resolução da lista de exercicios de Processo de Poisson

Prévia do material em texto

Stochastic Models in OR: J. Lee Assignment 6 Solution Notes
Problem 1. Let T denote the time in hours until the next train arrives; and so T is uniform on (0, 1).
Conditioned on T , X is Poisson with mean 7T (and also variance 7T ), since passengers arrive at a rate of 7
per hour. Since T is Uniform(0,1), we know that E(T ) = (0 + 1)/2 = 1/2 and Var(T ) = (1− 0)2/12 = 1/12.
(a) E(X) = E(E(X|T )) = E(7T ) = 7/2.
(b) E(X|T ) = 7T and Var(X|T ) = 7T. By the conditional variance formula:
Var(X) = E(Var(X|T )) + Var(E(X|T )) = E(7T ) + Var(7T ) = 7E(T ) + 49Var(T ) = 7
2
+
49
12
=
91
12
.
Problem 2. Let T be the total time you spend in the system. Let Ti be the time you spend being served
by server i. Let W1 be the time you spend waiting for server 1 to be free (since you arrive at a time
that server 1 is busy with a customer). Let W2 be the time you spend waiting for server 2 to be free
(after you have completed service at server 1). Then, T = W1 + T1 + W2 + T2. Our goal is to compute
E(T ) = E(W1) + E(T1) + E(W2) + E(T2). Now, W1 is exponential(µ1), by the memoryless property of
exponential (the remaining service for the customer at server 1 is exponential). Thus, E(W1) = 1/µ1.
E(T1) = 1/µ1 and E(T2) = 1/µ2, since we are told that the service time Ti is exponential(µi). Finally, to
compute E(W2) we condition on the event F = “you finish your service at server 1 before the customer at
server 2 is done” (in which case you must wait an exponential(µ2) amount of time for server 2 to complete
the service before you can start at server 2). We get:
E(W2) = E(W2|F )P (F ) + E(W2|F c)P (F c) = 1
µ2
· µ1
µ1 + µ2
+ 0 · µ2
µ1 + µ2
=
1
µ2
· µ1
µ1 + µ2
.
Thus, in total, E(T ) = 2/µ1 + (1/µ2)[1 + µ1/(µ1 + µ2)].
Problem 3. The rate of the Poisson precess is λ = (1/10)minutes−1 (6 hour−1). Let NC(t) denote the
number of cars up until time t (in minutes); NC(t) is a Poisson process of rate λC = λ/3 = 1/30. Let NT (t)
denote the number of trucks up until time t (in minutes); NT (t) is a Poisson process of rate λT = 2λ/3 = 1/15.
(a)
P (at least 8 cars arrived in [12:03, 12:07] | 10 cars arrived during the lunch hour)
=
10∑
i=8
(
10
i
)(
4
60
)i(
56
60
)10−i
.
(b)
P (at least 2 trucks arrived in [12:10, 12:30] | 10 cars arrived during the lunch hour)
= P (at least 2 trucks arrived in [12:10, 12:30])
= P (NT (30)−NT (10) ≥ 2) = 1− e−20/15 (20/15)
0
0!
− e−20/15 (20/15)
1
1!
,
since the car process is independent of the truck process.
(c) Var(NT (140)) = 140(1/15).
(d) Assume now that vehicles arrive more frequently during the lunch hour: the average interarrival time is
5 minutes during the lunch hour (and 10 minutes otherwise). Then, the integral of the λT (t) function
over the interval [11:00, 1:20] is 60(1/15) + 60(2/15) + 20(1/15) = 200/15, so we get a variance of
200/15 for the number of trucks that arrive in [11:00, 1:20].
Problem 4. Let N(t) be the number of rainfalls during the time interval [0, t], where t is measured in days,
and time 0 is the present. N(t) defines a Poisson process with rate 0.2 day−1.
(a) Since the present level is just below 5000 units, and the level decreases by exactly 1000 units per day
(constant, non-random), the only way for the reservoir to be empty after 5 days is if there are no
rainfalls in the time interval [0,5]. This is simply P (N(5) = 0) = e−(.2)5 ((.2)5)
0
0! = e
−1, since N(5) is
Poisson ((.2)5) = Poisson (1).
(b) How can the reservoir be empty some time during [0,10]? There are two possibilities: (i). event E =
“no rainfalls during [0,10]”, or (ii). event F = “exactly one rainfall during [0,5], that rainfall amount
(call it Y ) is 5000 units (not 8000 units), and no rainfalls during [5,10]”. We want to compute P (E∪F )
and we know that E and F are disjoint, so
P (E∪F ) = P (E)+P (F ) = P (N(10) = 0)+P (N(5) = 1, Y = 5000, N(10)−N(5) = 0) = e−2+e−1(.8)e−1,
where we have used independence and stationary increments to write P (N(5) = 1, Y = 5000, N(10)−
N(5) = 0) = P (N(5) = 1)P (Y = 5000)P (N(5) = 0).