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随机过程-遍历定理和平稳分布

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随机过程-遍历定理和平稳分布 This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 1 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full ...

随机过程-遍历定理和平稳分布
This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 1 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 1 of 31 H®e>ŒÆ2009?a¬ïÄ) /Hený¢�«ã4Ò�¿2009.10.13-2009.12: 4-7(�: 1-4!); 10+15-17+19-20(o:1-4!) 2009.11.21-2010.01: 22-23(8: 5-8!) ‘ÅL§ / Stochastic Processes ///\\\ììì http://math.carleton.ca/∼tangjs URL Reµhttp://here.is/tangjs E-mail: jiashant@yahoo.ca HHH®®®eee>>>ŒŒŒÆÆÆ nnnÆÆÆ��� ÚÚÚOOOXXX This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 2 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 2 of 31 � á 8 ¹  Ch 1 VÇØÄ:  Ch 2 ‘ÅL§Ä�Vg  Ch 3 ÑtL§  Ch 4 ê�‰Åó  Ch 5 ëYëêê�‰Åó  Ch 6 ²­‘ÅL§  Ch 7 ²­L§�Ì©Û � áµ4gu. ‘ÅL§9ÙA^(1n‡).p��˜Ñ‡�, 2004.7 ëÖµ(1)²À.‘ÅL§n؆A^. ˜uŒÆч�§2005c8� (2)Û(É�È.‘ÅL§(1˜‡).¥IÚOч�2005c8�(15g 0. KdóäkH{5§ …4©Ù휋푗 ´§|⎧⎨⎩ 휋푗 = 푠∑ 푖=1 휋푖푝푖푗, 푗 = 1, 2, ⋅ ⋅ ⋅ , 푠, 휋푗 > 0, 푠∑ 푗=1 휋푗 = 1. �˜). This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 9 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 9 of 31 ~(p98/72/86, ex 4.15) This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 10 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 10 of 31 ∙Ϗ�푗 �~ˆž§4lim푛→∞ 푝(푛)푖푗 ؘ½3§=¦ 3kŒU†푖k'§ÏdÒ�Ä푝 (푛푑) 푖푗 (푑 ≥ 1G�푗�±Ï)± 91푛 ∑푛 푘=1 푝 (푘) 푖푗 �4" . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . P 푓푖푗(푟) = ∞∑ 푚=0 푓 (푚푑+푟) 푖푗 , 0 ≤ 푟 ≤ 푑− 1. §L«Ÿ:d푖Ñu§3ž푛 = 푟( mod 푑)Äg�ˆ푗�Vǧw , 푑−1∑ 푟=1 푓푖푗(푟) = ∞∑ 푚=0 푑−1∑ 푟=0 푓 (푚푑+푟) 푖푗 = ∞∑ 푚=0 푓 (푚) 푖푗 = 푓푖푗. This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 11 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 11 of 31 Theorem 48X푗 �~ˆ§±Ï푑,Ké?¿푖90 ≤ 푟 ≤ 푑− 1k lim 푛→∞ 푝 (푛푑+푟) 푖푗 = 푓푖푗(푟) 푑 휇푗 . Proof. Ϗ�푛 ∕= 0( mod 푑)ž§k푝(푛)푗푗 = 0,l 푝 (푛푑+푟) 푖푗 = 푛푑+푟∑ 푣=0 푓 (푣) 푖푗 푝 (푛푑+푟−푣) 푗푗 = 푛∑ 푚=0 푓 (푚푑+푟) 푖푗 푝 (푛−푚)푑 푗푗 , u´§éu1 ≤ 푁 ≤ 푛,k 푁∑ 푚=0 푓 (푚푑+푟) 푖푗 푝 (푛−푚)푑 푗푗 ≤ 푝(푛푑+푟)푖푗 ≤ 푁∑ 푚=0 푓 (푚푑+푟) 푖푗 푝 (푛−푚)푑 푗푗 + ∞∑ 푚=푁+1 푓 (푚푑+푟) 푖푗 . 3þª¥k�½푁 ,,�-푛→∞,2-푁 →∞,�� 푓푖푗(푟) 푑 휇푗 ≤ lim 푛→∞ 푝 (푛푑+푟) 푖푗 ≤ 푓푖푗(푟) 푑 휇푗 . A~µe푑 = 1§K... This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 12 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 12 of 31 ∙Ϗ∑푛푘=1 푝(푘)푗푗 L«g푗 Ñu§3푛ڃSˆ£�푗 �²þg ê§�1푛 ∑푛 푘=1 푝 (푘) 푖푗 L«ü žmS2£�푗�²þgê¶, §휇푗 L«l푗 Ñu2ˆ£�푗 �²þžm§Ïd 1휇푗 L«l푗 Ñuü ž mS2ˆ£�푗 �²þgꧤ±ATk 1 푛 푛∑ 푘=1 푝 (푘) 푖푗 ≈ 1 휇푗 ∙XJŸ:l푖Ñu§K‡Äl푖ÑuUÄ�ˆ푗 �œ¹§= ‡Ä푓푖푗 �Œ� . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Theorem 49é?¿G�푖, 푗,k lim 푛→∞ 1 푛 푛∑ 푘=1 푝 (푘) 푖푗 = { 0, X푗š~ˆ½"~ˆ, 푓푖푗 휇푗 , X푗�~ˆ This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 13 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 13 of 31 Lemma 50b�k푑‡ê�{푎푛푑+푠}, 푠 = 0, 1, 2, ⋅ ⋅ ⋅ , 푑 − 1, Xéz˜ ‡푠,Ñklim푛→∞ 푎푛푑+푠 = 푏푠,K lim 푛→∞ 1 푛 푛∑ 푘=1 푎푘 = 1 푑 푑−1∑ 푠=0 푏푠. Proof of Theorem 49. X푗 š~ˆ½"~ˆ§K(Ø´�" X푗 �~ˆ!k±Ï푑, dÚnŒ-푎푛푑+푠 = 푝 (푛푑+푠) 푖푗 , Ï d푏푠 = 푓푖푗(푠) 푑 휇푗 ,l k lim 푛→∞ 1 푛 푛∑ 푘=1 푝 (푘) 푖푗 = 1 푑 푑−1∑ 푠=0 푓푖푗(푠) 푑 휇푗 = 1 휇푗 푑−1∑ 푠=0 푓푖푗(푠) = 푓푖푗 휇푗 . This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 14 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 14 of 31 4.3.2 ²²²­­­©©©ÙÙÙ Definition 51 (p88/65/75, Def 4.11)�{푋(푛), 푛 = 0, 1, ⋅ ⋅ ⋅ }àgê ¼ó, =£VÇÝ (푝푖푗), ek˜‡VÇ©Ù푣푗, 푗 ∈ 푆, =푣푗 ≥ 0 … ∑ 푗∈푆 푣푗 = 1é∀푗 ∈ 푆 ÷v 푣푗 = ∑ 푖∈푆 푣푖푝푖푗. K¡VÇ©Ù푣푗, 푗 ∈ 푆 ê¼ó�˜‡²­©Ù. This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 15 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 15 of 31 Theorem 52�ê¼ó�G�˜mk©)푆 = 퐷 ∪ 퐶1 ∪ 퐶2 ∪ ⋅ ⋅ ⋅ . -푆 △ = 푄 ∪ 퐻 , Ù¥푄 dš~ˆ(63)G�8퐷 9퐶1, 퐶2, ⋅ ⋅ ⋅ ¥ �"~ˆG�8퐶˜훽1, 퐶˜훽2, ⋅ ⋅ ⋅ �N|¤, =푄 = 퐷 ∪ (∪푗퐶˜훽푗). , §퐻 = ∪훼퐶˜훼. z‡퐶˜훼Ä��~ˆ48,K (1)²­©ÙØ3�¿‡^‡퐻 = ∅. (2) ²­©Ù˜3�¿‡^‡k˜‡Ä��~ˆ4 8퐶˜훼. (3)²­©ÙkÁõ‡�¿‡^‡–�kü‡퐶˜훼. (4)kê¼ó�²­©Ùð3" This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 16 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 16 of 31 ~. (p93/69/79, ex 4.18) This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 17 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 17 of 31 ²­©ÙÚ²þˆ£žm�'X Theorem 53 1. (p89/66/75, Thm 4.16) 2. (p91/67/77, Cor 3) This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 18 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 18 of 31 ~. (p92/68/77, ex 4.16) This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 19 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 19 of 31 4.4 iii\\\êêê¼¼¼óóó(p//80) M/G/1 Queueing System For this model, we suppose that the arrival process is a Poisson process 퐻(푡) with the parameter 휆 and the service time is a general dis- tribution 퐵(푥) with mean 1휇 and the second moment 휇2 <∞. In this model, if we let 푁(푡) denote the number of customers in the system at time 푡, then 푁(푡) is not a Markov process. However, if