Currently I have the following code:
call = []
diff = []
def results(S0, K, T, r, sigma, k, N, M, Iteration):
for i in range(1, Iteration):
S0 = float(S0)
d1 = (log(S0 / K) + (r + 0.5 * sigma ** 2) * T) / (sigma * sqrt(T))
d2 = (log(S0 / K) + (r - 0.5 * sigma ** 2) * T) / (sigma * sqrt(T))
call1 = (S0 * stats.norm.cdf(d1, 0.0, 1.0) - K * exp(-r * T) * stats.norm.cdf(d2, 0.0, 1.0))
call.append(call1)
dilution = N/(N +k*M)
Value_2 = Value_1 + call*M
diff1 = Value_1 - Value_2 == 0
diff.append(diff1)
return call
print(results(100,100,1,0.1,0.2,1,100,10, 1000))
I am trying to make make iterations so that the program find value of "call" that gives minimum value of "Value_1 - Value_2) based on the number of iterations. Can you please, advise me how to advance the code? Specifically, I dont know how to code - "return me the output of a "call" such that "Value_1 - Value_2" is minimum that is based on the number of iterations"
Related
I am trying to place evenly-spaced markers/dots on a quadratic curve drawn with HTML Canvas API. Found a nice article explaining how the paths are calculated in the first place, at determining coordinates on canvas curve.
There is a formula, at the end, to calculate the angle:
function getQuadraticAngle(t, sx, sy, cp1x, cp1y, ex, ey) {
var dx = 2*(1-t)*(cp1x-sx) + 2*t*(ex-cp1x);
var dy = 2*(1-t)*(cp1y-sy) + 2*t*(ey-cp1y);
return Math.PI / 2 - Math.atan2(dx, dy);
}
The x/y pairs that we pass, are the current position, the control point and the end position of the curve - exactly what is needed to pass to the canvas context, and t is a value from 0 to 1. Unless I somehow misunderstood the referenced article.
I want to do something very similar - place my markers over the distance specified s, rather than use t. This means, unless I am mistaken, that I need to calculate the length of the "curved path" and from there, I could probably use the above formula.
I found a solution for the length in JavaScript at length of quadratic curve. The formula is similar to:
.
And added the below function:
function quadraticBezierLength(x1, y1, x2, y2, x3, y3) {
let a, b, c, d, e, u, a1, e1, c1, d1, u1, v1x, v1y;
v1x = x2 * 2;
v1y = y2 * 2;
d = x1 - v1x + x3;
d1 = y1 - v1y + y3;
e = v1x - 2 * x1;
e1 = v1y - 2 * y1;
c1 = a = 4 * (d * d + d1 * d1);
c1 += b = 4 * (d * e + d1 * e1);
c1 += c = e * e + e1 * e1;
c1 = 2 * Math.sqrt(c1);
a1 = 2 * a * (u = Math.sqrt(a));
u1 = b / u;
a = 4 * c * a - b * b;
c = 2 * Math.sqrt(c);
return (
(a1 * c1 + u * b * (c1 - c) + a * Math.log((2 * u + u1 + c1) / (u1 + c))) /
(4 * a1)
);
}
Now, I am trying to space markers evenly. I thought that making "entropy" smooth - dividing the total length by the step length would result in the n markers, so going using the 1/nth step over t would do the trick. However, this does not work. The correlation between t and distance on the curve is not linear.
How do I solve the equation "backwards" - knowing the control point, the start, and the length of the curved path, calculate the end-point?
Not sure I fully understand what you mean by "space markers evenly" but I do have some code that I did with curves and markers that maybe can help you ...
Run the code below, it should output a canvas like this:
function drawBezierCurve(p0, p1, p2, p3) {
distance = 0
last = null
for (let t = 0; t <= 1; t += 0.0001) {
const x = Math.pow(1 - t, 3) * p0[0] + 3 * Math.pow(1 - t, 2) * t * p1[0] + 3 * (1 - t) * Math.pow(t, 2) * p2[0] + Math.pow(t, 3) * p3[0];
const y = Math.pow(1 - t, 3) * p0[1] + 3 * Math.pow(1 - t, 2) * t * p1[1] + 3 * (1 - t) * Math.pow(t, 2) * p2[1] + Math.pow(t, 3) * p3[1];
ctx.lineTo(x, y);
if (last) {
distance += Math.sqrt((x - last[0]) ** 2 + (y - last[1]) ** 2)
if (distance >= 30) {
ctx.rect(x - 1, y - 1, 2, 2);
distance = 0
}
}
last = [x, y]
}
}
const canvas = document.getElementById('canvas');
const ctx = canvas.getContext('2d');
ctx.beginPath();
drawBezierCurve([0, 0], [40, 300], [200, -90], [300, 150]);
ctx.stroke();
<canvas id="canvas" width=300 height=150></canvas>
I created the drawBezierCurve function, there I'm using a parametric equation of a bezier curve and then I use lineTo to draw between points, and also we get a distance between points, the points are very close so my thinking is OK to use the Pythagorean theorem to calculate the distance, and the markers are just little rectangles.
