Issue getting decision variable to optimize properly in Pyomo












0














I have a simple multi-period optimization problem I am working on using pyomo. The goal of the model is to determine which hours a power plant should be on or off based on the Spark Spread (Power Price - Gas Price * Heat Rate + Variable Costs) for that hour. The Spark spread can be negative, which would indicate that the plant should be off, or positive which means the plant should be running.



Currently the results are a showing that the plant just gets turned on and runs despite the spark spread being negative.



How can I get the plant to turn on and off at each time step, given the spark spread for that hour?



I am sure this is a fairly simple solution, but I am very new to pyomo and optimization problems, so any guidance and help would be much appreciated.



gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

priceDict = dict(enumerate(power_price))
gasDict = dict(enumerate(gas_price))

m = en.ConcreteModel()

m.Time = en.RangeSet(0, len(power_price)-1)
m.powerPrice = en.Param(m.Time, initialize=priceDict)
m.gasPrice = en.Param(m.Time, initialize=gasDict)

m.generation = en.Var(m.Time, bounds=(0,800),
initialize=0)
m.spark = en.Var(m.Time,initialize=0)
m.heatRate = en.Var(m.Time,initialize=7)
m.vom = en.Var(m.Time,initialize=2)

m.max_gen = en.Param(initialize=800)


def Obj_fn(m):
return sum((m.spark[i]*m.generation[i]) for i in m.Time)
m.total_cost = en.Objective(rule=Obj_fn,sense=en.maximize)

# 7 is the heat rate of the plant
def spark_rule(m,i):
return (m.spark[i] == m.powerPrice[i]-(m.gasPrice[i]*7+m.vom[i]))
m.hourly_spark = en.Constraint(m.Time,rule=spark_rule)

def generation_rule(m,i):
return (0<=m.generation[i]<=m.max_gen)
m.t_generation_rule = en.Constraint(m.Time, rule=generation_rule)


opt = SolverFactory("clp",executable='C:\clp.exe')
results = opt.solve(m)


The output of the model is currently:



Time Generation     Spark Spread
1 0 6.57
2 800 4.13
3 800 2.27
4 800 0.73
5 800 1.53
6 800 0.31
7 800 -0.33
8 800 -0.84
9 800 -0.48
10 800 -1.19
11 800 1.33
12 800 2.1
13 800 3.78
14 800 6.86
15 800 10.18
16 800 12.13
17 800 9.03









share|improve this question
























  • Not sure I follow, but don´t you want to optimize sum((max(m.spark[i, 0)]*m.generation[i]) for i in m.Time) so that when spark is negative the plant should not run -> not add generation?
    – juvian
    Nov 12 '18 at 18:47










  • Well the model should evaluate the spark spread at each time step and decide if the 'plant' should be generating in that hour or off in that hour. My thinking was that by maximizing the sum of generation * spark spread, the optimization would decide when generation should be 0 or 800
    – mgeoffb
    Nov 12 '18 at 19:30










  • Could you add the values of the current solution it is showing?
    – juvian
    Nov 12 '18 at 21:28










  • This looks like a nonconvex QP. They are not so easy to solve. You need a global solver for that. I believe CLP only supports convex QPs.
    – Erwin Kalvelagen
    Nov 13 '18 at 6:00










  • Thanks for the thoughts @ErwinKalvelagen, I may try and reformulate the problem so I can use a solver such as GLPK/Clp.
    – mgeoffb
    Nov 13 '18 at 22:06
















0














I have a simple multi-period optimization problem I am working on using pyomo. The goal of the model is to determine which hours a power plant should be on or off based on the Spark Spread (Power Price - Gas Price * Heat Rate + Variable Costs) for that hour. The Spark spread can be negative, which would indicate that the plant should be off, or positive which means the plant should be running.



Currently the results are a showing that the plant just gets turned on and runs despite the spark spread being negative.



