Optimal Service Capacity in a Multiple-server
Queueing System: A Game Theory Approach
Wai-Ki Ching
∗
Sin-Man Choi†
Min Huang‡
1 December 2008
Abstract
The study of economic behavior of service providers in a competition environment is an
important and interesting research issue. A two-server service network has been proposed
in Kalai et al. [10] for this purpose. Their model aims at studying the role and impact of
service capacity in capturing larger market share so as to maximize the long-run expected
profit. A Markovian queueing system of two servers is used in their model and analysis.
They formulate the problem as a two-person strategic game and analyze the equilibrium
solutions. The main aim of this paper is to extend the results of the two-server queueing
model to the case of a general multiple-server queueing model. Here we will focus on the
case when the queueing system is stable.
Keywords: Capacity Allocation, Competition, Markovian Queueing Systems, Nash
Equilibrium.
1
Introduction
The problem of finding the optimal strategy and control policy of a queueing system
is a traditional mathematical problem and has been well studied in the literature, see for
instance [2, 8, 9, 10, 11, 16]. In an optimal control problem, it usually involves making
decisions on system parameters such as the system service capacity and number of servers
in the system under a specified cost structure (convex or concave). Here service capacity
is an important competitive factor in the design of a system, for example, in the areas
of telecommunication networks [6], data transmission systems [10] and Vendor-Managed
∗
Advanced Modeling and Applied Computing Laboratory, Department of Mathematics, The University of
Hong Kong, Pokfulam Road, Hong Kong. E-mail: [email protected]. Research supported in part by RGC
Grants, HKU CRCG Grants, Hung Hing Ying Physical Sciences Research Fund and HKU Strategic Research
Theme Fund on Computational Sciences. The preliminary version of the paper has been accepted for publication
in the proceedings of the COMPLEX2009 conference, Shanghai, China.
†
Advanced Modeling and Applied Computing Laboratory, Department of Mathematics, The University of
Hong Kong, Pokfulam Road, Hong Kong. E-mail: [email protected]
‡
College of Information Science and Engineering, Northeastern University, Shenyang, 110004, China. Key
Laboratory of Integrated Automation of Process Industry, Ministry of Education, Shenyang, 110004, China.
Email: [email protected]
1
Inventory (VMI) system [3, 15]. In particular, the current development in supply chain
management emphasizes the coordination and integration of inventory and transportation
logistics [4, 17]. VMI is a supply chain initiative where the distributor is responsible for
all decisions regarding the selection of retailers or agents. This creates a competitive
environment for the agents and retailers to compete in the market [13].
Regarding the service capacity, Kalai et al. [10] studied a strategic game of two
servers competing for their market shares through determining their service capacities. A
Markovian queueing system of two servers is used in their model and analysis. Markovian
queueing systems are popular tools for modeling service systems as they are mathematically tractable [6, 7] when compared to non-Markovian queueing systems. The problem is
then analyzed using game theory [14]. Game theory is a popular and promising approach
[1, 5] for the captured problem. They classified the Nash equilibria into three different
cases concerning the cost function and the revenue per customer. The waiting time is
finite in one of these cases and there is a unique symmetric equilibrium. Although their
model is simple, it brings in two important concepts. The first one is the “competitive
game of servers” and the second one is the “market share of a server in a multi-server
facility”. Furthermore, they also report that when the marginal cost of providing service is “high”, there is a unique symmetric equilibrium and the total service capacity is
less than the mean demand rate. In such a case, each server actually behaves as if it
were a monopolist. Competition therefore has no effect and this leads to an undesirable
situation. On the other hand, when the marginal cost of providing service is “low”, a
unique symmetric equilibrium exists and the total service capacity is greater than the
mean demand rate.
The remainder of the paper is structured as follows. In Section 2, we will give a
brief review on the two-server queueing system discussed in [10]. We present the general
multiple-server queueing model and our analysis on the system performance in Section 3.
Finally concluding remarks are given to address further research issues in Section 4.
2
A Review on the Two-Server Queueing Model
The service system studied by Kalai et al. [10] consists of two independently operated
servers. Customers arrive according to a Poisson process of rate λ and the service times
are assumed to follow the exponential distribution. Each of the server i operates independently and determines its own service capacity µi so as to maximize its own profit.
The cost to operate at service capacity µ is c(µ). Here the operating cost function c(.) is
assumed to be an increasing and strictly convex function, i.e., both c′ (µ) and c′′ (µ) are
positive and an example of such a function is c(µ) = µ2 .
The servers earn a fixed amount R for each unit of service rendered. The queueing
system consists of a single First-In-First-Out queue. If a customer arrives when both
servers are idle, he/she is assigned to either server with equal likelihood. No server is
2
allowed to be idle when at least one customer in the system. If a customer arrives when
one server is idle and the other is busy, he/she will be assigned to the idle server. In the
followings, we give a brief review on the queueing models discussed in [10].
2.1
The System Steady-state Probability Distribution
If Server i (i = 1, 2) chooses service capacity µi and such that
µ1 + µ2 > λ
(1)
the system has a steady-state probability distribution. We remark that condition (1) is
a necessary and sufficient condition for an Markovian queueing system to be stable or
to have steady-state probability distribution. Let Pn be the probability that there are n
customers in the system; P10 be the probability that Server 1 is busy and Server 2 is idle;
P01 be the probability that Server 2 is busy and Server 1 is idle. By studying the balanced
equations of the queueing system, the following results are obtained:
P0 =
and
P10 =
1−ρ
1−ρ+
λP0
2µ1
where
ρ=
(2)
λ(µ1 +µ2 )
2µ1 µ2
and P01 =
λP0
2µ2
λ
(µ1 + µ2 )
(3)
(4)
is the system load. Moreover, one also has
P1 = P10 + P01
(5)
and
Pn = ρn−1 P1
2.2
n = 2, 3, . . . .
(6)
The Market Share
Computing the market share of Server i is equivalent to computing the mean number
of customers per time unit that enter service with Server i. Using the results in Section
2.1.1, if µ1 + µ2 > λ, then the mean number of customers per time unit that enter service
with Server 1 is
λ
(7)
P0 + P01 λ + P3 µ1 + P4 µ1 + . . .
2
and that with Server 2 is
P0
λ
+ P10 λ + P3 µ2 + P4 µ2 + . . . .
2
3
(8)
We then divide these two expressions by the mean number of customers per time unit
that enter service, i.e., λ, to obtain the market share of Server i. Thus the fraction of all
customers served by Server i(i = 1, 2), is given by
αi (µ1 , µ2) =
2.3
λµ2i + µ1 µ2 (µ1 + µ2 )
.
λ(µ1 + µ2 )2 + 2µ1 µ2 (µ1 + µ2 − λ)
(9)
The Profit Function
Given the market shares of the servers in Section 2.1.2, the profit function πi (µ1 , µ2 )
of Server i ∈ {1, 2}, the expected profit per time unit earned by Server i, is then given by
(
Rλαi (µ1 , µ2 ) − c(µi ) if µ1 + µ2 > λ
πi (µ1 , µ2 ) =
(10)
Rµi − c(µi )
if µ1 + µ2 ≤ λ.
Here c(µ) is the cost per time unit of providing service at capacity µ and R is the revenue
that the server earns for each customer served.
2.4
The Equilibrium
Kalai et al. [10] considered the situation as a two-person strategic game and found
that finite waiting times exist at equilibrium if and only if
λ
R
′
c
< .
(11)
2
2
Moreover, if this condition is satisfied, then a unique equilibrium exists in which both
servers select the same service capacity µc = µ1 = µ2 , such that
c′ (µc ) =
3
Rλ2
.
2µc (2µc + λ)
(12)
The General Multiple-server Queueing System
In this section, we extend the two-server queueing system studied in [10] to a n-server
queueing system. The arrival process of customers is assumed to be a Poisson process.
In this queueing system, arriving customers wait in a single First-In-First-Out (FIFO)
queue if all servers are busy. No server is allowed to be idle when there is at least one
customer in the queueing system. If a customer arrives when more than one server is idle,
the customer is assigned to any of the idle servers with equal likelihood. Once a server
completes the service of a customer, the first customer in the queue, if any, is assigned
to the server. Each server i may choose its own service capacity µi , and its service time
follows the exponential distribution with mean 1/µi. The servers earn a revenue of R for
each customer served, and each of them incurs a cost of c(µ) to operate at service capacity
µ.
4
In the following subsections, we present some important properties of the multipleserver queueing system through the propositions. The proofs of the propositions can be
found in the Appendix.
3.1
The Steady-state Probability Distribution of the Queueing System
Given the service capacities µ1 , . . . , µn and the mean demand rate λ, suppose
n
X
µi > λ.
i=1
This condition is to guarantee that the queueing system is stable and the system steadystate probability distribution exists. We would like to obtain the steady-state probability
distribution of the number of customers in the system. Let us give the following definitions.
Let Pi be the steady-state probability of having i customers in the system, where i =
0, 1, 2, . . .. Also let Ps , where s = (s1 , s2 , . . . , sn ) and si = 0 or 1, be the steady-state
probability of having si customers at Server i. We note that by definition
Pk =
X
Ps
for k = 0, 1, . . . , n.
(13)
{s|s1 +...+sn =k}
We establish the balanced equations governing the steady-state probability distribution. The equations can be obtained by equating the incoming rate and outgoing rate at
P
each of the state. For si = 0, 1 and ni=1 si 6= n, we have
(
X
µi + λ)P(s1 ,s2 ,...,sn ) =
{i|si =1}
X
µi P(s−i ,si =1) +
{i|si =0}
X
{i|si =1}
λP(s−i ,si =0)
.
|{j|sj = 0}| + 1
(14)
where (s−i , s′i ) denotes (s1 , . . . , si−1 , s′i , si+1 , . . . , sn ). When si = 0 for all i (14) gives
λP(0,0,...,0) = µ1 P(1,0,...,0) + µ2 P(0,1,0,...,0) + · · · + µn P(0,...,0,1)
(15)
For the states with at least n customers we have
(
n
X
µi + λ)P(1,1,...,1) =
i=1
and
(
n
X
i=1
n
X
µi Pn+1 +
i=1
n
X
λP(s−i =1,si =0)
(16)
i=1
n
X
µi + λ)Pk = (
µi )Pk+1 + λPk−1 for k = n + 1, n + 2, . . . .
(17)
i=1
We note that these two equations together are equivalent to
(
n
X
i=1
n
X
µi + λ)Pk = (
µi )Pk+1 + λPk−1 for k = n, n + 1, . . . , .
i=1
5
(18)
We also have the normalization equation
∞
X
Pi = 1.
(19)
i=0
It can be shown by direct verification that the solution is given by
Proposition 1. We have
P(s1 ,s2 ,...,sn ) =
and
(n − k)!λk P0
Q
n! {i|si =1} µi
where
Pk = ρk−n Pn
k = s1 + s2 + . . . + sn > 0
(20)
for k > n
(21)
and
P0 =
1+
n−1
X
(n − k)!λk (
k=1
P
i1 <i2 <...<in−k
µi1 µi2 . . . µin−k )
n!µ1 µ2 . . . µn
1
λn
+
1 − ρ n!µ1 µ2 . . . µn
!−1
.
(22)
The steady-state probability distribution describes the long-run behavior of the system.
Each of these probabilities Pk represents the long-run proportion of time that there are
k customers in the system. They are essential in studying how each server determines
its strategy to maximize its long-run profit. In the next subsection, we will write the
market share of each server in terms of these probabilities and obtain an expression for
the market share.
3.2
The Market Share
We derive the market share of each server from the steady-state distribution. We note
P
that when nj=1 µj ≤ λ, i.e., customers arrive at least as fast as the servers can serve them.
the steady-state probability distribution does not exist and the queue is infinite. In this
case, each server receives customers at its service capacity in the long run. Otherwise,
Pn
j=1 µj > λ and all customers will be served. Each server only receives a fraction of the
arriving customers, at a rate lower than its service capacity. The server’s profit is thus
affected by the fraction of all customers it serves, i.e. its market share.
When k(1 ≤ k ≤ n) servers are idle, customers arrive at a rate of λ and an arriving
customer is served by any one of the k idle servers with equal likelihood. Each of these
idle servers therefore receives customers at a rate of λ/k. On the other hand, when all
servers are busy with at least one customer waiting in the system, each of the busy servers
i receives a new customer when it completes the service for a customer, i.e. at a rate of
its service capacity µi .
To obtain the market share, we find the expected value of the server’s rate of receiving
customers in different states of the systems, taking expectation over the steady-state
6
probabilities. In the following, we give the formula for the market share for an individual
server.
Proposition 2. If
n
X
µj > λ,
j=1
the market share of Server i, αi (µ1 , µ2 , . . . , µn ) is given by




n−1
X
X
ρ 
n−1
n−k−1 


)
+
λ
(
k!λ
µ
µ
.
.
.
µ
µi 
j
j
j
1
2
k

1−ρ 
k=0
j1 <j2 <...<jk ,jp 6=i ∀p
!
.
n
n
X
X
λ
k!λn−k
µj 1 µj 2 . . . µj k +
1−ρ
j <j <...<j
k=1
1
2
(23)
k
We note that when µi → ∞, we have the market share of Server i
αi (µ1 , µ2 , . . . , µn ) i = 1, 2, . . . , n
tends to the following limit
n−1
X
k=0
n−1
X
k=0

k!λn−k−1 
X
j1 <j2 <...<jk ,jp 6=i∀p

X
(k + 1)!λn−k−1 

µj 1 µj 2 . . . µj k 
j1 <j2 <...<jk ,jp 6=i∀p
.
µj 1 µj 2 . . . µj k 
As we focus on the case when the mean demand rate is less than the total service
rate, the market share is directly tied to the profit of a server. Before formulating the
profit function of a server, we state the following two propositions related to the partial
derivatives of the market share αi with respect to µi. These will be useful in determining
the Nash equilibrium of the system when we considered the system as a n-player strategic
game.
Proposition 3. Suppose that
n
X
µj > λ
j=1
then
∂αi (µ1 , µ2 , . . . , µn )
> 0.
∂µi
Furthermore, when µi → ∞, we have
∂αi (µ1 , . . . , µn )
→ 0.
∂µi
7
(24)
Proposition 4. Suppose that
n
X
µj > λ,
j=1
then
∂ 2 αi (µ1 , µ2, . . . , µn )
< 0.
∂µ2i
(25)
Propositions 3 and 4 together mean that the market share αi is increasing and concave
with respect to µi (i = 1, 2, . . . , n).
3.3
The Profit Function
Here we proceed to find out the profit function of an individual server, which represents
the server’s profit per time unit in the long run. There are two cases to be considered.
When
n
X
µj > λ,
j=1
Server i receives customers at a rate of λαi (µ1 , µ2 , . . . , µn ). When
n
X
µj ≤ λ,
j=1
Server i receives customer at a rate of µi . In both cases, Server i incurs a cost of c(µi).
Therefore, the profit function of Server i takes a similar form as that in [10] and is given
by

n
X



µj > λ
Rλα
(µ
,
µ
,
.
.
.
,
µ
)
−
c(µ
)
if
i
1
2
n
i


j=1
(26)
πi (µ1 , µ2, . . . , µn ) =
n
X



if
µj ≤ λ.

