queuing theory in continuous time assignment 4 k due:...

ECEN 750

Each problem is worth 10 points

Assignment 4 Spring 2016

Due: April 18

1. Exercise 8.1 of Communication Networks by Srikant and Ying.






1

Chapter 8

Queuing Theory in Continuous Time

8.1 Problems

Exercise 8.1 Prove Result 7.2.1 (the sum of two independent Poisson process is a Poisson process)and Result 7.2.2 (K random processes generated from a Poisson process are K independent Poissonprocess).

Solution 1. Consider two independent Poisson processes N1(t) (with parameter �1) and N2(t)(with parameter �2), and compute the following probability Pr (N1(t) +N2(t)�N1(s)�N2(s) = K) .

Pr (N1(t) +N2(t)�N1(s)�N2(s) = K)

=K�

n=0

Pr (N1(t)�N1(s) = n) Pr (N2(t)�N2(s) = K � n)

=K�

n=0

e��1(t�s)(�1(t� s))n

n!

e��2(t�s)(�2(t� s))K�n

(K � n)!

=K�

n=0

e�(�1+�2)(t�s)�n1�

K�n2 (t� s)K

n!(K � n)!

= e�(�1+�2)(t�s)(t� s)KK�

n=0

�n1�

K�n2

n!(K � n)!

= e�(�1+�2)(t�s)(t� s)K(�1 + �2)K

K!

=e�(�1+�2)(t�s)((�1 + �2)(t� s))K

K!.

81

82 CHAPTER 8. QUEUING THEORY IN CONTINUOUS TIME

2. Denote the K random processes by Ni for i = 1, · · · ,K, and the Poisson process by N(t).

Pr (N1(t)�N1(s) = n1, N2(t)�N2(s) = n2, · · · , NK(t)�NK(s) = nK)

= Pr

⌥N(t)�N(s) =

K✏

i=1

ni

�⇣

i

⇧ �Kj=i ni

ni

⌃pnii

=e��(t�s)(�(t� s))

�Ki=1 ni

⇤�Ki=1 ni

⌅!

(�K

i=1 ni)!

(�

i ni!)

⇣

i

pnii

=⇣

i

e��pi(t�s)(�pi(t� s))ni

ni!.

�

Exercise 8.2 Consider an M/M/s/k queue with arrival rate � and mean service time 1/µ. Findthe steady-state distribution of this queueing system. Are there conditions on � and µ for thesteady-state distribution to exist?

Solution By local balance equation, we have

⇥i+1 =�

µ(i+ 1)⇥i =

⇤

i+ 1⇥i 0 ⇥ i ⇥ s,

⇥i+1 =�

µs⇥i =

⇤

s⇥i s < i ⇥ k,

which leads to

⇥i =⇤i

i!⇥0 0 ⇥ i ⇥ s,

⇥s+j =⇤⇤s

⌅j⇥s =

⇤⇤s

⌅j ⇤s

s!⇥0 s < s+ j ⇥ k.

To normalize the distribution we have

1 =k✏

i=0

⇥i =

⌥s✏

i=0

⇤i

i!

�⇥0 +

↵k�s✏

j=1

⇤⇤s

⌅j ⇤s

s!

⌦

�⇥0

=

⌥s✏

i=0

⇤i

i!

�⇥0 +

⇤s

s!

⇥s

⇤1��⇥s

⇥k�s⌅

1� ⇥s

⇥0

=

⌥s✏

i=0

⇤i

i!

�⇥0 +

⇤s+1

s!

⇤1��⇥s

⇥k�s⌅

s� ⇤⇥0.

Thus,

⇥0 =1⇤�s

i=0⇥i

i!

⌅+ ⇥s+1

s!(s�⇥)

⇤1��⇥s

⇥k�s⌅

⇥i =⇤i

i!⇥0 1 ⇥ i ⇥ s,

⇥s+j =⇤⇤s

⌅j ⇤s

s!⇥0 s < s+ j ⇥ k.

