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spring2009-final

# spring2009-final - ISyE 3232 Section A Monday April 27 8am...

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ISyE 3232 Spring 2009 Section A, Monday, April 27 8am Professor Jim Dai Final No notes, no books, and no calculators are allowed. Georgia Tech student honor code applies to this test. Solution to this test will be posted by 9pm on Tuesday. Please do not discuss this test with any students who are not in the classroom until after that time. The worksheet has a total of 6 pages. 1. (15 points) Assume that call arrival to a call center follows a non-homogeneous Pois- son process. The call center opens from 9am to 5pm. During the first hour, the arrival rate increases linearly from 0 at 9am to 60 calls per hour at 10am. After 10am, the arrival rate is constant at 60 calls per hour. (a) Plot the arrival rate function λ ( t ) as a function of time t ; indicate clearly the time unit used. (b) Find the probability that exactly 5 calls have arrived by 9:10am. (c) What is the probability that the 1st call arrives after 9:20am? (d) What is the probability that there are exactly one call between 11:00am and 11:05am and at least two calls between 11:03am and 11:06am? Solution. (a) If we use the time units of mins: 0 t 60 min s λ ( t ) = 1 60 t 60 t 480 min s λ ( t ) = 1 (b) The arrival process N(t) during 9am and 9:10 am is a non-homogeneous arrival process, the average rate is 1/12 call per min. So we can view N(9am,9:10am) as possion random variable with mean 10 / 12 = 5 / 6call. The mean of poisson random variable can also be calculated through: 10 Z 0 λ ( t ) dt = 10 Z 0 1 60 tdt = 1 60 · 1 2 t 2 | t =10 t =0 = 1 60 · 1 2 · 100 = 5 6 So the probability of exactly 5 calls during 9am and 9:10 am should be P ( N (9 am, 9 : 10 am ) = 5) = ( 5 6 ) 5 5! e - 5 6 1

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(c) It’s equivalent to show the probability of there’s no arrival between 9am to 9:20am. Similar as part b,the average rate is 1/6 call per min, thus we can view N(9am,9:20am) as possion random variable with mean 10/3 call. Thus the required probability should be P ( N (9 am, 9 : 20 am ) = 0) = e - 10 3 (d) the rate after 11 am is always 1 call per min. We are calculating the intersec- tion of two events. There’re two possibilities of the arrival time of the one call in 11am-11:05 am: either it arrives between 11am-11:03m,or it arrives between 11:03am-11:05am. P(there are exactly one call between 11:00am and 11:05am and at least two calls between 11:03am and 11:06am) =P(exactly one call in 11am-11:03am and no call in 11:03-11:05 am and at least two calls between 11:05am and 11:06am)+P(no call in 11am-11:03am and one call in 11:03am-11:05am and at least one call in 11:05-11:06am) Denote N ( x ) as possion random variable with mean x. By independent incre- ment property of non-homogenous possion process, we know the above identity
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