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constraints working no or trivial solution found

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%% Cell type:markdown id: tags:
Formulieren der Optimierungsgleichung in pymoo
%% Cell type:markdown id: tags:
Es gilt die Kontinuitätsgleichung:
$ \Sigma \dot{V}_k(t) = O$
und die aus der Topologie resultierende Inzidenzmatrix $A_i$
sowie die aus dem Pumpenkennfeld folgende Beziehung:
$\Delta p=\alpha_1 Q^2+\alpha_2 Q n+\alpha_3 n^2 : n \in \{0\} \cup [n_{\mathrm{min}},n_{\mathrm{max}}] $
$P=\beta_1 Q^3+\beta_2 Q^2 n+\beta_3 Q n^2+\beta_4n^3+\beta_5$
und die beziehung für den Druckverlust an den Ventilen:
$\Delta p_{\mathrm{loss}} = - \frac{1}{2} \varrho \zeta \left(\frac{Q}{A}\right)^2 = -l Q^2 :l\in [l_{\mathrm{min}}:\infty )$
nun soll für einen Gegebenen Volumenstrom $Q$ eine Optimale Drehzahl bestimmt werden, welche die Pumpenlesitung minimiert.
$$
\begin{align*}
\mathrm{min} \sum_{p \in \mathcal{P}} Po_{p} \\
Q_{p,i} \geq \sum_{strang} Q_v + \sum_{strang} Q_p \\
Q_p , n\epsilon [n_{min},n_{max}] \\
\overrightarrow{n} = (1,n,n^2,n^3)^T \\
min P = A \overrightarrow{n} \\
-n\leq n_{min} \\
n\leq n_{max}
\end{align*}
$$
Förderhöhe als constraint continuität fomulieren pro strang
%% Cell type:code id: tags:
``` python
!pip install pyomo
```
%% Output
Defaulting to user installation because normal site-packages is not writeable
Requirement already satisfied: pyomo in c:\users\steinmann\appdata\roaming\python\python312\site-packages (6.8.2)
Requirement already satisfied: ply in c:\users\steinmann\appdata\roaming\python\python312\site-packages (from pyomo) (3.11)
[notice] A new release of pip is available: 24.2 -> 25.0
[notice] To update, run: C:\Program Files\Python312\python.exe -m pip install --upgrade pip
%% Cell type:code id: tags:
``` python
#Pump-Powercurve and Pump-Hightcurve
import regression_own
(LR_H,LR_P)=regression_own.regress_pump()
```
%% Output
R^20.9998289611292903
R^20.9994449560888792
%% Cell type:code id: tags:
``` python
#Graph constroctor
#Alle Ventile sind direkt mit der Quelle/Senke Verbunden
import multiDiGraph as gr
nodes =['source','pump1','pump2','valveA','valveB','valveC']
graph = gr.construct_graph('source',('source','pump1',0.),('pump1','pump2',0.),('pump2','valveA',0.),('pump2','valveB',0.),
('pump1','valveC',0.),('valveA','source',4.),('valveB','source',4.),('valveC','source',4.))
#ist das notwendig?!?
for node in graph.nodes:
#definieren der Drehzahl für jede Pumpe im graphen
#inizieren des Durchflusses für jedes Ventil im Graphen
if 'pump' in node:
graph.nodes[node]['n']=750/3600
else:
graph.nodes[node]['n']=None
graph.nodes[node]['flow']=0.
if 'valve' in node:
graph.nodes[node]['flow']= graph[node]['source'][0]['weight']
for node in graph.nodes:
#Berechnen des Durchflusses im Knoten
if 'valve' in node:
continue
for inF in graph.predecessors(node):
graph.nodes[node]['flow'] += graph[inF][node][0]['weight']
#Berechnen des Durchflusses der abgehenden Kanten
tempF=graph.nodes[node]['flow']
SC=0
for outF in graph.successors(node):
if 'valve' in outF:
graph[node][outF][0]['weight']=graph.nodes[outF]['flow']
tempF=tempF - graph.nodes[outF]['flow']
else:
SC+=1
for outF in graph.successors(node):
if SC!=0. and not'valve' in outF:
graph[node][outF][0]['weight']=tempF/SC
else:continue
```
%% Output
%% Cell type:code id: tags:
``` python
import networkx as nx
Mtrx= nx.incidence_matrix(graph,nodes,oriented=True)
```
%% Cell type:code id: tags:
``` python
import networkx as nx
def create_dict(GR:nx.multidigraph):
data={None:{'nodes':{},
'pumps':{},
'valves':{},
}
}
for node in GR.nodes:
data[None]['nodes'][node]=None
data[None]['Q'][node]=GR.nodes[node]['flow']
if 'pump' in node:
data[None]['pumps'][node]=None
data[None]['n'][node]=0.