2.

3.


2. Denote the K random processes by Ni for i = 1, · · · ,K, and the Poisson process by N(t).

Pr (N1(t)�N1(s) = n1, N2(t)�N2(s) = n2, · · · , NK(t)�NK(s) = nK)

= Pr

⌥N(t)�N(s) =

K✏

i=1

ni

�⇣

i

⇧ �Kj=i ni

ni

⌃pnii

=e��(t�s)(�(t� s))

�Ki=1 ni

⇤�Ki=1 ni

⌅!

(�K

i=1 ni)!

(�

i ni!)

⇣

i

pnii

=⇣

i

e��pi(t�s)(�pi(t� s))ni

ni!.

�

Exercise 8.2 Consider an M/M/s/k queue with arrival rate � and mean service time 1/µ. Findthe steady-state distribution of this queueing system. Are there conditions on � and µ for thesteady-state distribution to exist?

Solution By local balance equation, we have

⇥i+1 =�

µ(i+ 1)⇥i =

⇤

i+ 1⇥i 0 ⇥ i ⇥ s,

⇥i+1 =�

µs⇥i =

⇤

s⇥i s < i ⇥ k,

which leads to

⇥i =⇤i

i!⇥0 0 ⇥ i ⇥ s,

⇥s+j =⇤⇤s

⌅j⇥s =

⇤⇤s

⌅j ⇤s

s!⇥0 s < s+ j ⇥ k.

To normalize the distribution we have

1 =k✏

i=0

⇥i =

⌥s✏

i=0

⇤i

i!

�⇥0 +

↵k�s✏

j=1

⇤⇤s

⌅j ⇤s

s!

⌦

�⇥0

=

⌥s✏

i=0

⇤i

i!

�⇥0 +

⇤s

s!

⇥s

⇤1��⇥s

⇥k�s⌅

1� ⇥s

⇥0

=

⌥s✏

i=0

⇤i

i!

�⇥0 +

⇤s+1

s!

⇤1��⇥s

⇥k�s⌅

s� ⇤⇥0.

Thus,

⇥0 =1⇤�s

i=0⇥i

i!

⌅+ ⇥s+1

s!(s�⇥)

⇤1��⇥s

⇥k�s⌅

⇥i =⇤i

i!⇥0 1 ⇥ i ⇥ s,

⇥s+j =⇤⇤s

⌅j ⇤s

s!⇥0 s < s+ j ⇥ k.

8.1. PROBLEMS 83

There is no condition on ⇤ and µ that the distribution always exists. The existence of the stationarydistribution depends on s and k.

�

Exercise 8.3 In this exercise, we will show a central-limit-theorem-like result for the M/M/s/sloss model when the number of servers and the tra⇥c intensity are both large and nearly equal toeach other.

Let B(s, ⌅) denote the Erlang-B formula for the blocking probability in an M/M/s/s system.Let s = ⇧⌅+ �

⌥⌅.⌃ Show that

lim⌅⇥⇤

⌥⌅B(s, ⌅) = ⇧(�)/�(�),

where ⇧(.) and �(.) are the pdf and cdf of the standard normal random variable N(0, 1). Use thefollowing facts to show the above result:

• Central limit theorem for Poisson random variables: Let p(k, ⌅) = e�⌅⌅k/k!. Then,

lim⌅⇥⇤

⌅⌅+�⌃⌅⇧⇤

k=0

p(k, ⌅) = �(�).

• Stirling’s formula: limn⇥⇤n!⌃

2⇤n(ne )n = 1.

• lim⌅⇥⇤ e�⌃⌅(1+�/

⌥⌅)�⌅ = e�

2/2. (Prove this fact. You don’t have to prove the two previousfacts.)

