Papoulis Solutions Manual 4th Edition

Solutions Manual to accompany

Probability, Random Variables and Stochastic Processes Fourth Edition

Athanasios Papoulis Polytechnic University

S. Unnikrishna Pillai Polytechnic University

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 not be reproduced in any other form or for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. www.mhhe.com

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(n2 1) p2(1 p)n;2 ;

;

;

;

;

;

3.2

;

;

;

;

P (doublesix) = 16 16 = 361 

P (\double six atleast three times in n trials00)



= 1 50 0 = 0:162 ;

1 0
35 50 36 36

3.6 (a)

p1 = 1

(b) 1

;

(c) 1





;

;

6

;

1 
35 49 36 36

;

18 1

1 6

;





50 2

1 2
35 48 36 36

6 ;

5 = 0:665 6

 ! 11 12 1 5

1

6

 ! 17

18

5 6

12 5

50 1

5 6

6

= 0:619

 ! 2 16 ;

18 2

1 6

5 6

= 0:597

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 4 n < 1 16 < n < 17:3 n = 17 ;

;

;



17
33 1 3 = 0:432 P (net gain does not exceed $1) = 50 17 4 4 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 2 n) < 5 13:3 < n < 20 ;

;

;



n
50;n X 50 1 3 = 0:349 P (net gain does not exceed $5) = 19 n = 14 n 4 4 P (net gain or loss exceeds $5) = 1 0:349 = 0:651 ;

3 3.8 Dene 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) = nr pr qn;r P (B ) = p
 P (C ) = nr 11 pr;1 qn;r We need ) = P (BC ) = P (B ) P (C ) = r : P (B A) = PP(AB (A) P (A) P (A) n ;

;

; ;

j


 3.9 There are 52 13 ways of selecting 13 cards out of 52 cards. The of ways to select 13 cards of any suit (out of 13 cards) equals
number 13 = 1. Four such (mutually exclusive) suits give the total number 13 of favorable outcomes to be 4. Thus the desired probability is given by 4  ! = 6:3 10;12 52 13 

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 = qp Mk 1p 8
k n o q q 1 > k
p=q > M 1 ( ) > p 1 p q p < => > M1 kp
p=q > : ;

;

;

;

6

;

This gives

Ni =

i;1 X k=0

Mk+1

8
> > > M1 + > > < => > > > > :

1

p q ;

i;1
k X q

p

k=0

;

i p q
p = q 6

;

iM1 i(i2p 1)
;

p=q

;

where we have used No = 0. Similarly Na+b = 0 gives

q=p : M1 + p 1 q = pa + qb 1 1 (q=p )a+b ;

Thus

8 > > > < Ni = >> > :




;

;

;

a + b 1 (q=pa)+i b i p q 1 (q=p) p q
p = q i(a + b i)
p=q ;




;

;

;

;

;

6

5 which gives for i = a 8 > > > < Na = >> > : 8 > > > < => > > :

a a + b 1 (q=pa)+a b p q 1 (q=p) p q
p = q ab
p=q a + b 1 (p=qa)+b b
p = q: b 2p 1 2p 1 1 (p=q) ab
p=q ;

;

;




;

;

;

;




;

;

;

6

6

6 3.11

Pn = pPn+
+ qPn;

Arguing as in (3.43), we get the corresponding iteration equation

Pn = Pn+
+ qPn; and proceed as in Example 3.15. 3.12 Suppose one best on k = 1
2
Then







6.




2 p1 = P (k appears on one dice) = 31 16 56

2
 p2 = P (k appear on two dice) = 32 16 65
3 p3 = P (k appear on all the tree dice) = 61


3 p0 = P (k appear none) = 65

Thus, we get Net gain = 2p1 + 3p2 + 4p3 p0 = 0:343: ;

215 Chapter 15 15.1 The chain represented by 

0

1/2 1/2

   1/2 P =  

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

  

1 , k = 0, 1, 2, · · · m. m+1 15.3 This is the “success runs” problem discussed in Example 15-11 and 15-23. From Example 15-23, we get lim P {xn = ek } =

n→∞

ui+1 = pi,i+1 ui =

1 uo ui = i+1 (i + 1)!

so that from (15-206) ∞ X

k=1

uk = u 0

∞ X 1

k=1

k!

= e · u0 = 1

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 lim P [xn = 0|x0 = m] =

n→∞

(1)

(2)

(2) = P [{x(1) n = 0|x0 = 1} {xn = 0|x0 = 1}

=

Qm

k=1

(k)

T

P [x(k) n = 0|x0 = 1]

T

(m)

= 0|x0 · · · {x(m) n

= 1}]

= π0m

15.5 From (15-288)-(15-289), P (z) = p0 + p1 z + p2 z 2 ,

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 + p2 z 2

= p0 − (p0 + p2 )z + p2 z 2

= (z − 1)(p2 z − 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

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 X

yk

k=1

where yk are i.i.d random variables with common moment generating function P (z) so that (see (15-287)-(15-289)) P 0 (1) = E{yk } = µ. Thus

E{xn+1 |xn } = E{ = E{ = E{

Similarly

P xn

yk |xn = m}

Pm

yk } = mE{yk } = xn µ

k=1

Pm

k=1 k=1

yk |xn = m}

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 . Define wn =

xn . µn

This gives E{wn } = 1.

Dividing both sider of (i) with µn+r we get E{

xn xn xn+r r |x = x} = µ · = = wn n µn+r µn+r µn

(ii) (iii)

218 or E{wn+r |wn =

x ∆ = w} = wn µn

which gives E{wn+r |wn } = wn , the desired result. 15.7 sn = x 1 + x 2 + · · · + x n 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+1 = i|xn = j} P {xn = j} P {xn+1 = i} qj pji = qi = p∗ij ,

P {xn = j|xn+1 = i} =

(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 = pij qi

(ii)

or, for a time-reversible chain we get qj pji = qi pij .

(iii)

219 Thus using (ii) we obtain by direct substitution pij pjk pki =

³q

j qi

pji

´ µ

qk p qj kj

= pik pkj pji ,

¶³

qi p ´ qk ik

the desired result. 15.9 (a) It is given that A = AT , (aij = aji ) and aij > 0. Define the ith row sum X aik > 0, i = 1, 2, · · · ri = k

and let

aij aij pij = P = . ri k aik

Then

a a a pji = X ji = rjij = rijj ajm m

(i)

a = rrji riji = rrji pij

or ri pij = rj pji . Hence X

ri pij =

X

rj pji = rj

since X i

pji =

pji = rj ,

(ii)

i

i

i

X

P

i

aji

rj

=

rj = 1. rj

Notice that (ii) satisfies the steady state probability distribution equation (15-167) with qi = c r i , i = 1, 2, · · · where c is given by c

X i

ri =

X i

1 qi = 1 =⇒ c = P

i ri

1 =P P i

j

aij

.

220 Thus

ri j aij qi = P =P P >0 i ri i j aij P

(iii)

represents the stationary probability distribution of the chain. With (iii) in (i) we get qi pji = pij qj or qj pji pij = = p∗ij qi 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 + = δij +

P

P

k

wik mkj ,

ei , ej ∈ T

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  (rj − 1)(r 6−i − 1)    , j≤i   (p − q)(r 6 − 1)  mij =  (ri − 1)(r 6−i − rj−i )   , j ≥ i.    (p − q)(r 6 − 1)

15.11 If a stochastic matrix A = (aij ), aij > 0 corresponds to the twostep transition matrix of a Markov chain, then there must exist another stochastic matrix P such that A = P 2,

P = (pij )

221 where pij > 0,

X

pij = 1,

j

and this may not be always possible. For example in a two state chain, let   α 1−α  P =   1−β β so that



A = P2 =  

α2 + (1 − α)(1 − β) (α + β)(1 − β)

(α + β)(1 − α)

β 2 + (1 − α)(1 − β)



 .

This gives the sum of this its diagonal entries to be a11 + a22 = α2 + 2(1 − α)(1 − β) + β 2 = (α + β)2 − 2(α + β) + 2

= 1 + (α + β − 1)

2

(i)

≥ 1.

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 (15167),and hence we get uk =

X j

uj pjk =

N X

j=0

uj

Ã

!

N k N −k p j qj . k

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 Examples 15A-1, 15A-2 and 15A-3. The eigenvalues in general satisfy the equation X (k) (k) pij xj = λk xi , k = 0, 1, 2, · · · N j

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 P

G(s) =

N X

pij sj

(i)

j=0

where from (15A-7) pij = =

{Ai (z)}j {B N −i (z)}N −j {Ai (z) B N −i (z)}N

coefficient of sj z N in {Ai (sz) B N −i (z)} {Ai (z) B N −i (z)}N

(ii)

Substituting (ii) in (i) we get the compact expression G(s) =

{Ai (sz) B N −i (z)}N . {Ai (z) B N −i (z)}N

(iii)

Differentiating G(s) with respect to s we obtain G0 (s) =

N X

Pij j sj−1

j=0

=

{iAi−1 (sz) A0 (sz)z B N −i (z)}N {Ai (z) B N −i (z)}N

=i·

{Ai−1 (sz) A0 (sz) B N −i (z)}N −1 . {Ai (z) B N −i (z)}N

(iv)

223 Letting s = 1 in the above expression we get N X

{Ai−1 (z) A0 (z) B N −i (z)}N −1 G (1) = . pij j = i {Ai (z) B N −i (z)}N j=0 0

(v)

In the special case when A(z) = B(z), Eq.(v) reduces to N X

pij j = λ1 i

(vi)

j=0

where

{AN −1 (z) A0 (z)}N −1 . {AN (z)}N Notice that (vi) can be written as

(vii)

λ1 =

P x 1 = λ 1 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

Ã

!

2N 2p q N pN −1 N − 1 Ã ! = 1. = 2N N N q p N

Thus N j=0 pij j = i and from (15-224) these chains represent Martingales. (Similarly for Examples 15A-2 and 15A-3 as well). To determine the remaining eigenvalues we differentiate G0 (s) once more. This gives P

G (s) = 00

N X

j=0

pij j(j − 1) sj−2 00

{i(i − 1)Ai−2 (sz)[A0 (sz)]2 z B N −i (z) + iAi−1 (sz) A (sz)z B N −i (z)}N −1 = . {Ai (z) B N −i (z)}N 2

{i Ai−2 (sz) B N −i (z)[(i − 1) (A0 (sz)) + A(sz) A00 (sz)]}N −2 . = {Ai (z) B N −i (z)}N

224 With s = 1, and A(z) = B(z), the above expression simplifies to N X

j=0

pij j(j − 1) = λ2 i(i − 1) + iµ2

(viii)

where λ2 =

{AN −2 (z) [A0 (z)]2 }N −2 {AN (z)}N

µ2 =

{AN −1 (z) A00 (z)}N −2 . {AN (z)}N

and

Eq. (viii) can be rewritten as N X

j=0

pij j 2 = λ2 i2 + (polynomial in i of degree ≤ 1)

and in general repeating this procedure it follows that (show this) N X

j=0

pij j k = λk ik + (polynomial in i of degree ≤ k − 1)

(ix)

where λk =

{AN −k (z) [A0 (z)]k }N −k , {AN (z)}N

k = 1, 2, · · · N.

(x)

Equations (viii)–(x) motivate to consider the identities P q k = λ k 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

225 obtain the desired eigenvalues to be λk =

{(q + pz)2(N −k) [2p(q + pz)]k }N −k {(q + pz)2N }N

= 2 k pk

= 2k

Ã

{(q + pz)2N −k }N −k {(q + pz)2N }N } !

2N − k N −! k à , 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 {eλ(N −k)(z−1) λk eλk(z−1) }N −k λk = {eλN (z−1) }N =

λk {eλN z }N −k λk (λN )N −k /(N − k)! = λN z (λN )N /N ! {e }N

1 N! = 1− N = (N − k)! N k ³

´³

µ

¶

2 ··· 1 − k − 1 , 1− N N ´

k = 1, 2, · · · N

(c) The transition probabilities in this case follow from Example 15A-3 (page 766-767) with q A(z) = B(z) = . 1 − pz Thus {1/(1 − pz)N +k }N −k λk = p k {1/(1 − pz)N }N = (−1)k

Ã

−(N + k) N − k! Ã −N N

!

Ã

!

2N − 1 N −k! =Ã , 2N − 1 N

r = 2, 3, · · · N

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 +

X

pik mk = 1 + pi,i+1 mi+1 + pi,i−1 mi−1

k∈T

= 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 = pq Mk − p1 = pq     

=   

·

³ ´2 ³ ´k−1 M1 − p1 1 + pq + pq + · · · + pq ³ ´k q M − 1 n1 − ( q )k o , p 6= q 1 p p p−q M1 − kp , p=q

³ ´k

¸

227 This gives mi =

i−1 X

Mk+1

k=0

 i−1 ³ ´ ³ ´ X  q k− i ,  1  M +  1  p p − q p−q  k=0

=

p 6= q

 i(i − 1)   iM1 − 2p , p=q    ³ ´ 1 − (q/p)i  i , p= 1  − p− M + 6 q  1  p−q q  1 − q/p

