Papoulis%20 solutions

1. Solutions Manual
to accompany
Probability,
Random Variables
and
Stochastic Processes
Fourth Edition
Athanasios Papoulis
Polytechnic University
S. Unnikrishna Pillai
Polytechnic University

2. Solutions Manual to accompany
PROBABILITY, RANDOM VARIABLES AND STOCHASTIC PROCESSES, FOURTH EDITION
ATHANASIOS PAPOULIS
Published by McGraw-Hill Higher Education, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas,
New York, NY 10020. Copyright © 2002 by The McGraw-Hill Companies, Inc. All rights reserved.
The contents, or parts thereof, may be reproduced in print form solely for classroom use with PROBABILITY, RANDOM
VARIABLES AND STOCHASTIC PROCESSES, FOURTH EDITION, provided such reproductions bear copyright notice, but may
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www.mhhe.com

11. 1
Problem Solutions for Chapter 3
3.1 (a) P(A occurs atleast twice in n trials)
= 1 ; P(A never occurs in n trials); P(A occurs once in n trials)
= 1 ; (1 ; p)n ; np(1 ; p)n;1
(b) P(A occurs atleast thrice in n trials)
= 1 ; P(A never occurs in n trials) ; P(A occurs once in n trials)
;P(A occurs twice in n trials)
= 1 ; (1 ; p)n ; np(1 ; p)n;1 ;
n(n; 1)
2 p2(1 ; p)n;2
3.2
P(doublesix) = 1
6
j
1
6
= 1
36
P(double six atleast three times in n trials00)
= 1 ;
50
0
1
36
0 35
36
50
;
50
1
1
36
35
36
49
;
50
2
1
36
2 35
36
48
= 0:162
3.6 (a)
p1 = 1 ;
5
6
6
= 0:665
(b)
1 ;
5
6
12
;
12
1
!
1
6
5
6
11
= 0:619
(c)
1 ;
5
6
18
;
18
1
!
1
6
5
6
17
;
18
2
!
1
6
2 5
6
16
= 0:597

12. 2
3.7 (a) Let n represent the number of wins required in 50 games so that
the net gain or loss does not exceed $1. This gives the net gain to be
;1 < n;
50 ; n
4 < 1
16 < n < 17:3
n = 17
P(net gain does not exceed $1) = 50
17
1
4
17 3
4
33
= 0:432
P(net gain or loss exceeds $1) = 1 ; 0:432 = 0:568
(b) Let n represent the number of wins required so that the net gain
or loss does not exceed $5. This gives
;5 < n;
(50 ; n)
2 < 5
13:3 < n < 20
P(net gain does not exceed $5) =
X19
n = 14
50
n 1
4
n 3
4
50;n
= 0:349
P(net gain or loss exceeds $5) = 1 ; 0:349 = 0:651

13. 3
3.8 De ne the events
A= r successes in n Bernoulli trials"
B=success at the ith Bernoulli trial"
C=r; 1 successes in the remaining n; 1 Bernoulli trials excluding
the ith trial"
P(A) = n
r pr qn;r
P(B) = p
P(C) = n; 1
r ; 1 pr;1 qn;r
We need
P(BjA) = P(AB)
P(A) = P(BC)
P(A) = P(B)P(C)
P(A) = r
n:
3.9 There are 52
13 ways of selecting 13 cards out of 52 cards. The
number of ways to select 13 cards of any suit (out of 13 cards) equals
13
13 = 1. Four such (mutually exclusive) suits give the total number
of favorable outcomes to be 4. Thus the desired probability is given by
4
52
13
! = 6:3 j 10;12

14. 4
3.10 Using the hint, we obtain
p(Nk+1 ; Nk) = q(Nk ; Nk;1) ; 1
Let
Mk+1 = Nk+1 ; Nk
so that the above iteration gives
Mk+1 = q
p Mk ;
1
p
=
8
>>><
>>>:
q
p
k
M1 ;
1
p ; q
n
1 ; (q
p)k
o
p 6= q
M1 ;
kp p = q
This gives
Ni =
i;1X
k=0
Mk+1
=
8
>>>>><
>>>>>:
M1 + 1
p; q
i;1X
k=0
q
p
k
;
i
p ; q p 6= q
iM1 ;
i(i; 1)
2p p = q
where we have used No = 0. Similarly Na+b = 0 gives
M1 + 1
p; q = a+ b
p ; q
1 ; q=p
1 ; (q=p)a+b :
Thus
Ni =
8
>>><
>>>:
a+ b
p ; q
1 ; (q=p)i
1 ; (q=p)a+b ;
i
p; q p 6= q
i(a+ b; i) p = q

15. 5
which gives for i = a
Na =
8
>>><
>>>:
a+ b
p; q
1 ; (q=p)a
1 ; (q=p)a+b ;
a
p; q p 6= q
ab p = q
=
8
>>><
>>>:
b
2p; 1 ;
a+ b
2p; 1
1 ; (p=q)b
1 ; (p=q)a+b p 6= q:
ab p = q

16. 6
3.11
Pn = pPn+ + qPn;a
Arguing as in (3.43), we get the corresponding iteration equation
Pn = Pn+ + qPn;a
and proceed as in Example 3.15.
3.12 Suppose one best on k = 1 2 6.
Then
p1 = P(k appears on one dice) = 3
1
1
6
5
6
2
p2 = P(k appear on two dice) = 3
2
1
6
2 5
6
p3 = P(k appear on all the tree dice) = 1
6
3
p0 = P(k appear none) = 5
6
3
Thus, we get
Net gain = 2p1 + 3p2 + 4p3 ; p0 = 0:343:

217. 215
Chapter 15
15.1 The chain represented by
P =







0 1/2 1/2
1/2 0 1/2
1/2 1/2 0







is irreducible and aperiodic.
The second chain is also irreducible and aperiodic.
The third chain has two aperiodic closed sets {e1, e2} and {e3, e4}
and a transient state e5.
15.2 Note that both the row sums and column sums are unity in this
case. Hence P represents a doubly stochastic matrix here, and
Pn
=
1
m + 1