we describe the model by the pair (푛, 푥) where 푛 denotes the number of customers in the system and 푥 the service time already received by the customer in service. We thus need a two-dimensional state description. The first dimension is discrete, but the other one is continuous and this essentially complicates the analysis. This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 20 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 20 of 31 However, if we look at the system just after departures (introduced at 1953 by D.G. Kendall), then the state description can be simplified to 푛 only, because 푥 = 0 for the new customer (if any) in service. The stochastic process of the first marginal is a Markov chain. This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 21 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 21 of 31 Imbedded Markov Chain Theorem 54 Let 푋푛 be the number of customers left in the system just after the 푛-th customer leaving the system, then {푋푛, 푛 ≥ 0} is a Markov chain with the state space 퐸 = {0, 1, 2, ⋅ ⋅ ⋅ } and the transition probabil- ity matrix given by 푃 = ⎛⎜⎜⎜⎜⎝ 푎0 푎1 푎2 ⋅ ⋅ ⋅ 푎0 푎1 푎2 ⋅ ⋅ ⋅ 푎0 푎1 ⋅ ⋅ ⋅ . . . . . . ⎞⎟⎟⎟⎟⎠ (1) where 푎푖 = ∫ ∞ 0 (휆푡)푖 푖! 푒−휆푡푑퐵(푡). (2) is the probability of that there are 푖 customers arrived during one service period. Proof. Let 푉푛 denote the service time of the 푛-th customer and he This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 22 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 22 of 31 left at the epoch 푇푛, 푛 ≥ 1. During the service time of the 푛-th customer, there are 푌푛 customers arrived. If we define 휀(푥) = { 1, if 푥 > 0, 0, if 푥 ≤ 0, then we have 푋푛+1 = 푋푛 − 휀(푋푛) + 푌푛+1, 푛 ≥ 1. we should note that if 푋푛 > 0, then the service of the 푛 + 1-th customer starts at 푇푛 and ends at 푇푛 + 푉푛+1 and 푌푛+1 = 퐻(푇푛 + 푉푛+1)−퐻(푇푛); if 푋푛 = 0, then the 푛+ 1-th customer is not in the system, let the customer comes at푅, then the service of the 푛+1-th customer starts at푅 and ends at 푅 + 푉푛+1, and 푌푛+1 = 퐻(푅 + 푉푛+1) − 퐻(푅). From the property of Poisson process, we know that the distribution of 푌푛+1 are the same no matter what situation it is. (1). Becasue of the independent relationship between 푌푛+1 and 푋푛, This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 23 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 23 of 31 for any non-negative integers 푘1, 푘2, ⋅ ⋅ ⋅ , 푘푛+1, we have 푃 ( 푋푛+1 = 푘푛+1∣푋푛 = 푘푛, 푋푛−1 = 푘푛−1, ⋅ ⋅ ⋅ , 푋1 = 푘1) = 푃 (푋푛 − 휀(푋푛) + 푌푛+1 = 푘푛+1∣푋푛 = 푘푛, 푋푛−1 = 푘푛−1, ⋅ ⋅ ⋅ , 푋1 = 푘1) = 푃 (푘푛 − 휀(푘푛) + 푌푛+1 = 푘푛+1∣푋푛 = 푘푛, 푋푛−1 = 푘푛−1, ⋅ ⋅ ⋅ , 푋1 = 푘1) = 푃 (푌푛+1 = 푘푛+1 − 푘푛 + 휀(푘푛)∣푋푛 = 푘푛, 푋푛−1 = 푘푛−1, ⋅ ⋅ ⋅ , 푋1 = 푘1) = 푃 (푌푛+1 = 푘푛+1 − 푘푛 + 휀(푘푛)) = 푃 (푋푛+1 = 푘푛+1∣푋푛 = 푘푛) Therefore, {푋푛, 푛 ≥ 0} is a Markov chain. (2). Since 푌1, 푌2, ⋅ ⋅ ⋅ are i.i.d. Let 푎푖 △ = 푃 (푌1 = 푖) = 푃 (퐻(푉1) = 푖) = ∫∞ 0 푃 (퐻(푉1) = 푖∣푉1 = 푡)푑퐵(푡) = ∫∞ 0 (휆푡)푖 푖! 푒 −휆푡푑퐵(푡). From (1), we know that if 푖 = 0, then 푝(푛, 푖;푛 + 1, 푗) = 푃 (푋푛+1 = 푗∣푋푛 = 푖) = 푃 (푌푛+1 = 푗) = 푎푗. If 0 < 푖 ≤ 푗 + 1, then 푝(푛, 푗;푛+1, 푗) = 푃 (푋푛+1 = 푗∣푋푛 = 푖) = 푃 (푌푛+1 = 푗− 푖+1) = 푎푗−푖+1. This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 24 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 24 of 31 If 푖 > 푗 + 1, then 푝(푛, 푖;푛 + 1, 푗) = 푃 (푋푛+1 = 푗∣푋푛 = 푖) = 푃 (푌푛+1 = 푗 − 푖 + 1) = 0. This indicates that 푝(푛, 푖;푛 + 1, 푗) is independent of 푛, which im- plies that {푋푛, 푛 ≥ 0} is homogeneous Markov chain. This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 25 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 25 of 31 Stationary Distribution of the Imbedded Markov Chain Theorem 55 For the M/G/1 model, the imbedded Markov chain {푋푛, 푛 ≥ 0} is irreducible and aperiod. {푋푛} is positive recurrent if and only 휌 = 휆휇 < 1. Moreover, if 휌 = 1, then {푋푛} is null recurrent; if 휌 > 1, then {푋푛} is transient. Theorem 56 Denote the limiting distribution of 푁푑푘 by {푑푛}∞푛=0 and the limiting random variable by푁푑. So 푑푛 = 푃 (푁푑 = 푛) = lim푘→∞ 푃 (푁푑푘 = 푛). If we define 푃푁푑(푧) = ∑∞ 푛=0 푑푛푧 푛, then 푃푁푑(푧) = (1− 휌)퐵˜(휆− 휆푧)(1− 푧) 퐵˜(휆− 휆푧)− 푧 . (3) where 퐵˜(훼) = ∫∞ 0 푒 −훼푡푑퐵(푡). This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 26 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 26 of 31 Proof. The limiting probabilities 푑푛, which we know are equal to 푝푛, satisfy the equilibrium equations 푑푛 = 푑푛+1푎0 + 푑푛푎1 + ⋅ ⋅ ⋅ + 푑1푎푛 + 푑0푎푛 = ∑푛 푘=0 푑푛+1−푘푎푘 + 푑0푎푛 푛 = 0, 1, ⋅ ⋅ ⋅ To solve the equilibrium equations we will use the generating function approach. Let us introduce the probability generating functions 푃퐴(푧) = ∞∑ 푛=0 푎푛푧 푛, which are defined for all 푧 ≤ 1. Multiplying the above equation by This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 27 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 27 of 31 푧푛 and summing over all 푛 leads to 푃푁푑(푧) = ∑∞ 푛=0 푑푛푧 푛 = ∑∞ 푛=0 ( ∑푛 푘=0 