 Rµi − c(µi )
j=1
Each of the servers aims to maximize its long-run profit when determining its service capacity. Therefore, how a server’s profit changes with its service capacity (when
other servers’ capacities remain unchanged) is important in characterizing the server’s
decision. By proposition 3 and 4, we readily obtain the following proposition describing
the properties of the profit function πi with respect to µi.
Proposition 5. For i = 1, 2, . . . , n, for each fixed λ > 0 and µj > 0 for j 6= i, the
function πi (µ1 , µ2 , . . . , µn ) is continuous and strictly concave in µi .
The continuity and concavity of the profit function ensure that the first-order condition
is a sufficient condition for a value of µi to maximize the profit function.
8
3.4
The Nash Equilibrium of the Queueing System
Since servers’ decisions of their service capacities would affect the profit of each other,
we model the situation as an n-player strategic game, in which each server i simultaneously chooses its service capacity µi to maximize its profit πi . Here we discuss the Nash
equilibrium of the system. In the two-server model in [10], a unique symmetric equilibrium is found in the case when the total demand rate is less than the total service rate.
In our analysis, we will show that, similar to the two-server case, when the marginal cost
is low enough, there is a unique equilibrium, in which all servers choose the same service
capacities. In the following, we will first look at how the profit of Server i changes with
its service capacity when all other servers choose the same service capacities.
Proposition 6. For µc > λ/n,



λ
∂
1 − n−1
αi (µ1 , µ2 , . . . , µn )
= 2 2
∂µi
n µc 
X

µ1 =µ2 =...=µn =µc

λn−1
(k + 1)!
k=0
n−1
k
!




k
λn−k−1µc
(27)
which is decreasing in µc . Also, we have
lim
µc →(λ/n)+
and
n−1
∂
αi (µ1 , µ2 , . . . , µn )
=
∂µi
nλ
µ1 =µ2 =...=µn =µc
∂
lim
αi (µ1 , µ2 , . . . , µn )
= 0.
µc →∞ ∂µi
µ1 =µ2 =...=µn =µc
It should be noted that proposition 6 implies that for
µc > λ/n,
we have
(28)
n−1
∂
αi (µ1 , µ2, . . . , µn )
.
<
∂µi
nλ
µ1 =µ2 =...=µn =µc
We also note that the partial derivative in proposition 6 gives the marginal benefit
Server i gets by unilaterally deviating from a service capacity µc commonly chosen by all
servers.
The following proposition gives the Nash equilibrium of the game, which represents
the decision of the servers on their service capacities in the long run.
Proposition 7. If (n − 1)R/n > c′ (λ/n) then there is a unique equilibrium where
µ1 = µ2 = . . . = µn = µc
9
(29)
and µc is the unique solution that satisfies µc > λ/n and
∂
αi (µ1 , µ2 , . . . , µn )
= c′ (µc ),
Rλ
∂µi
µ1 =µ2 =...=µn =µc
(30)
i.e.,
R
λ
nµc

2 

1 −

n−1
X


λn−1
n−1
k
(k + 1)!
k=0
!


 = c′ (µc ).

k
(31)
λn−k−1 µc
If (n − 1)R/n ≤ c′ (λ/n) then the system has no equilibrium in which the expected
waiting time is finite.
We note that from the proposition, we have µc > λ/n and so the expected waiting
times are finite. This means that we know that if the marginal cost of serving 1/n of all
customers is less than (n − 1)/n of the revenue received per customer, there is a unique
symmetric equilibrium with finite waiting times.
For equation (30) to hold, it means that the marginal benefit Server i gets by unilaterally deviating from a service capacity µc commonly chosen by all servers must be equal to
the marginal cost to do so. In this case, Server i does not benefit from changing its service
capacity. Mathematically, the first-order condition for πi holds. From the concavity of πi
obtained in proposition 5, we know that choosing µc as the service capacity maximizes
the profit for Server i.
Since the servers share the same cost function and the same profit function with respect
to their own service capacities, the condition for which the marginal benefit equals the
marginal cost is identical for all servers when they choose the same service capacities.
The proposition asserts that there is only one value of µc which satisfies the condition,
and that this symmetric equilibrium is the unique equilibrium of the system.
This proposition shows that, given the arrival rate of customer λ, the number of servers
n and the revenue per customer R, all servers will choose the same service capacity given
by equation (31) in the long run if the condition
λ
(n − 1)R
> c′ ( )
n
n
(32)
is satisfied. The proposition is useful for determining the minimum value of revenue per
customer R for which the system will have a finite-waiting time equilibrium.
When n = 2, Propositions 6 and 7 reduce to the results in [10]. It is worth noting
that as n increases, (n − 1)R/n increases and c′ (λ/n) decreases. Therefore, the minimum
value of R required for the existence of a finite waiting-time equilibrium decreases as
n increases. An increase in the number of servers causes competition to become more
intense. Thus the minimum revenue per customer needed to achieve an equilibrium with
finite waiting times becomes lower.
10
3.5
A Numerical Example on Three-Server Queueing System
In this subsection, we present a numerical example for a three-server service system,
i.e., n = 3. Here we assume the cost function takes the following form:
c(µ) = µ2
(33)
and the condition for the queueing system to be stable holds, i.e.
µ1 + µ2 + µ3 > λ.
(34)
We note that c′ (µ) > 0 and c′′ (µ) > 0 for µ > 0. So c(µ) is strictly increasing and strictly
convex.
We first give the steady-state probability distribution of the system. The following
result comes from Proposition 1 in Section 3.1.1. We have
P0 =
1−ρ
(1 − ρ) 1 +
P(0,0,1) =
P(0,1,1) =
λP0
,
3µ3
λ2 P0
,
6µ2 µ3
λ(µ1 µ2 +µ1 µ3 +µ2 µ3 )
2µ1 µ2 µ3
P(0,1,0) =
P(1,0,1) =
λP0
,
3µ2
λ2 P0
,
6µ1 µ3
+
λ2 (µ1 +µ2 +µ3 )
6µ1 µ2 µ3
P(1,0,0) =
,
λP0
,
3µ1
P(1,1,0) =
λ2 P0
,
6µ1 µ2
and
Pk = ρk−2 P2
for k > 2
where
P2 = P(0,1,1) + P(1,0,1) + P(1,1,0) .
Moreover, we have
αi (µ1 , µ2 , µ3 ) =
h
µi λ2 + λ(µj + µl ) + 2µj µl +
λ3
µi +µj +µl −λ
i