How can I get the plant to turn on and off at each time step, given the spark spread for that hour?



I am sure this is a fairly simple solution, but I am very new to pyomo and optimization problems, so any guidance and help would be much appreciated.



gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

priceDict = dict(enumerate(power_price))
gasDict = dict(enumerate(gas_price))

m = en.ConcreteModel()

m.Time = en.RangeSet(0, len(power_price)-1)
m.powerPrice = en.Param(m.Time, initialize=priceDict)
m.gasPrice = en.Param(m.Time, initialize=gasDict)

m.generation = en.Var(m.Time, bounds=(0,800),
initialize=0)
m.spark = en.Var(m.Time,initialize=0)
m.heatRate = en.Var(m.Time,initialize=7)
m.vom = en.Var(m.Time,initialize=2)

m.max_gen = en.Param(initialize=800)


def Obj_fn(m):
return sum((m.spark[i]*m.generation[i]) for i in m.Time)
m.total_cost = en.Objective(rule=Obj_fn,sense=en.maximize)

# 7 is the heat rate of the plant
def spark_rule(m,i):
return (m.spark[i] == m.powerPrice[i]-(m.gasPrice[i]*7+m.vom[i]))
m.hourly_spark = en.Constraint(m.Time,rule=spark_rule)

def generation_rule(m,i):
return (0<=m.generation[i]<=m.max_gen)
m.t_generation_rule = en.Constraint(m.Time, rule=generation_rule)


opt = SolverFactory("clp",executable='C:\clp.exe')
results = opt.solve(m)


The output of the model is currently:



Time Generation     Spark Spread
1 0 6.57
2 800 4.13
3 800 2.27
4 800 0.73
5 800 1.53
6 800 0.31
7 800 -0.33
8 800 -0.84
9 800 -0.48
10 800 -1.19
11 800 1.33
12 800 2.1
13 800 3.78
14 800 6.86
15 800 10.18
16 800 12.13
17 800 9.03









share|improve this question
























  • Not sure I follow, but don´t you want to optimize sum((max(m.spark[i, 0)]*m.generation[i]) for i in m.Time) so that when spark is negative the plant should not run -> not add generation?
    – juvian
    Nov 12 '18 at 18:47










  • Well the model should evaluate the spark spread at each time step and decide if the 'plant' should be generating in that hour or off in that hour. My thinking was that by maximizing the sum of generation * spark spread, the optimization would decide when generation should be 0 or 800
    – mgeoffb
    Nov 12 '18 at 19:30










  • Could you add the values of the current solution it is showing?
    – juvian
    Nov 12 '18 at 21:28










  • This looks like a nonconvex QP. They are not so easy to solve. You need a global solver for that. I believe CLP only supports convex QPs.
    – Erwin Kalvelagen
    Nov 13 '18 at 6:00










  • Thanks for the thoughts @ErwinKalvelagen, I may try and reformulate the problem so I can use a solver such as GLPK/Clp.
    – mgeoffb
    Nov 13 '18 at 22:06














0












0








0







I have a simple multi-period optimization problem I am working on using pyomo. The goal of the model is to determine which hours a power plant should be on or off based on the Spark Spread (Power Price - Gas Price * Heat Rate + Variable Costs) for that hour. The Spark spread can be negative, which would indicate that the plant should be off, or positive which means the plant should be running.



Currently the results are a showing that the plant just gets turned on and runs despite the spark spread being negative.



How can I get the plant to turn on and off at each time step, given the spark spread for that hour?



