{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapt 10c - 10.5 Poisson Process 353 restaurant will go to...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
10.5 Poisson Process 353 restaurant will go to refuel at the gas station before going back to the highway with a probability of 0.25. What is the probability that exactly 10 cars have been refueled at the gas station within a particular two-hour period? Solution The process that governs car arrivals at the gas station is Poisson with a rate of λ G = p λ = ( 0 . 25 )( 12 ) = 3 cars per hour. Thus, if K represents the number of cars that arrive at the gas station within 2 hours, the probability that K = 10 cars is given by P [ K = 10 ] = ( 2 λ G ) 10 10 ! e 2 λ G = 6 10 10 ! e 6 = 0 . 0413 trianglesolid 10.5.11 Random Incidence Consider a Poisson process X ( t ) in which events (or arrivals) occur at times T 0 = 0 , T 1 , T 2 ,... . Let the interarrival times Y k be defined as follows: Y 1 = T 1 T 0 Y 2 = T 2 T 1 ... Y k = T k T k 1 These interarrival times are illustrated in Figure 10.6, where A k denotes the k th arrival. The Poisson process belongs to a class of random processes called renewal processes that have the property that the Y k are mutually independent and iden- tically distributed. For the Poisson process with mean arrival rate λ , the Y k are exponentially distributed with mean 1 , as discussed earlier. Consider the following problem in connection with the Y k . Assume the T k are the points in time that buses arrive at a bus stop. A passenger arrives at the bus stop at a random time and wants to know how long he or she will wait until the next bus arrival. This problem is usually referred to as the random incidence problem , since the subject (or passenger in this example) is incident to the process Figure 10.6 Interarrival Times of a Poisson Process
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
354 Chapter 10 Some Models of Random Processes Figure 10.7 Random Incidence at a random time. Let R be the random variable that denotes the time from the moment the passenger arrived until the next bus arrival. R is referred to as the residual life of the Poisson process. Also, let W denote the length of the interar- rival gap that the passenger entered by random incidence. Figure 10.7 illustrates the random incidence problem. Let f Y ( y ) denote the PDF of the interarrival times; let f W ( w ) denote the PDF of W , the gap entered by random incidence; and let f R ( r ) denote the PDF of the residual life, R . The probability that the random arrival occurs in a gap of length between w and w + dw can be assumed to be directly proportional to the length w of the gap and relative occurrence f Y ( w ) dw of such gaps. That is, f W ( w ) dw = β wf Y ( w ) dw where β is the constant of proportionality. Thus, f W ( w ) = β wf Y ( w ) . Since f W ( w ) is a PDF, we have that integraldisplay −∞ f W ( w ) dw = 1 = β integraldisplay −∞ wf Y ( w ) dw = β E [ Y ] Thus, β = 1 / E [ Y ] , and we obtain f W ( w ) = wf Y ( w ) E [ Y ] The expected value of W is given by E [ W ] = E bracketleftbig Y 2 bracketrightbig / E [ Y ] . This result applies to all renewal processes. For a Poisson process, Y is exponentially distributed with E [ Y ] = 1 and E bracketleftbig Y 2 bracketrightbig = 2 2 . Thus, for a Poisson process we obtain f W ( w ) = λ wf Y ( w ) = λ 2 we λ w w 0 E [ W ] = 2
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern