X(t) = B(t), t 0,

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1 IEOR 4106: Introduction to Operations Research: Stochastic Models Spring 2007, Professor Whitt, Final Exam Chapters 4-7 and 10 in Ross, Wednesday, May 9, 1:10pm-4:00pm Open Book: but only the Ross textbook, the CTMC Notes, the class notes related to Chapters 7 and 10, and three 8 11 pages of notes Honor Code: Students are expected to behave honorably when taking exams. After completing the test, please testify to your adherence by writing the following on your bluebook and signing your name: I have neither given nor received help while taking this test. Justify your answers; show your work. 1. The Halasa-Yildirim Consulting Company Goes Public! (25 points) Ranwa Halasa and Burcu Yildirim have been so successful with their Halasa-Yildirim Consulting Company, that their company has gone public, and is now listed on the New York Stock Exchange under the HY symbol. Suppose that a share of HY stock initially costs $80 per share. Suppose that the HY stock share price over time (measured in years) is modelled as the stochastic process {X(t) : t 0}, where X(t) = B(t), t 0, with {B(t) : t 0} being standard Brownian motion, having E[B(t)] = 0 and V ar(b(t)) = t for t 0. (a) What is the probability that the HY stock price is greater than $100 after 2 years? ( X(4) E[X(4)] P(X(4) > 100) = P SD(X(4)) > ) = P(N(0,1) > 2) = (b) What is the conditional expected HY stock price after 1 year, given that the HY stock price is $130 per share after 5 years? See (10.4) in Section The conditional mean lies on the line connecting the specified values at t = 0 and t = 5. Hence E[X(1) X(5) = 130] = 90. (c) What is the approximate conditional probability that the HY stock price after 1 year exceeds 94.5, given that the HY stock price is $130 per share after 5 years? (Hint: ) From Section 10.1, we know the conditional distribution is normal, with the mean determined in part (a). The variance is given in (10.4) as well, except we need to adjust for the variance parameter σ = 5. We get that the variance is 25(1)(4)/5 = 20, so that the standard

2 deviation is That means that 94.5 is approximately one standard deviation above the mean. So the probability is approximately P(N(0, 1) > 1) (d) What is the probability that the HY stock price rises to $100 per share before it drops to $70 per share? 1/3 using the linear martingale along with the optional stopping theorem. See Section 5 of the martingale lecture notes. (e) What is the expected length of time until the HY stock price first either rises to $100 per share or drops to $70 per share? If T is that random time, then E[T] = 8, using the quadratic martingale along with the optional stopping theorem. See Section 5 of the martingale lecture notes. 2. The Tello-Tang Tattoo Parlor (25 points) Juan Tello and Jianzhou Tang have opened the Tello-Tang Tattoo Parlor near campus, specializing in probability tattoos, including images of Pascal s triangle, Galton s quincunx and, their best seller - Gauss s bell curve. The two proprietors each work on one customer at a time. There is a waiting room, which can accommodate two people in addition to the two in service. Suppose that potential customer s come to the Tello-Tang Tattoo Parlor according to a Poisson process at constant rate of 2 per hour. Suppose that potential customers arriving when the waiting room is full (when there are two customers in service and two others waiting) leave without getting a tattoo, and without affecting future arrivals. Suppose that waiting customers have limited patience, so that they may leave without receiving service if they have waited too long. Suppose that the times successive customers are willing to wait before starting service are independent random variables, each with an exponential distribution having a mean of 1 hour. The proprietors are very good at their work, so it takes only an average of one hour to give customers their desired tattoo. However, the required times are random, since some tattoos are much more complicated and ornate than others. Suppose that these service times are independent and identically distributed random variables, each exponentially distributed with a mean of 1 hour. (a) What kind of stochastic process describes the number of customers in the Tello-Tang Tattoo Parlor over time, and what are its parameters? It is a birth-and-death process. The parameters are the birth rates λ j = 2 for j = 0,1,2,3 and the death rates µ j = j for j = 1,2,3,4. An interesting observation is that the queueing system is equivalent to an M/M/4 system since the individual rate of abandonment is equal to the service rate. 2

