Aufgabe Makro 1 KE 2 Aufgabe 9/5
A. Total Differenzieren:
(1) S * di = I * di
(2) 0 = L * dP + P * L[Y] * dY
(3) dY = Y[N] * dN
(4) dN = dNs[/COLOR]
(5) dNs = W/P * da + a * d(W/P)[/COLOR]
(6) d(W/P) = Y[NN] * dN
B. Gleichungen (4) und (5) durch Einsetzen von dNs und dN in die Gleichungen (3) und (6) eliminieren:
(1) S * di = I * di
(2) 0 = L * dP + P * L[Y] * dY
(3) dY = Y[N] * (W/P * da + a * d(W/P))[/COLOR]
(4) ---
(5) ---
(6) d(W/P) = Y[NN] * (W/P * da + a * d(W/P))
C. Gleichung (3) durch Einsetzen von dY in die Gleichung (2) eliminieren:
(1) S * di = I * di
(2) 0 = L * dP + P * L[Y] * (Y[N] * (W/P * da + a * d(W/P)))
(3) ---
(4) ---
(5) ---
(6) d(W/P) = Y[NN] * (W/P * da + a * d(W/P))
D. Die Verbleibenden Gleichungen ausmultiplizieren
(1) S * di = I * di
(2) 0 = L * dP + P * L[Y] * Y[N] * W/P * da + P * L[Y] * Y[N] * a * d(W/P)
(6) d(W/P) = Y[NN] * W/P * da + Y[NN] * a * d(W/P)
E. Gleichungen umstellen für die Matrixschreibweise
(1) (S - I)) * di = 0
(2) L * dP + P * L[Y] * Y[N] * a * d(W/P) = - P * L[Y] * Y[N] * W/P * da
(6) (1 - Y[NN] * a) * d(W/P) = Y[NN] * W/P * da
F. Matrixschreibweise: 3x3 Matrix A * x = z
A =
S - I ... + 0 ...........................+ 0
0 .............. + L .......................... + P * L[Y] * Y[N] * a
0 .............. + P * L[Y] * Y[N] * a ..+ 1 - Y[NN] * a
x = (di, dP, d(W/P))
z = (0, - P * L[Y] * Y[N] * W/P * da, Y[NN] * W/P * da)
G. Determinanten
det(A) = (S - I) * L * (1 - Y[NN] * a)
det(d(W/P)) = (S - I) * L * Y[NN] * W/P * da
H. Multiplikator d(W/P))/da
d(W/P)
= det(d(W/P)) / det(A)
= (S - I) * L * Y[NN] * W/P * da / ((S - I) * L * (1 - Y[NN] * a))
= Y[NN] * W/P * da / (1 - Y[NN] * a)[/COLOR]
= Y[NN] * Y[N] * da / (1 - Y[NN] * a) .......// Beachte: (6) W/P[/COLOR] = Y[N]
Ergebnis: d(W/P))/da = Y[NN] * Y[N] / (1 - Y[NN] * a)
Liebe Grüße