1 1 · · · 1 1
1 1 · · · 1 1
...
...
...
...
...
1 1 · · · 1 1











limn→∞
P{xn = ek} =
1
m + 1
, k = 0, 1, 2, · · · m.
15.3 This is the “success runs” problem discussed in Example 15-11
and 15-23. From Example 15-23, we get
ui+1 = pi,i+1ui =
1
i + 1
ui =
uo
(i + 1)!
so that from (15-206)
∞
k=1
uk = u0
∞
k=1
1
k!
= e · u0 = 1

218. 216
gives u0 = 1/e and the steady state probabilities are given by
uk =
1/e
k!
, k = 1, 2, · · ·
15.4 If the zeroth generation has size m, then the overall process may
be considered as the sum of m independent and identically distributed
branching processes x(k)
n , k = 1, 2, · · · m, each corresponding to unity
size at the zeroth generation. Hence if π0 represents the probability of
extinction for any one of these individual processes, then the overall
probability of extinction is given by
limn→∞
P[xn = 0|x0 = m] =
= P[{x(1)
n = 0|x
(1)
0 = 1} {x(2)
n = 0|x
(2)
0 = 1} · · · {x(m)
n = 0|x
(m)
0 = 1}]
= m
k=1 P[x(k)
n = 0|x
(k)
0 = 1]
= πm
0
15.5 From (15-288)-(15-289),
P(z) = p0 + p1z + p2z2
, since pk = 0, k ≥ 3.
Also p0 + p1 + p2 = 1, and from (15-307) the extinction probability is
given by sloving the equation
P(z) = z.
Notice that
P(z) − z = p0 − (1 − p1)z + p2z2
= p0 − (p0 + p2)z + p2z2
= (z − 1)(p2z − p0)
and hence the two roots of the equation P(z) = z are given by
z1 = 1, z2 =
p0
p2
.
Thus if p2 < p0, then z2 > 1 and hence the smallest positive root of
P(z) = z is 1, and it represents the probability of extinction. It follows

219. 217
that such a tribe which does not produce offspring in abundence is
bound to extinct.
15.6 Define the branching process {xn}
xn+1 =
xn
k=1
yk
where yk are i.i.d random variables with common moment generating
function P(z) so that (see (15-287)-(15-289))
P (1) = E{yk} = µ.
Thus
E{xn+1|xn} = E{ xn
k=1 yk|xn = m}
= E{ m
k=1 yk|xn = m}
= E{ m
k=1 yk} = mE{yk} = xn µ
Similarly
E{xn+2|xn} = E{E{xn+2|xn+1, xn}}
= E{E{xn+2|xn+1}|xn}
= E{µxn+1|xn} = µ2
xn
and in general we obtain
E{xn+r|xn} = µr
xn. (i)
Also from (15-310)-(15-311)
E{xn} = µn
. (ii)
Define
wn =
xn
µn
. (iii)
This gives
E{wn} = 1.
Dividing both sider of (i) with µn+r
we get
E{
xn+r
µn+r
|xn = x} = µr
·
xn
µn+r
=
xn
µn
= wn

220. 218
or
E{wn+r|wn =
x
µn
∆
= w} = wn
which gives
E{wn+r|wn} = wn,
the desired result.
15.7
sn = x1 + x2 + · · · + xn
where xn are i.i.d. random variables. We have
sn+1 = sn + xn+1
so that
E{sn+1|sn} = E{sn + xn+1|sn} = sn + E{xn+1} = sn.
Hence {sn} represents a Martingale.
15.8 (a) From Bayes’ theorem
P{xn = j|xn+1 = i} =
P{xn+1 = i|xn = j} P{xn = j}
P{xn+1 = i}
=
qj pji
qi
= p∗
ij,
(i)
where we have assumed the chain to be in steady state.
(b) Notice that time-reversibility is equivalent to
p∗
ij = pij
and using (i) this gives
p∗
ij =
qj pji
qi
= pij (ii)
or, for a time-reversible chain we get
qj pji = qi pij. (iii)

221. 219
Thus using (ii) we obtain by direct substitution
pij pjk pki =
qj
qi
pji
qk
qj
pkj
qi
qk
pik
= pik pkj pji,
the desired result.
15.9 (a) It is given that A = AT
, (aij = aji) and aij > 0. Define the ith
row sum
ri =
k
aik > 0, i = 1, 2, · · ·
and let
pij =
aij
k aik
=
aij
ri
.
Then
pji =
aji
m
ajm
=
aji
rj
=
aij
rj
= ri
rj
aij
ri
= ri
rj
pij
(i)
or
ri pij = rj pji.
Hence
i
ri pij =
i
rj pji = rj
i
pji = rj, (ii)
since
i
pji = i aji
rj
=
rj
rj
= 1.
Notice that (ii) satisfies the steady state probability distribution equa-
tion (15-167) with
qi = c ri, i = 1, 2, · · ·
where c is given by
c
i
ri =
i
qi = 1 =⇒ c =
1
i ri
=
1
i j aij
.

222. 220
Thus
qi =
ri
i ri
= j aij
i j aij
> 0 (iii)
represents the stationary probability distribution of the chain.
With (iii) in (i) we get
pji =
qi
qj
pij
or
pij =
qj pji
qi
= p∗
ij
and hence the chain is time-reversible.
15.10 (a) M = (mij) is given by
M = (I − W)−1
or
(I − W)M = I
M = I + WM
which gives
mij = δij + k wik mkj, ei, ej ∈ T
= δij + k pik mkj, ei, ej ∈ T
(b) The general case is solved in pages 743-744. From page 744,
with N = 6 (2 absorbing states; 5 transcient states), and with r = p/q
we obtain
mij =



(rj
− 1)(r6−i
− 1)
(p − q)(r6
− 1)
, j ≤ i
(ri
− 1)(r6−i
− rj−i
)
(p − q)(r6
− 1)
, j ≥ i.
15.11 If a stochastic matrix A = (aij), aij > 0 corresponds to the two-
step transition matrix of a Markov chain, then there must exist another
stochastic matrix P such that
A = P2
, P = (pij)

