SOLUTIONS MANUAL FOR ANALYSIS OF QUEUES METHODS AND … · 2019-12-20 · SOLUTIONS MANUAL FOR ANALYSIS OF QUEUES METHODS AND APPLICATIONS 1ST EDITION GAUTAM SOLUTIONS SOLUTIONS MANUAL

SOLUTIONS MANUAL FOR ANALYSIS OF QUEUES METHODS AND APPLICATIONS 1ST EDITION GAUTAM

SOLUTIONS


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2 Solutions to Exercises in Chapter 2

1. The variance of the number of customers in the system in steady state for the M/M/s/K queue is

K∑j=0

pjj2 − L2

where

p0 =

[s∑

n=0

{1

n!(λ/µ)n

}+

(λ/µ)s

s!

K∑n=s+1

ρn−s

]−1with ρ = λ/(sµ) and for all j ∈ {0, 1, . . . ,K},

pj =

1j!

(λµ

)jp0 if j < s,

1s!sj−s

(λµ

)jp0 if s ≤ j ≤ K,

L =p0(λ/µ)sρ

s!(1− ρ)2[1− ρK−s − (K − s)ρK−s(1− ρ)] +

λ(1− pK)

µ.

2. The graphs of pK and W for various values of K (x-axis) are plotted in the figure below for the two

cases s = 10 (left) and s = 5 (right).

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 5 10 15 20 25 30 35

pK

0

0.1

0.2

0.3

0.4

0.5

0.6

0 5 10 15 20 25 30

pK

0.78

0.8

0.82

0.84

0.86

0.88

0.9

0.92

0.94

0.96

0.98

0 5 10 15 20 25 30 35

W

0

0.5

1

1.5

2

2.5

3

3.5

4

0 5 10 15 20 25 30

W

5



Clearly pK decreases with K and W increases with K. Notice that as K increases for the case s = 10

where ρ = λ/(sµ) < 1 pK and W converge to the M/M/10 values. However when s = 5, since ρ > 1,

as K increases W shoots off to infinity and pK converges to a non-zero point.

3. The generating function for the number of customers in the system in steady state is ψ(z) = 1−ρ1−ρz .

Using that L(2) = ψ′′(1) = 2ρ2

(1−ρ)2 . Likewise since the waiting time in steady state is according to

exp(µ− λ), we have W2 = 2(µ−λ)2 (by adding the variance and the square of the mean). Substituting

ρ = λ/µ, we get the result.

4. Without loss of generality assume that a service has just completed. Let X denote the time of the

next arrival and let Y denote the service time of the next customer. We would like to obtain the CDF

of U , the time of the next departure. Define F (x) = P (U ≤ x). Also, let Z be a random variable such

that Z = 0 if there are no customers in the system currently and Z = 1 otherwise. Then we have

F̃ (s) = E[e−sU ] = E[e−sU |Z = 0]P (Z = 0) + E[e−sU |Z = 1]P (Z = 1)

= E[e−s(X+Y )](1− ρ) + E[e−sY ]ρ

=

(λ

λ+ s

)(µ

µ+ s

)(1− λ

µ

)+

(µ

µ+ s

)(λ

µ

)=

λ

λ+ s

Inverting the LST, F (x) = P (U ≤ x) = 1− e−λt or U ∼ exp(λ).

5. The system can be modeled as a CTMC with rate diagram:

λ

0 1 2 3 4 5 6

λ λ λ λ λ λ

(1-q)µ (1-q)µ (1-q)µ (1-q)µ (1-q)µ (1-q)µ (1-q)µ

Since the above is equivalent to an M/M/1 queue with PP (λ) arrivals and exp((1− q)µ) service, the

stability condition is λ/[(1− q)µ] < 1. Also L = λ/[(1− q)µ− λ] and W = 1/[(1− q)µ− λ].

6. The problem can be stated as

Minimize C(µ) = cµ+hλ

µ− λSubject to µ > λ

Solving the problem using one of the NLP (non-linear programming) techniques, one can show µ∗ =

λ+√

hλc and C(µ∗) = cλ+ 2

√hλc.

6



7. Consider the following cases:

Case (i) Single M/M/2 system with PP (λ) arrivals and exp(µ) service for each server. The waiting

time in this case is W1 = 4µ/(4µ2 − λ2).

