I am trying to use Gekko to optimize (dis)charging of a battery energy storage system. Electricity prices per hour EP
, energy production from solar panels PV
, and energy demand Dem
are considered over the entire horizon (0-24h) to minize total costs TC
. Arbitrage should take place as the battery is (dis)charged (Pbat_ch
& Pbat_dis
) to/from the grid (Pgrid_in
& Pgrid_out
) at the optimal moments.
As opposed to most of the examples online, the problem is not formulated as a state-space model, but mostly relies on exogenous data for price, consumption and production. 3 specific issues with reference to Gurobi are outlined below, the entire code which results in the following error, can be found at the bottom of this post.
Exception: @error: Inequality Definition
invalid inequalities: z > x < y
at0x0000016c6b214040>
STOPPING . . .
PowerGridOut
and PowerGridIn = m.MV(...)
) at specific timesteps in this manner ([t]).m.Obj(sum(ElectricityPrice[t]*PowerGridOut[t] - ElectricityPrice[t]*PowerGridIn[t]) for t in range(25))
Is this also possible in Gekko, or should this summation be recast as an integral? Is the following code correct?
ElectricityPrice = m.Param([..])
.
.
.
TotalCosts = m.integral(ElectricityPrice*(PowerGridOut - PowerGridIn))
m.Obj(TotalCosts)
m.options.IMODE = 6
m.solve()
m.addConstrs(SoC[t+1] == (SoC[t] - ((1/(DischargeEfficiency*BatteryCapacity)) * (PowerBattery
Discharge[t+1]) * Delta_t - ChargeEfficiency/BatteryCapacity * (PowerBatteryCharge[t+1]) * Delta_t)) for t in range(24))
Based on a question on stackoverflow regarding a similar problem, I have reformulated this in a continuous manner as:
m.Equation(SoC.dt() == SoC - 1/(DischargeEfficiency*BatteryCapacity) * Pbattdis - (ChargeEfficiency/BatteryCapacity) * Pbattch)
Demand[t]
& PV[t]
are exogenous vectors, while the other variables are m.MV()
:m.Equation(((Demand[t] + Pbat_ch + Pgrid_in) == (PV[t] + Pgrid_out + Pbat_dis)) for t in range(25))
Unfortunately, all of this has not worked so far. I would highly appreciate it if someone could give me some hints. Ideally I would like to formulate both the objective function and constraints in discrete terms.
entire code
m = GEKKO()
# horizon
m.time = list(range(0,25))
# data vectors
EP = m.Param(list(Eprice))
Dem = m.Param(list(demand))
PV = m.Param(list(production))
# constants
bat_cap = 13.5
ch_eff = 0.94
dis_eff = 0.94
# manipulated variables
Pbat_ch = m.MV(lb=0, ub=4)
Pbat_ch.DCOST = 0
Pbat_ch.STATUS = 1
Pbat_dis = m.MV(lb=0, ub=4)
Pbat_dis.DCOST = 0
Pbat_dis.STATUS = 1
Pgrid_in = m.MV(lb=0, ub=3)
Pgrid_in.DCOST = 0
Pgrid_in.STATUS = 1
Pgrid_out = m.MV(lb=0, ub=3)
Pgrid_out.DCOST = 0
Pgrid_out.STATUS = 1
#State of Charge Battery
SoC = m.Var(value=0.5, lb=0.2, ub=1)
#Battery Balance
m.Equation(SoC.dt() == SoC - 1/(dis_eff*bat_cap) * Pbat_dis - (ch_eff/bat_cap) * Pbat_ch)
#Energy Balance
m.Equation(((Dem[t] + Pbat_ch + Pgrid_in) == (PV[t] + Pbat_dis + Pgrid_out)) for t in range(0,25))
#Objective
TC = m.Var()
m.Equation(TC == sum(EP[t]*(Pgrid_out-Pgrid_in) for t in range(0,25)))
m.Obj(TC)
m.options.IMODE=6
m.options.NODES=3
m.options.SOLVER=3
m.solve()
Nice application! You can either write out all of your discrete equations yourself with m.options.IMODE=3
or else let Gekko manage the time dimension for you. When you include an objective or constraint, it applies them to all of the time points that you specify. With m.options.IMODE=6
, there is no need to add the set indices in Gekko such as [t]
. Here is a simplified model:
from gekko import GEKKO
import numpy as np
m = GEKKO()
# horizon
m.time = np.linspace(0,3,4)
# data vectors
EP = m.Param([0.1,0.05,0.2,0.25])
Dem = m.Param([10,12,9,8])
PV = m.Param([10,11,8,10])
# constants
bat_cap = 13.5
ch_eff = 0.94
dis_eff = 0.94
# manipulated variables
Pbat_ch = m.MV(lb=0, ub=4)
Pbat_ch.DCOST = 0
Pbat_ch.STATUS = 1
Pbat_dis = m.MV(lb=0, ub=4)
Pbat_dis.DCOST = 0
Pbat_dis.STATUS = 1
Pgrid_in = m.MV(lb=0, ub=3)
Pgrid_in.DCOST = 0
Pgrid_in.STATUS = 1
Pgrid_out = m.MV(lb=0, ub=3)
Pgrid_out.DCOST = 0
Pgrid_out.STATUS = 1
#State of Charge Battery
SoC = m.Var(value=0.5, lb=0.2, ub=1)
#Battery Balance
m.Equation(bat_cap * SoC.dt() == -dis_eff*Pbat_dis + ch_eff*Pbat_ch)
#Energy Balance
m.Equation(Dem + Pbat_ch + Pgrid_in == PV + Pbat_dis + Pgrid_out)
#Objective
m.Minimize(EP*Pgrid_in)
# sell power at 90% of purchase (in) price
m.Maximize(0.9*EP*Pgrid_out)
m.options.IMODE=6
m.options.NODES=3
m.options.SOLVER=3
m.solve()
I modified your differential equation to remove SoC
from the right hand side, otherwise you'll get an exponential increase. The energy balance differential equation is Accumulation=In-Out
. Here is some additional code to visualize the solution.
import matplotlib.pyplot as plt
plt.subplot(3,1,1)
plt.plot(m.time,SoC.value,'b--',label='State of Charge')
plt.ylabel('SoC')
plt.legend()
plt.subplot(3,1,2)
plt.plot(m.time,Dem.value,'r--',label='Demand')
plt.plot(m.time,PV.value,'k:',label='PV Production')
plt.legend()
plt.subplot(3,1,3)
plt.plot(m.time,Pbat_ch.value,'g--',label='Battery Charge')
plt.plot(m.time,Pbat_dis.value,'r:',label='Battery Discharge')
plt.plot(m.time,Pgrid_in.value,'k--',label='Grid Power In')
plt.plot(m.time,Pgrid_in.value,':',color='orange',label='Grid Power Out')
plt.ylabel('Power')
plt.legend()
plt.xlabel('Time')
plt.show()