The following code generates numpy 2D lists of r and E values for the specified intervals.
r = np.linspace(3, 14, 10)
E = np.linspace(0.05, 0.75, 10)
r, E = np.meshgrid(r, E)
I am then using the following nested loop to generate output from the function ionisationGamma for each r and E interval value.
for ridx in trange(len(r)):
z = []
for cidx in range(len(r[ridx])):
z.append(ionisationGamma(r[ridx][cidx], E[ridx][cidx]))
Z.append(z)
Z = np.array(Z)
This loop gives me a 2D numpy array Z, which is my output and I am using it for a 3D graph. The problem with it is: it is taking ~6 hours to generate the output for all these intervals as there are so many values due to np.meshgrid. I have just discovered multi-threading in Python and wanted to know how I can implement this by using it. Any help is appreciated.
See below code for ionisationGamma
def ionisationGamma(r, E):
I = complex(0.1, 1.0)
a_soft = 1.0
omega = 0.057
beta = 0.0
dt = 0.1
steps = 10000
Nintervals = 60
N = 3000
xmin = float(-300)
xmax = -xmin
x = [0.0]*N
dx = (xmax - xmin) / (N - 1)
L = dx * N
dk = 2 * M_PI / L
propagator = None
in_, out_, psi0 = None, None, None
in_ = [complex(0.,0.)] * N
psi0 = [complex(0.,0.)] * N
out_ = [[complex(0.,0.)]*N for i in range(steps+1)]
overlap = exp(-r) * (1 + r + (1 / 3) * pow(r, 2))
normC = 1 / (sqrt(2 * (1 + overlap)))
gammai = 0.5
qi = 0.0 + (r / 2)
pi = 0.0
gammai1 = 0.5
gammai2 = 0.5
qi1 = 0.0 - (r / 2)
qi2 = 0.0 + (r / 2)
pi1 = 0.0
pi2 = 0.0
# split initial wavepacket
for i in range(N):
x[i] = xmin + i * dx
out_[0][i] = (normC) * ((pow(gammai1 / M_PI, 1. / 4.) * exp(complex(-(gammai1 / 2.) * pow(x[i] - qi1, 2.), pi1 * (x[i] - qi1)))) + (pow(gammai2 / M_PI, 1. / 4.) * exp(complex(-(gammai2 / 2.) * pow(x[i] - qi2, 2.), pi2 * (x[i] - qi2)))))
in_[i] = (normC) * ((pow(gammai1 / M_PI, 1. / 4.) * exp(complex(-(gammai1 / 2.) * pow(x[i] - qi1, 2.), pi1 * (x[i] - qi1)))) + (pow(gammai2 / M_PI, 1. / 4.) * exp(complex(-(gammai2 / 2.) * pow(x[i] - qi2, 2.), pi2 * (x[i] - qi2)))))
psi0[i] = in_[i]
for l in range(1, steps+1):
for i in range(N):
propagator = exp(complex(0, -potential(x[i], omega, beta, a_soft, r, E, dt, l) * dt / 2.))
in_[i] = propagator * in_[i];
in_ = np.fft.fft(in_, N)
for i in range(N):
k = dk * float(i if i < N / 2 else i - N)
propagator = exp(complex(0, -dt * pow(k, 2) / (2.)))
in_[i] = propagator * in_[i]
in_ = np.fft.ifft(in_, N)
for i in range(N):
propagator = exp(complex(0, -potential(x[i], omega, beta, a_soft, r, E, dt, l) * dt / 2.))
in_[i] = propagator * in_[i]
out_[l][i] = in_[i]
initialGammaCentre = 0.0
finalGammaCentre = 0.0
for i in range(500, 2500 +1):
initialGammaCentre += pow(abs(out_[0][i]), 2) * dx
finalGammaCentre += pow(abs(out_[steps][i]), 2) * dx
ionisationGamma = finalGammaCentre / initialGammaCentre
return ionisationGamma
def potential(x, omega, beta, a_soft, r, E, dt, l):
V = (-1. / sqrt((x - (r / 2)) * (x - (r / 2)) + a_soft * a_soft)) + ((-1. / sqrt((x + (r / 2)) * (x + (r / 2)) + a_soft * a_soft))) + E * x
return V
Since the question is about how to use multiprocessing, the following code will work:
import multiprocessing as mp
if __name__ == '__main__':
with mp.Pool(processes=16) as pool:
Z = pool.starmap(ionisationGamma, arguments)
Z = np.array(Z)
Where the arguments are:
arguments = list()
for ridx in range(len(r)):
for cidx in range(len(r[ridx])):
arguments.append((r[ridx][cidx], E[ridx][cidx]))
I am using starmap instead of map, since you have multiple arguments that you want to unpack. This will divide the arguments iterable over multiple cores, using the ionisationGamma function and the final result will be ordered.