=

iM1 −

   

i(i − 1) 2p ,

p=q

where we have used mo = 0. Similarly ma+b = 0 gives M1 + Thus

1 a+b 1 − q/p = · . p−q p − q 1 − (q/p)a+b

 i  a + b 1 − (q/p) − i , p 6= q    p−q · 1 − (q/p)a+b p − q mi =    i(a + b − i), p=q 

which gives for i = a ma

 a  a + b · 1 − (q/p) − a , p =  6 q   p − q 1 − (q/p)a+b p−q =   ab, p=q   b  b − a + b · 1 − (p/q) , p 6= q    a+b

=

by writing

2p − 1

  

2p − 1 1 − (p/q) ab,

p=q

(q/p)a − (q/p)a+b 1 − (p/q)b 1 − (q/p)a =1− =1− 1 − (q/p)a+b 1 − (q/p)a+b 1 − (p/q)a+b

(see also problem 3-10).

228 Chapter 16 16.1 Use (16-132) with r = 1. This gives

pn =

ρn p , n ≤ 1 n! 0 ρn p 0 , 1 < n ≤ m

      

= ρ n p0 , Thus

m X

pn = p 0

m X

0≤n≤m ρn = p 0

n=0

n=0

=⇒

p0 =

(1 − ρm+1 ) =1 1−ρ

1−ρ 1 − ρm+1

and hence pn =

1−ρ n ρ , 1 − ρn+1

0 ≤ n ≤ m,

ρ 6= 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 X

P {X = k} P {Y = n − k}

k=0

=

n X

(i) (1 − ρ)ρk (1 − ρ)ρn−k

k=0

= (n + 1)(1 − ρ)2 ρn , where ρ = λ/µ.

n = 0, 1, 2, · · ·

229 0

(b) When the two queues are merged, the new input rate λ = λ + λ = 2λ. Thus from (16-102)      

pn =     

0

(2ρ)n (λ /µ)n p 0 = n! n! p0 , n < 2 0

22 ( λ )n p = 2ρn p , 0 0 2! 2µ

n ≥ 2.

Hence ∞ X

pk = p0 (1 + 2ρ + 2

∞ X

ρk )

k=2

k−0

2 = p0 (1 + 2ρ + 12ρ − ρ) = 1 p−0 ρ ((1 + 2ρ) (1 − ρ) + 2ρ2 ) = 1 p−0 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) 0

E{X} = L1 =

∞ X

n=2

(n − 1) pn

230 where pn is an in (16-88). Thus 0

L1 =

∞ X

(n − 1)(1 − ρ) ρn

n=2

= (1 − ρ) ρ

2

∞ X

(n − 1) ρn−2

n=2

= (1 − ρ) ρ

2

∞ X

(iv) kρ

k−1

k=1

1 ρ2 = (1 − ρ) ρ = . (1 − ρ)2 (1 − ρ) 2

Since n1 (t) = X + Y we have L1 = E{n1 (t)} = E{X} + E{Y } 0

= 2L1 =

2ρ2 1−ρ

(v)

For L2 we can use (16-106)-(16-107) with r = 2. Using (iii), this gives ρ L2 = p r (1 − ρ)2 =2

(1 − ρ) ρ2 2 ρ3 ρ = 1 + ρ (1 − ρ)2 1 − ρ2

2 ρ2 = 1−ρ

Ã

ρ 1+ρ

!

(vi)

< L1

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, λi,i+1 = (m − i) λ,

λ0,1 = µ λi,i−1 = iµ

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 Ã

! Ã

m pi = i

λ λ+µ

!i Ã

µ λ+µ

!m−i

,

i = 0, 1, 2, · · · , m

!n

p0 , n < m

16.4 (a) In this case        

pn =       

where

λ λ λ ··· = µ1 µ1 µ1

λ µ1

λ λ λ λ λ ··· ··· p0 , n ≥ m µ1 µ1 µ1 µ2 µ2   

= 

∞ X

Ã

ρn1 p0 ,

n<m

ρm−1 ρ2n−m+1 p0 , n ≥ m, 1

pn = p 0

n=0

"m−1 X

ρk1 +

= p0

∞ X

n=0

k=0

"

ρm−1 ρ2 1

ρm 1

ρ2 ρm−1 1

1− + 1 − ρ1 1 − ρ2

#

ρn2

#

=1

232 gives p0 =

Ã

ρ2 ρm−1 1 − ρm 1 1 + 1 − ρ1 1 − ρ2

!−1

.

(b) L =

∞ X

n pn

n=0

= p0

"m−1 X

n ρn1 +

= p 0  ρ1

m−1 X n=0

Ãm−1 X



Ã

d = p 0  ρ1 dρ1

d = p 0  ρ1 dρ1 = p0

n ρn−1 + ρ1 1



"

n ρm−1 1

ρ2n−m+1

n=m

n=0



∞ X

Ã

ρ1 ρ2

!