Case (ii) Two identical M/M/1 queues with PP (λ/2) arrivals and exp(µ) service for each server. The

waiting time in this case is W2 = 2/(2µ−λ). Note that the stability condition for both cases is 2µ > λ.

Try the following:

W1 −W2 = 4µ/(4µ2 − λ2)− 2/(2µ− λ)

=−2λ

(2µ− λ)(2µ+ λ)< 0

Thus W1 < W2 and therefore it is better to go with case (i), the single queue system.

8. Let X(t) = number of customers in the system. {X(t), t ≥ 0} is a birth and death process with

parameters λi =

{λ1 + λ2 0 ≤ i < Kλ2 i ≥ K and µi = µ, i ≥ 1.

Let ρj =

(λ1 + λ2

µ

)j0 ≤ j < K(

λ2µ

)j−K (λ1 + λ2

µ

)Kj ≥ K.

The system is stable when

∞∑i=0

ρi =

K−1∑i=0

(λ1 + λ2

µ

)i+

∞∑i=K

(λ2µ

)i−K (λ1 + λ2

µ

)K< ∞ ⇔ λ2 < µ.

Assuming stability, p0 = 1

/(1− ((λ1 + λ2)/µ)K

1− (λ1 + λ2)/µ+

(λ1 + λ2

µ

)K1

1− λ2/µ

)and pj = ρjp0, j ≥ 1.

Due to PASTA we have the probability an arbitrary arrival of type 2 sees j customers ahead as pj .

Hence assuming FCFS we have

E[W2] =

∞∑j=0

j + 1

µpj =

p0µ

[1− (K + 1)(ρ1 + ρ2)K +K(ρ1 + ρ2)K+1

(1− ρ1 − ρ2)2+

[K(1− ρ2) + 1](ρ1 + ρ2)K

(1− ρ2)2

].

Again due to PASTA the probability a potential arrival of type 1 sees j customers ahead is pj . But

the probability a type 1 arrival enters is

K−1∑j=0

pj = 1 − p0(ρ1 + ρ2)K

1− ρ2. Thus the probability a type 2

“entry” sees j customers ahead is pj/[1 − p0 (ρ1+ρ2)K

1−ρ2 ] for j = 0, 1, . . . ,K − 1. Again assuming FCFS

we have

E[W1] =

∑K−1j=0

j+1µ pj

1− p0 (ρ1+ρ2)K

1−ρ2

=1

µ

[1− (K + 1)(ρ1 + ρ2)K +K(ρ1 + ρ2)K+1

(1− ρ1 − ρ2)[1− (ρ1 + ρ2)K ]

].

9. The following is the rate diagram of the birth and death process.

λ

0 1 2 3 4 5 6

λ λ λ λ λ λ

µ1 µ1 µ1 µ2 µ2 µ2 µ2

7



10. The fraction of time the server is busy is ρ, i.e. λ/µ. Hence E[B]E[B]+E[I] = λ

µ , although this is intuitive

the result comes from alternating renewal processes result. From the above we have

E[B]/E[I] =ρ

1− ρ=

λ

µ− λ.

Since the idle times are according to exp(λ) we have E(B) = 1µ−λ .

11. Let X(t) =

{j if there are j customers in the system, j = 0, 2, 3, . . .

(1, i) if there is 1 customer in the system at server i, i = 1, 2

{X(t), t ≥ 0} is a CTMC with rate diagram

λ0

1,1

2 3 4 5 6

λλ λ λ λ λ

λµc1 µc2

λµc1

1,2µc2 µ(c1+c2) µ(c1+c2) µ(c1+c2) µ(c1+c2) µ(c1+c2)

From the balance equations λp0 = c1µp1,1 +c2µp1,2, (λ+c1µ)p1,1 = λp0 +c2µp2, (λ+c2µ)p1,2 = c1µp2

and λpk = (c1 + c2)µpk+1, k ≥ 2, we have

p0 =c1c2µ

2(2λ+ (c1 + c2)µ)

λ2(λ+ c2µ)p2, p1,1 =

c2µ(λ+ (c1 + c2)µ)

λ(λ+ c2µ)p2,

p1,2 =c1µ

λ+ c2µp2 and pk =

(λ

(c1 + c2)µ

)k−2p2, k ≥ 3.