I am sure this is a fairly simple solution, but I am very new to pyomo and optimization problems, so any guidance and help would be much appreciated.



gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

priceDict = dict(enumerate(power_price))
gasDict = dict(enumerate(gas_price))

m = en.ConcreteModel()

m.Time = en.RangeSet(0, len(power_price)-1)
m.powerPrice = en.Param(m.Time, initialize=priceDict)
m.gasPrice = en.Param(m.Time, initialize=gasDict)

m.generation = en.Var(m.Time, bounds=(0,800),
initialize=0)
m.spark = en.Var(m.Time,initialize=0)
m.heatRate = en.Var(m.Time,initialize=7)
m.vom = en.Var(m.Time,initialize=2)

m.max_gen = en.Param(initialize=800)


def Obj_fn(m):
return sum((m.spark[i]*m.generation[i]) for i in m.Time)
m.total_cost = en.Objective(rule=Obj_fn,sense=en.maximize)

# 7 is the heat rate of the plant
def spark_rule(m,i):
return (m.spark[i] == m.powerPrice[i]-(m.gasPrice[i]*7+m.vom[i]))
m.hourly_spark = en.Constraint(m.Time,rule=spark_rule)

def generation_rule(m,i):
return (0<=m.generation[i]<=m.max_gen)
m.t_generation_rule = en.Constraint(m.Time, rule=generation_rule)


opt = SolverFactory("clp",executable='C:\clp.exe')
results = opt.solve(m)


The output of the model is currently:



Time Generation     Spark Spread
1 0 6.57
2 800 4.13
3 800 2.27
4 800 0.73
5 800 1.53
6 800 0.31
7 800 -0.33
8 800 -0.84
9 800 -0.48
10 800 -1.19
11 800 1.33
12 800 2.1
13 800 3.78
14 800 6.86
15 800 10.18
16 800 12.13
17 800 9.03









share|improve this question















I have a simple multi-period optimization problem I am working on using pyomo. The goal of the model is to determine which hours a power plant should be on or off based on the Spark Spread (Power Price - Gas Price * Heat Rate + Variable Costs) for that hour. The Spark spread can be negative, which would indicate that the plant should be off, or positive which means the plant should be running.



Currently the results are a showing that the plant just gets turned on and runs despite the spark spread being negative.



How can I get the plant to turn on and off at each time step, given the spark spread for that hour?



I am sure this is a fairly simple solution, but I am very new to pyomo and optimization problems, so any guidance and help would be much appreciated.



gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

priceDict = dict(enumerate(power_price))
gasDict = dict(enumerate(gas_price))

m = en.ConcreteModel()

m.Time = en.RangeSet(0, len(power_price)-1)
m.powerPrice = en.Param(m.Time, initialize=priceDict)
m.gasPrice = en.Param(m.Time, initialize=gasDict)

m.generation = en.Var(m.Time, bounds=(0,800),
initialize=0)
m.spark = en.Var(m.Time,initialize=0)
m.heatRate = en.Var(m.Time,initialize=7)
m.vom = en.Var(m.Time,initialize=2)

m.max_gen = en.Param(initialize=800)


def Obj_fn(m):
return sum((m.spark[i]*m.generation[i]) for i in m.Time)
m.total_cost = en.Objective(rule=Obj_fn,sense=en.maximize)

# 7 is the heat rate of the plant
def spark_rule(m,i):
return (m.spark[i] == m.powerPrice[i]-(m.gasPrice[i]*7+m.vom[i]))
m.hourly_spark = en.Constraint(m.Time,rule=spark_rule)

def generation_rule(m,i):
return (0<=m.generation[i]<=m.max_gen)
m.t_generation_rule = en.Constraint(m.Time, rule=generation_rule)


opt = SolverFactory("clp",executable='C:\clp.exe')
results = opt.solve(m)


The output of the model is currently:



Time Generation     Spark Spread
1 0 6.57
2 800 4.13
3 800 2.27
4 800 0.73
5 800 1.53
6 800 0.31
7 800 -0.33
8 800 -0.84
9 800 -0.48
10 800 -1.19
11 800 1.33
12 800 2.1
13 800 3.78
14 800 6.86
15 800 10.18
16 800 12.13
17 800 9.03






python optimization pyomo glpk clp






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share|improve this question








edited Nov 12 '18 at 22:19

























asked Nov 12 '18 at 18:21









mgeoffb

11




11












  • Not sure I follow, but don´t you want to optimize sum((max(m.spark[i, 0)]*m.generation[i]) for i in m.Time) so that when spark is negative the plant should not run -> not add generation?
    – juvian
    Nov 12 '18 at 18:47