Solution In the following, we prove the equation lim⌅⇥⇤

er⌃⌅(1 + r/

⌥⌅)�⌅ = er

2/2

log⌅er

⌃⌅(1 + r/

⌥⌅)�⌅

⇧= r

⌥⌅� ⌅ log(1 + r/

⌥⌅)

= r⌥⌅� ⌅

�r⌥⌅� r2

2⌅+ o(

1

⌅)

⇥

= r2/2 + o(1) ⇤ r2/2 as ⌅ ⇤ ⌅

Thus, er⌃⌅(1 + r/

⌥⌅)�⌅ ⇤ er

2/2 as ⌅ ⇤ ⌅.

Next, we prove that the equation lim⌅⇥⇤

es�⌅

⌃s/⌅(s/⌅)s

= e�r2/2, where s = ⇧⌅+ r⌥⌅⌃.

Suppose s = ⌅+ r⌥⌅� ⇥, 0 ⇥ ⇥ < 1, notice that

es�⌅

⌃s/⌅(s/⌅)s

=

⌅er

⌃⌅(1 + r/

⌥⌅)�⌅

⇧ � e�⇥(1 + r/⌥⌅)⌅

(1 + r/⌥⌅� ⇥/⌅)⌅

⇥ ⌅(1 + r/

⌥⌅� ⇥/⌅)⇥�1/2

⇧ ⌅(1 + r/

⌥⌅� ⇥/⌅)�r

⌃⌅⇧

4.

8.1. PROBLEMS 85

Q⇥(z) =⇥(0)A⇥(z)(z � 1) + ⇥(0)A(z)

z �A(z)� ⇥(0)A(z)(z � 1)(1�A⇥(z))

(z �A(z))2.

According to L. Hopital’s rule

⇥(0)A⇥(z)(z � 1) + ⇥(0)A(z)

z �A(z)|z=1 =

2⇥(0)⇤

1� ⇤,

and

⇥(0)A(z)(z � 1)(1�A⇥(z))

(z �A(z))2|z=1 =

2⇥(0)⇤(1� ⇤)� ⇥(0)�2E[S2]

2(1� ⇤)2.

Note that ⇥(0) = 1� ⇤, we therefore have L = Q⇥(1) = ⇤+ �2E[S2]2(1�⇥) .

�

Exercise 8.5 Consider an M/GI/1 queue with two classes of customers, with the class i arrivalrate, mean service time, variance of the service time being �i, 1/µi and ⌅2

i , respectively. Let usassume that the service discipline is non-preemptive priority with higher priority to class 1 packets,i.e., when the server becomes free, the first customer in the class 1 queue begins service. However, aclass 2 customer undergoing service cannot be interrupted even if a class 1 customer arrives (hence,the terminology “non-preemptive” priority). Show that the average waiting time in the queue fora class i customer is given by

W iq =

⇤2i=1 �i

�1/µ2

i + ⌅2i

⇥

2(1� ⇤i�1)(1�⇤i

j=1 ⇤j),

where ⇤0 = 0 and ⇤i = �i/µi, i = 1, 2.Hint: Note the a class 2 customer has to wait for the service completion of all class 2 and class

1 customers that arrived prior to it, the service completion of all class 1 customers that arrivedwhile the class 2 customer is waiting in the queue and the remaining service time of the customerin service when the class 2 customer arrived.

Solution First, we note that

W (1)q =

L(1)q

µ1+R

W (2)q =

L(2)q

µ2+

L̃(1)q

µ1+

L̂(1)q

µ1+R.

where R is the expected remaining service time of the packet in the service; L(1)q is the mean number

of customers of class 1 in the queue; L̃(1)q is the average number of class 1 customers seen by a class

2 customer, L̂(1)q is the average number of class 1 customers that arrive while the class 2 customer


is waiting; L(2)q is the average number of class 2 customers in the queue seen by a class 2 arrival.

