Totale Differential von der Übungsaufgabe 4 b) von Makro 2 (Kurseinheit 2)

Makro II KE 2 Kapitel 1 Aufgabe 4 b)

A. Totales Differenzial

(1) dY = C[Y] * dY + I[i] * di + dG + NX[Y] * dY + NX[q] * dq
(2) 0 = P * L[Y] * dY + P * L[i] * di
(3) P * NX[Y] * dY + P * NX[q] * dq = 0
(4) dq = Pa/P * de


[B]B. Gleichung (4) durch Einsetzen von dq = Pa/P * de in Gleichung (1) und (3) eliminieren[/B]

(1) dY = C[Y] * dY + I[i] * di + dG + NX[Y] * dY + NX[q] * Pa/P * de
(2) 0 = P * L[Y] * dY + P * L[i] * di
(3) P * NX[Y] * dY + P * NX[q] * Pa/P * de = 0


[B]C. Gleichungen umstellen für die Matrixschreibweise[/B]

(1) (C[Y] + NX[Y] - 1) * dY + I[i] * di + NX[q] * Pa/P * de = -dG
(2) P * L[Y] * dY + P * L[i] * di = 0
(3) P * NX[Y] * dY + NX[q] * Pa * de = 0


[B]D. Matrixschreibweise A * x = z[/B]

A =

(C[Y] + NX[Y] - 1)...+ I[i] ...........+ NX[q] * Pa/P
P * L[Y].............+ P * L[i] .......+ 0
P * NX[Y]............+ 0 ..............+ NX[q] * Pa

x = (dY, di, de)

z = (-dG, 0, 0)


[B]E. Determinanten (Mit Sarrus-Regel)[/B]

det(A)
= (C[Y] + NX[Y] - 1) * P * L[i] * NX[q] * Pa - P * NX[Y] * P * L[i] * NX[q] * Pa/P - NX[q] * Pa * P * L[Y] * I[i]
= P * Pa * L[i] * NX[q] * (C[Y] + NX[Y] - 1 - NX[Y] - L[Y] * I[i] / L[i])
= P * Pa * L[i] * NX[q] * (C[Y] - 1 - I[i])

det(de)
= P * NX[Y] * P * L[i] * dG
= P^2 * NX[Y] * L[i] * dG


[B]F. Lösung für de [/B]

de
= det(de) / det(A)
= P^2 * NX[Y] * L[i] * dG / (P * Pa * L[i] * NX[q] * (C[Y] - L[Y] * I[i] / L[i])
= P * NX[Y] * dG / (Pa * NX[q] * (C[Y] - 1 - I[i])


[B]G. Multiplikator de/dG[/B]

de/dG = P * NX[Y] / (Pa * NX[q] * (C[Y] - 1 - L[Y] * I[i] / L[i])

Falls P = Pa (wie in der Lösung angenommen):

de/dG
= NX[Y] / (NX[q] * (C[Y] - 1 - L[Y] * I[i] / L[i])
= -NX[Y] / (NX[q] * (1 - C[Y] + L[Y] * I[i] / L[i])

Liebe Grüße