223. 221
where
pij > 0,
j
pij = 1,
and this may not be always possible. For example in a two state chain,
let
P =



α 1 − α
1 − β β



so that
A = P2
=



α2
+ (1 − α)(1 − β) (α + β)(1 − α)
(α + β)(1 − β) β2
+ (1 − α)(1 − β)


 .
This gives the sum of this its diagonal entries to be
a11 + a22 = α2
+ 2(1 − α)(1 − β) + β2
= (α + β)2
− 2(α + β) + 2
= 1 + (α + β − 1)2
≥ 1.
(i)
Hence condition (i) necessary. Since 0 < α < 1, 0 < β < 1, we also
get 1 < a11 + a22 ≤ 2. Futher, the condition (i) is also sufficient in the
2 × 2 case, since a11 + a22 > 1, gives
(α + β − 1)2
= a11 + a22 − 1 > 0
and hence
α + β = 1 ±
√
a11 + a22 − 1
and this equation may be solved for all admissible set of values 0 <
α < 1 and 0 < β < 1.
15.12 In this case the chain is irreducible and aperiodic and there are
no absorption states. The steady state distribution {uk} satisfies (15-
167),and hence we get
uk =
j
uj pjk =
N
j=0
uj
N
k
pk
j qN−k
j .

224. 222
Then if α > 0 and β > 0 then “fixation to pure genes” does not occur.
15.13 The transition probabilities in all these cases are given by (page
765) (15A-7) for specific values of A(z) = B(z) as shown in Exam-
ples 15A-1, 15A-2 and 15A-3. The eigenvalues in general satisfy the
equation
j
pij x
(k)
j = λk x
(k)
i , k = 0, 1, 2, · · · N
and trivially j pij = 1 for all i implies λ0 = 1 is an eigenvalue in all
cases.
However to determine the remaining eigenvalues we can exploit the
relation in (15A-7). From there the corresponding conditional moment
generating function in (15-291) is given by
G(s) =
N
j=0
pij sj
(i)
where from (15A-7)
pij =
{Ai
(z)}j {BN−i
(z)}N−j
{Ai
(z) BN−i
(z)}N
=
coefficient of sj
zN
in {Ai
(sz) BN−i
(z)}
{Ai
(z) BN−i
(z)}N
(ii)
Substituting (ii) in (i) we get the compact expression
G(s) =
{Ai
(sz) BN−i
(z)}N
{Ai
(z) BN−i
(z)}N
. (iii)
Differentiating G(s) with respect to s we obtain
G (s) =
N
j=0
Pij j sj−1
=
{iAi−1
(sz) A (sz)z BN−i
(z)}N
{Ai
(z) BN−i
(z)}N
= i ·
{Ai−1
(sz) A (sz) BN−i
(z)}N−1
{Ai
(z) BN−i
(z)}N
.
(iv)

225. 223
Letting s = 1 in the above expression we get
G (1) =
N
j=0
pij j = i
{Ai−1
(z) A (z) BN−i
(z)}N−1
{Ai(z) BN−i(z)}N
. (v)
In the special case when A(z) = B(z), Eq.(v) reduces to
N
j=0
pij j = λ1 i (vi)
where
λ1 =
{AN−1
(z) A (z)}N−1
{AN (z)}N
. (vii)
Notice that (vi) can be written as
Px1 = λ1 x1, x1 = [0, 1, 2, · · · N]T
and by direct computation with A(z) = B(z) = (q + pz)2
(Example
15A-1) we obtain
λ1 =
{(q + pz)2(N−1)
2p(q + pz)}N
{(q + pz)2N
}N
=
2p{(q + pz)2N−1
}N−1
{(q + pz)2N
}N
=
2p
2N
N − 1
qN
pN−1
2N
N
qN
pN
= 1.
Thus N
j=0 pij j = i and from (15-224) these chains represent Martin-
gales. (Similarly for Examples 15A-2 and 15A-3 as well).
To determine the remaining eigenvalues we differentiate G (s) once
more. This gives
G (s) =
N
j=0
pij j(j − 1) sj−2
=
{i(i − 1)Ai−2
(sz)[A (sz)]2
z BN−i
(z) + iAi−1
(sz) A (sz)z BN−i
(z)}N−1
{Ai
(z) BN−i
(z)}N
=
{i Ai−2
(sz) BN−i
(z)[(i − 1) (A (sz))
2
+ A(sz) A (sz)]}N−2
{Ai
(z) BN−i
(z)}N
.
.

226. 224
With s = 1, and A(z) = B(z), the above expression simplifies to
N
j=0
pij j(j − 1) = λ2 i(i − 1) + iµ2 (viii)
where
λ2 =
{AN−2
(z) [A (z)]2
}N−2
{AN (z)}N
and
µ2 =
{AN−1
(z) A (z)}N−2
{AN (z)}N
.
Eq. (viii) can be rewritten as
N
j=0
pij j2
= λ2 i2
+ (polynomial in i of degree ≤ 1)
and in general repeating this procedure it follows that (show this)
N
j=0
pij jk
= λk ik
+ (polynomial in i of degree ≤ k − 1) (ix)
where
λk =
{AN−k
(z) [A (z)]k
}N−k
{AN (z)}N
, k = 1, 2, · · · N. (x)
Equations (viii)–(x) motivate to consider the identities
P qk = λk qk (xi)
where qk are polynomials in i of degree ≤ k, and by proper choice of
constants they can be chosen in that form. It follows that λk, k =
1, 2, · · · N given by (ix) represent the desired eigenvalues.
(a) The transition probabilities in this case follow from Example 15A-1
(page 765-766) with A(z) = B(z) = (q + pz)2
. Thus using (ix) we