λ2 (µi + µj + µl ) + 2λ(µi µj + µi µl + µj µl ) + 6µi µj µl +
where j, l ∈ {1, 2, 3} and i, j, l are distinct.
Now we have
2λ(2λ + 3µc )
∂
αi (µ1 , µ2 , µ3)
=
.
∂µi
9µc (λ2 + 4µc λ + 6µ2c )
µ1 =µ2 =µ3 =µc
λ3 (µi +µj +µl )
µi +µj +µl −λ
.
If 2R
> c′ (λ/n) = 2λ
i.e., R > λ then there is a symmetric equilibrium where µ1 = µ2 =
3
3
µ3 = µc and µc is the unique solution that satisfies
2λ2 (2λ + 3µc )
λ
R = c′ (µc ) = 2µc
and
µc >
2
2
3
9µc (λ + 4µc λ + 6µc )
11
i.e.,
54µ4c + 36λµ3c + 9λ2 µ2c − 3Rλ2 µc − 2Rλ3 = 0.
4
Concluding Remarks
In this paper, we extend the analytical results of the two-server queueing model in
[10] to the case of a general multiple-server queueing model. To extend our study to the
incentive aspect of the queue system is our future work.
A service system of two servers coordinated by one central agency was studied by
Gilbert and Weng [11]. The principal-agent relationship [12] between the central agency
and the servers was studied, from the principal’s perspective. It is of interest whether
the allocation policy with a separate queue or that with a common queue would allow the
coordinator to control waiting times at a lower cost. Again customers arrive according to
a Poisson process and the service times are assumed follow an exponential distribution.
Each of the server operates independently and determine their own service capacities so as
to maximize their individual profits. The coordinating agency determine a fixed amount
R to compensate the servers per unit of service rendered. The coordinating agency then
has to determine the minimum value of R needed to maintain expected system time below
a given level.
It was found that the servers have a weaker incentives to increase their service capacities in common queue systems than in separate queue systems. They conclude that in
some cases, the competition incentive effects can more than offset the risk-pooling benefits of a common queue. In these cases the separate queue allocation scheme then has
advantages over common queue allocation scheme. In particular, cases with small permissible waiting times or not severe diseconomies on increasing capacity favor the separate
queue system. The queueing system discussed here corresponds to the common queue
allocation policy with N servers. Therefore the results obtained here are ready to apply
to generalize the models and conclusions addressed in [11]. We will extend the model in
[11] by allowing the number of servers to be more than two.
5
Appendix
5.1
Proof of Proposition 1
We note that
X
s1 +s2 +...+sn =k
P(s1 ,s2 ,...,sn )
P
Q
(n − k)!λk
s1 +s2 +...+sn =k
{i|si =0} µi
·
P0
=
n!
µ1 µ2 . . . µn
"
#
P
(n − k)!λk ( i1 <i2 <...<in−k µi1 µi2 . . . µin−k )
=
P0 .
n!µ1 µ2 . . . µn
We will then verify that this solution satisfies (14) and (17).
12
Recall that Equation (14) we have for si = 0, 1 and
X
(
X
µi + λ)P(s1 ,s2 ,...,sn ) =
{i|si =1}
Pn
i=1
µi P(s−i ,si =1) +
{i|si =0}
X
X
{i|si =1}
So the left-hand-side is
(
si 6= n,
µi + λ)
{i|si =1}
The right-hand-side is
λP(s−i ,si =0)
.
|{j|sj = 0}| + 1
(n − k)!λk P0
Q
.
n! {i|si =1} µi
(n − k + 1)!λk−1P0
λ
(n − k − 1)!λk+1 P0 P
Q
Q
+ {i|si =1}
n!µi {j|sj =1} µj
n − k + 1 n! {j|sj =1 and j6=i} µj
P
(n − k − 1)!λk+1P0 P
(n − k)!λk P0
Q
= {i|si =0}
+ {i|si =1} µi Q
n! {j|sj =1} µj
n! {j|sj =1} µj
k
k+1
(n − k)!λ P0 P
(n − k − 1)!λ P0
Q
+( Q
) {i|si =1} µi
= (n − k)
n! {j|sj =1} µj
n! {j|sj =1} µj
(n − k)!λk P0
(n − k)!λk P0 P
=λ Q
+( Q
) {i|si =1} µi
n! {j|sj =1} µj
n! {j|sj =1} µj
P
(n − k)!λk P0
= (λ + {i|si =1} µi ) Q
n! {j|sj =1} µj
P
{i|si =0} µi
which equals the left-hand-side. For Equation (17)
(
n
X
i=1
n
X
µi + λ)Pk = (
µi )Pk+1 + λPk−1
for k = n, n + 1, . . . .
i=1
When k = n, the left-hand-side is
(
n
X
µi + λ)
i=1
Note that
Pn−1
n!
λn P
Qn 0
j=1 µj
.
P
λn−1 ( ni=1 µi )
=
P0 .
n!µ1 µ2 . . . µn
Therefore the right-hand-side is
(
n
X
i=1
P
n Pn
n
X
λ ( i=1 µi )
λn−1 ( ni=1 µi )
µi + λ)Pn
P0 = λPn +
P0 = (
µi )ρPn + λ
n!µ1 µ2 . . . µn
n!µ1 µ2 . . . µn
i=1
which is equal to the left-hand-side.
When k > n, the left-hand-side is
(
n
X
µi + λ)ρk−n Pn .
i=1
13
And the right-hand-side is
(
n
X
i=1
n
n
X
X
µi )ρk−n+1 Pn + λρk−n−1Pn = λρk−n Pn + (
µi)ρk−n Pn = (
µi + λ)ρk−n Pn
i=1
i=1
which is equal to the left-hand-side. The expression of P0 can be readily obtained from
the normalization equation substituting the above expressions.
5.2
Proof of Proposition 2
The mean number of jobs per time unit that enter service with Server i is given by
Qi =
=
=
=
=
=
λPs
+ µi (Pn+1 + Pn+2 + . . .)
|{j|sj = 0}|
P
λ
ρ
(n − k)!λk P0
λn P0
Q
(
)
+
µ
i
s|si =0
|{j|sj = 0}| n! {i|si =1} µi
n!µ1 µ2 . . . µn 1 − ρ
P
ρ
µ i λn P0
(n − k − 1)!λk+1 P0
Q
(
)
+
s|si =0
n! {i|si =1} µi
n!µ1 µ2 . . . µn 1 − ρ
P
Pn−1
λµi P0
ρ
k
n−1
(
)
k=0 (n − k − 1)!λ (
j1 <j2 <...<jn−k−1 ,jp 6=i∀p µj1 µj2 . . . µjn−k−1 ) + λ
n!µ1µ2 . . . µn
1−ρ
P
Pn−1
ρ
k
n−1
)
λµi
(
k=0 (n − k − 1)!λ (
j1 <j2 <...<jn−k−1 ,jp 6=i∀p µj1 µj2 . . . µjn−k−1 ) + λ
1−ρ
−1
Pn−1
P
1
k
n
× n!µ1 µ2 . . . µn + k=1 (n − k)!λ ( i1 <i2 <...<in−k µi1 µi2 . . . µin−k ) +
λ µi )
1−ρ
−1
P
Pn−1
λn
k
(n
−
k)!λ
(
×
µ
µ
.
.
.
µ
)
+
in−k
k=0
i1 <i2 <...<in−k i1 i2
1−ρ
Pn−1
P
ρ
n−k−1
n−1
λµi
( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk ) + λ (
)
k=0 k!λ
1
−
ρ
−1
P
Pn
λn
n−k
( i1 <i2 <...<ik µi1 µi2 . . . µik ) +
×
.
k=1 k!λ
1−ρ
P
s|si =0
The market share of Server i is
αi (µ1 , µ2 , . . . , µn ) = Qi /
n
X
j=1
We note that
n
X
Qi = λ,
i=1
14
Qj .
P
since ni=1 Qi is the mean number of jobs per time unit that enter service with any server.
Thus the market share of Server i is
Qi
λ
Pn−1
P
ρ
n−k−1
n−1
µi
( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk ) + λ (
)
k=0 k!λ
1−ρ
=
.
P
Pn
λn
n−k (
k!λ
µ
µ
.
.
.
µ
)
+
jk
k=1
j1 <j2 <...<jk j1 j2
1−ρ
αi (µ1 , µ2 , . . . , µn ) =
5.3
Proof of Proposition 3
We consider the partial derivative of αi (λ, µ1 , . . . , µn ) with respect to µi . Here we
assume that λ and µj (j 6= i) are given and fixed, and write αi (µi ) for short. Let

