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lect0421 - IEOR 4106 Introduction to Operations Research...

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Unformatted text preview: IEOR 4106: Introduction to Operations Research: Stochastic Models Spring 2009, Professor Whitt Topics for Discussion: Monday, April 21 Renewal Theory: Patterns 0. The Pattern Problem: see Section 7.9 Consider successive independent flips of a biased coin. On each flip, the coin comes up heads (H) with probability p or tails (T) with probability q = 1- p , where 0 < p < 1. A given segment of finitely many consecutive outcomes is called a pattern . The pattern is said to occur at flip n if the pattern is completed at flip n . For example, the pattern A ≡ HTHTHT occurs at flips 8 and 10 in the sequence TTHTHTHTHTTTTHHHT ... and at no other times among the first 17 flips. Note a key property of this example: The two occurrences of the pattern overlap: The final portion of the pattern ending at trial 8 serves as the initial portion of the pattern ending at trial 10. It should be clear that this framework generalizes: In particular, we could have more than two outcomes in each trial. Each trial could be the outcome of the roll of two dice. Then there would be 11 possible outcomes on each trial: 2 , 3 ,..., 12; e.g., with the probability of 4 being 3 / 36 and the probability of 11 being 2 / 36 in each trial. We would then want 11 different symbols to use for each trial. However, we will discuss the special case of only two possible outcomes on each trial. 1. The Main Problem Let N A be the number of trials (coin tosses) until the pattern A appears for the first time . Our main goal is to develop a relatively simple method for calculating the mean E [ N A ] for any pattern A . We will be drawing on renewal theory (Chapter 7). The key observation is that renewal theory applies here. Let S n be the trial where a specified pattern appears for the n th time, for n ≥ 1. Let X n ≡ S n- S n- 1 , where S ≡ 0. The key observation is that X 1 ,X 2 ,X 3 , ··· are independent random variables. In general, X 1 has a different distribution from X n for n ≥ 2, but { X n : n ≥ 2 } is a sequence of independent and identically distributed random variables. Since X 1 has a different distribution from X n for n ≥ 2, this is called a delayed renewal process . In particular, let N ( n ) be the number of times the pattern appears in the first n trials. Then { N ( n ) : n ≥ } is a discrete-time delayed renewal counting process, having N (0) ≡ 0. We will exploit this renewal structure. 2. Divide and Conquer Once Again The main idea is to approach this goal of calculating E [ N A ] indirectly, exploiting the divide-and-conquer strategy. To do so, we consider a more general concept: Let N A → B be the number of trials (coin tosses) to go from pattern A to pattern B . In other words, we assume that pattern A has just occurred at some trial, say trial j . (That means that pattern A is completed on trial j .) We then ask how many additional trials are needed before pattern B next appears. (If pattern B is completed on trial j + n , then we say N A → B = n . A key thing to....
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lect0421 - IEOR 4106 Introduction to Operations Research...

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