227. 225
obtain the desired eigenvalues to be
λk =
{(q + pz)2(N−k)
[2p(q + pz)]k
}N−k
{(q + pz)2N
}N
= 2k
pk {(q + pz)2N−k
}N−k
{(q + pz)2N
}N }
= 2k
2N − k
N − k
2N
N
, k = 1, 2, · · · N.
(b) The transition probabilities in this case follows from Example 15A-2
(page 766) with
A(z) = B(z) = eλ(z−1)
and hence
λk =
{eλ(N−k)(z−1)
λk
eλk(z−1)
}N−k
{eλN(z−1)
}N
=
λk
{eλNz
}N−k
{eλNz
}N
=
λk
(λN)N−k
/(N − k)!
(λN)N
/N!
= N!
(N − k)! Nk = 1 − 1
N 1 − 2
N · · · 1 − k − 1
N , k = 1, 2, · · · N
(c) The transition probabilities in this case follow from Example 15A-3
(page 766-767) with
A(z) = B(z) =
q
1 − pz
.
Thus
λk = pk {1/(1 − pz)N+k
}N−k
{1/(1 − pz)N
}N
= (−1)k
−(N + k)
N − k
−N
N
=
2N − 1
N − k
2N − 1
N
, r = 2, 3, · · · N

228. 226
15.14 From (15-240), the mean time to absorption vector is given by
m = (I − W)−1
E, E = [1, 1, · · · 1]T
,
where
Wik = pjk, j, k = 1, 2, · · · N − 1,
with pjk as given in (15-30) and (15-31) respectively.
15.15 The mean time to absorption satisfies (15-240). From there
mi = 1 +
k∈T
pik mk = 1 + pi,i+1 mi+1 + pi,i−1 mi−1
= 1 + p mi+1 + q mi−1,
or
mk = 1 + p mk+1 + q mk−1.
This gives
p (mk+1 − mk) = q (mk − mk−1) − 1
Let
Mk+1 = mk+1 − mk
so that the above iteration gives
Mk+1 = q
p Mk − 1
p
= q
p
k
M1 − 1
p 1 + q
p + q
p
2
+ · · · + q
p
k−1
=



q
p
k
M1 − 1
p − q 1 − (q
p)k
, p = q
M1 − k
p, p = q

229. 227
This gives
mi =
i−1
k=0
Mk+1
=



M1 + 1
p − q
i−1
k=0
q
p
k
− i
p − q, p = q
iM1 −
i(i − 1)
2p , p = q
=



M1 + 1
p − q
1 − (q/p)i
1 − q/p
− i
p − q, p = q
iM1 −
i(i − 1)
2p , p = q
where we have used mo = 0. Similarly ma+b = 0 gives
M1 +
1
p − q
=
a + b
p − q
·
1 − q/p
1 − (q/p)a+b
.
Thus
mi =



a + b
p − q ·
1 − (q/p)i
1 − (q/p)a+b − i
p − q, p = q
i(a + b − i), p = q
which gives for i = a
ma =



a + b
p − q ·
1 − (q/p)a
1 − (q/p)a+b − a
p − q, p = q
ab, p = q
=



b
2p − 1 − a + b
2p − 1 ·
1 − (p/q)b
1 − (p/q)a+b , p = q
ab, p = q
by writing
1 − (q/p)a
1 − (q/p)a+b = 1 −
(q/p)a
− (q/p)a+b
1 − (q/p)a+b = 1 −
1 − (p/q)b
1 − (p/q)a+b
(see also problem 3-10).

230. 228
Chapter 16
16.1 Use (16-132) with r = 1. This gives
pn =



ρn
n! p0, n ≤ 1
ρn
p0, 1 < n ≤ m
= ρn
p0, 0 ≤ n ≤ m
Thus
m
n=0
pn = p0
m
n=0
ρn
= p0
(1 − ρm+1
)
1 − ρ
= 1
=⇒ p0 =
1 − ρ
1 − ρm+1
and hence
pn =
1 − ρ
1 − ρn+1
ρn
, 0 ≤ n ≤ m, ρ = 1
and lim ρ → 1, we get
pn =
1
m + 1
, ρ = 1.
16.2 (a) Let n1(t) = X + Y , where X and Y represent the two queues.
Then
pn = P{n1(t) = n} = P{X + Y = n}
=
n
k=0
P{X = k} P{Y = n − k}
=
n
k=0
(1 − ρ)ρk
(1 − ρ)ρn−k
= (n + 1)(1 − ρ)2
ρn
, n = 0, 1, 2, · · ·
(i)
where ρ = λ/µ.

231. 229
(b) When the two queues are merged, the new input rate λ =
λ + λ = 2λ. Thus from (16-102)
pn =



(λ /µ)n
n! p0 =
(2ρ)n
n! p0, n < 2
22
2! ( λ
2µ)n
p0 = 2ρn
p0, n ≥ 2.
Hence
∞
k−0
pk = p0(1 + 2ρ + 2
∞
k=2
ρk
)
= p0(1 + 2ρ + 2ρ2
1 − ρ)
= p0
1 − ρ((1 + 2ρ) (1 − ρ) + 2ρ2
)
= p0
1 − ρ (1 + ρ) = 1
=⇒ p0 =
1 − ρ
1 + ρ
, (ρ = λ/µ). (ii)
Thus
pn =



2 (1 − ρ) ρn
/(1 + ρ), n ≤ 1
(1 − ρ)/(1 + ρ), n = 0
(iii)
(c) For an M/M/1 queue the average number of items waiting is
given by (use (16-106) with r = 1)
E{X} = L1 =
∞
n=2
(n − 1) pn