  • Well the model should evaluate the spark spread at each time step and decide if the 'plant' should be generating in that hour or off in that hour. My thinking was that by maximizing the sum of generation * spark spread, the optimization would decide when generation should be 0 or 800
    – mgeoffb
    Nov 12 '18 at 19:30










  • Could you add the values of the current solution it is showing?
    – juvian
    Nov 12 '18 at 21:28










  • This looks like a nonconvex QP. They are not so easy to solve. You need a global solver for that. I believe CLP only supports convex QPs.
    – Erwin Kalvelagen
    Nov 13 '18 at 6:00










  • Thanks for the thoughts @ErwinKalvelagen, I may try and reformulate the problem so I can use a solver such as GLPK/Clp.
    – mgeoffb
    Nov 13 '18 at 22:06


















  • Not sure I follow, but don´t you want to optimize sum((max(m.spark[i, 0)]*m.generation[i]) for i in m.Time) so that when spark is negative the plant should not run -> not add generation?
    – juvian
    Nov 12 '18 at 18:47










  • Well the model should evaluate the spark spread at each time step and decide if the 'plant' should be generating in that hour or off in that hour. My thinking was that by maximizing the sum of generation * spark spread, the optimization would decide when generation should be 0 or 800
    – mgeoffb
    Nov 12 '18 at 19:30










  • Could you add the values of the current solution it is showing?
    – juvian
    Nov 12 '18 at 21:28










  • This looks like a nonconvex QP. They are not so easy to solve. You need a global solver for that. I believe CLP only supports convex QPs.
    – Erwin Kalvelagen
    Nov 13 '18 at 6:00










  • Thanks for the thoughts @ErwinKalvelagen, I may try and reformulate the problem so I can use a solver such as GLPK/Clp.
    – mgeoffb
    Nov 13 '18 at 22:06
















Not sure I follow, but don´t you want to optimize sum((max(m.spark[i, 0)]*m.generation[i]) for i in m.Time) so that when spark is negative the plant should not run -> not add generation?
– juvian
Nov 12 '18 at 18:47




Not sure I follow, but don´t you want to optimize sum((max(m.spark[i, 0)]*m.generation[i]) for i in m.Time) so that when spark is negative the plant should not run -> not add generation?
– juvian
Nov 12 '18 at 18:47












Well the model should evaluate the spark spread at each time step and decide if the 'plant' should be generating in that hour or off in that hour. My thinking was that by maximizing the sum of generation * spark spread, the optimization would decide when generation should be 0 or 800
– mgeoffb
Nov 12 '18 at 19:30




Well the model should evaluate the spark spread at each time step and decide if the 'plant' should be generating in that hour or off in that hour. My thinking was that by maximizing the sum of generation * spark spread, the optimization would decide when generation should be 0 or 800
– mgeoffb
Nov 12 '18 at 19:30












Could you add the values of the current solution it is showing?
– juvian
Nov 12 '18 at 21:28




Could you add the values of the current solution it is showing?
– juvian
Nov 12 '18 at 21:28












This looks like a nonconvex QP. They are not so easy to solve. You need a global solver for that. I believe CLP only supports convex QPs.
– Erwin Kalvelagen
Nov 13 '18 at 6:00




This looks like a nonconvex QP. They are not so easy to solve. You need a global solver for that. I believe CLP only supports convex QPs.
– Erwin Kalvelagen
Nov 13 '18 at 6:00












Thanks for the thoughts @ErwinKalvelagen, I may try and reformulate the problem so I can use a solver such as GLPK/Clp.
– mgeoffb
Nov 13 '18 at 22:06




Thanks for the thoughts @ErwinKalvelagen, I may try and reformulate the problem so I can use a solver such as GLPK/Clp.
– mgeoffb
Nov 13 '18 at 22:06












1 Answer
1






active

oldest

votes


















0














I might be wrong here, but I think you actually meant to define heatRate and vom as parameters, rather than variables.