By the PASTA property and Little’s Law, we have,

L(1)q = �1W

(1)q

L(2)q = �2W

(2)q

L̃(1)q = L(1)

q = �1W(1)q

L̂(1)q = �1W

(2)q

Thus, we get

W (1)q = �1W

(1)q /µ1 +R

W (2)q = ⇥2W

(2)q + ⇥1W

(1)q + ⇥1W

(2)q +R

=⇥1W

(1)q +R

1� ⇥1 � ⇥2

=

R�11��1

+R

1� ⇥1 � ⇥2

Solve the equations above, we get

W (1)q =

R

1� ⇥1

W (2)q =

R

(1� ⇥1)(1� ⇥1 � ⇥2)

Now we o�er the value of R. Let Ni(t) be the number of class i customers that have been servedin [0, t], ski be the service time of the ith class k customer. Then

R = limt⇥⇤

1

t

⇤

⇧N1(t)⌥

i=1

s21i2

+

N2(t)⌥

i=1

s22i2

⌅

⌃

= limt⇥⇤

⇤

⇧N1(t)

t

N1(t)⌥

i=1

s21i2N1(t)

+N2(t)

t

N2(t)⌥

i=1

s21i2N2(t)

⌅

⌃

= �1E[s21]

2+ �2

E[s22]

2

=�1

2

�1

µ21

+ ⇤21

⇥+

�2

2

�1

µ22

+ ⇤22

⇥.

.

�

Exercise 8.6 For a GI/GI/1 queue with � < µ, show that

Wq ⇥�(⇤2

a + ⇤2s)

2(1� ⇥),

5.


2. Now consider a system consisting of two independent M/M/1 queues with mean service timeequal to 1/µ in each queue. Assume that packets arrive to this system according to a Poissonprocess of rate 2�. When a packet arrives it joins queue 1 with probability 1/2 and queue 2with probability 1/2. Compute the expected steady-state waiting time in this system.

Note: The mean waiting time in part (1) is smaller because it has a single queue so that noserver is idle when there is work to be done.

Solution 1. The local balance equation for this system is

µ⇥1 = 2�⇥0

2µ⇥i = 2�⇥i�1 i ⇥ 2.

Let ⇤ = �/µ. We have

⇥1 =2�

µ⇥0 = 2⇤⇥0

⇥i = ⇤i�1⇥1 = 2⇤i⇥0 i ⇥ 2.

Since�

i ⇥=1, we have ⇥0 =1��1+� . Therefore, the expected queue length (the number of packets

in the queue) is

⇥

i

i⇥i+2 =⇥

i

2i⇤i+2⇥0 = 2⇥0⇤2⇥

i

i⇤i = 2⇥0⇤3

(1� ⇤)2,

and the expected waiting time is

Wq =Lq

2�=

⇤3

�(1� ⇤).

2. Because the arrival rate of the system is 2� and probability packets enter each queue is 1/2,arrival rate to each queue is �, and the waiting time of each queue is

Wq =⇤3

�(1� ⇤),

which is larger than the waiting time of the system considered in question (1).

�

Exercise 8.8 Consider the M/G/s/s loss model, where the service-time distribution is Erlang withK stages, each with mean 1/(Kµ). Show that the blocking probability depends only on the meanof the service-time distribution 1/µ and not on the number of stages.

Hint: Proceed as in the insensitivity proof for M/GI/1-PS queue.Note: this result can be extended to mixtures of Erlang distributions and more generally, to any

service-time distribution, to prove the insensitivity of the model to the distribution of service timebeyond the mean.

6.

8.1. PROBLEMS 91

3.

�n =⌅

{ni}:�

i ni=n

�(n1, · · · , nK)

=e�⇥⇥n

n!

�K⌅

i=1

piµi

⇥n

=e�⇥⇥n

n!.

�

Exercise 8.10 Let X be a CTMC over a state-space S and suppose that X is time-reversible insteady-state. Now consider a CTMC Y which is a restriction of X to the state space A, whereA ⇥ S. By restriction, we mean that Y takes values only in A, has the same transition rates as Xbut transitions out of A are not allowed. Show that Y is also time-reversible in steady-state andits steady-state distribution is the same as the steady-state distribution of X restricted to A butrescaled to add up to 1. This result is called the truncation theorem.