232. 230
where pn is an in (16-88). Thus
L1 =
∞
n=2
(n − 1)(1 − ρ) ρn
= (1 − ρ) ρ2
∞
n=2
(n − 1) ρn−2
= (1 − ρ) ρ2
∞
k=1
k ρk−1
= (1 − ρ) ρ2 1
(1 − ρ)2
=
ρ2
(1 − ρ)
.
(iv)
Since n1(t) = X + Y we have
L1 = E{n1(t)} = E{X} + E{Y }
= 2L1 =
2ρ2
1 − ρ
(v)
For L2 we can use (16-106)-(16-107) with r = 2. Using (iii), this
gives
L2 = pr
ρ
(1 − ρ)2
= 2
(1 − ρ) ρ2
1 + ρ
ρ
(1 − ρ)2
=
2 ρ3
1 − ρ2
=
2 ρ2
1 − ρ
ρ
1 + ρ
< L1
(vi)
From (vi), a single queue configuration is more efficient then two
separate queues.
16.3 The only non-zero probabilities of this process are
λ0,0 = −λ0 = −mλ, λ0,1 = µ
λi,i+1 = (m − i) λ, λi,i−1 = iµ

233. 231
λi,i = [(m − i) λ + iµ], i = 1, 2, · · · , m − 1
λm,m = −λm,m−1 = −mµ.
Substituting these into (16-63) text, we get
m λ p0 = µ p1 (i)
[(m−i)λ+iµ] pi = (m−i+1) pi−1 +(i+1) µ pi+1, i = 1, 2, · · · , m−1
(ii)
and
m µ pm = λ pm−1. (iii)
Solving (i)-(iii) we get
pi =
m
i
λ
λ + µ
i
µ
λ + µ
m−i
, i = 0, 1, 2, · · · , m
16.4 (a) In this case
pn =



λ
µ1
λ
µ1
· · ·
λ
µ1
=
λ
µ1
n
p0, n < m
λ
µ1
λ
µ1
· · ·
λ
µ1
λ
µ2
· · ·
λ
µ2
p0, n ≥ m
=



ρn
1 p0, n < m
ρm−1
1 ρn−m+1
2 p0, n ≥ m,
where
∞
n=0
pn = p0
m−1
k=0
ρk
1 + ρm−1
1 ρ2
∞
n=0
ρn
2
= p0
1 − ρm
1
1 − ρ1
+
ρ2ρm−1
1
1 − ρ2
= 1

234. 232
gives
p0 =
1 − ρm
1
1 − ρ1
+
ρ2ρm−1
1
1 − ρ2
−1
.
(b)
L =
∞
n=0
n pn
= p0
m−1
n=0
n ρn
1 +
∞
n=m
n ρm−1
1 ρn−m+1
2
= p0

ρ1
m−1
n=0
n ρn−1
1 + ρ1
ρ1
ρ2
m−2 ∞
n=m
n ρn−1
2


= p0

ρ1
d
dρ1
m−1
n=0
ρn
1 + ρ1
ρ1
ρ2
m−2
d
dρ2
∞
n=m
ρn
2


= p0

ρ1
d
dρ1
1 − ρm
1
1 − ρ
+ ρ1
ρ1
ρ2
m−2
d
dρ
ρm
1 − ρ


= p0
ρ1[1 + (m − 1)ρm
1 − mρm−1
1 ]
(1 − ρ1)2
+
ρ2 ρm−1
1 + [m − (m − 1)ρ2]
(1 − ρ2)2
.
16.5 In this case
λi =



λ, j < r
pλ, j ≥ r
µi =



jµ, j < r
rµ, j ≥ r.
Using (16-73)-(16-74), this gives
pn =



(λ/µ)n
n! p0, n < r
(λ/µ)r
r! (pλ/rµ)n−r
, n ≥ r.

235. 233
16.6
P{w > t} =
m−1
n=r
pn P(w > t|n)
=
m−1
n=r
pn (1 − Fw(t|n)) = pr
λ
rµ
n−r
(1 − Fw(t|n))
fw(t|n) = e−γµt (γµ)n−r+1
tn−r
(n − r)!)
(see 16.116)
and
Fw(t|n) = 1 −
n−r
k=0
(γµt)k
k!
e−γµt
(see 4.)
so that
1 − Fw(t|n) =
n−r
k=0
(γµt)k
k!
e−γµt
P{w > t} =
m−1
n=r
pr
λ
γµ
n−r n−r
k=0
(γµt)k
k! e−γµt
=
m−r−1
i=0
pr ρi
i
k=0
(γµt)k
k! e−γµt
, n − r = i
= pr e−γµt
m−r−1
k=0
ρk
k
i=0
(γµt)i
i!
=
m−r−1
k=0
k
i=0
=
m−r−1
i=0
m−r−1
k=i
P{w > t} = pr e−γµt
m−r−1
i=0
(γµt)i
i!
m−r−1
k=i
ρk
= pr
1 − ρ e−γµt
m−r−1
i=0
(γµt)i
i! (ρi
− ρm−r
), ρ = λ/γµ.

236. 234
Note that m → ∞ =⇒ M/M/r/m =⇒ M/M/r and
P(w > t) = pr
1 − ρ e−γµt
∞
i=0
(γµρt)i
i!
= pr
1 − ρ e−γµ(1−ρ)t
t > 0.
and it agrees with (16.119)
16.7 (a) Use the hints
(b)
−
∞
n=1
(λ + µ) pn zn
+
µ
z
∞
n=1
pn+1 zn+1
+ λ
∞
n=1
n
k=1
pn−k ck zn
= 0
−(ρ + 1) (P(z) − p0) +
µ
z
(P(z) − p0 − p1z) + λ
∞
k=1
ckzk
m=0
pm zm
= 0
which gives
P(z)[1 − z − ρz (1 − C(z))] = p0(1 − z)
or
P(z) =
p0(1 − z)
1 − z − ρz (1 − C(z))
.
1 = P(1) =
−p0
−1 − ρ + ρz C (z) + ρC(z)
=
−p0
−1 + ρC (1)
=⇒ p0 = 1 − ρ0, ρ0 = ρC (1).
Let
D(z) =
1 − C(z)
1 − z
.
Then
P(z) =
1 − ρL
1 − ρzD(z)
.
(c) This gives
P (z) =
(1 − ρc)
(1 − ρzD(z))2 (ρD(z) + ρzD (z))

