316 IOE Fall 2007 Homework 5 Solution
Due December 4, 2007
1. (25 points) Cars cross a certain point in the highway in accordance with a Poisson process with rate = 4 per minute. (a) (5 points) If Kate blindly runs across the highway, then what is the probability that she will be uninjured if the amount of time that it takes her to cross the road is s seconds? (Assume that if she is on the highway when a car passes by, then she will be injured.) (b) (5 points) Compute part (a) for s = 2, 5, 10, 20. (c) (10 points) Now assume that Kate is agile enough to escape from a single car, but if she encounters two or more cars while attempting to cross the road, then she will be injured. What is the probability that she will be unhurt if it takes her s seconds to cross? (d) (5 points) Compute part (c) for s = 5, 10, 20, 30. SOLUTION: Let N(t) = the number of cars that cross the road by time t, where t is minutes. (a) (5 points) P (Kate is uninjured) = P (N (s) = 0) = e- 60 = e- 15 (b) (5 points) 2 s s = 2: P(Kate is uninjured) = e-4 60 = e- 15 = 0.87517. s 1 s = 5: P(Kate is uninjured) = e-4 60 = e- 3 = 0.71653. s 2 s = 10: P(Kate is uninjured) = e-4 60 = e- 3 = 0.513417. 4 s s = 20: P(Kate is uninjured) = e-4 60 = e- 3 = 0.263597. (c) (10 points) P (Kate is uninjured) = P (N (t) = 0 or N (t) = 1) = e-t + te-t s s s = e- 60 + e- 60 60 s s s = e-4 60 + 4 e-4 60 60 (d) (5 points) s s s P(Kate is uninjured) = e-4 60 + 4 60 e-4 60 s s s s = = = = 5: P(Kate is uninjured) = 4 e- 3 = 0.955375 3 2 10: P(Kate is uninjured) = 5 e- 3 = 0.855695 3 4 20: P(Kate is uninjured) = 7 e- 3 = 0.61506 3 30: P(Kate is uninjured) = 3 e-2 = 0.406
1 s s
1
2. (10 points) A certain scientific theory supposes that mistakes in cell division occur according to a Poisson process with rate 2.5 per year, and that an individual dies when 196 such mistakes have occurred. Assuming this theory, find (a) (5 points) the mean lifetime of an individual, (b) (5 points) the variance of the lifetime of an individual. SOLUTION: Let Xi be the time between the (i - 1)st and the ith mistake, where X0 = 0. Since mistakes occur according to a Poisson process with rate 2.5 per year, the time between mistakes is exponentially distributed with rate = 2.5 per year. (a) (5 points) We are looking for E
196 i=1 Xi
. Since each Xi exponentially distributed
1 2.5
with parameter = 2.5 per year, E[Xi ] =
= 0.4 for all i. Since the mistakes occur
196 i=1 Xi
independently by virtue of the Poisson process, E 78.4 years. Thus, V
196 i=1 Xi
=
196 i=1 E[Xi ] 1 2.52
= 196(0.4) =
(b) (5 points) Following the same reasoning as in part (a), V [Xi ] = =
196 i=1 V
= 0.16 for all i.
[Xi ] = 196(0.16) = 31.36 years.
3. (20 points) In a certain system, a customer must first be served by server 1 and then by server 2. The service times at server i are exponential with rate i , i = 1, 2. An arrival finding server 1 busy waits in line for that server. Upon completion of service at server 1, a customer either enters service with server 2 if that server is free or else remains with server 1 (blocking any other customers from entering service) until server 2 is freed. Customers depart the system after being served by server 2. Suppose that when you arrive there is one customer in the system and that customer is being served by server 1. What is the expected total time you spend in the system? SOLUTION: The total time you spend in the system can be broken into four components: waiting time at server 1 (call this X1 ), service time at server 1 (X2 ), waiting time at server 2 (X3 ), and service time at server 2 (X4 ). Let T be the total time. Then T = 4 Xi so , i=1 that E[T ] = 4 E[Xi ]. i=1
1 1 First, because of the memoryless distribution, E[X1 ] = 1 . Clearly, E[X2 ] = 1 and 1 E[X4 ] = 2 . As for X3 , we can calculate it by conditioning on whether you (at server 1) or the other customer (at server 2) finishes service first. If you finish first, then by the 1 memoryless property, the expected waiting time is 2 . If the other customer finishes first, then you will have no waiting time. The probability that you finish first is 11 2 , whereas the + probability that the other customer finishes first is 12 2 . We can now give E[X3 ] as follows. +
E[X3 ] = E[X3 |you finish first]P (you finish first) +E[X3 |you finish second]P (you finish second) 1 1 2 = +0 2 1 + 2 1 + 2 1 1 = 2 1 + 2
2
Putting all of the facts together, we find that E[T ] = 2 + 1 1 2 1+ 1 1 + 2 .
