SPLecture18

# SPLecture18 - Lecture 18 Mariana Olvera-Cravioto Columbia...

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Lecture 18 Mariana Olvera-Cravioto Columbia University [email protected] April 6th, 2015 IEOR 4106, Intro to OR: Stochastic Models Lecture 18 1/18

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Regenerative processes I Let { X ( t ) : t 0 } be a regenerative process with regeneration points { T i } i 0 . I Note that the times between regeneration points, i.e., X i = T i - T i - 1 , are i.i.d. random variables, and therefore form a renewal process. I Suppose that when the process is in state j it earns a reward at rate 1 per unit of time. I Define I ( s ) = ( 1 , if X ( s ) = j, 0 , if X ( s ) 6 = j. I The total reward earned by time t is then Z t 0 I ( s ) ds. IEOR 4106, Intro to OR: Stochastic Models Lecture 18 2/18
Regenerative processes... continued I It follows from what we know about renewal processes that lim t →∞ 1 t Z t 0 I ( s ) ds = E [ reward by time T 1 ] E [ T 1 ] . IEOR 4106, Intro to OR: Stochastic Models Lecture 18 3/18

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Regenerative processes... continued I It follows from what we know about renewal processes that lim t →∞ 1 t Z t 0 I ( s ) ds = E [ reward by time T 1 ] E [ T 1 ] . I In this case, the average reward per unit of time is the same as the proportion of time spent at time j . I Proposition: For a regenerative process, the long-run proportion of time spent at state j , i.e., lim t →∞ 1 t Z t 0 I ( s ) dx, satisfies lim t →∞ 1 t Z t 0 I ( s ) dx = E [ amount in time spent in j during a cycle ] E [ length of a cycle ] . IEOR 4106, Intro to OR: Stochastic Models Lecture 18 3/18
A finer result I Suppose that T 1 is a continuous random variable. I Using an advanced result called the Key Renewal Theorem we obtain lim t →∞ P ( X ( t ) = j ) = E [ amount in time spent in j during a cycle ] E [ length of a cycle ] . IEOR 4106, Intro to OR: Stochastic Models Lecture 18 4/18

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Example I Suppose that { X ( t ) : t 0 } is a continuous-time Markov chain such that X (0) = k . I The sequence of times at which X ( t ) returns to state k , i.e., T 0 = 0 , T i = inf { t > T i - 1 : X ( t ) = k } , i 1 , constitute regeneration points. IEOR 4106, Intro to OR: Stochastic Models Lecture 18 5/18
Example I Suppose that { X ( t ) : t 0 } is a continuous-time Markov chain such that X (0) = k . I The sequence of times at which X ( t ) returns to state k , i.e., T 0 = 0 , T i = inf { t > T i - 1 : X ( t ) = k } , i 1 , constitute regeneration points. I It follows that the long-run proportion of time spent in state j is E [ time spent in j during a k - k cycle ] E [ length of a k - k cycle ] . I If we set j = k and let μ kk denote the expected return time to state k , then E [ time spent in k during a k - k cycle ] E [ length of a k - k cycle ] = 1 k μ k . IEOR 4106, Intro to OR: Stochastic Models Lecture 18 5/18

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Example: The GI/GI/1 queue I Consider a single-server queueing system with an infinite waiting area. I Customers arrive according to a renewal process { N ( t ) : t 0 } . I Service times of customers are i.i.d. according to some general distribution F . I Customers are served in the order in which they arrive. IEOR 4106, Intro to OR: Stochastic Models Lecture 18 6/18
Example: The GI/GI/1 queue I Consider a single-server queueing system with an infinite waiting area.

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• Summer '12
• Stochastic
• Probability theory, Pon, Queueing theory, stochastic models

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