Solution Note that �xqxy = qyx�y for any y ⇤= x. Now for any x ⇥ A, we have �xqxyZ = qyx�y

Z forany y ⇤= x and y, x ⇥ A, where Z =

⇤x⇥A �x so that

⇤x⇥A �x/Z = 1.

�

Exercise 8.11 Consider the following variation of the M/M/s/s loss model. Consider a singlelink of capacity 50 with two call classes. Each call from each class requests one unit of capacity onarrival. Assume both call classes have holding times that are exponentially distributed with unitmean. The call arrival processes are independent Poisson processes, with a mean arrival rate of40 for Class 1 calls and a mean arrival rate of 20 for Class 2 calls. Class 2 calls are blocked fromentering (and lost) if the available capacity is less than or equal to 5. Class 1 calls are blocked andlost only if there is no available capacity. Thus, this is a loss model with some priority given toClass 1 calls. This is an example of the trunk reservation introduced in Section ??.

1. Compute the blocking probabilities of Class 1 and Class 2 calls in this system.

Hint: Consider the Markov chain describing the total number of calls in the system.

2. What are the expected numbers of Class 1 and Class 2 calls in the system in steady-state?

Hint: Use Little’s law.

Solution 1. The Markov chain is shown in Figure 8.1. We first use the following recursion tocompute �i :

�i+1 =60�ii+ 1

, i � 44

�i+1 =40�ii+ 1

, 45 � i � 49.


� � � ��

��

Figure 8.1: The Markov chain for the system

We then obtain

PB1 = ⇥50 = 0.0697

PB2 = ⇥45 + ⇥46 + · · ·+ ⇥50 = 0.6576.

2. Using the PASTA property and Little’s law, we obtain

E[n1] = (1� ⇥50)40 = 37.212

E[n2] = (1� PB2)20 = 6.848.

�

Exercise 8.12 Consider calls accessing a link with time-varying capacity. The call holding timesare independent and exponentially distributed with mean 1/µ and each call requires 1 unit ofbandwidth from the link. The link has an available capacity of nc0 in the time interval [0, ⇤1], anavailable capacity of nc1 in [⇤1, ⇤2] and an available capacity of nc2 in [⇤2,⇤), where ⇤2 > ⇤1 > 0.We assume that nc0 > nc1 > nc2, and n > 0. Suppose that there are n� calls in progress at timet = 0 (assume � ⌅ (0, c0)) is an integer) and no further calls ever arrive at the system. Let N(t)denote the number of calls in progress at time t. Thus, the bandwidth requested by the calls attime t is N(t). We are interested in estimating the probability that the required bandwidth everexceeds the available capacity, up to a logarithmic equivalence. To this end, prove the followingfact for appropriate functions Ii(x) :

limn⇥⇤

1

nlog Pr (there exists a t ⌅ [0,⇤) such that N(t) ⇥ c(t)) = �min

1,2Ii(ci),

where c(t) is the available capacity at time t. Explicitly compute the functions Ii(x).

Solution Define

Xi = I{Call i is on at time t1}Yi = I{Call i is on at time t2}.

So

Xi =

�0, w. p. 1� e�µt1

1, w. p. e�µt1 ,

and

Yi =

�0, w. p. 1� e�µt2

1, w. p. e�µt2 .

8.1. PROBLEMS 93

Let q be the probability that the required bandwidth exceeds the available capacity, then

q = Pr

nX

i=1

Xi � nc1

,nX

i=1

Yi � nc2

!

.

So

max{Pr

nX

i=1

Xi � nc1

!

,Pr

nX

i=1

Yi � nc2

!

} q Pr

nX

i=1

Xi � nc1

!

+ Pr

nX

i=1

Yi � nc2

!

.

By the Cherno↵ bound and taking log dividing by n as n ! 1, both LHS and RHS is�mini=1,2 Ii(ci) <where I

1

(x) is the rate function of X and I2

(x) is the rate function of Y.

