0003614 0003670
C COMPUTATION, Is 97
X1= 0.9441F 01 Pis 0.1842E
0 0 0 3*14 0003670 0003750
C rOMPuT AT T ON vI= 58
X1 s 0.9656F 01 "1 = 0.1792 =
0003*70 0003750 0004100
C COMPUTATION, Is 59
X1 s 0.9890F 01 Pl= 0 .1777E
0003750 0004100 0004250
c computation,r= 30
Xl= 0.1032c 02 Pis 0.12?3E
0004100 0004250 0004535
C COMPUTATION,1= 31
Xl= 0.1083F 02 Pl= 0.88A3E
0004250 0004535 0005573
01 °5=.0.70i6F 01 P3s 0.1"45c 01 Vis 0.1895F 01 P1N = 0.R459F 01
0P01040 2002350 0001170 0001*07 0001400 0001^73 0403304
01 P2=.0.3636c 01 P3 = 0.160"= 01 Yl« 0.1876F 01 niN = 0.B220C 01
0 0 0136 4 5002722 0 0 0l2?0 00 01400 0001373 0001421 0403223
01 P2=.0.3638C 01 P3= 0.1629C 01 Vis 0.1865F 01 niN= O.R440F 01
0001353 5002722 0001232 0001373 0001421 0001444 0403300
01 P2 = »0•3638P 01 P3= 0.1590C 01 Vis 0.1918? 01 HTN= 0.B689F B1
0001373 2002722 0001 213 0001421 0001444 0001463 0403363
01 P?s.0.3652c 01 P3 = 0.159AF 01 Vl= 0.1964F 01 DJNs 0.B90BF 01
000136? 2002714 0001215 0001444 0001463 0001477 0403440
01 P2==0.3639C 01 P3a 0.1614C 01 Yls 0.2001E 01 D!N= 0.9138F 1)1
000136? 20027 23 0001255 00014 63 0001477 0001511 04 0 3517
01 P2 = = 0 • 3640F 01 P3= 0.1 * 6 3 F 01 Yls 0.?030F 01 PINs 0.9372F I)1
0001335 2002723 0001250 0001477 0001511 0001514 0403577
01 P3s.n.3037c 01 P3s 0.1074F 01 Vis 0.2055E 01 P1N= 0.955RF 01
0001327 2002334 0000667 0001511 000l5l« 0001535 0403645
01 P2s-0.2184c 01 P3s 0.777fi= 00 Yls 0 .2063F 01 tlTNs 0.9978F 01
0000764 20015 77 0000476 0001514 0 0 01535 0001601 0 403773
00 P2s=0•1748F 01 P3s 0.6787F 00 Yls 0.7103F 01 HlNs 0.1054F 02
0000552 2001314 00004?5 0001535 0001601 0001630 0404157
C =
0.13338217F 02 0.12761O05F 02 0 .125B6081F 02 0 .13514041E 02 0.11970696F 02 0.10759463E 0?.
0.10583638= 02 0.10593407F 02 0.1036R742F 02 0.9B705739E 01 0.93626373F 01 0.8O52380OE 01
0.82393162F 01 0.73504273E 01 0.73113553F 01 0.75457875E 01 0.78485958F 01 0.80732601E 03
0.R4053724F 01 0.84590«64E 01 0.82197802F 01 0.843O560*E 01 0.B6B86447F 01 0.89084249E 01
0.9137O731F 01 0.93724054F 01 0.95579975F 01 0.997802206 01 0.10544567F 02 0.11501R31E 0?
L COMPUTATION,[s 2
X1s 0.4000F 01 Pis 0.1864£ 01 P2s-0.23?0F 01 P3s 0.3?61F-00 Yls 0. [}INs 0.11515 02
0001463 0001414 0001550 0001373 2001667 0000205 0000000 0000477 0000351 0404465
L COMPUTATION,Is 3
XI= 0.3814F 01 P1= 0.13ft7E 01 P2=.0.1750C 01 P3s 0.2760F-00 Yls 0 .78001= 00 DINs 0.3736F 01
0001414 000155 0 0001630 0001057 2001315 0000161 0000*77 000 0351 00 0 0136 04 01375
D-14
-------
xeo
BINARY
LIBRARY ROUTINE"? REOUFSTfD
• CI .FL .PT
LOaiiing map
UNIlSeD CORF- 13052 THRU 17400
*MAIN<
MOLDA
ROhUN
.TO
• OP
. WTB
• EM
00 BO
.OE
YFIX
FLOAT
FX 1 T
ENTRY
07566
07566
00330
«;topc
07566
10007
10 035
10126
10233
07610
V
10252
10567
10234
12314
12321
10527
AT
12472
12472
12442
13000
13002
12712
.ER
12562 12566
HUMP
13000
13002
RQT J M
0 7566
10035
STRTC
075*6
1(1(112
SETir
07566
07777
MRST
075*6
07767
.CO
10126
10136
.PT
10156
10156
.C\M
10252
10256
.CVO
10252
10262
. TPN
12314
12405
• TPNW
12314
12404
XFIX
12520
12520
.TP
12542
12552
=0 FACT0RS-
0.005
204,75204,
,75204.75409.51409,
,50409.50409
.50409
"53,
33.