The normalizing equation is p0 + p1,1 + p1,2 + p2

∞∑j=0

(λ

(c1 + c2)µ

)j= 1. Thus, the CTMC is stable

when λ < (c1 + c2)µ and p2 =

[(c1 + c2)(λ2µ+ c1c2µ

3) + c2λµ2(3c1 + c2)

λ2(λ+ c2µ)+

(c1 + c2)µ

(c1 + c2)µ− λ

]−1.

12. First for L, our conjecture is that system 1 has a higher number in the system than system 2, i.e.

λ2

sµ(sµ− λ)+

λ

sµ<p0

(λµ

)sλsµ

s!(sµ− λ)2+λ

µ,

where p0 =

[s−1∑n=0

{1

n!(λ/µ)n

}+

(λ/µ)s

s!

1

1− λ/(sµ)

]−1. Rearranging the terms appropriately, the

above is true if the following is true

− 1

s!

(λ

µ

)ssµ

sµ− λ< (s− 1− λ/µ)/p0.

Rewriting the above inequality, we need the following to be true in order for the conjecture to be true:

(s− 1− λ/µ)

s−1∑n=0

{1

n!(λ/µ)n

}+

(λ/µ)s

(s− 1)!> 0

which is true because the LHS can be written as

(s− 1) +

s−1∑n=1

{1

(n− 1)!(λ/µ)n

[s− 1

n− 1

]}

8



that is a positive quantity.

Now for the Lq values, the following steps indicate that system 1 is better (i.e. lower Lq):

[1− λ/(sµ)]

s−1∑n=0

{1

n!(λ/µ)n

}> 0

(1− λ/(sµ))

[s−1∑n=0

{1

n!(λ/µ)n

}+

(λ/µ)s

s!

1

1− λ/(sµ)

]>

(λ/µ)s

s!

1− λ/(sµ) > p0(λ/µ)s

s!

λ2

sµ(sµ− λ)>

p0

(λµ

)sλsµ

s!(sµ− λ)2

13. Let X(t) = number of failed machines. {X(t), t ≥ 0} is a CTMC with rate diagram

2λ

0 1 2

λ

µ µ

and steady state distribution p0 =µ2

µ2 + 2λµ+ 2λ2, p1 =

2λµ

µ2 + 2λµ+ 2λ2and

p2 =2λ2

µ2 + 2λµ+ 2λ2. The long run rate at which the system earns profits = 2rp0+rp1−c2λp0−cλp1 =

2µ(λ+ µ)(r − cλ)

µ2 + 2λµ+ 2λ2.

14. To use the formula for L, first calculate, E[X], E[X2] and ρ.

(a) E[X] = 1/(1− q), E[X2] = (1 + q)/(1− q)2 and L = λ/[(1− q)(µ− qµ− λ]

(b) E[X] = K, E[X2] = K2 and L = λK(K + 1)/[2(µ− λK)]

(c) E[X] = (K + 1)/2, E[X2] = (2K + 1)(K + 1)/6 and L = λ(K + 2)(K + 1)/[3(2µ−−λ− λK)]

(d) E[X] = θ/(1− e−θ), E[X2] = θ(θ + 1)/(1− e−θ) and L = 0.5λθ(θ + 2)/[µ(1− e−θ)− λθ]

15. The number of customers in the system process, {X(t), t ≥ 0} is a CTMC on S = {0, 1, 2, . . .} with

rate matrix Q =

−λ λ(1− q) λq(1− q) λq2(1− q) λq3(1− q) · · ·µ −(λ+ µ) λ(1− q) λq(1− q) λq2(1− q) · · ·0 2µ −(λ+ 2µ) λ(1− q) λq(1− q) · · ·0 0 3µ −(λ+ 3µ) λ(1− q) · · ·...

. . .. . .

.

Multiplying the balance equations by zk and summing

9



z0[ λp0 = µp1 ]z1[ (λ+ µ)p1 = 2µp2 + λ(1− q)p0 ]z2[ (λ+ 2µ)p2 = 3µp3 + λ(1− q)q1 + λq(1− q)p0 ]z3[ (λ+ 3µ)p3 = 4µp4 + λ(1− q)q2 + λq(1− q)p1 + λq2(1− q)p0 ]

.... . .