This leads to a weird problem because the plant will naturally use its max power whenever the "spark" price is positive, and go to 0 whenever the spark price is negative. I guess you'll add more constraints later on.



If heatRate and vom are fixed, then you can redefine the problem in the following way:



from pyomo import environ as pe

gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

priceDict = dict(enumerate(power_price))
gasDict = dict(enumerate(gas_price))

m = pe.ConcreteModel()

m.Time = pe.RangeSet(0, len(power_price)-1)

# this are all input parameters
m.powerPrice = pe.Param(m.Time, initialize=priceDict)
m.gasPrice = pe.Param(m.Time, initialize=gasDict)
m.vom = pe.Param(default=7)
m.heatRate = pe.Param(default=2)
m.maxGen = pe.Param(default=800)

# this is a "dependent" parameter
m.spark = pe.Param(m.Time,
initialize = lambda m,t: m.powerPrice[t]-(m.gasPrice[t]*7+m.vom)
)

# this is the only variable
m.generation = pe.Var(m.Time,
initialize=0,
bounds = (0, m.maxGen)
)

def Obj_fn(m):
return sum((m.spark[t]*m.generation[t]) for t in m.Time)

m.total_cost = pe.Objective(rule=Obj_fn,sense=pe.maximize)





share|improve this answer





















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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    I might be wrong here, but I think you actually meant to define heatRate and vom as parameters, rather than variables.



    This leads to a weird problem because the plant will naturally use its max power whenever the "spark" price is positive, and go to 0 whenever the spark price is negative. I guess you'll add more constraints later on.



    If heatRate and vom are fixed, then you can redefine the problem in the following way:



    from pyomo import environ as pe

    gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
    power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

    priceDict = dict(enumerate(power_price))
    gasDict = dict(enumerate(gas_price))

    m = pe.ConcreteModel()

    m.Time = pe.RangeSet(0, len(power_price)-1)

    # this are all input parameters
    m.powerPrice = pe.Param(m.Time, initialize=priceDict)
    m.gasPrice = pe.Param(m.Time, initialize=gasDict)
    m.vom = pe.Param(default=7)
    m.heatRate = pe.Param(default=2)
    m.maxGen = pe.Param(default=800)

    # this is a "dependent" parameter
    m.spark = pe.Param(m.Time,
    initialize = lambda m,t: m.powerPrice[t]-(m.gasPrice[t]*7+m.vom)
    )

    # this is the only variable
    m.generation = pe.Var(m.Time,
    initialize=0,
    bounds = (0, m.maxGen)
    )

    def Obj_fn(m):
    return sum((m.spark[t]*m.generation[t]) for t in m.Time)

    m.total_cost = pe.Objective(rule=Obj_fn,sense=pe.maximize)





    share|improve this answer


























      0














      I might be wrong here, but I think you actually meant to define heatRate and vom as parameters, rather than variables.



      This leads to a weird problem because the plant will naturally use its max power whenever the "spark" price is positive, and go to 0 whenever the spark price is negative. I guess you'll add more constraints later on.