ρn1 + ρ1

n=0

1 − ρm 1 1−ρ

!

!m−2

Ã

+ ρ1

ρ1 ρ2

Ã

#

∞ X

n=m

!m−2

ρ1 ρ2



 n ρn−1 2



∞ d X ρn  dρ2 n=m 2

!m−2

d dρ

Ã

!

ρm  1−ρ

m−1 ρ1 [1 + (m − 1)ρm ] ρ2 ρm−1 + [m − (m − 1)ρ2 ] 1 1 − mρ1 + . 2 (1 − ρ1 ) (1 − ρ2 )2

16.5 In this case   

λi =  

λ,

j
pλ, j ≥ r

  

µi =  

jµ, j < r rµ, j ≥ r.

Using (16-73)-(16-74), this gives     

pn =    

(λ/µ)n n
#

233 16.6 P {w > t} =

m−1 X

pn P (w > t|n)

n=r

=

m−1 X

X

pn (1 − Fw (t|n)) =

n=r

fw (t|n) = e−γµt

µ

pr

(γµ)n−r+1 tn−r (n − r)!)

λ rµ

¶n−r

(1 − Fw (t|n))

(see 16.116)

and Fw (t|n) = 1 −

n−r X

k=0

(γµt)k −γµt e k!

(see 4.)

so that 1 − Fw (t|n) =

n−r X

k=0

P {w > t} =

m−1 X

pr

n=r

=

m−r−1 X

µ

λ γµ

p r ρi

i=0

¶n−r n−r X (γµt)k

m−r−1 X

(γµt)k −γµt , k! e ρk

k=0

=

m−r−1 k X X k=0

P {w > t} = pr e−γµt

i=0

m−r−1 X i=0

= 1 p−r ρ e−γµt

=

k!

k=0

i X

k=0

= pr e−γµt

(γµt)k −γµt e k! e−γµt n−r =i

k X (γµt)i

i!

i=0

m−r−1 X m−r−1 X i=0

k=i

m−r−1 (γµt)i X k ρ i!

m−r−1 X i=0

k=i

(γµt)i i m−r ), i! (ρ − ρ

ρ = λ/γµ.

234 Note that m → ∞ =⇒ M/M/r/m =⇒ M/M/r and P (w > t) = 1 p−r ρ e−γµt

∞ X (γµρt)i i=0

= 1 p−r ρ e−γµ(1−ρ)t

i!

t > 0.

and it agrees with (16.119) 16.7 (a) Use the hints (b) −

∞ X

(λ + µ) pn z n +

n=1

n ∞ X ∞ X µ X pn−k ck z n = 0 pn+1 z n+1 + λ z n=1 n=1 k=1

∞ X X µ −(ρ + 1) (P (z) − p0 ) + (P (z) − p0 − p1 z) + λ ck z k pm z m = 0 z m=0 k=1

which gives

P (z)[1 − z − ρz (1 − C(z))] = p0 (1 − z) or

p0 (1 − z) . 1 − z − ρz (1 − C(z)) −p0 −p0 1 = P (1) = = 0 −1 − ρ + ρz C (z) + ρC(z) −1 + ρC 0 (1) P (z) =

=⇒ p0 = 1 − ρ0 , Let D(z) = Then P (z) =

ρ0 = ρC 0 (1).

1 − C(z) . 1−z

1 − ρL . 1 − ρzD(z)

(c) This gives P 0 (z) =

(1 − ρc ) (ρD(z) + ρzD 0 (z)) (1 − ρzD(z))2

235 (1 − ρc ) ρ (D(1) + D 0 (1)) (1 − ρc )2 1 = (C 0 (1) + D 0 (1)) (1 − ρc )

L = P 0 (1) =

C 0 (1) = E(x) D(z) = D0 (z) = =

1 − C(z) 1−z

(1 − z) (−C 0 (z)) − (1 − C(z)) (−1) (1 − z)2 1 − C(z) − (1 − z)C 0 (z) (1 − z)2

By L-Hopital’s Rule D0 (1) = limz→1

−C 0 (z) − (−1)C 0 (z) − (1 − z)C 00 (z) −2(1 − z)

C 00 (z) 2 P E(X 2 ) − E(X) = 1/2 k(k − 1) Ck = 2

= limz→1 = 1/2C 00 (z) =

L=

ρ (E(X) + E(X 2 )) . 2 (1 − ρE(X))

(d) C(z)z m P (z) = D(z) =

E(X) = m

1−ρ P k 1−ρ m k=1 z

X 1 − z m m−1 = zk 1−z k=0

E(X) = m, L=

E(X 2 ) = m2

ρ(m + m2 ) 2(1 − ρm)

236 (e) C(z) = P (z) =

qz 1 − Pz

1 − ρ0 , 1 − ρzD(z)

C(z) =

qz 1 − pz

qz 1 − 1−P 1 − P z − (1 − P )z 1−z 1 1 − C(z) (z) = = = = D(z) = 1−z 1−z (1 − z)(1 − P z) (1 − z)(1 − P z) 1 − Pz

P (z) =

(1 − ρ0 )(1 − pz) (1 − ρ0 )(1 − pz) = 1 − pz − ρz 1 − (p + ρ)z

C 0 (1) =

(1 − pz)q − qz(−p) q 1 = = (1 − P z)2 q2 q

1 − C(z) 1−z D(1) = C 0 (1) 1 − ρc L = P 0 (1) = (ρ · C 0 (1) + ρ · D 0 (1)) (1 − ρc )2 D(z) =

D0 (z) =

−(1 − z)C 0 (z) − (1 − C(z)) (ρ − 1) 1 − C(z) − (1 − z)C 0 (1) = (1 − z)2 (1 − z)2

limz→1 D0 (z) = limz→1 =

−C 0 (z) − (−1)C 0 (z) − (1 − z)C 00 (z) 2(1 − z)

−(1 − z)C 00 (z) ρ00 (z) = 2 −2(1 − z)

C 00 (1) 2 Ã ! 2 1 ρ (E(X ) − E(X)) ρE(X) + ρE(X 2 ) L= ρE(X) + = . (1 − ρc ) 2 2(1 − ρc ) D0 (1) =

16.8 (a) Use the hints. (b) −

∞ X

n=1

(λ + µ) pn z n +

∞ ∞ X µ X n+m pn−1 z n−1 = 0 p z + λz n+m z n n=1 n=1

237 or 1 −(1 + ρ) (P (z) − p0 ) + m z

Ã

P (z) −

m X

pk z

k

k=0

!

+ ρzP (z) = 0

which gives h

i

P (z) ρ z m+1 − (ρ + 1) z m + 1 = or P (z) =

m X

m X

pk z k − p0 (1 + ρ) z m

k=0

pk z k − p0 (1 + ρ) z m

k=0 ρ z m+1

− (ρ +

1) z m

+1

=

N (z) . M (z)

(i)

(c) Consider the denominator polynomial M (z) in (i) given by M (z) = ρ z m+1 − (1 + ρ) z m + 1 = f (z) + g(z) where f (z) = −(1 + ρ) z m , g(z) = 1 + ρ z m+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)

238 Using (ii) and (iii) in (i) we get a N (z) = . M (z) z − z0

P (z) =

But P (1) = 1 gives a = 1 − z0 or z0 − 1 P (z) = z0 − z 1 = 1− z0 µ

µ

=⇒ pn = 1 −

1 z0

¶ X ∞

(z/z0 )n

n=0

¶n

¶ µ

1 z0

∞ X

n pn =

= (1 − r) r n ,

n≥0

(iv)

where r = 1/z0 . (d) Average system size L=

n=0

r . 1−r

16.9 (a) Use the hints in the previous problem. (b) −

∞ X

(λ + µ) pn z n + µ

n=m

∞ X

pn+m z n + λ

n=m

Ã

−(1 + ρ) P (z) −

m−1 X

pk z

k

k=0

Ã

+ρ z P (z) −

!

2m−1 X 1 pk z k + m P (z) − z k=0

m−2 X

!

pk z k = 0.

h

i

P (z) ρ z m+1 − (ρ + 1) z m + 1 = (1 − z m ) or (1 − z m ) P (z) =

m−1 X

pn−1 z n

n=m

Ã

k=0

After some simplifications we get

∞ X

k=0

pk z k

k=0

pk z k

ρ z m+1 − (ρ + 1) z m + 1

m−1 X

(z0 − 1) =

m−1 X

zk

k=0

m (z0 − z)

!