if 'valve' in node:
data[None]['valves'][node]=None
return data
```
%% Cell type:markdown id: tags:
Durchfluss aus Incidenzmatrix beerechnen
Zeilen = knoten
Spalten = kanten
Summe pro knoten = 0
.Q_valve <= .Q
.Q['pumps']==.Q**2['successors']
$-l Q^2 = \alpha_1 Q^2+\alpha_2 Q n+\alpha_3 n^2$
%% Cell type:code id: tags:
``` python
#defining abstract modell for given Network
import pyomo.environ as pyo
from pyomo.dataportal import DataPortal
import numpy as np
from sklearn.linear_model import LinearRegression
modell = pyo.AbstractModel()
#notwendige Mengen zur Berechnung der Constraints
modell.nodes = pyo.Set()
modell.pumps = pyo.Set()
modell.valves = pyo.Set()
modell.Q_valve=pyo.Param(modell.valves)
#Optimierungsvariable
modell.n = pyo.Var(modell.pumps,bounds=(750/3600,1))
modell.Q = pyo.Var(modell.nodes)
#Objective
def PumpPower(modell):
return sum(np.dot(
np.array(
[modell.Q[i]**3,(modell.Q[i]**2)*modell.n[i],modell.Q[i]*modell.n[i]**2,modell.n[i]**3]
),LR_P.coef_
) for i in modell.pumps)
modell.Power_Objective = pyo.Objective(rule=PumpPower,sense=pyo.minimize)
#expressions for constraints:
def PumpFlow(modell,pump):
return np.dot(np.array([modell.Q[pump]**2,modell.n[pump]*modell.Q[pump],modell.n[pump]**2]),LR_H.coef_)
def Pump_delivery_req(modell,pump):
return PumpFlow(modell,pump) + (pyo.summation(modell.Q,index=graph.successors(pump))**2)==0.
return PumpFlow(modell,pump) ==pyo.summation(modell.Q,index=graph.successors(pump))
def valve_req_rule(modell,valve):
return pyo.summation(modell.Q,index=graph.predecessors(valve))>=modell.Q_valve[valve]
return modell.Q[valve]>=modell.Q_valve[valve]
#modell.Flow_Objective = pyo.Objective(modell.pumps,rule=Flow_req,sense=pyo.minimize)
#Constaints
def continuityRule(modell,node):
return pyo.summation(modell.Q, index=graph.predecessors(node))==pyo.summation(modell.Q, index=graph.successors(node))
#Objective
def PumpPower(modell):
return sum(np.dot(
np.array(
[modell.Q[i]**3,(modell.Q[i]**2)*modell.n[i],modell.Q[i]*modell.n[i]**2,modell.n[i]**3]
),LR_P.coef_
) for i in modell.pumps)
modell.Power_Objective = pyo.Objective(rule=PumpPower,sense=pyo.minimize)
#alternative
def continuityRule2(modell,node):
return 0.==sum(graph[node][i][0]['weight'] for i in graph[node])
#continuity adjustment for change in hight needed
#construction of test Data dictionairy missing
TestData={
None:{
'nodes':[key for key in graph.nodes.keys()],
'pumps':[key for key in graph.nodes.keys() if 'pump' in key],
'valves':[key for key in graph.nodes.keys() if 'valve' in key],
'Q_valve':{'valveA':4.,'valveB':4.,'valveC':4.},
'Q_valve':{'valveA':1.,'valveB':1.,'valveC':2.},
}
}
print(TestData)
```
%% Output
{None: {'nodes': ['source', 'pump1', 'pump2', 'valveA', 'valveB', 'valveC'], 'pumps': ['pump1', 'pump2'], 'valves': ['valveA', 'valveB', 'valveC'], 'Q_valve': {'valveA': 4.0, 'valveB': 4.0, 'valveC': 4.0}}}
{None: {'nodes': ['source', 'pump1', 'pump2', 'valveA', 'valveB', 'valveC'], 'pumps': ['pump1', 'pump2'], 'valves': ['valveA', 'valveB', 'valveC'], 'Q_valve': {'valveA': 1.0, 'valveB': 1.0, 'valveC': 2.0}}}
%% Cell type:markdown id: tags:
Frage: gibt es nur eine Lösung für Drehzahl?