3 (b) What is the steady-state (or long-run limiting) probability that there are three customers in the Tello-Tang Tattoo Parlor, with two in service and one waiting? We define the state of the birth-death process to be the number of customers in the parlor so that the steady-state probability that there are three customers in the parlor is α 3 = r i=1 r, i where r i = λ 0 λ i 1 µ 1 µ i. Plugging in our birth and death rates we get α 3 = 4/21. (c) What is the long-run abandonment rate? Since the total abandonment rate is 1 when the system is in state 3 and 2 when the system is in state 4, the total long-run abandonment rate is α 3 + 2α 4. Computing α 4 just as we computed α 3 in part (b) we get α 4 = 2/21. Therefore, our total abandonment rate is 4/ /21 = 8/21. (d) Suppose that new arrivals are not admitted after 7:00 pm, but all customers in the system at that time are served. Suppose that the Tello-Tang Tattoo Parlor is full at 7:00 pm. What is the expected remaining length of time until all four customers present at 7:00 pm are served? The time in question can be represented as the sum of four independent exponential random variables, the first three each with mean 1/2 and the last with mean 1 because there will only be one in service at the end. So the overall mean is 5/2 = 2.5. (e) What is the variance of the remaining time in part (d) above? The variance of an exponential is the square of the mean. The variance of a sum of independent random variables is the sum of the variances. Hence the overall variance is 3 (1/4) + 1 = 7/4 = Ishan s Mood (25 points) Suppose that Ishan is either cheerful (C) or glum (G). Suppose that his mood (which is always one of these two) changes randomly over time, with his mood on any given day depending on his mood on the previous two days. If he was cheerful on both the previous two days, then the probability that he will be cheerful on the current day is 0.8; if he was glum on both the previous two days, then the probability that he will be cheerful on the current day is 0.5; if he was first cheerful and then glum on the previous two days, then the probability that he will be cheerful on the current day is 0.5; if he was first glum and then cheerful on the previous two days, then the probability that he will be cheerful on the current day is

4 (a) Is it possible to construct a Markov chain that, along with a specification of appropriate initial conditions, can be used to describe the probability distribution of Ishan s mood over time? If so, construct the Markov chain. This problem is the same as the weather Markov chain in Example 4.4. We need to let states be the pairs of moods in successive days: (C,C),(C,G),(G,C),(G, G). The transition probability matrix for this Markov chain is P = where the rows and columns are indexed by the states in the order (C,C),(C,G),(G,C),(G,G). (b) Suppose that Ishan has been cheerful for five days in a row. What is the probability that he also remains cheerful every day for the next three days? By the Markov property, the fact that Ishan was cheerful before yesterday has no effect on his mood the next three days. Here we simply want the probability of transitioning from (C,C) to (C,C) three times, which is.8 3 =.512. (c) Suppose that Ishan has been cheerful for five days in a row. What is the probability that his mood changes every day for the next four days? Similarly, here we want the probability of making the transitions (C,C) to (C,G), (C,G) to (G,C), (G,C) to (C,G), and (C,G) to (G,C) over the next four days which is =.04. (d) Suppose that Ishan has been cheerful for five days in a row. What is the expected number of days until he has again been cheerful for at least two consecutive days? (Note that event could possibly happen in only one more day.) Again, Ishan s mood before yesterday is irrelevant. Note that we can not use our theory of patterns since Ishan s mood on consecutive days is not independent. Let T (C,C),(C,C) be the number of days until Ishan has again been cheerful for at least two days. We condition on X 1, the state of the Markov chain on the next step: E[T (C,C),(C,C) ] =E[T (C,C),(C,C) X 1 = (C,C)]P (C,C),(C,C) + E[T (C,C),(C,C) X 1 = (C,G)]P (C,C),(C,G) =P (C,C),(C,C) + (1 + E[T (C,G),(C,C) ])P (C,C),(C,G) The expectation in the second term can be computed using the fundamental matrices. If we make (C, C) an absorbing state, then we are interested in the expected time until absorption given the Markov chain begins in the transient state (C, G). Our modified transition probability 4