푑푛+1−푘푎푘 + 푑0푎푛) 푧 푛 = 푧−1 ∑∞ 푛=0 ∑푛 푘=0 푑푛+1−푘푧 푛+1−푘푎푘푧푘 + ∑∞ 푛=0 푑0푎푛푧 푛 = 푧−1 ∑∞ 푘=0 ∑∞ 푛=푘 푑푛+1−푘푧 푛+1−푘푎푘푧푘 + 푑0푃퐴(푧) = 푧−1 ∑∞ 푘=0 푎푘푧푘 ∑∞ 푛=푘 푑푛+1−푘푧 푛+1−푘 + 푑0푃퐴(푧) = 푧−1푃퐴(푧)(푃푁푑(푧)− 푑0) + 푑0푃퐴(푧). Hence we find 푃푁푑(푧) = 푑0푃퐴(푧)(1− 푧−1) 1− 푧−1푃퐴(푧) . To determine the probability 푑0 we note that 푑0 is equal to 푝0, which is the fraction of the system is empty. Hence 푑0 = 푝0 = 1 − 휌. So, by multiplying numerator and denominator by −푧 we obtain 푃푁푑(푧) = (1− 휌)푃퐴(푧)(1− 푧) 푃퐴(푧)− 푧 . By using the definition of 푎푖, the generating function 푃퐴(푧) can be This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 28 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 28 of 31 rewritten as 푃퐴(푧) = ∑∞ 푛=0 ∫∞ 0 (휆푡)푛 푛! 푒 −휆푡푧푛푑퐵(푡) = ∫∞ 0 ∑∞ 푛=0 (휆푡푧)푛 푛! 푒 −휆푡푑퐵(푡) = ∫∞ 0 푒 −(휆−휆푧)푡푑퐵(푡) = 퐵˜(휆− 휆푧) This completes the proof. This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 29 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 29 of 31 Example 57 M/M/1 queue For exponential service time with mean 1/휇 we have 퐵˜(훼) = 휇 휇 + 훼 . Thus 푆˜(훼) = (1−휌) 휇휇+훼훼 휆 휇휇+훼+훼−휆 = (1−휌)휇훼휆휇+(훼−휆)(휇+훼) = (1−휌)휇훼 (휇−휆)훼+훼2 = 휇(1−휌) 휇(1−휌)+훼. Hence 푆 is exponentila distribution with parameter 휇(1− 휌), i.e. 퐹푆(푡) = 푃 (푆 ≤ 푡) = 1− 푒−휇(1−휌)푡. 푡 ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G/M/1 Queueing System to be added here... This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 30 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 30 of 31 Exercise 4.12:page //85 ²L{ü�OŽL²G�1Ú2´š~ˆG�§G�6Ú7´˜ ‡~ˆ48"G�3§4Ú5�‡E,˜ §yÄG�4§dG� =£ãŒ 푓 (1) 44 = 0 푓 (2) 44 = 0.4 ∗ 0.4 + 0.6 ∗ 0.5 푓 (3) 44 = 0.4 ∗ 0.6 ∗ 0.4 + 0.6 ∗ 0.3 ∗ 0.5 + 0.6 ∗ 0.2 ∗ 0.4 푓 (4) 44 = 0.4 ∗ 0.62 ∗ 0.4 + 0.6 ∗ 0.32 ∗ 0.5 + 0.6 ∗ 0.3 ∗ 0.2 ∗ 0.4 + 0.6 ∗ 0.2 ∗ 0.6 ∗ 0.4 = 0.4 ∗ 0.62 ∗ 0.4 + 0.6 ∗ 0.32 ∗ 0.5 + 0.6 ∗ 0.2 ∗ (0.3 + 0.6) ∗ 0.4 푓 (푛) 44 = 0.4 ∗ 0.6푛−2 ∗ 0.4 + 0.6 ∗ 0.3푛−2 ∗ 0.5 + 0.6 ∗ 0.2 ∗ ( 푛−3∑ 푘=0 0.3푘 ∗ 0.6푛−3−푘 ) ∗ 0.4 This document is created by Jiashan TANG with NUPT on November 2, 2009: jiashant@yahoo.ca; http://math.carleton.ca/∼tangjs All right reserved. ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 31 of 31 ∙First ∙Prev ∙Next ∙Last ∙Go Back ∙Full Screen ∙Close ∙Quit ∙ Page 31 of 31 푓44 = ∞∑ 푛=1 푓 (푛) 44 = 0.4 ∗ 0.4 + 0.6 ∗ 0.5 + ∞∑ 푛=3 ( 0.4 ∗ 0.6푛−2 ∗ 0.4 + 0.6 ∗ 0.3푛−2 ∗ 0.5) + ∞∑ 푛=3 ( 0.6 ∗ 0.2 ∗ 0.4 ∗ ( 푛−3∑ 푘=0 0.3푘 ∗ 0.6푛
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