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\deflang1031\pard\ri4\plain\f4\fs48\cf0\b Warteschlangen
\par \plain\f4\fs24\cf0 Prof. Dr. D\'f6rte Haftendorn, Dez. 04 Version vom 12.12.04\plain\f6\fs44\cf1\b
\par \plain\f4\fs32\cf0\b lam Personen kommen im Mittel in Zeiteinheit Z an, z.B. 3 pro 1/4-Stunde.
\par my Personen werden im Mittel pro Zeiteinheit Z bedient, z.B. 4 pro 1/4-Stunde.\plain\f6\fs44\cf1\b
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f6\fs32\cf1\b {\pntext\f1\'b7\tab}delete lam, my:
\par {\pntext\f1\'b7\tab}lam:=3:my:=4:
\par {\pntext\f1\'b7\tab}rho:=lam/my //rho hei\'dft auch Auslastung
\par \pard\li50\ri6\plain\f6\fs32\cf2\b\protect {\pict\wmetafile8\picw707\pich1208\picwgoal400\pichgoal684
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}\plain\f6\fs32\cf2\b\protect
\par \pard\li50\ri2\plain\f6\fs32\cf2\b\protect
\par \pard\ri4\plain\f4\fs32\cf0\b \'dcbergangsmatrix\plain\f6\fs44\cf1\b
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f6\fs32\cf1\b {\pntext\f1\'b7\tab}A:=matrix([[1-lam,lam, 0,0],
\par \pard\li600\ri1\fi-300\plain\f6\fs32\cf1\b [my,1-lam-my,lam,0],
\par [0,my,1-lam-my,lam],
\par [0, 0,my,1-lam-my]])
\par \pard\li50\ri6\plain\f6\fs32\cf2\b\protect {\pict\wmetafile8\picw4198\pich2152\picwgoal2379\pichgoal1220
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}\plain\f6\fs32\cf2\b\protect
\par \pard\li50\ri2\plain\f6\fs32\cf2\b\protect
\par \pard\ri4\plain\f4\fs24\cf0 Eigentlich m\'fcsste es stets lam*h, my*h hei\'dfen mit einem h, das dann sp\'e4ter immer kleiner gemacht werden
\par kann. Bei dem Problem, die Grenzverteilung zu bestimmen, f\'e4llt h dann aber heraus. Darum
\par wird auf das Mitziehen in den Termen verzichtet.\plain\f6\fs44\cf1\b
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f6\fs44\cf1\b {\pntext\f1\'b7\tab}A-A^0
\par \pard\li50\ri6\plain\f6\fs32\cf2\b\protect {\pict\wmetafile8\picw4198\pich2152\picwgoal2379\pichgoal1220
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}\plain\f6\fs32\cf2\b\protect
\par \pard\li50\ri2\plain\f6\fs32\cf2\b\protect
\par \pard\ri4\plain\f4\fs28\cf0 Beginnt man das Gleichungssystem vA=v zu l\'f6sen, so
\par l\'e4sst man v1 zun\'e4chst in allen Gleichungen stehen .
\par Man merkt: v2= rho*v1 und v(n+1) =rho*v(n), also eine geometrische Folge mit Faktor rho.
\par v1 bestimmt man nun durch die Bedingung, dass v1+v2+v3+.....=1
\par sein muss, denn v muss stochastische Vektor sein, eine Verteilung beschreiben.
\par Wegen der Summe der geometrische Reihe f\'fchr das zu v1=1-rho.
\par Die Summe - und damit der Grenzwerteilung- existiert nur f\'fcr
\par \plain\f4\fs44\cf0 rho<1, also lam