5 matrix is P = , With the rows and columns of the matrix indexed just as in the original transition probability matrix. Using the notation for fundamental matrices introduced in class, we have Q = so that and N = (I Q) 1 = , M (C,G) = N (C,G),(C,G) + N (C,G),(G,C) + N (C,G),(G,G) = 15. Plugging M (C,G) into our original expression for E[T (C,C),(C,C) ] gives us E[T (C,C),(C,C) ] =.8 + (1 + 15).2 = 4. (e) What is the long-run proportion of days that Ishan is cheerful? First we must compute the stationary distribution of the Markov chain. Solving π = πp and i π i = 1, we find that π i =.25 for each state i. Ishan is cheerful on the current day when the state of the Markov chain is (C,C) or (G,C). Therefore, the long-run proportion of days Ishan is cheerful is π (C,C) + π (G,C) = 1/2. 4. Valerie Lau s Life as a Traffic Cop (25 points) Valerie Lau has decided to become a traffic cop. Her life has settled into a happy routine: Her day is spent cruising (driving around) in a police car looking for traffic violators. When she finds a traffic violator, she stops the violator and writes up a ticket. Then she goes back to cruising, and so forth, throughout the day. Suppose that successive cruising times (before catching a traffic violator) are independent and identically distributed exponential random variables with a mean of 15 minutes. Suppose that 90% of violators are stopped for speeding and given a $80 fine, while the remaining 10% of violators are stopped for more serious offenses, for which one half of these fines are for $100 and the other half are for $500. Suppose that the time required to write up a speeding ticket is uniformly distributed between 5 minutes and 15 minutes. Suppose that the time required to write up a ticket for a more serious offense is uniformly distributed between 20 minutes and 40 minutes. (a) What is the long-run rate at which Valerie stops traffic violators? 5

6 This problem is about renewal theory, which is covered by Chapter 7. Let C be the cruising time and let T be the ticket time. Let X = C +T. Then the number N(t) of completed tickets written over the interval [0, t] is a renewal counting process with inter-renewal time distributed as X. Note that E[X] = E[C] + E[T] = 15 + (0.9 10) + (0.1 30) = = 27 So the rate is 1 per 27 minutes or 1/27 per minute or 60/27 = per hour. We can apply Proposition 7.1 (b) What is the long-run proportion of Valerie s working day spent cruising (as opposed to writing up tickets)? The successive cruising and ticket-writing periods form an alternating renewal process; see Section Thus the long-run proportion of time spent cruising is E[C]/(E[C] + E[T]) = 15/27 = 5/9. (c) What is the long-run rate at which Valerie earns money? Use the renewal-reward theorem: The long-run reward rate is the expected earnings per cycle divided by the expected length of the cycle, and so is ( )/27 = 102/27 = 34/9 = (d) Let Y be the time remaining after 4 : 00 pm some day until Valerie next completes writing up a ticket and is ready to start cruising again. What is the approximate expected value E[Y ]? (Hint: This expected value coincides with the long-run average remaining time until Valerie next completes writing up a ticket.) Notice that we are looking at a renewal process at an arbitrary time, which we regard as being in equilibrium. Thus Y has the equilibrium residual lifetime (or excess) distribution; e.g., see Example That example only refers to the average excess, but actually it applies to this equilibrium excess. Hence, E[Y ] = E[X2 ] 2E[X] = E[(C + T)2 ] V ar(c) + V ar(t) + (27)2 = 2E[C + T] = V ar(t) so we have reduced the problem to one of calculating V ar(t). We show how that can be done: We have E[T] = 12 while E[T 2 ] = 0.9E[U[5,15] 2 ] + 0.1E[U[20,40] 2 ]. However, V ar(u[5,15]) = V ar(5 + 10U[0,1] = V ar(10u[0,1] = 100/12, and Hence V ar(u[20,40]) = V ar( U[0,1] = V ar(20u[0,1]) = 400/12. ( ) ( ) E[T 2 ] = = =

7 and V ar(t) = E[T 2 ] (E[T]) 2 = = so that, finally (assuming no arithmetic errors), E[Y ] = So the expected remaining time is minutes. =