This is a follow-up question from my previous post on linear interpolation while solving ODEs in GEKKO. Basically, I have a parameter k as a linear function of time, k(t), like:
and I want to include this parameter to be automatically interpolated during ODE integration.
Below is the code for running the simulation:
from gekko import GEKKOimport numpy as npkt = np.array([0,0.2,0.45,0.75,1.0])kv = np.array([0.5,1.0,0.2,1.2,1.0])m = GEKKO()m.time = np.linspace(0, 1, 11)# include kt time points in m.timem.time = np.concatenate((m.time, kt))m.time = np.sort(m.time)m.time = np.unique(m.time)k = np.interp(m.time, kt, kv)k = m.Param(value=k)x = m.Var(value=0)m.Equation(x.dt() == k)m.options.IMODE = 4m.solve(disp=False)import matplotlib.pyplot as pltplt.figure(figsize=(5,2.5))plt.plot(kt,kv,'bo-',label='k data')plt.plot(m.time,k.value,'rx-',label='k interp')plt.plot(m.time,x.value,'k-',label='x')plt.plot(); plt.legend(); plt.grid()##plt.tight_layout(); plt.savefig('p.png',dpi=300)plt.show()My follow-up question is that I now want to calculate k(t), i.e., optimization, to achieve some objective function.
So, I modified the above code to change k from Param to MV and switched to IMODE = 6.
from gekko import GEKKOimport numpy as npkt = np.array([0,0.2,0.45,0.75,1.0])kv = np.array([0.5,1.0,0.2,1.2,1.0])m = GEKKO()m.time = np.linspace(0, 1, 11)# include kt time points in m.timem.time = np.concatenate((m.time, kt))m.time = np.sort(m.time)m.time = np.unique(m.time)k = np.interp(m.time, kt, kv)k = m.MV(value=k,lb = 0, ub = 2,fixed_initial=False)k.STATUS = 1x = m.Var(value=0)m.Equation(x.dt() == k)m.options.IMODE = 6m.Minimize(x)m.solve(disp=False)import matplotlib.pyplot as pltplt.figure(figsize=(5,2.5))plt.plot(kt,kv,'bo-',label='k data')plt.plot(m.time,k.value,'rx-',label='k interp')plt.plot(m.time,x.value,'k-',label='x')plt.plot(); plt.legend(); plt.grid()plt.show()The thing is that the above code calculates k for 13 points corresponding to the length of m.time. However, I want the solver to calculate k for only 5 points corresponding to the original length of kv. The reason is that, in the actual problem, only 5 points are enough for optimziation, and so calculating 13 points is too expensive. This is a cartoon example, so the reason may not be obvious, but in the real problem, it is 15 points for kv vs 150 points for m.time (the ODE is very stiff), and so that is really important.
Another approach I tried is force k to have 5 points:
from gekko import GEKKOimport numpy as npkt = np.array([0,0.2,0.45,0.75,1.0])kv = np.array([0.5,1.0,0.2,1.2,1.0])m = GEKKO()m.time = np.linspace(0, 1, 11)k = m.MV(value=kv, lb = 0, ub = 2,fixed_initial=False)k.STATUS = 1x = m.Var(value=0)m.Equation(x.dt() == k)m.options.IMODE = 6m.Minimize(x)m.solve(disp=False)import matplotlib.pyplot as pltplt.figure(figsize=(5,2.5))plt.plot(kt,kv,'bo-',label='k data')plt.plot(m.time,k.value,'rx-',label='k interp')plt.plot(m.time,x.value,'k-',label='x')plt.plot(); plt.legend(); plt.grid()plt.show()But got an error: Data arrays must have the same length, and match time discretization in dynamic problems, which is understandable.
Really appreciate if someone could help solve this problem.
Thank you!