Start up costs & min/max runtimes

examples/oemof_0.2/min_max_runtimes/data.csv

@@ -0,0 +1,25 @@
+timestep,demand_el
+1,0
+2,0
+3,0
+4,0
+5,1
+6,0
+7,0
+8,0
+9,0
+10,0
+11,1
+12,0
+13,0
+14,0
+15,0
+16,1
+17,0
+18,0
+19,0
+20,1
+21,0
+22,0
+23,0
+24,0

examples/oemof_0.2/min_max_runtimes/min_max_runtimes.py

@@ -0,0 +1,89 @@
+# -*- coding: utf-8 -*-
+"""
+General description
+-------------------
+Example that illustrates how to model min and max runtimes.
+
+Installation requirements
+-------------------------
+This example requires the version 0.2.1 of oemof. Install by:
+
+    pip install 'oemof>=0.2.1,<0.3'
+
+"""
+
+import pandas as pd
+import oemof.solph as solph
+from oemof.network import Node
+from oemof.outputlib import processing, views
+try:
+    import matplotlib.pyplot as plt
+except ImportError:
+    plt = None
+
+
+# read sequence data
+file_name = 'data'
+data = pd.read_csv(file_name + '.csv', sep=",")
+
+# select periods
+periods = len(data)-1
+
+# create an energy system
+idx = pd.date_range('1/1/2017', periods=periods, freq='H')
+es = solph.EnergySystem(timeindex=idx)
+Node.registry = es
+
+# power bus and components
+bel = solph.Bus(label='bel')
+
+demand_el = solph.Sink(
+    label='demand_el',
+    inputs={bel: solph.Flow(
+        fixed=True, actual_value=data['demand_el'], nominal_value=10)})
+
+dummy_el = solph.Sink(
+    label='dummy_el',
+    inputs={bel: solph.Flow(variable_costs=10)})
+
+pp1 = solph.Source(
+    label='cheap_plant_min_down_constraints',
+    outputs={
+        bel: solph.Flow(
+            nominal_value=10, min=0.5, max=1.0, variable_costs=10,
+            nonconvex=solph.NonConvex(
+                minimum_downtime=4, initial_status=0))})
+
+pp2 = solph.Source(
+    label='expensive_plant_min_up_constraints',
+    outputs={
+        bel: solph.Flow(
+            nominal_value=10, min=0.5, max=1.0, variable_costs=10,
+            nonconvex=solph.NonConvex(
+                minimum_uptime=2, initial_status=1))})
+
+# create an optimization problem and solve it
+om = solph.Model(es)
+
+# debugging
+#om.write('problem.lp', io_options={'symbolic_solver_labels': True})
+
+# solve model
+om.solve(solver='cbc', solve_kwargs={'tee': True})
+
+# create result object
+results = processing.results(om)
+
+# plot data
+if plt is not None:
+    # plot electrical bus
+    data = views.node(results, 'bel')['sequences']
+    data[[(('bel', 'demand_el'), 'flow'), (('bel', 'dummy_el'), 'flow')]] *= -1
+    exclude = ['dummy_el', 'status']
+    columns = [c for c in data.columns
+               if not any(s in c[0] or s in c[1] for s in exclude)]
+    data = data[columns]
+    ax = data.plot(kind='line', drawstyle='steps-post', grid=True, rot=0)
+    ax.set_xlabel('Hour')
+    ax.set_ylabel('P (MW)')
+    plt.show()

examples/oemof_0.2/start_and_shutdown_costs/data.csv

@@ -0,0 +1,25 @@
+timestep,demand_el
+1,0
+2,0
+3,0
+4,1
+5,1
+6,1
+7,0
+8,0
+9,1
+10,1
+11,1
+12,0
+13,0
+14,1
+15,1
+16,1
+17,0
+18,0
+19,1
+20,1
+21,1
+22,0
+23,0
+24,0

examples/oemof_0.2/start_and_shutdown_costs/startup_shutdown.py

@@ -0,0 +1,89 @@
+# -*- coding: utf-8 -*-
+"""
+General description
+-------------------
+Example that illustrates how to model startup and shutdown costs attributed
+to a binary flow.
+
+Installation requirements
+-------------------------
+This example requires the version 0.2.1 of oemof. Install by:
+
+    pip install 'oemof>=0.2.1,<0.3'
+
+"""
+
+import pandas as pd
+import oemof.solph as solph
+from oemof.network import Node
+from oemof.outputlib import processing, views
+try:
+    import matplotlib.pyplot as plt
+except ImportError:
+    plt = None
+
+
+# read sequence data
+file_name = 'data'
+data = pd.read_csv(file_name + '.csv', sep=",")
+
+# select periods
+periods = len(data)-1
+
+# create an energy system
+idx = pd.date_range('1/1/2017', periods=periods, freq='H')
+es = solph.EnergySystem(timeindex=idx)
+Node.registry = es
+
+# power bus and components
+bel = solph.Bus(label='bel')
+
+demand_el = solph.Sink(
+    label='demand_el',
+    inputs={bel: solph.Flow(
+        fixed=True, actual_value=data['demand_el'], nominal_value=10)})
+
+# pp1 and pp2 are competing to serve overall 12 units load at lowest cost
+# summed costs for pp1 = 12 * 10 * 10.25 = 1230
+# summed costs for pp2 = 4*5 + 4*5 + 12 * 10 * 10 = 1240
+# => pp1 serves the load despite of higher variable costs since
+#    the start and shutdown costs of pp2 change its marginal costs
+pp1 = solph.Source(
+    label='power_plant1',
+    outputs={bel: solph.Flow(nominal_value=10, variable_costs=10.25)})
+
+# shutdown costs only work in combination with a minimum load
+# since otherwise the status variable is "allowed" to be active i.e.
+# it permanently has a value of one which does not allow to set the shutdown
+# variable which is set to one if the status variable changes from one to zero
+pp2 = solph.Source(
+    label='power_plant2',
+    outputs={
+        bel: solph.Flow(
+            nominal_value=10, min=0.5, max=1.0, variable_costs=10,
+            nonconvex=solph.NonConvex(startup_costs=5, shutdown_costs=5))})
+
+# create an optimization problem and solve it
+om = solph.Model(es)
+
+# debugging
+#om.write('problem.lp', io_options={'symbolic_solver_labels': True})
+
+# solve model
+om.solve(solver='cbc', solve_kwargs={'tee': True})
+
+# create result object
+results = processing.results(om)
+
+# plot data
+if plt is not None:
+    # plot electrical bus
+    data = views.node(results, 'bel')['sequences']
+    data[(('bel', 'demand_el'), 'flow')] *= -1
+    columns = [c for c in data.columns
+               if not any(s in c for s in ['status', 'startup', 'shutdown'])]
+    data = data[columns]
+    ax = data.plot(kind='line', drawstyle='steps-post', grid=True, rot=0)
+    ax.set_xlabel('Hour')
+    ax.set_ylabel('P (MW)')
+    plt.show()