237. 235
L = P (1) =
(1 − ρc)
(1 − ρc)2 ρ (D(1) + D (1))
= 1
(1 − ρc)
(C (1) + D (1))
C (1) = E(x)
D(z) =
1 − C(z)
1 − z
D (z) =
(1 − z) (−C (z)) − (1 − C(z)) (−1)
(1 − z)2
=
1 − C(z) − (1 − z)C (z)
(1 − z)2
By L-Hopital’s Rule
D (1) = limz→1
−C (z) − (−1)C (z) − (1 − z)C (z)
−2(1 − z)
= limz→1 = 1/2C (z) =
C (z)
2
= 1/2 k(k − 1) Ck =
E(X2
) − E(X)
2
L =
ρ (E(X) + E(X2
))
2 (1 − ρE(X))
.
(d)
C(z)zm
E(X) = m
P(z) =
1 − ρ
1 − ρ m
k=1 zk
D(z) =
1 − zm
1 − z
=
m−1
k=0
zk
E(X) = m, E(X2
) = m2
L =
ρ(m + m2
)
2(1 − ρm)

238. 236
(e)
C(z) =
qz
1 − Pz
P(z) =
1 − ρ0
1 − ρzD(z)
, C(z) =
qz
1 − pz
D(z) =
1 − C(z)
1 − z
=
1 − qz
1−P(z)
1 − z
=
1 − Pz − (1 − P)z
(1 − z)(1 − Pz)
=
1 − z
(1 − z)(1 − Pz)
=
1
1 − Pz
P(z) =
(1 − ρ0)(1 − pz)
1 − pz − ρz
=
(1 − ρ0)(1 − pz)
1 − (p + ρ)z
C (1) =
(1 − pz)q − qz(−p)
(1 − Pz)2
=
q
q2
=
1
q
D(z) =
1 − C(z)
1 − z
D(1) = C (1)
L = P (1) =
1 − ρc
(1 − ρc)2
(ρ · C (1) + ρ · D (1))
D (z) =
−(1 − z)C (z) − (1 − C(z)) (ρ − 1)
(1 − z)2
=
1 − C(z) − (1 − z)C (1)
(1 − z)2
limz→1 D (z) = limz→1
−C (z) − (−1)C (z) − (1 − z)C (z)
2(1 − z)
=
−(1 − z)C (z)
−2(1 − z)
=
ρ (z)
2
D (1) =
C (1)
2
L =
1
(1 − ρc)
ρE(X) +
ρ (E(X2
) − E(X))
2
=
ρE(X) + ρE(X2
)
2(1 − ρc)
.
16.8 (a) Use the hints.
(b)
−
∞
n=1
(λ + µ) pn zn
+
µ
zn
∞
n=1
pn+m zn+m
+ λz
∞
n=1
pn−1 zn−1
= 0

239. 237
or
−(1 + ρ) (P(z) − p0) +
1
zm
P(z) −
m
k=0
pk zk
+ ρzP(z) = 0
which gives
P(z) ρ zm+1
− (ρ + 1) zm
+ 1 =
m
k=0
pk zk
− p0 (1 + ρ) zm
or
P(z) =
m
k=0
pk zk
− p0 (1 + ρ) zm
ρ zm+1 − (ρ + 1) zm + 1
=
N(z)
M(z)
. (i)
(c) Consider the denominator polynomial M(z) in (i) given by
M(z) = ρ zm+1
− (1 + ρ) zm
+ 1 = f(z) + g(z)
where
f(z) = −(1 + ρ) zm
,
g(z) = 1 + ρ zm+1
.
Notice that |f(z)| > |g(z)| in a circle defined by |z| = 1 + ε, ε > 0.
Hence by Rouche’s Theorem f(z) and f(z)+g(z) have the same number
of zeros inside the unit circle (|z| = 1 + ε). But f(z) has m zeros inside
the unit circle. Hence f(z) + g(z) = M(z) also has m zeros inside the
unit circle. Hence
M(z) = M1(z) (z − z0) (ii)
where |z0| > 1 and M1(z) is a polynomial of degree m whose zeros are
all inside or on the unit circle. But the moment generating function
P(z) is analytic inside and on the unit circle. Hence all the m zeros
of M(z) that are inside or on the unit circle must cancel out with the
zeros of the numerator polynomial of P(z). Hence
N(z) = M1(z) a. (iii)

240. 238
Using (ii) and (iii) in (i) we get
P(z) =
N(z)
M(z)
=
a
z − z0
.
But P(1) = 1 gives a = 1 − z0
or
P(z) =
z0 − 1
z0 − z
= 1 −
1
z0
∞
n=0
(z/z0)n
=⇒ pn = 1 −
1
z0
1
z0
n
= (1 − r) rn
, n ≥ 0 (iv)
where r = 1/z0.
(d) Average system size
L =
∞
n=0
n pn =
r
1 − r
.
16.9 (a) Use the hints in the previous problem.
(b)
−
∞
n=m
(λ + µ) pn zn
+ µ
∞
n=m
pn+m zn
+ λ
∞
n=m
pn−1 zn
−(1 + ρ) P(z) −
m−1
k=0
pk zk
+
1
zm
P(z) −
2m−1
k=0
pk zk
+ρ z P(z) −
m−2
k=0
pk zk
= 0.
After some simplifications we get
P(z) ρ zm+1
− (ρ + 1) zm
+ 1 = (1 − zm
)
m−1
k=0
pk zk
or
P(z) =
(1 − zm
)
m−1
k=0
pk zk
ρ zm+1 − (ρ + 1) zm + 1
=
(z0 − 1)
m−1
k=0
zk
m (z0 − z)

241. 239
where we have made use of Rouche’s theerem and P(z) ≡ 1 as in
problem 16-8.
(c)
P(z) =
∞
n=0
pn zn
=
1 − r
m
m−1
k=0
zk
1 − rz
gives
pn =



(1 + r + · · · + rk
) p0, k ≤ m − 1
rn−m+1
(1 + r + · · · + rm−1
) p0, k ≥ m
where
p0 =
1 − r
m
, r =
1
z0
.
Finally
L =
∞
n=0
n pn = Pn(1).
But
P (z) =
1 − r
m
m−1
k=1
k zk−1
(1 − rz) −
m−1
k=0
zk
(−r)
(1 − rz)2
so that
L = P (1) =
1 − r
m
m − 1 + r
(1 − r)2
=
m − (1 − r)
m (1 − r)
=
1
1 − r
−
1
m
.
16.10 Proceeding as in (16-212),
ψA(u) =
∞
0
e−uτ
dA(τ)
=
λ m
u + λ m z
m
.