33.
33.
33.
33.
S31
33.
33.
34,
35.
35.
55,
35.
36.
36.
36.
36.
571
37.
36.
36,
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34.
54,
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34.
34.
34 .
33.
53
33.
33.
33.
33.
33.
53,
S3.
32.
32.
33.
33.
54
34.
34.
34.
33.
33.
53,
33.
34.
34.
33.
32.
52
32.
32.
32.
32.
32.
53
33.
34.
34.
35.
35.
56
36.
36.
36.
36.
36.
59
41.
41.
41.
42.
43.
(4
43.
42.
41.
40.
42.
44
45.
47.
43.
44 .
45.
45.
46.
47.
48.
49.
50.
50.
50.
50.
49.
49.
50.
50.
51.
50.
52.
52.
53.
54.
55.
56.
56.
56.
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37.
57.
58.
58.
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59.
60.
*1.
52.
62.
62.
63.
64 .
64.
65.
66.
67.
68.
61.
68.
67.
67.
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66.
66.
67.
67.
67.
68.
61.
68.
69.
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70.
71.
70.
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72.
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74.
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73.
73.
73.
73.
74 .
74.
75.
75.
75.
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76.
77.
77.
77.
77.
77.
77.
77.
7B.
71.
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78.
78.
79.
79.
79.
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82.
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no.
79.
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D-15
-------
«D.
70 .
70 .
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/a .
78.
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78 .
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77.
76.
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4160 .
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3820.
8350.
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4150 ¦
5000.
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3390.
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daily
FLOW
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=
432000000
400039000
708480000
730080000
518400000
6T0720000
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552960000
464832000
390521000
803520000
578*80000
902P80000
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2760
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16. 15.
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410400000
440640000
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648000000 .
704160000 .
432000000.
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548*40000.
611712000.
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346464000 .
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321408000 .
301536000 .
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0
D-18
INITIAL
CONDITI QMS[C
ANn L IN 15
e=
0.
l.s
4.00
Cs
13.40
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r=
13.30
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l.'s
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13.00
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f s
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Cs
12.80
l.s
5.40
£ S
12.50
l.s
6,00
c s
12.20
l_S
6.90
C s
11.90
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7,60
r 3
11.60
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8.30
n s
11.20
l.s
9.40
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10.20
l.s
12.20
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9.60
l.s
11. on
e=
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8.40
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12.40
-------
n.438095?4E HI
0.U072039E 02
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0 . 6695° 70 7E 01
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DAY 3 ¦Ls
0.33015873? 01
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0.36483516F 01
0.28376068F 01
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0.77ft06B3PE 01
0.64?24*64E 01
0.12A93529E 02
0.11013431F 02
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0.97533577F 01
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0 .1 0788767C 02
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-------
r=
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L =
12.30
r.c
8.60
l_ =
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c=
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L =
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r=
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r=
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rs
10.30
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c=
1 0.50
L =
9.50
Cs
10.80
l.s
9.10
Cs
11.00
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8.70
c=
11.30
L =
8.30
Cs
11.50
l.s
7.80
Cs
11.90
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12.40
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6,40
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0.
L =
4,00
S
AT T ON CONSTANTS
ST
T T ON
1 .
ALPHAs
0.97
PHIs 0.0 3
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0.53000000F
08
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0.41320000F-08
ST
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0.90
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ALPHAS
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ALPHa's
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20 .
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A= 0
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0.00600000
14,56299999 -0.3729B000 0.004667R0
DAY 1 ,L
0.43614164F
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0.4791208RE
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0.11658120F
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D-20
C =
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0.11570207F
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(1. 369BH27P
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DAY 2 ,L=
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02 0.12«74237F 02 0
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02 0 .10857143E 02
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0.951R9255F
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02 0.11804640E 02
02 0.9714?B57E 01
01 0.83174603E 01
01 0.97142B57E 01
02 0.11951160E 02
01 0 ."46544566F 01 0.48302808F 01 0.50061050E 01 0.444932A4F 01 0 .41 465201E: Jl
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