4. (15 points) Let {N (t), t 0} be a Poisson process with rate . Let Sn denote the time of the nth event. (a) (5 points) Find E[S6 ]. (b) (5 points) Find E[S6 |N (2) = 3]. (c) (5 points) Find E[N (6) - N (3)|N (2) = 4]. SOLUTION: (a) (5 points) E[S6 ] =
6 3
(b) (5 points) E[S6 |N (2) = 3] = 2 + E[Time for 3 more events] = 2 +
(c) (5 points) E[N (6) - N (3)|N (2) = 4] = E[N (6) - N (3)] = 3 (by the independent increments property) 5. (25 points) Cars pass a point on the highway at a Poisson rate of one per minute. If 7 percent of the cars on the road are hybrid (that is, there is a probability of 0.07 that any passing car is a hybrid car), then (a) (5 points) what is the probability that at least one hybrid car passes by during an hour? (b) (5 points) given that ten hybrids have passed by in an hour, what is the expected number of non-hybrid cars to have passed by in that time? (c) (7 points) given that ten hybrids have passed by in an hour, what is the expected number of cars to have passed by in that time? (d) (8 points) if 70 cars have passed by in an hour, what is the probability that five of them were hybrids? SOLUTION: (a) (5 points) Each passing car is either a hybrid or not a hybrid. Let N (t) be the number of cars passing the point in t time units. Let N1 (t) and N2 (t) denote the number of hybrid and non-hybrid cars that pass the point in t time units. Then N (t) = N1 (t) + N2 (t). It is important to note that both N1 (t) and N2 (t) are Poisson processes with rates 1 = .07 60 = 4.2 and 2 = .93 60 = 55.8 cars per hour, respectively. Given that N1 (t) is a Poisson process with rate 4.2 per hour, P (N1 (1) 1) = 1 - P (N1 (1) = 0) = 1 - e-4.2 = .9850. (b) (5 points) The number of non-hybrid cars to pass in an hour is independent of the number of hybrid cars to pass in an hour, so E[N2 (1)|N1 (1) = 10] = E[N2 (1)] = 55.8. (c) (7 points) From the previous question, we expect 55.8 non-hybrid cars to have passed. Also, we know that 10 hybrid cars have passed, so that we expect 65.8 cars total to have passed. 3
(d) (8 points) If N (1) = 70 and N1 (1) = 5, then N2 (1) = 65. So P (N1 (1) = 5|N (1) = 70) = = = = P (N1 (1) = 5 N (1) = 70) P (N (1) = 70 P (N1 (1) = 5 N2 (1) = 65) P (N (1) = 70) P (N1 (1) = 5) P (N2 (1) = 65) P (N (1) = 70)
e-4.2 (4.2)5 e-55.8 (55.8)65 5! 65! -60 (60)70 e 70!
= 0.1819 6. (5 points) Cars pass a certain street location according to a Poisson process with rate . A woman who wants to cross the street at that location waits until she can see that no cars will come by in the next T time units. Find the probability that her waiting time is 0. SOLUTION: N (t), the number of cars to come by in t time units, is a Poisson random variable with rate t. Her waiting time is 0 only if no cars come in the next T time units. So P (N (T ) = 0) = e-T .
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