n−1
X
A(µi ) = µi 
k!λn−k−1(
k=0
and
B(µi ) =
n
X
k=1
X
µj1 µj2 . . . µjk ) + λn−1 (
j1 <j2 <...<jk ,jp 6=i∀p
X
k!λn−k (
µj 1 µj 2 . . . µj k ) +
j1 <j2 <...<jk
so we have
Ci =
n−1
X
and
Di =
n−1
X
k=0
then
Pn
k=1 k!λ
n−k
(
P
j1 <...<jk
X
(k + 1)!λn−k−1(
k=0
ρ 
)
1−ρ
λn
1−ρ
A(µi )
.
B(µi )
αi (µi ) =
Let

µj 1 µj 2 . . . µj k )
j1 <j2 <...<jk ,jp 6=i∀p
X
k!λn−k−1 (
µj 1 µj 2 . . . µj k )
j1 <j2 <...<jk ,jp 6=i∀p
P
Pn−1
n−k
( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk )
k=1 k!λ
P
P
+ nk=1 k!λn−k µi ( j1 <j2 <...<jk−1 ,jp 6=i∀p µj1 µj2 . . . µjk−1 )
P
Pn−1
k!λn−k−1 ( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk ) − λn
= λ k=0
Pn−1
P
+µi k=0
(k + 1)!λn−k−1( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk )
= λDi + µi Ci − λn .
µj 1 µj 2 . . . µj k ) =
15
We also note that both Ci and Di are positive, and are constant with respect to µi . Now
we have

n
λ
ρ

n−1

)] = µi Di + P
A(µi ) = µi [Di + λ (


µj − λ
1−ρ



n

λ
λn+1


= λDi + µi Ci + P
 B(µi ) = λDi + µi Ci − λn +
1−ρ
µj − λ
P
n
λ
(
µ
−
λ
−
µ
)