239 where we have made use of Rouche’s theerem and P (z) ≡ 1 as in problem 16-8. (c) m−1 X

zk 1 − r k=0 pn z n = P (z) = m 1 − rz n=0 ∞ X

gives pn =

    

where

(1 + r + · · · + r k ) p0 ,

k ≤m−1

rn−m+1 (1 + r + · · · + r m−1 ) p0 , k ≥ m 1−r , m

p0 =

r=

1 . z0

Finally L=

∞ X

0

n pn = Pn (1).

n=0

But 1−r P (z) = m 0

µ

¶

m−1 X

k z k−1 (1 − rz) −

m−1 X

z k (−r)

k=0

k=1

(1 − rz)2

so that 1−r m−1+r m − (1 − r) = 2 m (1 − r) m (1 − r) 1 1 = − . 1−r m 0

L = P (1) =

16.10 Proceeding as in (16-212), ∞

ψA (u) =

Z

=

Ã

0

e−uτ dA(τ )

λm u + λmz

!m

.

240 This gives B(z) = ψA (ψ(1 − z)) =

!m

Ã

λm µ (1 − z) + λ m



m

1  = 1 1 + ρ (1 − z) =

Ã

ρ (1 + ρ) − z

!m

,

(i)

ρ=

λ . mµ

Thus the equation B(z) = z for π0 reduce to Ã

ρ (1 + ρ) − z

!m

=z

or ρ = z 1/m , (1 + ρ) − z 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

241 which is equivalent to 1 + ρ (1 − z) = z −1/m .

(i)

Let x = z 1/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 k dB(t) X di δ(t − Ti ) = dt i=1

and this Laplace transform equals Φs (s) =

k X

di e−s Ti

(i)

i=1

substituting (i) into (15.219), we get A(z) = Φs (λ (1 − z)) =

k X

di e−λ Ti (1−z)

k X

di e−λ Ti eλ Ti z

i=1

=

i=1

=

k X

di e

i=1

−λ Ti

∞ X (λ T

j=0

i)

j!

j

zj

=

∞ X

aj z j .

j=0

Hence aj =

k X i=1

di e−λ Ti

(λ Ti )j , j!

j = 0, 1, 2, · · · .

(i)

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 − a 0 ,

cn = 1 −

n X

ak ,

n≥1

k=0 (m)

and let {ck } 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)

n ∞ X X

(m)

ak cn−k .

m=0 k=0

(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 =

=

Z

∞ 0

              

(λt)k dBi (t) k! Z ∞ (λt)k µ1 λk e−λt µ1 e−µ1 t dt = , i=0 k! (λ + µ1 )k+1 0

e−λt

Z

∞ 0

e−λt

µ2 λk (λt)k µ2 e−µ2 t dt = , i≥1 k! (λ + µ2 )k+1

(i)

243 This gives

Ai (z) =

∞ X

k=0

1 , i=0 1 + ρ1 (1 − z) 1 , i≥1 1 + ρ2 (1 − z)

      

ak,i z k =      

(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 X

qi aj−i+1,i

(iii)

i=1

and (see(15.212)) Q(z) =

∞ X

qj z j

j=0

= q0

∞ X

aj,0 z j +

∞ X

qi aj−i+1,i

j=0

j=0

= q0 A0 (z) +

∞ X i=1

qi z i

(iv) ∞ X

am,i z m z −1

m=0

= q0 A0 (z) + (Q(z) − q0 ) A1 (z)/z where (see (ii)) A0 (z) =

1 1 + ρ1 (1 − z)

(v)

A1 (z) =

1 . 1 + ρ2 (1 − z)

(vi)

q0 (z A0 (z) − A1 (z)) . z − A1 (z)

(vii)

and

From (iv) Q(z) =

244 Since h

Q(1) = 1 = =

0

0

q0 A0 (1) + A0 (1) − A1 (1) 0

1 − A1 (1)

i

q0 (1 + ρ1 − ρ2 ) 1 − ρ2

we obtain q0 =

1 − ρ2 . 1 + ρ 1 − ρ2

(viii)

Substituting (viii) into (vii) we can rewrite Q(z) as Q(z) = (1 − ρ2 ) =

Ã

(1 − z) A1 (z) 1 − z A0 (z)/A1 (z) 1 · A1 (z) − z 1 + ρ 1 − ρ2 1−z

1 − ρ2 1 − ρ2 z

!

1− 1 1 + ρ 1 − ρ2 1 −

ρ2 1+ρ1 ρ1 1+ρ1

z z

= Q1 (z) Q2 (z) (ix) where Q1 (z) =

∞ X 1 − ρ2 ρk2 z k = (1 − ρ2 ) 1 − ρ2 z k=0

and 1 Q2 (z) = 1 + ρ 1 − ρ2

Ã

ρ2 1− z 1 + ρ1

!

