I'm new to both the Julia Language and its optimization package JuMP. I'm trying to solve a really simple optimization problem, the objective is the minimization of the fixed costs for a facility location problem
o = Model(with_optimizer(GLPK.Optimizer))
#variable(o, x[i = 1:size(candidates, 1)], Bin) #creating variables for each possible location
#variable(o, ϑ[i = 1:size(demand, 2)] >= 0) #creating one variable for each demand scenario (in this case only one scenario is considered)
#objective(o, Min, dot(x,cost_facility) * (1 + M) + sum(capacities .* x .* TH .* lab_cost_plant) + 1/size(demand,2) * sum(ϑ))
#constraint(o, 30000x[1] + 20000x[2] + 30000x[3] + 20000x[4] >= 1500.0)
#constraint(o, ϑ[1] + 1.6317502375004835e10x[1] + 1.0878334916669891e10x[2] + 1.6318862646433956e10x[3] >= 2.9076517866671824e9)
JuMP.optimize!(o)
st = MOI.get(o, MOI.TerminationStatus())
#info "Status $st"
And I get the following result:
┌ Info: Status INFEASIBLE
└ # Main In[37]:3
I couldn't see why such a problem could be infesible, considering that those two constraints were the only ones present. So, I tried to modify them to understand what was wrong and it turned out that by substituting the second constraint with an equality constraint (and keeping everything else unchanged):
#constraint(o, ϑ[1] + 1.6317502375004835e10x[1] + 1.0878334916669891e10x[2] + 1.6318862646433956e10x[3] == 2.9076517866671824e9)
An optimal solution is found:
┌ Info: Status OPTIMAL
└ # Main In[43]:3
I couldn't find any explanation for that, shouldn't the first problem be feasible too? Is there any mistake in the code? Thank you in advance for your help
candidate_plant = ["Roma", "London", "Berlin", "Milano"]
candidate_whs = ["Munich", "Glasgow","Madrid", "Lione"]
capacity_plant = [30000, 20000, 30000, 20000]
capacity_whs = [400000, 300000, 400000, 300000]
cost_plant = [174739, 293932, 174739, 293932]
cost_whs = [124739, 213932, 124739, 213932]
demand = [10000; 10000; 10000]
TH = 20
M = 0.3
lab_cost_plant = 8.64
candidates = vcat(candidate_plant, candidate_whs)
capacities = vcat(capacity_plant, capacity_whs)
cost_facility = vcat(cost_plant, cost_whs)
This is the input data used
Related
I'm new to Julia programming I managed to solve some 1st order DDE (Delay Differential Equations) and ODE. I now need to solve a second order delay differential equation but I didn't manage to find documentation about that (I previously used DifferentialEquations.jl).
The equation (where F is a function and τ the delay):
How can I do this?
Here is my code using the given information, it seems that the system stay at rest which is incorrect. I probably did something wrong.
function bc_model(du,u,h,p,t)
# [ u'(t), u''(t) ] = [ u[1], -u[1] + F(ud[0],u[0]) ] // off by one in julia A[0] -> A[1]
γ,σ,Q = p
ud = h(p, t-σ)[1]
du = [u[2], + Q^2*(γ/Q*tanh(ud)-u[1]) - u[2]]
end
u0 = [0.1, 0]
h(p, t) = u0
lags = [σ,0]
tspan = (0.0,σ*100.0)
alg = MethodOfSteps(Tsit5())
p = (γ,σ,Q,ω0)
prob = DDEProblem(bc_model,u0,h,tspan,p; constant_lags=lags)
sol = solve(prob,alg)
plot(sol)
The code is in fact working! It seems that it is my normalization constants that are not consistent. Thank you!
You get a state space of dimension 2, containing u = [u(t),u'(t)]. Consequently the return vector of the right-side function is [u'(t),u''(t)]. Then if ud is the delayed state [u(t-τ),u'(t-τ)] the right side function can be formulated as
[ u'(t), u''(t) ] = [ u[1], -u[1] + F(ud[0],u[0]) ]
I would like to calculate RSI 14 in line with the tradingview chart.