Bsp. Optimierung nach Dezentraler Pumpe um modell zu prüfen
%% Cell type:code id: tags:
``` python
from pyomo.opt import SolverFactory
opt = pyo.SolverFactory('scipampl', executable=r'C:\Program Files\SCIPOptSuite 9.2.0\bin\scip.exe')
instance = modell.create_instance(TestData)
instance.Continuity_constaint=pyo.Constraint(instance.nodes, rule=continuityRule)
instance.Flow_constraint=pyo.Constraint(instance.valves,rule=valve_req_rule)
instance.pump_Flow_constraint=pyo.Constraint(instance.pumps,rule=Pump_delivery_req)
result=opt.solve(instance, tee=True)
print(result)
instance.n.pprint()
instance.Q.pprint()
```
%% Output
SCIP version 9.2.0 [precision: 8 byte] [memory: block] [mode: optimized] [LP solver: Soplex 7.1.2] [GitHash: 74cea9222e]
Copyright (c) 2002-2024 Zuse Institute Berlin (ZIB)
External libraries:
Soplex 7.1.2 Linear Programming Solver developed at Zuse Institute Berlin (soplex.zib.de) [GitHash: b040369c]
CppAD 20180000.0 Algorithmic Differentiation of C++ algorithms developed by B. Bell (github.com/coin-or/CppAD)
TinyCThread 1.2 small portable implementation of the C11 threads API (tinycthread.github.io)
MPIR 3.0.0 Multiple Precision Integers and Rationals Library developed by W. Hart (mpir.org)
ZIMPL 3.6.2 Zuse Institute Mathematical Programming Language developed by T. Koch (zimpl.zib.de)
AMPL/MP 690e9e7 AMPL .nl file reader library (github.com/ampl/mp)
PaPILO 2.4.0 parallel presolve for integer and linear optimization (github.com/scipopt/papilo) (built with TBB) [GitHash: 2d9fe29f]
Nauty 2.8.8 Computing Graph Automorphism Groups by Brendan D. McKay (users.cecs.anu.edu.au/~bdm/nauty)
sassy 1.1 Symmetry preprocessor by Markus Anders (github.com/markusa4/sassy)
Ipopt 3.14.16 Interior Point Optimizer developed by A. Waechter et.al. (github.com/coin-or/Ipopt)
user parameter file <scip.set> not found - using default parameters
read problem <C:\Users\STEINM~1\AppData\Local\Temp\tmpn32p48nz.pyomo.nl>
read problem <C:\Users\STEINM~1\AppData\Local\Temp\tmprv0ikbwh.pyomo.nl>
============
original problem has 9 variables (0 bin, 0 int, 0 impl, 9 cont) and 12 constraints
solve problem
=============
presolving:
(round 1, fast) 2 del vars, 6 del conss, 0 add conss, 5 chg bounds, 0 chg sides, 0 chg coeffs, 0 upgd conss, 0 impls, 0 clqs
(round 1, fast) 2 del vars, 6 del conss, 0 add conss, 12 chg bounds, 0 chg sides, 0 chg coeffs, 0 upgd conss, 0 impls, 0 clqs
presolving (2 rounds: 2 fast, 0 medium, 0 exhaustive):
2 deleted vars, 6 deleted constraints, 0 added constraints, 5 tightened bounds, 0 added holes, 0 changed sides, 0 changed coefficients
2 deleted vars, 6 deleted constraints, 0 added constraints, 16 tightened bounds, 0 added holes, 0 changed sides, 0 changed coefficients
0 implications, 0 cliques
presolving detected infeasibility
Presolving Time: 0.00
SCIP Status : problem is solved [infeasible]
Solving Time (sec) : 0.00
Solving Nodes : 0
Primal Bound : +1.00000000000000e+20 (0 solutions)
Dual Bound : +1.00000000000000e+20
Gap : 0.00 %
WARNING: Loading a SolverResults object with a warning status into
model.name="unknown";
- termination condition: infeasible
- message from solver: infeasible
Problem:
- Lower bound: -inf
Upper bound: inf
Number of objectives: 1
Number of constraints: 0
Number of variables: 0
Sense: unknown
Solver:
- Status: warning
Message: infeasible
Termination condition: infeasible
Id: 200
Error rc: 0
Time: 0.047808170318603516
Solution:
- number of solutions: 0
number of solutions displayed: 0
n : Size=2, Index=pumps
Key : Lower : Value : Upper : Fixed : Stale : Domain
pump1 : 0.20833333333333334 : None : 1 : False : True : Reals
pump2 : 0.20833333333333334 : None : 1 : False : True : Reals
Q : Size=6, Index=nodes
Key : Lower : Value : Upper : Fixed : Stale : Domain
pump1 : None : None : None : False : True : Reals
pump2 : None : None : None : False : True : Reals
source : None : None : None : False : True : Reals
valveA : None : None : None : False : True : Reals
valveB : None : None : None : False : True : Reals
valveC : None : None : None : False : True : Reals
......
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