242. 240
This gives
B(z) = ψA(ψ(1 − z))
=
λ m
µ (1 − z) + λ m
m
=


1
1 + 1
ρ
(1 − z)


m
=
ρ
(1 + ρ) − z
m
, ρ =
λ
m µ
.
(i)
Thus the equation B(z) = z for π0 reduce to
ρ
(1 + ρ) − z
m
= z
or
ρ
(1 + ρ) − z
= z1/m
,
which is the same as
ρ z−1/m
= (1 + ρ) − z (ii)
Let x = z−1/m
. Sustituting this into (ii) we get
ρ x = (1 + ρ) − x−m
or
ρ xm+1
− (1 + ρ) xm
+ 1 = 0 (iii)
16.11 From Example 16.7, Eq.(16-214), the characteristic equation for
Q(z) is given by (ρ = λ/m µ)
1 − z[1 + ρ (1 − z)]m
= 0

243. 241
which is equivalent to
1 + ρ (1 − z) = z−1/m
. (i)
Let x = z1/m
in this case, so that (i) reduces to
[(1 + ρ) − ρ xm
] x = 1
or the characteristic equation satisfies
ρ xm+1
− (1 + ρ) x + 1 = 0. (ii)
16.12 Here the service time distribution is given by
dB(t)
dt
=
k
i=1
di δ(t − Ti)
and this Laplace transform equals
Φs(s) =
k
i=1
di e−s Ti
(i)
substituting (i) into (15.219), we get
A(z) = Φs (λ (1 − z))
=
k
i=1
di e−λ Ti (1−z)
=
k
i=1
di e−λ Ti
eλ Ti z
=
k
i=1
di e−λ Ti
∞
j=0
(λ Ti)j zj
j!
=
∞
j=0
aj zj
.
Hence
aj =
k
i=1
di e−λ Ti
(λ Ti)j
j!
, j = 0, 1, 2, · · · . (i)

244. 242
To get an explicit formula for the steady state probabilities {qn}, we
can make use of the analysis in (16.194)-(16.204) for an M/G/1 queue.
From (16.203)-(16.204), let
c0 = 1 − a0, cn = 1 −
n
k=0
ak, n ≥ 1
and let {c
(m)
k } represent the m−fold convolution of the sequence {ck}
with itself. Then the steady-state probabilities are given by (16.203) as
qn = (1 − ρ)
∞
m=0
n
k=0
ak c
(m)
n−k.
(b) State-Dependent Service Distribution
Let Bi(t) represent the service-time distribution for those customers
entering the system, where the most recent departure left i customers
in the queue. In that case, (15.218) modifies to
ak,i = P{Ak|Bi}
where
Ak = ”k customers arrive during a service time”
and
Bi = ”i customers in the system at the most recent departure.”
This gives
ak,i =
∞
0
e−λt (λt)k
k!
dBi(t)
=



∞
0
e−λt (λt)k
k!
µ1 e−µ1t
dt =
µ1λk
(λ + µ1)k+1
, i = 0
∞
0
e−λt (λt)k
k!
µ2 e−µ2t
dt =
µ2λk
(λ + µ2)k+1
, i ≥ 1
(i)

245. 243
This gives
Ai(z) =
∞
k=0
ak,i zk
=



1
1 + ρ1(1 − z)
, i = 0
1
1 + ρ2(1 − z)
, i ≥ 1
(ii)
where ρ1 = λ/µ1, ρ2 = λ/µ2. Proceeding as in Example 15.24, the
steady state probabilities satisfy [(15.210) gets modified]
qj = q0 aj,0 +
j+1
i=1
qi aj−i+1,i (iii)
and (see(15.212))
Q(z) =
∞
j=0
qj zj
= q0
∞
j=0
aj,0 zj
+
∞
j=0
qi aj−i+1,i
= q0 A0(z) +
∞
i=1
qi zi
∞
m=0
am,i zm
z−1
= q0 A0(z) + (Q(z) − q0) A1(z)/z
(iv)
where (see (ii))
A0(z) =
1
1 + ρ1(1 − z)
(v)
and
A1(z) =
1
1 + ρ2(1 − z)
. (vi)
From (iv)
Q(z) =
q0 (z A0(z) − A1(z))
z − A1(z)
. (vii)

246. 244
Since
Q(1) = 1 =
q0 A0(1) + A0(1) − A1(1)
1 − A1(1)
=
q0 (1 + ρ1 − ρ2)
1 − ρ2
we obtain
q0 =
1 − ρ2
1 + ρ1 − ρ2
. (viii)
Substituting (viii) into (vii) we can rewrite Q(z) as
Q(z) = (1 − ρ2)
(1 − z) A1(z)
A1(z) − z
·
1
1 + ρ1 − ρ2
1 − z A0(z)/A1(z)
1 − z
=
1 − ρ2
1 − ρ2 z
1
1 + ρ1 − ρ2
1 − ρ2
1+ρ1
z
1 − ρ1
1+ρ1
z
= Q1(z) Q2(z)
(ix)
where
Q1(z) =
1 − ρ2
1 − ρ2 z
= (1 − ρ2)
∞
k=0
ρk
2 zk
and
Q2(z) =
1
1 + ρ1 − ρ2
1 −
ρ2
1 + ρ1
z
∞
i=0
ρ1
1 + ρ1
i
zi
.
Finally substituting. Q1(z) and Q2(z) into (ix) we obtain
qn = q0


n
i=0
ρ1
1 + ρ1
n−i
ρi
2 −
n−1
i=0
ρi+1
2
ρn−i−1
1
(1 + ρ1)n−i

 . n = 1, 2, · · ·
with q0 as in (viii).