According to there wiki this should be the solution:
https://www.tradingview.com/wiki/Talk:Relative_Strength_Index_(RSI)
I implemented this is in a object called RSI:
Calling within object RSI:
self.df['rsi1'] = self.calculate_RSI_method_1(self.df, period=self.period)
Implementation of the code the calculation:
def calculate_RSI_method_1(self, ohlc: pd.DataFrame, period: int = 14) -> pd.Series:
delta = ohlc["close"].diff()
ohlc['up'] = delta.copy()
ohlc['down'] = delta.copy()
ohlc['up'] = pd.to_numeric(ohlc['up'])
ohlc['down'] = pd.to_numeric(ohlc['down'])
ohlc['up'][ohlc['up'] < 0] = 0
ohlc['down'][ohlc['down'] > 0] = 0
# This one below is not correct, but why?
ohlc['_gain'] = ohlc['up'].ewm(com=(period - 1), min_periods=period).mean()
ohlc['_loss'] = ohlc['down'].abs().ewm(com=(period - 1), min_periods=period).mean()
ohlc['RS`'] = ohlc['_gain']/ohlc['_loss']
ohlc['rsi'] = pd.Series(100 - (100 / (1 + ohlc['RS`'])))
self.currentvalue = round(self.df['rsi'].iloc[-1], 8)
print (self.currentvalue)
self.exportspreadsheetfordebugging(ohlc, 'calculate_RSI_method_1', self.symbol)
I tested several other solution like e.g but non return a good value:
https://github.com/peerchemist/finta
https://gist.github.com/jmoz/1f93b264650376131ed65875782df386
Therefore I created a unittest based on :
https://school.stockcharts.com/doku.php?id=technical_indicators:relative_strength_index_rsi
I created an input file: (See excel image below)
and a output file: (See excel image below)
Running the unittest (unittest code not included here) should result in but is only checking the last value.
if result == 37.77295211:
log.info("Unit test 001 - PASSED")
return True
else:
log.error("Unit test 001 - NOT PASSED")
return False
But again I cannot pass the test.
I checked all values by help with excel.
So now i'm a little bit lost.
If I'm following this question:
Calculate RSI indicator from pandas DataFrame?
But this will not give any value in the gain.
a) How should the calculation be in order to align the unittest?
b) How should the calculation be in order to align with tradingview?
Here is a Python implementation of the current RSI indicator version in TradingView:
https://github.com/lukaszbinden/rsi_tradingview/blob/main/rsi.py
I had same issue in calculating RSI and the result was different from TradingView,
I have found RSI Step 2 formula described in InvestoPedia and I changed the code as below:
N = 14
close_price0 = float(klines[0][4])
gain_avg0 = loss_avg0 = close_price0
for kline in klines[1:]:
close_price = float(kline[4])
if close_price > close_price0:
gain = close_price - close_price0
loss = 0
else:
gain = 0
loss = close_price0 - close_price
close_price0 = close_price
gain_avg = (gain_avg0 * (N - 1) + gain) / N
loss_avg = (loss_avg0 * (N - 1) + loss) / N
rsi = 100 - 100 / (1 + gain_avg / loss_avg)
gain_avg0 = gain_avg
loss_avg0 = loss_avg
N is the number of period for calculating RSI (by default = 14)
the code is put in a loop to calculate all RSI values for a series.
For those who are experience the same.
My raw data contained ticks where the volume is zero. Filtering this OLHCV rows will directly give the good results.
I want to teach myself about solving PDEs with Julia and I am trying to solve the complex Ginzburg Landau equation (CGLE) with a pseudospectral method in Julia now. However, I struggle with it and I am a bit of ideas what to try.
The CGLE reads:
With Fourier transform and its inverse , I can transform into the spectral form:
This is for example also given in this old script I found (https://www.uni-muenster.de/Physik.TP/archive/fileadmin/lehre/NumMethoden/SoSe2009/Skript/script.pdf) From the same source I know, that alpha=1, beta=2 and initial conditions with small noise of order 0.01 around 0 should result in plane waves as solutions. Thats what I want to test first.