However, I do feel the need to say that the main solution is not really the multiprocessing but the original function code. In ionisationGamma you are using several times the slow python for loops. And it would benefit your code a lot if you could vectorize those operations.
A second observation is that you are using many of those loops separately and it would be nice if you could separate that one big function into multiple smaller functions. Then you can time every function individually and speed up those that are too slow.
I am trying to create a quadratic solver where it will tell me the values of x. It works flawlessly until one of the answers would be an imaginary number. How do I make it so the program doesn't crash every time and actually gives the correct answer in terms of i?
import math
print()
a = input("Coefficient of a: ")
a = float(a)
b = input("Coefficient of b: ")
b = float(b)
c = input("Coefficient of c: ")
c = float(c)
x_1 = (-b + math.sqrt(b ** 2 - (4 * a * c))) / 2 * a
x_2 = (-b - math.sqrt(b ** 2 - (4 * a * c))) / 2 * a
print()
print(f" X = {x_1} or {x_2}")
print()
You can use complex() instead of float():
a = input("Coefficient of a: ")
a = complex(a)
b = input("Coefficient of b: ")
b = complex(b)
c = input("Coefficient of c: ")
c = complex(c)
x_1 = (-b + (b ** 2 - (4 * a * c))**0.5) / 2 * a
x_2 = (-b - (b ** 2 - (4 * a * c))**0.5) / 2 * a
print()
print(f" X = {x_1} or {x_2}")
print()
Prints (for example):
Coefficient of a: 3+2j
Coefficient of b: 1
Coefficient of c: -2-1j
X = (3.1748906833227015+8.198076043446118j) or (-6.174890683322701-10.198076043446118j)
Check the value of b ** 2 - (4 * a * c). If it's negative, you have imaginary solutions.
So your code becomes:
import math
print()
a = input("Coefficient of a: ")
a = float(a)
b = input("Coefficient of b: ")
b = float(b)
c = input("Coefficient of c: ")
c = float(c)
delta = b ** 2 - (4 * a * c)
if delta >= 0:
x_1 = (-b + math.sqrt(b ** 2 - (4 * a * c))) / 2 * a
x_2 = (-b - math.sqrt(b ** 2 - (4 * a * c))) / 2 * a
print()
print(f" X = {x_1} or {x_2}")
print()
else:
print("The 2 solutions are imaginary:")
real = -b
imag = abs(math.sqrt(-delta) / (2 * a))
print(f" X = {real} + i*{imag} or {real} - i*{imag}")
you can use sympy this:example(Equation of the second and third degree and ...)
import sympy as sp
x = sp.Symbol("x")
print(sp.solve(x ** 4 - 1, x))
output:
[-1, 1, -I, I]
pip install sympy :https://pypi.org/project/sympy/
What I basically want, is comparing a timevalue (t1 and tuit)(in hours) to determine which method to use to calculate 'S' and 'k' in a function called 'stijghoogteverlaging'. Then a fitted curve can be made with those values.
I tried multiple things, like putting 'return s' underneath both s-methods.
if t1[i] < tuit:
s = Q / (4 * np.pi * k * D) * exp1(S * r**2 / (4 * k * D * t))
return s
else:
s = Q / (4 * np.pi * k * D) * ((exp1(S * r**2 / (4 * k * D * t))) - (exp1(S * r**2 / (4 * k * D * (t - tuit)))))
return s
But then I got a wrong fitted curve as can be seen in the image below.
Now I tried putting only one 'return s', but then it takes forever to calculate and I have to interrupt the kernel.