∞ X i=0

Ã

ρ1 1 + ρ1

!i

zi.

Finally substituting. Q1 (z) and Q2 (z) into (ix) we obtain 

qn = q 0 

n X i=0

Ã

ρ1 1 + ρ1

with q0 as in (viii).

!n−i

ρi2 −

n−1 X i=0

ρi+1 2



ρn−i−1 1  . n = 1, 2, · · · (1 + ρ1 )n−i

245 16.13 From (16-209), the Laplace transform of the waiting time distribution is given by Ψw (s) = =

1−ρ 1−λ

³

1−Φs (s) s

1−ρ ³

1 − ρµ

´

1−Φs (s) s

(i) ´.

Let Fr (t) = µ

Z

t 0

[1 − B(τ )]dτ

·

=µ t−

Z

t

(ii)

¸

B(τ )dτ .

0

represent the residual service time distribution. Then its Laplace transform is given by ΦF (s) = L {Fr (t)} = µ =µ

Ã

Ã

1 Φs (s) − s s

!

(iii)

!

1 − Φs (s) . s

Substituting (iii) into (i) we get Ψw (s) =

∞ X 1−ρ = (1 − ρ) [ρ ΦF (s)]n , 1 − ρ ΦF (s) n=0

|ΦF (s)| < 1.

(iv)

Taking inverse transform of (iv) we get Fw (t) = (1 − ρ)

∞ X

ρn Fr(n) (t),

n=0

where Fr(n) (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

246

0

ρ = A (1) > 1 where A(z) =

∞ X

ak z k

k=0

From Theorem 15.9 on Extinction probability (pages 759-760) it 0 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 substitution with xi = π0i we get ∞ X

j=1

pij xj =

∞ X

aj−i+1 π0j

(ii)

j=1

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 ∞ X

k=2−i

ak π0k+i−1 = π0i−1

∞ X

ak π0k

k=0

= π0i−1 π0 = π0i = xi

(iii)

since π0 satisfies (i). Thus if ρ > 1, the M/G/1 system is transient with probabilities σi = π0i . 16.15 (a) The transition probability matrix here is the truncated version of (15.34) given by

247



             P =            

a0 a1 a2 · · ·

·

am−2

1−

m−2 X

ak

     m−2 X  ak  1−   k=0   m−3  X  1− ak    k=0   ..   .   1 − (a0 + a1 )     k=0

a0 a1 a2 · · ·

·

am−2

0

·

am−3

a 0 a1 · · ·

.. .

.. .

.. .

.. .

.. .

0

0

0

· · · a0

a1

0

0

0

· · ·

a0

0



(i)

1 − a0

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 probabilities {qj∗ }m−1 j=0 satisfy qj∗

=

q0∗

aj +

j+1 X

qi∗ aj−i+1 ,

j = 0, 1, 2, · · · , m − 2

(ii)

i=1

and the normalization condition gives ∗ qm−1

=1−

m−2 X

qi∗ .

(iii)

i=0

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 {qj∗ }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 qj∗ = 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

248

c=

1 m−1 X

.

(v)

qi

i=0

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} =

∞ X

P {An ∩ Bk,n } =

n=k

∞ X

P {Bk,n |An }P (An ).

n=k

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 =

=

Z

Z

t 0

t 0

P (S > t − x|x = x)fx (x)dx

1Z t α(t) 1 (1 − B(τ )) dτ = [1 − B(t − x)] dx = t t 0 t

Thus Bk,n given An has a Binomial distribution, so that

P {Bk,n |An } =

à !

n k p (1 − pt )n−k , k t

k = 0, 1, 2, · · · n,

249 and P {X(t) = k} =

∞ X

e

−λt

n=k

= e−λt

(λt)n n n! k

à !Ã

α(t) t

!k µ

1Z t B(τ )dτ t 0

1Z t λt B(τ )dτ ∞ k X [λα(t)] t 0 k! (n − k)! n=k µ

h R

¶n−k

¶n−k

i

[λα(t)]k −λ t− 0t B(τ )dτ = e k! [λα(t)]k −λ R t [1−B(τ )]dτ e 0 = k! [λα(t)]k −λ α(t) = e , k = 0, 1, 2, · · · k! (i) (b) lim α(t) =

t→∞

Z

∞ 0

[1 − B(τ )]dτ (ii)

= E{s}

where we have made use of (5-52)-(5-53). Using (ii) in (i), we obtain lim P {x(t) = k} = e−ρ

t→∞

where ρ = λ E{s}.

ρk k!

(iii)