Following the very nice tutorial from Chris Rackauckas (https://youtu.be/okGybBmihOE), I tried to use ApproxFun and DifferentialEquations in the following way to solve this problem:
EDIT: I corrected two mistakes from the original post, a missing dot and minus sign, but the code is still not giving the correct results
EDIT2: Figured out that I computed the wavenumber k completely wrong
using ApproxFun
using DifferentialEquations
F = Fourier()
n = 512
L = 100
T = ApproxFun.plan_transform(F, n)
Ti = ApproxFun.plan_itransform(F, n)
x = collect(range(-L/2,stop=L/2, length=n))
k = points(F, n)
alpha = 1im
beta = 2im
u0 = 0.01*(rand(ComplexF64, n) .- 0.5)
Fu0 = T*u0
function cgle!(du, u, p, t)
a, b, k, T, Ti = p
invu = Ti*u
du .= (1.0 .- k.^2*(1.0 .+a)).*u .- T*( (1.0 .+b) .* (abs.(invu)).^2 .* invu)
end
pars = alpha, beta, k, T, Ti
prob = ODEProblem(cgle!, Fu0, (0.,50.), pars)
u = solve(prob, Rodas5(autodiff=false))
# plotting on a equidistant time stepping
t = collect(range(0, stop=50, length=1000))
sol = zeros(eltype(u),(n, length(t)))
for it in eachindex(t)
sol[:,it] = Ti*u(t[it])
end
IM = PyPlot.imshow(real.(sol))
cb = PyPlot.colorbar(IM, orientation="horizontal")
gcf()
(edited) I tried different solvers, as also recommended in the video, some apparently wont work for complex numbers, some do, but when I run this code it does not give the expected results. The solution remain very small in value and it wont result in the plane waves that actually should be the result. I also tested other intial conditions that should result in chaos, but those result in the same very small solutions as well. I also alternativly used an explicit Laplace Operator with ApproxFun, but the results are the same. My problem here, is that I am neither really an expert with PDE mathemitacaly, nor with their numerical treatment, so far I mainly worked with ODEs.
EDIT2 This now seems to work more or less. I am still wondering about some things though
How can I compute this on a specified domain like , I am seriously confused about how this works with ApproxFun, as far as I can see the wavenumbers k should be (2pi/L)*[-N/2+1 ; N/2 -1], but I am not so sure about how to do this with ApproxFun
https://codeinthehole.com/tutorial/coherent.html shows the different dynamic regimes / phase portrait of the equation. While I can reproduce some of them, some don't seem to work, like the Spatio-temporal intermittency
EDIT 3: I solved these issues by using FFTW directly instead of ApproxFun. In case somebody knows how to this with ApproxFun, I would still be interessted though. Below follows the code with FFTW (it is also a bit more optimized for performance)
begin
using FFTW
using DifferentialEquations
using PyPlot
end
begin
n = 512
L = 200
n2 = Int(n/2)
alpha = 2im
beta = 1im
x = range(-L/2,stop=L/2,length=n)
u0 = 0.01*(rand(ComplexF64, n) .- 0.5)
k = [0:n/2-1; 0; -n/2+1:-1] .*(2*pi/L);
k2 = k.*k
k2[n2 + 1] = (n2*(2*pi/L))^2
T = plan_fft(u0)
Ti = plan_ifft(T*u0)
LinOp = (1.0 .- k2.*(1.0 .+alpha))
Fu0 = T*u0
end
function cgle!(du, u, p, t)
LinOp, b, T, Ti = p
invu = Ti*u
du .= LinOp.*u .- T*( (1.0 .+b) .* (abs.(invu)).^2 .* invu)
end
pars = LinOp, beta, T, Ti
prob = ODEProblem(cgle!, Fu0, (0.,100.), pars)
#time u = solve(prob)
t = collect(range(0, stop=50, length=1000))
sol = zeros(eltype(u),(n, length(t)))
for it in eachindex(t)
sol[:,it] = Ti*u(t[it])
end
IM = PyPlot.imshow(abs.(sol))
cb = PyPlot.colorbar(IM, orientation="horizontal")
gcf()
EDIT 4: Rodas turned out to be a extremly slow solver for this case, just using the default works out nicely for me.
Any help is appreciated.
du = (1. .- k.^2*(1. .+(im*a))).*u + T*( (1. .+(im*b)) .* abs.(invu).^2 .* invu)
Notice that is replacing the pointer to du, not updating it. Use something like .= instead:
du .= (1. .- k.^2*(1. .+(im*a))).*u + T*( (1. .+(im*b)) .* abs.(invu).^2 .* invu)
Otherwise your derivative is just 0.