⇤

Exercise 8.13 Consider S ON-OFF sources which switch between an active (ON) state and aninactive (OFF) state. Let us suppose that the sources access a bottleneck node with transmissionrate C bits/sec., and that each source accesses this node via an access link whose transmission rateis 1 bit/sec. In other words, when the number of ON sources in the system is less than or equal to C,each source is served at rate 1. Assume C < S. When the number of ON sources exceeds C, there issome form of resource allocation mechanism that divides the available capacity C to all the sourcesequally. In other words, if there are N(t) active sources in the system at time t, each source is thatis ON is served at rate C/N(t) bits/sec. We assume that the OFF times are exponential with mean1/�. Once a source enters the ON state, it remains in this state till it receives an exponentiallydistributed amount of service whose mean is 1/µ bits.

1. Let N(t) be the number of active sources in the system at time t. Draw the state transitiondiagram of the Markov chain that describes the evolution of N(t).

2. Let ⇡i denote the steady-state probability that there are i active sources in the system. Writedown the equations needed to compute {⇡i}.

3. Derive an expression for the average amount of time spent in the ON state by a source, as afunction of {⇡i}.

Solution 1. The Markov chain is shown in Figure 8.2.

� � �

Figure 8.2: The Markov chain for the system

2.

⇡i+1

=(S � i)⇡i(i+ 1)µ

0 i C � 1

⇡i+1

=(S � i)⇡i

cµc i S � 1.


3. L =PS

i=1

i⇡i. The average arrival rate � is equal to the average departure rate, which is

µCX

i=0

i⇡i + cµSX

i=C+1

⇡,

and the average time spent in ON state is L/�.

⇤

Exercise 8.14 CSMA in continuous-time: Consider an ad hoc network whose scheduling algorithmoperates in continuous-time. Assume that if a link l interferes with the transmission of link j, thenj also interferes with l. Associate with each link l in the network a weight wl. Let S1

, S2

, · · · SK

be the independent sets in the interference graph of the network, i.e., each Sk represents a set oflinks that can be scheduled simultaneously without interference. We include the empty set in theset of independent sets. Consider the following MAC (medium access control algorithm): each linkcan be either active or inactive at each time instant. At time 0, assume that the set of all activelinks is an independent set. Further, at time 0, each link starts an exponentially distributed timerwith mean 1. When a link’s timer expires (call this link l), it checks to see if any other link in itsneighborhood (where the neighborhood is the set of other links with which it interferes) is active.If there is an active link in its neighborhood, then link l maintains its current state (which mustbe inactive), starts another timer and repeats the process. If no link in its neighborhood is active,then the link chooses be active with probability ewl/(1 + ewl), and chooses to be inactive withprobability 1/(1+ ewl). Then, it starts a new timer and repeats the process. Note that the state ofa link does not change between two of its consecutive timer expiration moments.

1. Let X(t) be the set of links that are active at time t. Note that the set of active links mustbe an independent set in the interference graph, and also note that X(t) is a CTMC. Findthe rate transition matrix of X(t).

2. Show that the steady-state distribution of X(t) is given by

⇡Sk=

Q

l2Skewl

Z=

eP

l2Skwl

Z,

where Z is a normalization constant and ⇡; = 1/Z.

Solution 1. Assume the state X(t) is the set of all “ON” links, and take values S1

, S2

, .., SK .In a su�ciently small interval, only one link changes it states. Assume link l is any linkwithout active neighbors. From Si to Sj ,there are two cases.

• One link l turns on. Sj = Si [ {l}, qij = ewl

1+ewl .

• One link l turns o↵. Sj = Si � {l}, qij = 1

1+ewl .

The rate transition matrix Q of X(t) is a matrix with entries qij . When Si and Sj is one ofthe two cases above, qij is defined accordingly. Otherwise, if j 6= i, qij = 0, while if i = j,qii = �P

j 6=i qij .

queuing theory in continuous time assignment 4 k due:...

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