λP (z) + zµP ′(z) = µP ′(z) + λ(1− q)zP (z)(1 + qz + q2z2 + · · · )

= µP ′(z) +λ(1− q)zP (z)

1− qz

By taking the logarithm and differentiating it is easy to show that P (z) =[

1−q1−qz

] λµq

is a solution to

the above differential equation.

16. In state {0, 0} the only scheduled event is a customer arrival, which occurs at rate λ, and is of type 1

with probability α and of type 2 with probability 1− α. Thus from state {0, 0}, the system moves to

state {1, 1} at rate αλ, and to state {1, 2} at rate (1− α)λ.

In state {i, j}, there are two scheduled events, an arrival and a departure. Arrivals are handled in the

method described above. Departures occur at rate µj , and thus the system goes to state {i− 1, 1} at

rate αµj and to state {i− 1, 2} at rate (1− α)µj . As all arrival and departure times are independent

and exponentially distributed, {X(t), Y (t), t ≥ 0} is a CTMC. The rate diagram is:

αλ 11 21 31

λ λ λ

00

12 22 32

µ1

λ λ λ

αµ1 αµ1 αµ1

βλ

βµ1αµ2 αµ2

βµ1

βµ2 βµ2 βµ2

µ2

The balance equations at the different nodes can be written as:

−λp00 + µ1p11 + µ2p12 = 0

−λp11 − µ1p11 + λαp00 + µ1αp21 + µ2αp22 = 0

−λp12 − µ2p12 + λβp00 + µ2βp22 + µ1αp21 = 0

...

For i ≥ 2, we can write

−λpi1 − µ1pi1 + λαpi−1,1 + µ1αpi+1,1 + µ2αpi+1,2 = 0.

10



Multiplying the above by zi and summing, we get

−λ(ψ1(z)−p11z)−µ1(ψ1(z)−p11z)+λzψ1(z)+µ1α

z(ψ1(z)−p11z−p21z2)+

µ2α

z(ψ2(z)−p12z−p22z2) = 0.

By suitably using the first 3 balance equations mentioned earlier, we get

[−λ− µ1 + λz + µ1α/z]ψ1(z) +µ2α

zψ2(z)− λα(1− z)p00 = 0.

Likewise, solving for nodes pi2 we get

[−λ− µ2 + λz + µ2β/z]ψ2(z) +µ1β

zψ1(z)− λβ(1− z)p00 = 0.

The previous two equations can be solved simultaneously to get

ψ1(z) =λα(λ(1− z) + µ2)

µ1µ2/z − λµ1(1− α/z)− λµ2(1− β/z)− λ2(1− z)p00

The expression for ψ2(z) is obtained by swapping α with β and µ1 with µ2. Using the relation

p00 + ψ1(1) + ψ2(1) = 1, p00 = 1− λβµ2− λα

µ1and the condition of stability corresponds to 0 < p00 < 1

which can be stated as λ(α/µ1 + β/µ2) < 1.

17. By multiplying the ith balance equation by zi for i = 0, 1, . . ., and summing we get

λΨ(z) =2µ

z[Ψ(z)− p0]− µp1.

Multiplying by z and rearranging the terms we get

Ψ(z) =2µp0 + µzp1

2µ− λz.

Since Ψ(1) = 1 we get substituting z = 1 in the above equation

2µp0 + µp1 = 2µ− λ.

But we also know from the first balance equation that λp0 = µp1. Hence by solving the two simulta-

neous equations we get p0 = 2µ−λ2µ+λ and p1 = λ(2µ−λ)

µ(2µ+λ)

18. Let the states of the retrial queue CTMC be (0, 0), (0, 1), (0, 2), (0, 3), . . ., (1, 0), (1, 1), (1, 2), (1, 3),

. . .

The balance equations from the arc cuts can be generalized as

λp1,i = (i+ 1)θp0,i+1 for i ≥ 0

(λ+ iθ)p0,i = µp1,i for i ≥ 0

11



The above equations can be solved in terms of p00 as follows:

p0,i+1 = p00

(λ

µ

)i+11

(i+ 1)!

i∏j=0

[λ

θ+ j

]for i ≥ 0

p1,i = p00θ

µ

(λ

µ

)i1

i!

i∏j=0

[λ

θ+ j

]for i ≥ 0

Since∞∑i=0

p0,i +

∞∑i=0

p1,i = 1,

we have

p00 +

∞∑i=0

{(i+ 1)θ

λ+ 1

}p0,i+1 = 1

p00

1 +

∞∑i=0

θ

λ

{λ

θ+ (i+ 1)

