ex9s

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MTH6141 Random Processes, Spring 2012

Solutions to Exercise Sheet 9

1. If d1 ≥ d2 then the interarrivial times are at least as long as the service timesand so the system will never contain more than 1 person (you could workout for which times it has 0 and which times it has 1 if you want). If d1 < d2

then the number of people in the queue will grow to infinity (again you couldwork out the number of people in the queue at any time).

Similarly, if d1 ≥ sd2 the system will never contain more than s people whileif d1 < sd2 the number of people in the queue will grow to infinity.

2. (a) Recall the result from lectures that for M (λ)/M (µ)/1 at equilibrium

we haveE(Q(t)) =

λ

µ − λ.

Here λ = 1 per minute, µ = 3/2 per minute (as the service time isdistributed Exp(3/2)) and so E(Q(t)) = 2.

(b) Halving the expected service time means doubling µ so we will nowhave λ = 1, µ = 3 and E(Q(t)) = 1/2.

(c) Part b) shows that this is not always true since halving the expectedservice time gave a factor 4 reduction in expected queue length. Ingeneral the formula for expected queue length shows that we will always

decrease the expected queue length by a factor of more that 2 (sinceλ

2µ−λ< λ

2(µ−λ)). However, if λ is much smaller than µ then the change

in queue length is close halving.

3. (a) The M (λ)/M (µ)/3 queueing system is a birth-death process with pa-rameters λ0 = λ1 = λ2 = · · · = λ, and µ1 = µ, µ2 = 2µ andµ3 = µ4 = µ5 = 3µ.

(b) We know for a birth-death process with strictly positive birth and deathparameters that

w j = λ0λ1 . . . λ j−1µ1µ2 . . . µ j

w0.

Substituting for λ0, µ1, etc.,

w1 =λ

µw0,

w j =λ2

2µ2

λ

3µ

j−2w0, for all j ≥ 2.

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For this to be an equilibrium distribution we must also have

j≥0 w j =1. That is

w0

1 +

λ

µ +

λ2

2µ2i≥0

λ

3µi

= 1.

This is possible iff λ/3µ < 1, as this is the condition for the infinite sumto converge. Thus, an equilibrium distribution exists iff λ < 3µ.

(c) Specialising to λ = 2 and µ = 1, we obtain

w0

1 + 2 + 2

i≥0

2

3

i= 1,

i.e., w0 = 19

. Also, w1 = 2w0 = 29

, and w j = 2(23

) j−2w0 = 29

(23

) j−2.

(d) i. We wantlimt→∞

P(Q(t) = 0)

which is just w0 = 19

.

ii. Here we want

1 − limt→∞

(P(Q(t) = 0) + P(Q(t) = 1) +P(Q(t) = 2))

which is just 1 − w0 − w1 − w2 = 1 − 19− 2

9− 2

9= 4

9.

iii. Finally, we wantlimt→∞

P(Q(t) = 4)

which is just w4 = 29

(23

)2 = 881

.

4. Let Y 1, Y 2, . . . , Y n denote the service times of the n customers who are in thesystem when the customer in question arrives and Y n+1 denote the servicetime of the arriving customer The arriving customer spends time T = Y 1 +Y 2 + · · ·+ Y n + Y n+1 in the system. We know that the S i are each distributedExp(µ), and so

E(T ) = nE(Y 1) =n + 1

µ.

Let T E the time a new customer spends in the system at equilibrium. Weuse the fact that

E(T E ) =n≥0

E(time in the system|Q(t) = n)P(Q(t) = n)

=n≥0

n + 1

µwn

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where wn is the equilibrium distribution, and we have used the first part of the question for the conditional expectation. Hence

E(T E ) = 1µ

n≥0

(n + 1)(1 − ρ)ρn

=ρ

µ

n≥0

n(1 − ρ)ρn−1 +1

µ

n≥0

(1 − ρ)ρn

=ρ

µ

1

1 − ρ+

1

µ

=1

µ(1 − ρ)

=1

λ − µ.

5. (a) In this case the interarrival times will not be independent. If manypeople have arrived recently then the waiting room is more likely to befull and so the next customer to arrive is more likely to leave in disgust.

(b) Using similar calculations to those in the lectures it can be shown thatthis is a birth-death process with λn = pnλ, µn = µ.

(c) This is the case when pn = 1 for 0 ≤ n ≤ k, pn = 0 for n ≥ k +1. (Thisis assuming the person being served is no longer in the waiting room.)

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