j
i

P
A′ (µi ) = Di +



( µj − λ)2



λn+1


 B ′ (µi ) = Ci − P
( µj − λ)2
Then we have
P
λn ( µ j − λ − µ i )
λn+1
P
Di +
B(µi)A (µi ) − A(µi )B (µi) =
λDi + µi Ci + P
(
µj − λ)2
µ
−
λ
j
n
n+1
λ
λ
−µi Di + P
Ci − P
µj − λ
( µj − λ)2
n 2
n+1
λ µ Ci
λ2n+1
2λ Di
P
− P i
+
= λDi2 + P
( µj − λ) ( µj − λ)2 ( µj − λ)2
′
and
′
2λn+1 Di
λn µ 2 C i
λ2n+1
P
λDi2 + P
− P i
+
∂
( µj − λ) ( µj − λ)2 ( µj − λ)2
.
αi (µi) =
2
∂µi
λn+1
λDi + µi Ci + P
µj − λ
(35)
We observe that
i2
P
n−k−1
k!λ
(
µ
µ
.
.
.
µ
)
jk
k=0
j1 <j2 <...<jk ,jp 6=i∀p j1 j2
Pn−1 Pl=n−k−1
k!
≥ λ[ k=0 l=0
l!λn−l−1 (k − l)!λn−k+l−1
l!(k − l)!
P
( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk )]
= λn C i .
λDi2 = λ
hP
n−1
Thus we have λDi2 − λn Ci ≥ 0. Assuming
P
j6=i
µj > λ,
λn µ 2 C i
λ2n+1
2λn+1 Di
P
− P i
+
B(µi )A′ (µi ) − A(µi )B ′ (µi) = λDi2 + P
( µj − λ) ( µj − λ)2 ( µj − λ)2
λn µ 2 C i
2λn+1 Di
λ2n+1
P
P
= (λDi2 − P i
)
+
+
( µj − λ)2
( µj − λ) ( µj − λ)2
n+1
λ2n+1
2λ Di
+ P
≥ (λDi2 − λn Ci ) + P
( µj − λ) ( µj − λ)2
≥ 0.
16
Assuming that
=
≥
=
=
=
≥
=
>
P
j6=i
µj ≤ λ but µi > λ −
P
j6=i
µj , so
P
j
µj > λ,
B(µi )A′ (µi ) − A(µi )B ′ (µi )
λn µ 2 C i
λ2n+1
2λn+1 Di
P
− P i
+
λDi2 + P
( µj − λ) ( µj − λ)2 ( µj − λ)2
2λn+1 Di
λ2n+1
λn µ 2 C i
P
P
)
+
+
(λn Ci − P i
2
( # µj − λ) ( µj − λ)2
"
P( µj − λ) 2
(µi + j6=i µj − λ) − µ2i
2λn+1 Di
λ2n+1
n
P
P
P
λ
C
+
+
i
( µj − λ)2
( µj − λ) ( µj − λ)2
P
P
P
λn
P
{2λ[µi − (λ − j6=i µj )]Di + λn+1 − [2µi − (λ − j6=i µj )](λ − j6=i µj )Ci }
2
( µj − λ)
P
P
P
λn
P
{2µi[λDi − λCi + ( j6=i µj )Ci ] + 2( j6=i µj )[λDi − λCi + ( j6=i µj )Ci ]
2
( µj − λ)
P
+λ2 (Ci − Di ) − ( j6=i µj )2 Ci + λn+1 }
P
P
P
P
λn
P
{2(λ − j6=i µj )[λDi − λCi + ( j6=i µj )Ci ] + 2( j6=i µj )[λDi − λCi + ( j6=i µj )Ci ]
2
( µj − λ)
P
P
+λ2 (Ci − Di ) − ( j6=i µj )2 Ci + λn+1 } since µi > (λ − j6=i µj )
P
P
P
λn
P
{λ[λDi − λCi + ( j6=i µj )Ci ] + [λ − ( j6=i µj )]( j6=i µj )Ci + λn+1 }
2
( µj − λ)
0
because
P
( j6=i µj )Ci =
≥
=
=
=
=
P
Pn−1
P
n−k−1
( j6=i µj )( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk )
k=0 (k + 1)!λ
Pn−1
P
n−k−1
(k + 1)( j1 <j2 <...<jk+1 ,jp 6=i∀p µj1 µj2 . . . µjk+1 )
k=0 (k + 1)!λ
P
Pn−2
n−k−1
(k + 1)( j1 <j2 <...<jk+1 ,jp 6=i∀p µj1 µj2 . . . µjk+1 )
k=0 (k + 1)!λ
P
Pn−1
n−k
( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk )
k=1 k · k!λ
P
Pn−1
k · k!λn−k−1 ( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk )
λ k=0
λ(Ci − Di )
i.e.
X
λDi − λCi + (
µj )Ci ≥ 0.
j6=i
Combining the two cases, we have
′
′
B(µi )A (µi ) − A(µi )B (µi ) > 0 if
n
X
µj > λ.
i=1
Since [B(µi )]2 > 0,
αi′ (µi )
B(µi )A′ (µi ) − A(µi )B ′ (µi )
=
>0
[B(µi )]2
17
Pn
for
i=1
µj > λ. We note that as µi → ∞,
∂αi (µ1 , . . . , µn )
→ 0,
∂µi
since the numerator approaches a constant
hP
i2
P
n−1
n−k−1
k!λ
(
)
µ
µ
.
.
.
µ
jk
j1 <j2 <...<jk ,jp 6=i∀p j1 j2
k=0
P
P
n−1
n−k−1
n
−λ
( j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 . . . µjk )
k=0 (k + 1)!λ
λDi2 − λn Ci = λ
and the denominator approaches +∞.
5.4
Proof of Proposition 4
We note that
∂ 2 A(µi )
∂µ2i B(µi )
∂ B(µi )A′ (µi ) − A(µi )B ′ (µi )
=
∂µi
(B(µi ))2
2
(B(µi ) )[B(µi )A′ (µi ) − A(µi )B ′ (µi)]′ − 2B(µi )B ′ (µi)[B(µi )A′ (µi) − A(µi )B ′ (µi )]
=
(B(µi))4
′
′
′
B(µi )[B(µi )A (µi ) − A(µi )B (µi )] − 2B ′ (µi )[B(µi )A′ (µi ) − A(µi )B ′ (µi )]
=
.
(B(µi ))3
Then
[B(µi )A′ (µi ) − A(µi )B ′ (µi )]′
′
n 2
2n+1
n+1
λ
µ
C
λ
2λ
D
i
i
i
= λDi2 + P
− P
+ P
( µj − λ) ( P
µj − λ)2 ( µj − λ)2
2λn µi ( µj − λ − µi )
2λ2n+1
2λn+1
P
P
D
−
C
−
= − P
i
i
( µj − λ)2
( µj − λ)3
( µj − λ)3
18
and we have
=
=
=
≤
≤
′
B(µ
A(µi
)B′ (µi )]′ − 2B ′ (µi )[B(µi )A′ (µi )P− A(µi )B ′ (µi )]
i )[B(µi )A (µi ) −n+1
2λn µi ( µj − λ − µi)
2λ2n+1
2λn+1
λ
P
D −
Ci − P
− P
λDi + µiCi + P
2 i
3
3
(
µ
−
λ)
(
µ
−
λ)
µ
−
λ
j
j
j
( µj − λ)
n+1
n+1
n 2
2n+1
λ
2λ Di
λ µ Ci
λ
−2 Ci − P
λDi2 + P
− P i
+ P
2
2
( µj − λ)
( µj − λ)
( µj − λ)2
( µj− λ) 2n+2
n+2
2n+2
n+2
2λ
2λ
2λ
4λ2n+2
2λ
2
+ P
Di + − P
− P
+ P
Di
− P
2
λ)2 ( µj − λ)P
( µj − λ)3 ( µj − λ)3 ( µj − λ)3
( µj −
2λn+1 µi
2λn+1 µi ( µj − λ − µi )
4λn+1
P
P
+ − P
Ci Di − 2λCiDi2
−
−
2
3
(
µ
−
λ)
(
µ
−
λ)
(
µ
−
λ)
j P
j
j
2λn µ2i
2λn µ2i ( µj − λ − µi )
P
+ P
Ci2
+ −
3
2
(
µ
−
λ)
(
µ
−
λ)
j
j
P
2λ2n+1 µi
2λ2n+1 µi ( µj − λ − µi )
λ2n+1
λ2n+1 µ2i
P
+ − P
Ci
−
−2 P
+2 P
− λ)3
( µj − λ)4
( µj − λ)2
( µj − λ)4
( µj3n+2
2λ3n+2
2λ
P
+
+ − P
( µj − λ)4 ( µj − λ)4
P
P
2λn+1
2
− P
{2(
µ
−
λ)
+
µ
[2(
µj − λ) − µi ]}Ci Di
j
i
( µj − λ)3
P
2n+1
n 3
2λ
[2(
µ − λ) + 2µi ]
λ
µ
i
P j
Ci ] −
Ci
−2Ci [λDi2 − P
3
( µj − λ)
( µj − λ)3
P
P
2λn+1
− P
{2( µj − λ)2 + µi [2( µj − λ) − µi ]}Ci Di
3
( µj − λ)
P
2λ2n+1 [2( µj − λ) + 2µi ]
2
n
P
Ci
−2Ci [λDi − λ Ci ] −
( µj − λ)3
P
0 assuming
j6=i µj > λ.
Since the denominator B(µi )3 > 0, we have αi′′ (µi ) < 0 if
19
P
j6=i
µj > λ.
Assuming
=
≤
=
=
=
=
where
P
j6=i
µj ≤ λ but µi > λ −
P
j6=i
µj , so
P
j
µj > λ,
B(µi)[B(µi )A′ (µi ) − A(µi )B ′ (µi)]′ − 2B ′ (µi )[B(µi)A′ (µi ) − A(µi )B ′ (µi)]
P
P
2λn+1
2
{2(
µ
−
λ)
+
µ
[2(
µj − λ) − µi]}Ci Di
− P
j
i
( µj − λ)3
P
2λ2n+1 [2( µj − λ) + 2µi ]
λn µ3i
2
P
Ci ] −
Ci
−2Ci [λDi − P
( µj − λ)3
( µj − λ)3
P
P
2λn+1
− P
{2( µj − λ)2 + µi[2( µj − λ) − µi]}Ci Di
3
( µj − λ)
P
2λ2n+1 [2( µj − λ) + 2µi]
λn µ3i
n
P
Ci ] −
Ci
−2Ci [λ Ci − P
( µj − λ)3
( µj − λ)3
n
P
P
2λ Ci
2
− P
{λD
[2[µ
−
(λ
−
µ
)]
+
µ
[µ
−
2(λ
−
i
i
j
i
i
j6
=
i
j6=i µj )]]
( µj − λ)3
P
P
P
−(λ − j6=i µj )Ci [3µ2i − 3µi(λ − j6=i µj ) + (λ − j6=i µj )2 ]
P
+λn+1[2( µj − λ) + 2µi ]}
P
P
2λn Ci
{λDi [3µ2i − 6µi(λ − j6=i µj ) + 2(λ − j6=i µj )2 ]
− P
3
( µj − λ)
P
P
P
−(λ − j6=i µj )Ci [3µ2i − 3µi(λ − j6=i µj ) + (λ − j6=i µj )2 ]
P
+λn+1[4µi − 2(λ − j6=i µj )]}
P
2λn Ci
{3[λDi − (λ − j6=i µj )Ci ]µ2i
− P
3
( µj − λ)
P
P
−[3[2λDi − (λ − j6=i µj )Ci ](λ − j6=i µj ) − 4λn+1 ]µi
P
P
P
+(λ − j6=i µj )[(2λDi − (λ − j6=i µj )Ci )(λ − j6=i µj ) − 2λn+1]}
2λn Ci
Gi (µi )
− P
( µj − λ)3
P
Gi (µi ) = 3[λDi − (λ − j6=i µj )Ci ]µ2i
P
P
−[3[2λDi − (λ − j6=i µj )Ci ](λ − j6=i µj ) − 4λn+1 ]µi
P
P
P
+(λ − j6=i µj )[(2λDi − (λ − j6=i µj )Ci )(λ − j6=i µj ) − 2λn+1 ].
In the following, we will prove that
Gi (µi ) > 0 for µi > λ −
X
µj .
j6=i
Before proceeding, we need the following result:
X
j1 <j2 <...<jk ,jp
µj
µj 1 µj 2
... k ≤
λ λ
λ
6 i∀p
=
n−1
k
! P
µj
(n − 1)λ
j6=i
k
(36)
P
for j6=i µj < λ and µj > 0 for j 6= i for some given λ > 0. The equality holds when
µj = µk for all j, k 6= i.
20
To prove this, we first consider the problem
X
max
j1 <j2 <...<jk ,jp
subject to
X
µj
µj 1 µj 2
... k
λ λ
λ
6 i∀p
=
µj = M and µj > 0 for j 6= i
j6=i
for some given M > 0. Using the Lagrangian method, we have at the maximum,