}(λ

µ

)i+11

(i+ 1)!

i∏j=0

[λ

θ+ j

] = 1

p00

1 +θ

λ

∞∑i=0

(λ

µ

)i+11

(i+ 1)!

i+1∏j=0

[λ

θ+ j

] = 1

p00

1 +

∞∑i=0

(λ

µ

)i+11

(i+ 1)!

i+1∏j=1

[λ

θ+ j

] = 1

p00

[(1− λ

µ

)−λθ−1]= 1

The last equation follows from

(1 + x)a = 1 + ax+ a(a− 1)x2/2! + a(a− 1)(a− 2)x3/3! + a(a− 1)(a− 2)(a− 3)x4/4! + . . .

which can be used to write

(1− x)−a = 1 + ax+ a(a+ 1)x2/2! + a(a+ 1)(a+ 2)x3/3! + a(a+ 1)(a+ 2)(a+ 3)x4/4! + . . .

19. If we let the occupancy limit to go from 4 to infinite, we have two independent queues: queue 1 would

be an M/M/1 queue with steady state probability of having j in the system being (1 − λ1/µ1)(λ1

µ1)j ,

and queue 2 would be an M/M/2 queue with steady state probability of having j in the system being

λ2(2µ2−λ2)µ2(2µ2+λ2)

( λ2

2µ2)j−1, when j > 0 and 2µ2−λ2

2µ2+λ2when j = 0. Also, the stochastic process corresponding

to the number in the system for the two queues (X1(t) and X2(t)) are reversible. Hence the joint

CTMC {(X1(t), X2(t)),−∞ < t < ∞} is reversible. That implies that the truncated process stated

in this question when the occupancy limit is 4 is also reversible. The state space of this truncated

12



reversible CTMC is {00, 01, 02, 03, 04, 10, 11, 12, 13, 20, 21, 22, 30, 31, 40}. Using the reversibility results

the steady state probability there are i customers in queue 1 and j in queue 2 is given by

pij =

R(λ1

µ1)i λ2

µ2( λ2

2µ2)j−1 if j > 0 and i+ j ≤ 4

R(λ1

µ1)i if j = 0 and i ≤ 4

0 otherwise

where R is given by

R =1

1 + λ1

µ1+

λ21

µ21

+λ31

µ31

+λ41

µ41

+ λ2

µ2+ λ1λ2

µ1µ2+

λ21λ2

µ21µ2

+λ31λ2

µ31µ2

+λ22

2µ22

+λ1λ2

2

2µ1µ22

+λ21λ

22

2µ21µ

22

+λ32

4µ32

+λ1λ3

2

4µ1µ32

+λ42

8µ42

.

20. The bivariate stochastic process {(X1(t), X2(t)), t ≥ 0} is a CTMC with state space {(0, 0), (1, 0), (0, 1), (1, 1), (2, 1), (1, 2), (2, 2), (3, 2), (2, 3), . . .}.

The rate diagram is:

λ

0,0

1,0 2,1

2,2

3,2

0 1

1,1

1 2 2 3

λ/2 λ/2 λ/2

λ/2 λ/2 λ/2

λ

λλ

µ µ

µ

2µ

2

µ

µ

2µ

20,1 1,2 2,3µ 2µ µ 2µ

To compute the steady-state probabilities it is critical to realize that pi,j = pj,i for all i and j due to

symmetry. Now performing arc cuts and solving the balance equations, one can obtain for n = 0, 1, . . .

p0,0 =2µ− λ2µ+ λ

pn,n+1 = pn+1,n =1

2

λ(2µ− λ)

µ(2µ+ λ)(λ

2µ)2n

pn+1,n+1 =λ(2µ− λ)

µ(2µ+ λ)(λ

2µ)2n+1

An alternative way to obtain the above result is to realize that {X1(t) + X2(t), t ≥ 0} is the queue

length process of an M/M/2 queue. Then using the steady state results of that we can obtain the

corresponding values here. It may also be possible to use reversibility argument to solve this.

21. The number of customers in the system at time t is a CTMC with generator matrix Q = [qij ] such

that

qij =

λ/2 if j = i+ 1 or j = i+ 2µ if j = 0 and i = 1 or j = i+ 2 and i > 10 otherwise.