data = read_csv("pompproef_data.csv", sep = ';')
pb1 = data.iloc[1:,1].values-1.87
pb2 = data.iloc[1:,2].values-1.86
t1 = data.iloc[1:,0].values / (60*24)
volume = 10/1000 #m3
duur = [128,136, 150, 137, 143, 141] #seconden
totaal = np.sum(duur)
debiet = (((len(duur) * volume)/totaal)) * (60*60*24) #m3/d
print(debiet)
print(t1)
print(pb1)
tuit = 15/(24*60)
D = 2.0
Q = debiet
def stijghoogteverlaging(t, k, S):
for i in range(len(t1)):
if t1[i] < tuit:
s = Q / (4 * np.pi * k * D) * exp1(S * r**2 / (4 * k * D * t))
else:
s = Q / (4 * np.pi * k * D) * ((exp1(S * r**2 / (4 * k * D * t))) - (exp1(S * r**2 / (4 * k * D * (t - tuit)))))
return s
r = 4.0 #afstand peilbuis1 tot put
poptpb1, pcovpb1 = curve_fit(stijghoogteverlaging, t1, pb1, p0=[100, 1e-25], maxfev = 10000000)
print('optimale waarde van k voor peilbuis1:', poptpb1[0])
print('optimale waarde van S voor peilbuis1:', poptpb1[1])
tijd = data.iloc[1:,0].values
t = np.linspace(0.00069*(24*60), 0.021*(24*60), 1000)
s1 = stijghoogteverlaging(t, poptpb1[0], poptpb1[1])
plt.plot(tijd, pb1, 'r.', label = 'Gemeten bij 4 meter')
plt.plot(t, s1, 'b', label = 'fitted bij 4 m')
Does anyone have a solution?
Used values for t1 and pb1:
Plot with a wrong fitted curve(time in minutes).
The function stijghoogteverlaging is performing a nonsense operation over and over:
def stijghoogteverlaging(t, k, S):
for i in range(len(t1)):
if t1[i] < tuit:
s = Q / (4 * np.pi * k * D) * exp1(S * r**2 / (4 * k * D * t))
else:
s = Q / (4 * np.pi * k * D) * ((exp1(S * r**2 / (4 * k * D * t))) - (exp1(S * r**2 / (4 * k * D * (t - tuit)))))
return s
You are iterating len(t1) times, and at each iteration, you are computing the full vectorized value of s each and every time. That means that you are computing len(t)**2 values per call, and using a Python for loop as your outer loop to do it. As a minor point, you are accessing the x-data as the global variable t1 instead of the local value t, which gets passed in.
Your function should probably look more like this:
def stijghoogteverlaging(t, k, S):
return np.where(t < tuit,
Q / (4 * np.pi * k * D) * exp1(S * r**2 / (4 * k * D * t)),
Q / (4 * np.pi * k * D) * ((exp1(S * r**2 / (4 * k * D * t))) - (exp1(S * r**2 / (4 * k * D * (t - tuit)))))
)
This computes len(t) * 2 values per call, not len(t)**2, and selects a value from the appropriate result for each value of t.
I keep getting the invalid procedure call or argument error on the definition of sigma2d line.
Any idea how to avoid this code error?
Private Sub CommandButton4_Click()
Application.Range("E19").value = ""
Application.Range("F19").value = ""
S0 = Application.Range("C5").value 'arithmetic average of underlying 1
K = Application.Range("C6").value 'strike
T = Application.Range("C10").value 'maturity
sigma = Application.Range("C8").value 'volatility
r = Application.Range("C8").value 'risk free rate
nsteps = Application.Range("C12").value 'no of timesteps
nsimulations = Application.Range("C13").value ' no of mc simulations
div = Application.Range("C9").value 'dividends
Randomize
Dim M1 As Double, M2 As Double, sigma2d As Double
Dim d1 As Double, d2 As Double, Nd1 As Double, Nd2 As Double
M1 = (Exp((r - div) * T) - 1) / (r - div) * T
v = (2 * Exp((2 * r) - (2 * div) + (sigma * sigma) * T)) * S0 * S0
w = (r - div + (sigma * sigma)) * (2 * r - 2 * q + (sigma * sigma)) * T * T
Z = 2 * S0 * S0 / ((r - div) * T * T)
y = (1 / 2 * (r - div) + sigma * sigma)
h = Exp((r - div) * T) / (r - div + (sigma * sigma))
M2 = (v / w) + Z * (y - h)
M3 = M1 * M1
sigma2d = Log(M2 / M3)
d1 = (Log(M1 / K) + (sigma2d * T) / 2) / sigma * Sqr(T)
d2 = d1 - sigma * Sqr(T)
callArith = Exp(-r * T) * (M1 * Nd1 - K * Nd2)
Application.Range("E19").value = Application.Max(ExactCall, 0)
Are you trying to do the log of a negative number? Set a breakpoint and check variables before that line. Maybe you have an error before that generating a negative.
First check the argument to the Log function is positive.
Failing that, it could be due to a missing reference in the project. This manifests itself in this curious way. Have a look at "Tools", "References" and see if there is one missing.
You can write sigma2d = Vba.Log(M2 / M3) instead but that's only really a short fix since missing references will cause you headaches elsewhere.
One more thing, why not create a function instead, passing in all the variables as function parameters? Your spreadsheet will be more stable if you do that.
(Also, at the end of your code, d1 definition is incorrect. You need brackets around sigma * Sqr(T)).
I think you need a pair of () or do "/T" as you are multiplying by T here:
M1 = (Exp((r - div) * T) - 1) / (r - div) * T