Solved Task 2 + 3

Schenk_Brandenberger_S9_Aufg1.py

@@ -0,0 +1,22 @@
+import numpy as np
+
+
+if __name__ == '__main__':
+
+    A = np.array([[1,0,2],
+                  [0,1,0],
+                  [1e-4,0,1e-4]])
+    b_exact = np.array([[1],[1],[0]])
+    b_approx = np.array([[1],[1],[1.67e-7]])
+    print("Inverse von A:\n", np.linalg.inv(A))
+    x_exact = np.linalg.solve(A,b_exact)
+    print(x_exact)
+    x_approx = np.linalg.solve(A,b_approx)
+    print(x_approx)
+    print(x_exact-x_approx)
+
+    print("Task 1 d:")
+
+    print(60003*(3e-7/3))
+    print((60003/(1-0.0060003))*((3e-7/3)+(1.67e-7/1)))
+    print(((0.01*(1-0.0060003))/60003)-(3e-7/3))

Schenk_Brandenberger_S9_Aufg2.py

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+import numpy as np
+from math import nan
+
+def Schenk_Brandenberger_S9_Aufg2(A, A_approx, b, b_approx):
+    cond_A = np.linalg.cond(A,np.inf)
+    norm_A_minus_A_approx = np.linalg.norm((A-A_approx),np.inf)
+    norm_A = np.linalg.norm(A,np.inf)
+    norm_b_minus_b_approx = np.linalg.norm((b-b_approx),np.inf)
+    norm_b = np.linalg.norm(b,np.inf)
+    x = np.linalg.solve(A,b)
+    x_approx = np.linalg.solve(A_approx,b_approx)
+    dxmax = (cond_A/(1-(cond_A*(norm_A_minus_A_approx/norm_A))))*((norm_A_minus_A_approx/norm_A)+(norm_b_minus_b_approx/norm_b))
+    dxobs = np.linalg.norm((x-x_approx),np.inf)/np.linalg.norm(x,np.inf)
+    if(cond_A*(norm_A_minus_A_approx/norm_A)) >= 1:
+        dxmax = nan
+    return [x, x_approx, dxmax, dxobs]
+
+
+
+if __name__ == '__main__':
+
+    # Werte aus Aufgabe 1
+    A = np.array([[20, 30, 10],
+                  [10, 17, 6],
+                  [2, 3, 2]])
+    A_approx = A - 0.1
+
+    b = np.array([[5720],
+                  [3300],
+                  [836]])
+    b_approx = b + 100
+    print(b_approx)
+
+    [x, x_approx, dxmax, dxobs] = Schenk_Brandenberger_S9_Aufg2(A, A_approx, b, b_approx)
+    print("Value x: ")
+    print(x)
+    print("X approx with error: ")
+    print(x_approx)
+    print("dx max: ", dxmax)
+    print("dx obs: ", dxobs)

Schenk_Brandenberger_S9_Aufg3.py

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+import numpy as np
+from Schenk_Brandenberger_S9_Aufg2 import *
+import matplotlib.pyplot as plt
+
+all_dxmax = np.array([])
+all_dxobs = np.array([])
+dxmax_dxobs_ratio = np.array([])
+index = np.arange(0,1000,1)
+
+for i in range(1000):
+    A = np.random.rand(100,100)
+    b = np.random.rand(100,1)
+    A_approx = A + np.random.rand(100,100)/1e5
+    b_approx = b + np.random.rand(100,1)/1e5
+    [x, x_approx, dxmax, dxobs] = Schenk_Brandenberger_S9_Aufg2(A, A_approx, b, b_approx)
+    all_dxmax = np.append(all_dxmax, [dxmax], axis=0)
+    all_dxobs = np.append(all_dxobs, [dxobs], axis=0)
+    dxmax_dxobs_ratio = np.append(dxmax_dxobs_ratio, [dxmax/dxobs], axis=0)
+
+
+plt.figure(1)
+plt.semilogy(all_dxmax)
+plt.semilogy(all_dxobs)
+plt.semilogy(dxmax_dxobs_ratio)
+plt.legend(["dxmax", "dxob", "dxmax/dxobs"])
+plt.grid()
+plt.xlabel("x")
+plt.ylabel("f(x)")
+plt.title("Aufgabe 3 Serie 9")
+plt.show()
+
+#Die obere Schranke dxmax liegt immer dxobs, deshalb ist sie sicher realistisch und korrekt
+#Jedoch liegt die obere Schranke immer etwa den Faktor 1e3 über dxobs