Balance equation at node 0 is

λp0 = µp1 + µp2.

Balance equation at node 1 is

(λ+ µ)p1 = µp3 + λ/2p0.

13



For node i > 1 the balance equations are

(λ+ µ)pi = µpi+2 + λ/2(pi−1 + pi−2).

The generating function of the number of customers in the system in steady state is

Ψ(z) =2µp0 + 2µp1z − 2µp1z

2 − 2µp0z2

2µ+ λz3 + λz4 − 2(λ+ µ)z2.

Now consider the special case λ = µ. Since Ψ(1) = 1 we have 2p1−4(p0 +p1) = −1. Also, since Ψ(z) is

a continuous and finite function of z for all finite z, whenever the demoninator is zero, the numerator

should also be zero. The only root of the denominator that would result in meaningful p0 and p1 values

is z = 0.6889. Using that and solving simultaneous equations for p0 and p2 we get p0 = 0.2039 and

p1 = 0.0921. Also L = 5 + 2(p0 + p1) = 5.592 and W = L3λ/2 = 3.728.

22. The number in the system can be modeled as a birth and death process with birth rates λ and death

rates in state i (for all i > 0) equal to µ + (i − 1)γ. Using that, say p0, p1, . . . are the steady state

probabilities there are 0, 1, . . . respectively in the system.

(a) Average departure rate after service is µ(1 − p0) and the average departure rate without service

is λ − µ(1 − p0), since the two must add to λ. Thus the fraction of arrivals that are served is

µ(1− p0)/λ.

(b) Using the probability of first failure property of exponential distribution, the customer in service

would complete before arriving customer abandons is µ/(µ+ γ). That would the probability that

this arriving customer would be served.

(c) Let Lq be the average number of customers in the queue (not counting any at the server) and it

can be computed as Lq = 0p0 + 0p1 + 1p2 + 2p3 + . . .. Notice that Lq includes customers that get

served and those that abandon without service. Thus the average time a customer that waits in

the queue before beginning service is Lq/[µ(1− p0)].

23. Writing down the balance equations we get λp0 = µp1; (µ + λ)p1 = 2µp2; (2µ + λ)p2 = 2µp3;

(2µ + λ)p3 = 2µp4; (2µ + λ)p4 = 2µp5 + λp0; (2µ + λ)p5 = 2µp6 + λp1; (2µ + λ)p6 = 2µp7 + λp2; . . .

which we multiply by z0, z1, z2, . . ., and sum to get

λP (z) + 2µP (z)− 2µp0 − µp1z =2µ

z(P (z)− p0)− µp1 + λz4P (z).

Using the first balance equation, we can substitute for p1 as λp0/µ. Then we can write

P (z) =2µ(1− z)p0 + λz(1− z)p0λz5 + 2µ− 2µz − λz

.

14



Using the fact that P (1) = 1, we can get p0 = 4−4ρ4+ρ where ρ = 2λ/µ. Substituting for p0, we get

P (z) =(1− z)(1− ρ)(16 + 4ρz)

(4 + ρ)(ρz5 + 4− 4z − ρz).

24. TRUE or FALSE?

(a) True. Since L = ρ/(1− ρ) and both queues have the same ρ.

(b) False. In steady state both queues have average departure rate of λ which is the arrival rate.

(c) False. Although the X(t) process is identical to an M/M/∞ queue, there is only one server. So

if X(t) = 5, then there are 4 in the queue. hence Lq 6= 0.

(d) True. The average sojourn time in the M [X]/M/1 queue is (N+12 ) 1

µ−Nλ which is larger than the

sojourn time for the M/M/1 queue which is 1/(µ−Nλ) for all N ≥ 1.

(e) True. The workload in the system at any time would not depend on the discipline. In addition,

due to PASTA the average workload as seen by an entering customer (which is equal to Wq of the

M/M/1/∞/FCFS queue) is the time-averaged workload which due to Ergodicity is the workload

seen at an arbitrary point in steady state.

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SOLUTIONS MANUAL FOR ANALYSIS OF QUEUES METHODS AND … · 2019-12-20 · SOLUTIONS MANUAL FOR ANALYSIS OF QUEUES METHODS AND APPLICATIONS 1ST EDITION GAUTAM SOLUTIONS SOLUTIONS MANUAL

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