      If heatRate and vom are fixed, then you can redefine the problem in the following way:



      from pyomo import environ as pe

      gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
      power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

      priceDict = dict(enumerate(power_price))
      gasDict = dict(enumerate(gas_price))

      m = pe.ConcreteModel()

      m.Time = pe.RangeSet(0, len(power_price)-1)

      # this are all input parameters
      m.powerPrice = pe.Param(m.Time, initialize=priceDict)
      m.gasPrice = pe.Param(m.Time, initialize=gasDict)
      m.vom = pe.Param(default=7)
      m.heatRate = pe.Param(default=2)
      m.maxGen = pe.Param(default=800)

      # this is a "dependent" parameter
      m.spark = pe.Param(m.Time,
      initialize = lambda m,t: m.powerPrice[t]-(m.gasPrice[t]*7+m.vom)
      )

      # this is the only variable
      m.generation = pe.Var(m.Time,
      initialize=0,
      bounds = (0, m.maxGen)
      )

      def Obj_fn(m):
      return sum((m.spark[t]*m.generation[t]) for t in m.Time)

      m.total_cost = pe.Objective(rule=Obj_fn,sense=pe.maximize)





      share|improve this answer
























        0












        0








        0






        I might be wrong here, but I think you actually meant to define heatRate and vom as parameters, rather than variables.



        This leads to a weird problem because the plant will naturally use its max power whenever the "spark" price is positive, and go to 0 whenever the spark price is negative. I guess you'll add more constraints later on.



        If heatRate and vom are fixed, then you can redefine the problem in the following way:



        from pyomo import environ as pe

        gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
        power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

        priceDict = dict(enumerate(power_price))
        gasDict = dict(enumerate(gas_price))

        m = pe.ConcreteModel()

        m.Time = pe.RangeSet(0, len(power_price)-1)

        # this are all input parameters
        m.powerPrice = pe.Param(m.Time, initialize=priceDict)
        m.gasPrice = pe.Param(m.Time, initialize=gasDict)
        m.vom = pe.Param(default=7)
        m.heatRate = pe.Param(default=2)
        m.maxGen = pe.Param(default=800)

        # this is a "dependent" parameter
        m.spark = pe.Param(m.Time,
        initialize = lambda m,t: m.powerPrice[t]-(m.gasPrice[t]*7+m.vom)
        )

        # this is the only variable
        m.generation = pe.Var(m.Time,
        initialize=0,
        bounds = (0, m.maxGen)
        )

        def Obj_fn(m):
        return sum((m.spark[t]*m.generation[t]) for t in m.Time)

        m.total_cost = pe.Objective(rule=Obj_fn,sense=pe.maximize)





        share|improve this answer












        I might be wrong here, but I think you actually meant to define heatRate and vom as parameters, rather than variables.



        This leads to a weird problem because the plant will naturally use its max power whenever the "spark" price is positive, and go to 0 whenever the spark price is negative. I guess you'll add more constraints later on.



        If heatRate and vom are fixed, then you can redefine the problem in the following way:



        from pyomo import environ as pe

        gas_price = [2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81,2.81]
        power_price = [26.24,23.8,21.94,20.4,21.2,19.98,19.34,18.83,19.19,18.48,21,21.77,23.45,26.53,29.85,31.8,28.7]

        priceDict = dict(enumerate(power_price))
        gasDict = dict(enumerate(gas_price))

        m = pe.ConcreteModel()

        m.Time = pe.RangeSet(0, len(power_price)-1)

        # this are all input parameters
        m.powerPrice = pe.Param(m.Time, initialize=priceDict)
        m.gasPrice = pe.Param(m.Time, initialize=gasDict)
        m.vom = pe.Param(default=7)
        m.heatRate = pe.Param(default=2)
        m.maxGen = pe.Param(default=800)

        # this is a "dependent" parameter
        m.spark = pe.Param(m.Time,
        initialize = lambda m,t: m.powerPrice[t]-(m.gasPrice[t]*7+m.vom)
        )

        # this is the only variable
        m.generation = pe.Var(m.Time,
        initialize=0,
        bounds = (0, m.maxGen)
        )

        def Obj_fn(m):
        return sum((m.spark[t]*m.generation[t]) for t in m.Time)

        m.total_cost = pe.Objective(rule=Obj_fn,sense=pe.maximize)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 21 '18 at 1:26









        Giorgio Balestrieri

        645




        645






























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