I am developing an adhoc multinomial logistic model using Julia.
It works fine (although I am sure it could be improved!)
I have written the likelihood function and use Optim to estimate the parameters and the standard errors.
I would like now to develop some robust estimates. Coming from R, I would be using the sandwich package. I did not find any equivalent in Julia. So I could develop something specific I guess.
For this I would need the value of the gradient for each observation (row). I do not find a way to do that using Optim. (When I use gradient!(func, x), I get the sum of the gradients across rows, which is not what I am looking for)
Is there a way to do that using OnceDifferentiable or TwiceDifferentiable ?
Alternatively, is there a package equivalent to R Sandwich that would have escaped from my google researches?
The code I have developed so far:
LLIK_MNL = function(init::Array{Float64})
b = init
u1 = X1*b
u2 = X2*b
u3 = X3*b
umax = max.(u1, u2, u3)
umin = min.(u1, u2, u3)
ucensor = (umax + umin)/2
exu1 = exp.(u1 - ucensor)
exu2 = exp.(u2 - ucensor)
exu3 = exp.(u3 - ucensor)
sexu = exu1 .+ exu2 .+ exu3
Pr=(choice1 .* exu1 + choice2 .* exu2 + choice3 .* exu3) ./ sexu
LL = sum(log.(Pr))
return -LL
end
func = TwiceDifferentiable(var -> LLIK_MNL(var), beta_ini)
opt = optimize(func, beta_ini)
est_MNL = Optim.minimizer(opt)
numerical_hessian = hessian!(func, est_MNL)
var_cov_matrix = inv(numerical_hessian)
temp = diag(var_cov_matrix)
t_stats = est_MNL ./ sqrt.(temp)
pp = 2 * cdf.(Normal(), -abs.(t_stats))
hcat(est_MNL, sqrt.(temp), t_stats, round.(pp, 4))
While solving a differential equation on satellite motion encountered this error:
dt <= dtmin. Aborting. If you would like to force continuation with dt=dtmin, set force_dtmin=true
Here is my code:
using JPLEphemeris
spk = SPK("some-path/de430.bsp")
jd = Dates.datetime2julian(DateTime(some-date))#date of the calculatinons
yyyy/mm/dd hh/mm/ss
jd2 = Dates.datetime2julian(DateTime(some-date))#date of the calculatinons
yyyy/mm/dd hh/mm/ss
println(jd)
println(jd2)
st_bar_sun = state(spk, 0, 10, jd)
st_bar_moon_earth = state(spk, 0, 3, jd)
st_bar_me_earth = state(spk, 3, 399, jd)
st_bar_me_moon = state(spk, 3, 301, jd)
moon_cord = st_bar_me_moon - st_bar_me_earth
a = st_bar_moon_earth + st_bar_me_earth
sun_cord = st_bar_sun - a
println(sputnik_cord)
sputnik_cord = [8.0,8.0,8.0,8.0,8.0,8.0,8.0]
moon_sputnik = sputnik_cord - moon_cord
sun_sputnic = sputnik_cord - sun_cord
Req = 6378137
J2 = 1.08262668E-3
GMe = 398600.4418E+9
GMm = 4.903E+12
GMs = 1.32712440018E+20
function f(dy,y,p,t)
re2=(y[1]^2 + y[2]^2 + y[3]^2)
re3=re2^(3/2)
rs3 = ((y[1]-sun_cord[1])^2 + (y[2]-sun_cord[2])^2 + (y[3]-sun_cord[3])^2)^(3/2)
rm3 = ((y[1]-moon_cord[1])^2 + (y[2]-moon_cord[2])^2 + (y[3]-moon_cord[3])^2)^(3/2)
w = 1 + 1.5J2*(Req*Req/re2)*(1 - 5y[3]*y[3]/re2)
w2 = 1 + 1.5J2*(Req*Req/re2)*(3 - 5y[3]*y[3]/re2)
dy[1] = y[4]
dy[2] = y[5]
dy[3] = y[6]
dy[4] = -GMe*y[1]*w/re3
dy[5] = -GMe*y[2]*w/re3
dy[6] = -GMe*y[3]*w2/re3
end
function f2(dy,y,p,t)
re2=(y[1]^2 + y[2]^2 + y[3]^2)
re3=re2^(3/2)
rs3 = ((y[1]-sun_cord[1])^2 + (y[2]-sun_cord[2])^2 + (y[3]-sun_cord[3])^2)^(3/2)
rm3 = ((y[1]-moon_cord[1])^2 + (y[2]-moon_cord[2])^2 + (y[3]-moon_cord[3])^2)^(3/2)
w = 1 + 1.5J2*(Req*Req/re2)*(1 - 5y[3]*y[3]/re2)
w2 = 1 + 1.5J2*(Req*Req/re2)*(3 - 5y[3]*y[3]/re2)
dy[1] = y[4]
dy[2] = y[5]
dy[3] = y[6]
dy[4] = -GMe*y[1]*w/re3 - GMm*y[1]/rm3 - GMs*y[1]/rs3