X
1


µj1 µj2 . . . µjk−1 + γ = 0 for l 6= i

 λk
j1 <j2 <...<jk−1 ,jp 6=i∀p,jp 6=l∀p
X


µj = M.


j6=i
The first equation gives µj = µk for all j 6= i, k 6= i. Then the second equation gives
µj =
M
n−1
for j 6= i.
We check that µj > 0 for all j 6= i.
The maximum of the objective function is therefore
X
j1 <j2 <...<jk ,jp 6=i∀p
M
(n − 1)λ
k
=
n−1
k
! P
µj
(n − 1)λ
j6=i
k
.
Since this is true for any M > 0, thus it is true for any values of µj ’s such that µj > 0 for
j 6= i. Hence we have inequality (36).
P
Now we will show that Gi (µi ) ≥ 0 for all µi > λ − j6=i µj . Note that
3[λDi − (λ −
X
µj )Ci ] > 0
j6=i
and so Gi (µi) is a quadratic function of µi with the minimum point attained at
P
P
3[2λDi − (λ − j6=i µj )Ci ](λ − j6=i µj ) − 4λn+1
P
µmin,i =
6[λD
−
(λ
−
µj )Ci ]
i
j6=iP
P
P
6[λDi − (λ − j6=i µj )Ci ](λ − j6=i µj ) + 3(λ − j6=i µj )2 Ci − 4λn+1
P
=
6[λD
−
(λ
−
i
j6=i µj )Ci ]
P
3(λ − j6=i µj )2 Ci − 4λn+1
P
P
= (λ − j6=i µj ) +
.
6[λDi − (λ − j6=i µj )Ci ]
21
Here we note that 6[λDi − (λ −
P
j6=i
µj )Ci ] > 0 as we obtained earlier, and
P
3(λ − j6=i µj )2 Ci − 4λn+1
P
P
Pn−1
= 3(λ − j6=iP
(k + 1)!λn−k−1 j1 <j2 <...<jk ,jp 6=i∀p µj1 µj2 ...µjk − 4λn+1
µj )2 k=0
P
µj 1 µj 2 µj k
j6=i µj 2 Pn−1
)
...
− 4λn+1
= 3λn+1 (1 −
k=0 (k + 1)!
j1 <j2 <...<jk ,jp 6=i∀p
λ
λ
λ
λ
! P
P
k
µ
µ
P
n
−
1
j
j
j6
=
i
j6
=
i
n−1
− 4λn+1
)2 k=0 (n − k)!
≤ 3λn+1 (1 −
λ
(n − 1)λ
k
where the inequality
P follows from (36).
j6=i µj
Now let γ =
, we have 0 < γ < 1.
λ
!
k
n−1
n−1
X
X
n−1 n−2
n−k k
γ
n−1
(k + 1)
=
·
...
γ
(k + 1)!
n−1
n−1 n−1
n−1
k
k=0
k=0
∞
X
≤
(k + 1)γ k
k=0
∞
X
4
(k + 1)γ k
3 k=0
4
1
=
·
3 (1 − γ)2
<
Thus we have
3(λ −
X
µj )2 Ci − 4λn+1 < 0
j6=i
P
and so µmin,i < λ − j6=i µj . Since Gi (.) is a quadratic function with minimum at µmin,i,
P
P
Gi (µi ) > Gi (λ − j6=i µj ) for all µi > λ − j6=i µj . Next,
Gi (λ −
But
P
j6=i
P
P
µj ) = 3[λDi − (λ − j6=i µj )Ci ](λ − j6=i µj )2
P
P
P
−[3[2λDi − (λ − j6=i µj )Ci](λ − j6=i µj ) − 4λn+1 ](λ − j6=i µj )
P
P
P
+(λ − j6=i µj )[(2λDi − (λ − j6=i µj )Ci )(λ − j6=i µj ) − 2λn+1]
P
P
P
= (λ − j6=i µj )[2λn+1 − (λ − j6=i µj )2 Ci − (λ − j6=i µj )λDi ]
2λn+1 − (λ −
P
µj )2 Ci − (λ − j6=i µj )λDi
Pn−1
P
µj µj µj
= 2λn+1 − λn+1 (1 − γ){ k=0
[(k + 1)(1 − γ) + 1]k!( j1 <j2 <...<jk ,jp 6=i∀p 1 2 ... k )}
λ λ
λ
P
j6=i µj
where γ =
and 0 < γ < 1
!
" λ
k #
P
γ
2
n−1
n−1
− k=0
[(k + 2) − γ(k + 1)]k!
≥ λn+1 (1 − λ)
1−λ
n−1
k
P
j6=i
≥ 0
22
since
!
k
γ
n−1
k=0 [(k + 2) − γ(k + 1)]k!
n−1
k
k
k
Pn−1
Pn−1
(n − 1)!
γ
1
(n − 1)!
=
− k=0 (k + 1)
γ k+1
k=0 (k + 2)
(n − k − 1)! n − 1
(n − k − 1)! n − 1
k
k−1
Pn
Pn−1
γ
1
(n − 1)!
(n − 1)!
− k=1 k
γk
=
k=0 (k + 2)
(n − k − 1)! n − 1 (n −k)! n − 1
Pn−1 (n − 1)!
n!
k+2
k
1
γk −
−
γn
= 2 + k=1
k−1
(n − k − 1)! (n − 1)
n−1 n−k
(n − 1)n−1
Pn−1
Pn−1 n − 2 n − 3
n−k+1 1
...
[(k + 2)(n − k) − k(n − 1)]
k=1
n−1n−1
n − 1 n− 1
n−2n−3
n−k+1 1
k(k + 1)
= 2+
...
2 n−
n−1 n−1
2
Pnn − 1k n − 1
≤ 2 k=0 γ
2
=
1−γ
≤ 2+
This gives
Gi (λ −
X
µj ) ≥ 0.
j6=i
Therefore we have
Gi (µi ) > Gi (λ −
X
µj ) ≥ 0 for µi > λ −
X
µj
j6=i
j6=i
which implies α′′ (µi ) < 0 as the denominator B(µi )3 > 0. Combining the two cases, we
P
have α′′ (µi ) < 0 if nj=1 µj > λ.
5.5
Proof of Proposition 5
For i = 1, 2, . . . , n, for each fixed λ > 0 and µj > 0 where j 6= i, Clearly πi (µi ) is
P
P
continuous for µi < λ − j6=i µj , as well as for µi > λ − j6=i µj . To show that πi (µi )
P
P
is continuous at the point µi = λ − j6=i µj , we note that, as µi approaches λ − j6=i µj
from above, πi (µi ) approaches the following limit:
Rλαi (µi )
i
hP
P
n−1
n−k−1
n−1 ρ
)
Rλµi
k!λ
(
µ
µ
.
.
.
µ
)
+
λ
(
jk
k=0
j1 <j2 <...<jk ,jp 6=i ∀p j1 j2
1−ρ
P
P
=
lim
n
P
n
λ
n−k (
µi →(λ− j6=i µj )+
k=1 k!λ
j1 <j2 <...<jk µj1 µj2 . . . µjk ) + 1−ρ
h
i
Pn−1
P
Rλµi (1 − ρ) k=0 k!λn−k−1 ( j1 <j2 <...<jk ,jp 6=i ∀p µj1 µj2 . . . µjk ) + λn−1 ρ
P
P
=
lim
P
µi →(λ− j6=i µj )+
(1 − ρ) nk=1 k!λn−k ( j1 <j2 <...<jk µj1 µj2 . . . µjk ) + λn
= Rµi .
lim
P
µi →(λ−
j6=i
µj )+
23
So we have
lim
P
µi →(λ−
j6=i
πi (µi ) = R(λ −
+
µj )
X
µj ) − c′ (λ −
X
µj ) =
j6=i
j6=i
lim
P
µi →(λ−
j6=i
µj )−
πi (µi ).
Thus πi (µi) is continuous for µi > 0.
To prove the concavity of πi , first note that c(.) is convex and so −c(.) is concave.
Also, it is proved in Proposition 4 that αi (µi) is concave in µi . Therefore, it remains to
prove that,
∂ ′
α (µi )
µi →(λ− j6=i µj
∂µi i
λn µ 2 C i
λ2n+1
2λn+1 Di
P
− P i
+
λDi2 + P
( µj − λ) ( µj − λ)2 ( µj − λ)2
Rλ
lim
2
P
µi →(λ− j6=i µj )+
λn+1
λDi + µi Ci + P
µj − λ
P
P
( µj − λ)2 λDi2 + ( µj − λ)2λn+1 Di − λn µ2i Ci + λ2n+1
Rλ
lim
P
P
µi →(λ− j6=i µj )+
[(λDi + µiCi )( µj − λ) + λn+1 ]2
λn+1 − µ2i Ci
R
lim
P
µi →(λ− j6=i µj )+
λn+1
R
lim
P
=
=
=
≤
)+
Rλ
i.e.,
lim
P
µi →(λ−
j6=i
µj
)+
∂
∂
πi (µi ) ≤
lim
πi (µi ).
P
−
µi →(λ− j6=i µj ) ∂µi
∂µi
So πi (µi ) is concave in µi for µi > 0. Therefore, the function πi (µ1 , µ2, . . . , µn ) is continuous and strictly concave in µi .
5.6
Proof of Proposition 6
Suppose that µ1 = µ2 = . . . = µn = µc , we have
Ci =
n−1
X
(k + 1)!
k=0
and
Di =
n−1
X
k=0
k!
n−1
k
n−1
k
24
!
!
λn−k−1 µkc
λn−k−1µkc
(37)
(38)
We then have
λDi + µc Ci =
=
=
=
=
=
!
n
−
1
(k + 1)!
λn−k−1µk+1
k!
λn−k µkc +
c
k
k=0
k=0
!
!
n
n−1
X
X
n−1
n−1
k!
λn−k µkc
k!
λn−k µkc +
k−1
k
k=1
k=0
"
!
!#
n−1
X
n
−
1
n
−
1
k!
+
λn−k µkc
λn + n!µnc +
k
k−1
k=1 !
n
X
n
λn +
k!
λn−k µkc
k
k=1
!
n
X
n−1
λn + nµc
(k − 1)!
λn−k µck−1
k−1
k=1
!
n−1
X
n
−
1
k!
λn−k−1µkc
λn + nµc
k
k=0
n−1
X
n−1
k
!
n−1
X
= nµc Di + λn
i.e.,
(nµc − λ)Di = µc Ci − λn .
Substituting µ1 = . . . = µn = µc in (35) we have
∂
αi (µ1 , µ2, . . . , µn )
∂µi
µ1 =...=µn =µc
n+1
λn µ2c Ci
λ2n+1
2λ
D
i
−
+
λDi2 +
(nµc − λ) (nµc − λ)2 (nµc − λ)2
=
2
λn+1
λDi + µc Ci +
nµc − λ
λDi2 (nµc − λ)2 + 2λn+1 Di (nµc − λ) − λn µ2c Ci + λ2n+1
=
[(λDi + µc Ci )(nµc − λ) + λn+1 ]2
λ(µc Ci − λn )2 + 2λn+1 (µc Ci − λn ) − λn µ2c Ci + λ2n+1
=
[(nµc Di + λn )(nµc − λ) + λn+1 ]2
Ci (Ci − λn−1 )
= λµ2c ·
2 4C 2
i
n µcn−1
λ
λ
1−
.
=
n2 µ2c
Ci
Substituting Ci with the expression in (37), we obtain (27).
As µc → λ/n,
!
n−1
X
1
n−1
(k + 1)!
λn−1 k
Ci →
n
k
k=0
25
so the above expression approaches





 1
1
1

! 
.
 = λ 1 − Pn−1
n
−
1
n
−
2
n
−
k
1 
n−1
...
k=0 (k + 1)
k=0 (k + 1)!
n
n
n
nk
k