247. 245
16.13 From (16-209), the Laplace transform of the waiting time distri-
bution is given by
Ψw(s) =
1 − ρ
1 − λ 1−Φs(s)
s
=
1 − ρ
1 − ρ µ 1−Φs(s)
s
.
(i)
Let
Fr(t) = µ
t
0
[1 − B(τ)]dτ
= µ t −
t
0
B(τ)dτ .
(ii)
represent the residual service time distribution. Then its Laplace
transform is given by
ΦF (s) = L {Fr(t)} = µ
1
s
−
Φs(s)
s
= µ
1 − Φs(s)
s
.
(iii)
Substituting (iii) into (i) we get
Ψw(s) =
1 − ρ
1 − ρ ΦF (s)
= (1 − ρ)
∞
n=0
[ρ ΦF (s)]n
, |ΦF (s)| < 1. (iv)
Taking inverse transform of (iv) we get
Fw(t) = (1 − ρ)
∞
n=0
ρn
F(n)
r (t),
where F(n)
r (t) is the nth
convolution of Fr(t) with itself.
16.14 Let ρ in (16.198) that represents the average number of customers
that arrive during any service period be greater than one. Notice that

248. 246
ρ = A (1) > 1
where
A(z) =
∞
k=0
ak zk
From Theorem 15.9 on Extinction probability (pages 759-760) it
follows that if ρ = A (1) > 1, the eqution
A(z) = z (i)
has a unique positive root π0 < 1. On the other hand, the transient
state probabilities {σi} satisfy the equation (15.236). By direct substi-
tution with xi = πi
0 we get
∞
j=1
pij xj =
∞
j=1
aj−i+1 πj
0 (ii)
where we have made use of pij = aj−i+1, i ≥ 1 in (15.33) for an
M/G/1 queue. Using k = j − i + 1 in (ii), it reduces to
∞
k=2−i
ak πk+i−1
0 = πi−1
0
∞
k=0
ak πk
0
= πi−1
0 π0 = πi
0 = xi (iii)
since π0 satisfies (i). Thus if ρ > 1, the M/G/1 system is transient
with probabilities σi = πi
0.
16.15 (a) The transition probability matrix here is the truncated version
of (15.34) given by

249. 247
P =



























a0 a1 a2 · · · · am−2 1 −
m−2
k=0
ak
a0 a1 a2 · · · · am−2 1 −
m−2
k=0
ak
0 a0 a1 · · · · am−3 1 −
m−3
k=0
ak
...
...
...
...
...
...
0 0 0 · · · a0 a1 1 − (a0 + a1)
0 0 0 · · · 0 a0 1 − a0



























(i)
and it corresponds to the upper left hand block matrix in (15.34)
followed by an mth
column that makes each row sum equal to unity.
(b) By direct sybstitution of (i) into (15-167), the steady state prob-
abilities {q∗
j }m−1
j=0 satisfy
q∗
j = q∗
0 aj +
j+1
i=1
q∗
i aj−i+1, j = 0, 1, 2, · · · , m − 2 (ii)
and the normalization condition gives
q∗
m−1 = 1 −
m−2
i=0
q∗
i . (iii)
Notice that (ii) in the same as the first m − 1 equations in (15-210)
for an M/G/1 queue. Hence the desired solution {q∗
j }m−1
j=0 must satisfy
the first m − 1 equations in (15-210) as well. Since the unique solution
set to (15.210) is given by {qj}∞
j=0 in (16.203), it follows that the desired
probabilities satisfy
q∗
j = c qj, j = 0, 1, 2, · · · , m − 1 (iv)
where {qj}m−1
j=0 are as in (16.203) for an M/G/1 queue. From (iii)
we also get the normalization constant c to be

250. 248
c =
1
m−1
i=0
qi
. (v)
16.16 (a) The event {X(t) = k} can occur in several mutually exclusive
ways, viz., in the interval (0, t), n customers arrive and k of them
continue their service beyond t. Let An = “n arrivals in (0, t)”, and
Bk,n =“exactly k services among the n arrivals continue beyond t”,
then by the theorem of total probability
P{X(t) = k} =
∞
n=k
P{An ∩ Bk,n} =
∞
n=k
P{Bk,n|An}P(An).
But P(An) = e−λt
(λt)n
/n!, and to evaluate P{Bk,n|An}, we argue as
follows: From (9.28), under the condition that there are n arrivals in
(0, t), the joint distribution of the arrival instants agrees with the joint
distribution of n independent random variables arranged in increasing
order and distributed uniformly in (0, t). Hence the probability that a
service time S does not terminate by t, given that its starting time x
has a uniform distribution in (0, t) is given by
pt =
t
0
P(S > t − x|x = x)fx(x)dx
=
t
0
[1 − B(t − x)]
1
t
dx =
1
t
t
0
(1 − B(τ)) dτ =
α(t)
t
Thus Bk,n given An has a Binomial distribution, so that
P{Bk,n|An} =
n
k
pk
t (1 − pt)n−k
, k = 0, 1, 2, · · · n,

251. 249
and
P{X(t) = k} =
∞
n=k
e−λt (λt)n
n!
n
k
α(t)
t
k
1
t
t
0
B(τ)dτ
n−k
= e−λt [λα(t)]k
k!
∞
n=k
λt
1
t
t
0
B(τ)dτ
n−k
(n − k)!
=
[λα(t)]k
k!
e
−λ t−
t
0
B(τ)dτ
=
[λα(t)]k
k!
e−λ
t
0
[1−B(τ)]dτ
=
[λα(t)]k
k!
e−λ α(t)
, k = 0, 1, 2, · · ·
(i)
(b)
lim
t→∞
α(t) =
∞
0
[1 − B(τ)]dτ
= E{s}
(ii)
where we have made use of (5-52)-(5-53). Using (ii) in (i), we obtain
lim
t→∞
P{x(t) = k} = e−ρ ρk
k!
(iii)
where ρ = λ E{s}.