dy[5] = -GMe*y[2]*w/re3 - GMm*y[2]/rm3 - GMs*y[2]/rs3
dy[6] = -GMe*y[3]*w2/re3- GMm*y[3]/rm3 - GMs*y[3]/rs3
end
y0 = sputnik_cord
jd=jd*86400
jd2=jd2*86400
using DifferentialEquations
prob = ODEProblem(f,y0,(jd,jd2))
sol = solve(prob,DP5(),abstol=1e-13,reltol=1e-13)
prob2 = ODEProblem(f2,y0,(jd,jd2))
sol2 = solve(prob2,DP5(),abstol=1e-13,reltol=1e-13)
println("Without SUN and MOON")
println(sol[end])
for i = (1:6)
println(sputnik_cord[i]-sol[end][i])
end
println("With SUN and MOON")
println(sol2[end])
What(except the values) can be a reason of this? It worked well before I added the terms with sun_coords and moon_coords in definition of dy[4], dy[5], dy[6] in function f2(As I suppose the function f1 works correctly).
There are two reasons this could be happening. For one, you could see this error because the model is unstable due to implementation issues. If you accidentally put something in wrong, the solution may be diverging to infinity and as it diverges the time steps shorten and it exists with this error.
Another thing that can happen is that your model might be stiff. This can happen if you have large time scale differences between different components. In that case, DP5(), an explicit Runge-Kutta method, is not an appropriate algorithm for this problem. Instead, you will want to look at something for stiff equations. I would recommend giving Rosenbrock23() a try: it's not the fastest but it's super stable and if the problem is integrable it'll handle it.
That's a very good way to diagnose these issues: try other integrators. Try Rosenbrock23(), CVODE_BDF(), radau(), dopri5(), Vern9(), etc. If none of these are working, then you will have just tested your algorithm with a mixture of the most well-tested algorithms (some of them Julia implementations, but others are just wrappers to standard classic C++ and Fortran methods) and this suggests that the issue is in your model formulation and not a peculiarity of a specific solver on this problem. Since I cannot run your example (you should make your example runnable, i.e. no extra files required, if you want me to test things out), I cannot be sure that your model implementation is correct and this would be a good way to find out.
My guess is that the model you have written down is not a good implementation because of floating point issues.
GMe = 398600.4418E+9
GMm = 4.903E+12
GMs = 1.32712440018E+20
these values have only precision 16 digits less than their prescribed value:
> eps(1.32712440018E+20)
16384.0
> 1.32712440018E+20 + 16383
1.3271244001800002e20
> 1.32712440018E+20 + 16380
1.3271244001800002e20
> 1.32712440018E+20 + 16000
1.3271244001800002e20
Notice the lack of precision below the machine epsilon for this value. Well, you're asking for
sol = solve(prob,DP5(),abstol=1e-13,reltol=1e-13)
precision to 1e-13 when it's difficult to be precise to 1e5 given the size of your constants. You need to adjust your units or utilize BigFloat numbers if you want this kind of precision on this problem. So what's likely going on is that the differential equation solvers are realizing that they are not hitting 1e-13 precision, shrinking the stepsize, and repeating this indefinitely (because they can never hit 1e-13 precision due to the size of the floating point numbers) until the stepsize is too small and it exits. If you change the units so that way the constants are more reasonable in size then you can fix this problem.