1
1 −
λ
Pn−1

But
Pn−1
k=0 (k
+ 1)
Pn−1
Qk n − i
n−k
n−1n−2
...
=
k=0 (k + 1)
i=1
n
n
n
n
Pn−1
Qk n − i
=
k=0 [n − (n − k − 1)]
i=1
n
Pn−1 Qk n − i
Q n−i
=
− (n − k − 1) ki=1
]
k=0 [n
i=1
n
n
Q
Qn−1 n − i
Pn−2 Qk n − i
n−i
− n k+1
] + n i=1
=
i=1
k=0 [n
i=1
n
n
n
= n.
Therefore as µc → λ/n,
d
dµc
"
∂
n−1
αi (µ1 , µ2 , . . . , µn )
.
→
∂µi
nλ
µ1 =...=µn =µc
#
λn−1
1
λ d
∂
1−
αi (µ1 , µ2 , . . . , µn )
=
∂µi
n2 dµc µ2c
Ci
µ1 =...=µn =µc
n−1
λ d
Ci − λ
=
2
n2 dµ
#
" c2 dCµc Ci
n−1
2 dCi
i
λ µc Ci dµc − (Ci − λ )(µc dµc + 2µc Ci )
=
n2
µ4c Ci2
#
"
dCi
n−1
n−1
λ 2(Ci − λ )Ci − λ µc dµc
= − 2
n
µ3c Ci2
#
" 2
dCi
n−1
λ 2Ci − λ (2Ci + µc dµc )
= − 2
.
n
µ3c Ci2
First we note that differentiating (37) with respect to µc and multiplying by µc we obtain
n−1
dCi X
=
(k + 1)!k
µc
dµc
k=0
So we have
n−1
dCi X
2Ci + µc
=
(k + 2)!
dµc
k=0
26
n−1
k
!
n−1
k
λn−k−1µkc .
!
λn−k−1 µkc .
i
)
2Ci2 − λn−1 (2Ci + µc dC
dµc
#2
"
!
"
Pn−1
Pn−1
n−1
n−k−1 k
n−1
= 2
(k
+
1)!
λ
µ
−
λ
c
k=0
k=0 (k + 2)!
k
n−1
k
!
#
λn−k−1µkc .
But
"
Pn−1
n−1
k
!
#2
λn−k−1µkc
!
!
Pn−1 Pm
n−1
n
−
1
λn−k−1µkc (m − k + 1)!
λn−m+k−1µm−k
2 m=0 k=0 (k + 1)!
c
k
m−k
Pn−1
Pm 2(k + 1)(m − k + 1) (n − 1)!
(n − 1)!
(n − m − 1)!
(m + 2)λ2n−m−2 µm
c
m=0
k=0
(n − m − 1)!
m+2
(n − k − 1)! (n − m + k − 1)!
Pn−1
P
(n − 1)!
2m
+
2
+
2k(m
−
k)
m
(m + 2)λ2n−m−2 µm
c
m=0
k=0
(n − m − 1)!
m+2
Pn−1
P
(n − 1)!
m
(m + 2)λ2n−m−2 µm
c
m=0
k=0 1
(n − m − 1)!
!
Pn−1
n−1
(m + 2)!
λn−m−1 µm
λn−1 m=0
c .
m
2
≥
=
≥
>
=
k=0 (k
+ 1)!
So we have
2Ci2 − λn−1 (2Ci + µc
and thus
d
dµc
5.7
"
dCi
)>0
dµc
#
∂
αi (µ1 , µ2 , . . . , µn )
< 0.
∂µi
µ1 =...=µn =µc
Proof of Proposition 7
The service capacities
µ1 = µ2 = . . . = µn = µc
is an symmetric equilibrium if and only if µc satisfies
∂
πi (µ1 , µ2 , . . . , µn )
=0
∂µi
µ1 =µ2 =...=µn =µc
for all i i.e.,
∂
Rλ
αi (µ1 , µ2 , . . . , µn )
= c′ (µc ).
∂µi
µ1 =µ2 =...=µn =µc
From Proposition 6 we have, for µc > λ/n,
∂
αi (µ1 , µ2 , . . . , µn )
∂µi
µ1 =µ2 =...=µn =µc
27
is decreasing in µc . Moreover, we have
Rλ
and
lim
µc →(λ/n)+
∂
(n − 1)R
αi (µ1 , µ2 , . . . , µn )
=
∂µi
n
µ1 =µ2 =...=µn =µc
∂
αi (µ1 , µ2 , . . . , µn )
= 0.
lim
µc →∞ ∂µi
µ1 =µ2 =...=µn =µc
Recall that c′ (µc ) is positive and increasing in µc . When
(n − 1)R
> c′ (λ/n)
n
equation (30) must have a unique solution of µc for µc > λ/n. Substituting (27) in (30)
gives (31). On the other hand, if
(n − 1)R
≤ c′ (λ/n)
n
(30) must have no solution of µc for which µc > λ/n.
It remains to prove that there is no equilibrium in which servers do not choose the
same service capacities. Suppose we have an equilibrium (µ1 , µ2 , . . . , µn ) where µi > µl .
P
First, suppose nj=1 µj > λ. For equilibrium we have

∂

 Rλ
αi (µ1 , µ2 , . . . , µn ) − c′ (µi ) = 0
∂µi
∂

 Rλ
αl (µ1 , µ2, . . . , µn ) − c′ (µl ) = 0.
∂µl
Therefore
∂
∂
Rλ
αi (µ1 , µ2 , . . . , µn ) −
αl (µ1 , µ2 , . . . , µn ) = c′ (µi ) − c′ (µl )
∂µi
∂µl
From Equation (35) we have
∂
∂
αi (µ1 , µ2 , . . . , µn ) −
αl (µ1 , µ2 , . . . , µn ) [B(µ1 , µ2 , . . . , µn )]2
∂µ
∂µ
i
l
2λn+1 Di
λn µ2i Ci
2λn+1 Dl
λn µ2l Cl
2
2
= λDi + P
− P
− λDl + P
− P
( µj − λ) ( µj − λ)2
( µj − λ) ( µj − λ)2
2λn+1 (Di − Dl ) λn (µ2i Ci − µ2l Cl )
P
P
−
= λ(Di2 − Dl2 ) +
( µj − λ)
( µj − λ)2
2λn+1 (Di − Dl ) λn (µ2i Ci − µ2l Cl )
P
P
= λ(Di − Dl )(Di + Dl ) +
−
( µj − λ)
( µj − λ)2
< 0
28
(39)
since
Pn−1
P
n−k−1
k=0 k!λ
j1 <...<jk ,jp 6=i µj1 µj2 . . . µjk
Pn−1
P
n−k−1
− k=0 k!λ
j1 <...<jk ,jp 6=l µj1 µj2 . . . µjk
hP
i
P
n−2
n−k−2
(k
+
1)!λ
= −(µi − µl )
µ
.
.
.
µ
µ
<0
j
j
j
1
2
k
k=0
j1 <...<jk ,jp 6=i,l
Di − D l =
P
Pn−1
(k + 1)!λn−k−1 j1 <...<jk ,jp 6=i µj1 µj2 . . . µjk
µ2i Ci − µ2l Cl = µ2i k=0
P
Pn−1
−µ2l k=0
(k + 1)!λn−k−1 j1 <...<jk ,jp 6=l µj1 µj2 . . . µjk
Pn−1
P
= µ2i µl k=1
(k + 1)!λn−k−1 j1 <...<jk−1 ,jp 6=i,l µj1 µj2 . . . µjk
P
Pn−1
−µi µ2l k=1
(k + 1)!λn−k−1 j1 <...<jk−1 ,jp 6=i,l µj1 µj2 . . . µjk
P
Pn−2
(k + 1)!λn−k−1 j1 <...<jk ,jp 6=i,l µj1 µj2 . . . µjk
+µ2i k=0
P
Pn−2
−µ2l k=0
(k + 1)!λn−k−1 j1 <...<jk ,jp 6=i,l µj1 µj2 . . . µjk
h
Pn−2
P
= (µi − µl ) µi µl k=0
(k + 2)!λn−k−2 j1 <...<jk ,jp 6=i,l µj1 µj2 . . . µjk
i
Pn−2
P
+ (µi + µl ) k=0
(k + 1)!λn−k−1 j1 <...<jk ,jp 6=i,l µj1 µj2 . . . µjk
> 0
The left-hand-side of (39) is negative while the right-hand-side is positive, by the convexity
P
of c(.). Therefore, equation (39) cannot be satisfied. If nj=1 µj < λ, then there must
exists i such that µi < λ/n. For equilibrium we have
R = c′ (µi )
(40)
However,
(n − 1)R
λ
< R.
(41)
c′ (µi) < c′ ( ) <
n
n
Therefore, equation (40) cannot be satisfied. We conclude that there is no Nash equilibrium in which servers do not choose the same service capacities.
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