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Unformatted text preview: 9, 6/19, 6/19, 4/19}. The probability of a
caller arriving to see all three lines busy equals π3 , so
4
15
λeﬀ = λ (1 − π3 ) = 1 1 −
=
per minute
19
19
Because there are three servers, there is no queue, and therefore the waiting time is zero.
Wq = 0 minute
2. This queuing system is an example of a CTMC called a birthdeath process with birth rates
λi = λ/(i + 1) and death rates µi = µ. The Rate diagram can be written as follows By writing the equations for the cuts, and setting η0 = 1, the following is a valid invariant
measure can be constructed.
n− 1
=0 λi
ηn = in
∀n = 1, 2, . . .
i=1 µi
Applying this formula to our instance yields, with ρ = λ/µ,
n− 1 i=0
ηn = n λ
i+1 i=1 Recalling the formula ex = ∞ n=0 µi = λn
ρn
=
µ n n!
n! xn /n!, yields
∞
ηn = n=0 ∞
ρn
= eρ
n!
n=0 Using the relationship between the stationary distribution and the invariant measure,
ηn
ρn e − ρ
πn = ∞
=
n!
( i=0 ηi ) Notice that πn is equal to the probability mass function for a Poisson random variable with
mean ρ, so the expected number customers in the long run is ρ. 3 3. (a) The generator is 1
− 12 1 15 0 0 0 1
12 0 0 3
− 20 1
12 0 2
15 − 13
60 1
12 0 2
15 − 13
60 0 0 2
15 0 0 0 1
12 2
− 15 (b) The rate diagram is given by Using the cut method generates the following equations
1
1 12 η0 = 15 η1 1 η = 2η
12 1
15 2
1
2 12 η2 = 15 η3 1
2
12 η3 = 15 η4 Substituting η0 = 1 gives η = {1, 1.25, 0.7813, 0.4883, 0.3052} with
The stationary distribution is given by the relation,
ηn
πn =
4 i=0 ηi 4 n=0 ηn = 3.8247. This calculation gives the stationary distribution π = {0.2615, 0.3268, 0.2043, 0.07979}. (c) The steady state probability the call is accepted is the complement of the probability that
the call will be transferred. A call is transferred if and only if four calls are currently in
the system.
P(Call Accepted) = 1 − P(Call Transferred)
= 1 − π4 = 0.07979
(d) A call does not have to wait if the an ambulance is currently available, or equivalently,
at most a single ambulance is responding to a call.
P(No Waiting) = P(At Most One Ambulance Responding)
= π0 + π1
= 0.5882
(e) Because there are two ambulances, the number of calls waiting is equal to the number in
the system minus two.
Nq = 1 π3 + 2 π4 = 0.2872 call
4 (f) The throughput of the system is the arrival rate times the probability that a call is
accepted.
1
λeﬀ = λ (1 − π4 ) =
(1 − 0.07979) = 0.07668 per minute
12
Using Little’s Law the average waiting time for a call to be dispached is
Wq = Nq
= 3.7458 minutes
λe ﬀ (g) Caller X will have an ambulance dispatched as soon as two ambulances ﬁnish responding
to current calls. Let N be a homogeneous Poisson process with rate 2/15 per minute, then
this probability is equivalent to having at least two service completions in 20 minutes.
P(Caller X has ambulance dispatched in less than 20 minutes)
= P(At least two service completions in 20 minutes)
= P(N (0, 20) ≥ 2) = 1 − P(N (0, 20) = 0) − P(N (0, 20) = 1)
2
2
2
= 1 − e−20· 15 − 20 · e−20· 15
15
= 0.7452 5...
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 Fall '07
 Billings

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