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Information Sciences xxx (2013) xxx–xxx

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An analysis of the migration rates for biogeography-basedoptimization

0020-0255/$ - see front matter � 2013 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.ins.2013.07.018

⇑ Corresponding author. Tel.: +86 15216706828.E-mail address: [email protected] (W. Guo).

Please cite this article in press as: W. Guo et al., An analysis of the migration rates for biogeography-based optimization, Inform. Sci.http://dx.doi.org/10.1016/j.ins.2013.07.018

Weian Guo ⇑, Lei Wang, Qidi WuSchool of Electronics and Information, Tongji University, Shanghai 201804, China

a r t i c l e i n f o

Article history:Received 17 February 2011Received in revised form 24 April 2013Accepted 27 July 2013Available online xxxx

Keywords:Biogeography-based optimizationEvolutionary algorithmMigration ratesTransition probability

a b s t r a c t

Biogeography-Based Optimization (BBO), inspired by the science of biogeography, is anovel population-based Evolutionary Algorithm (EA). For optimization problems, BBObuilds the matching mathematical model of the organism distribution. In this evolutionarymechanism, species migrating among islands can be considered as the information transi-tion among different solutions represented by habitats. Solutions are reassembled accord-ing to migration rates. However, so far, the migration models are generally designed byempirical studies. This leads to immature conclusions that are unreliable. To completethe previous works, this paper investigates transition probability matrices of BBO to clarifythat the transition probability of median number of species is not the only determinant fac-tor to influence performance. The impact of migration rates on BBO is mathematically dis-cussed, which is helpful to design migration models. Using numerical simulations, the BBOand several other classical evolutionary algorithms are compared. The simulations alsocomprehensively explain the effect of the BBO’s properties on its performance includingdimension, population size, and migration models. The results validate the theoreticalanalysis in this paper.

� 2013 Elsevier Inc. All rights reserved.

1. Introduction

Research on optimization has been very active during the last decades in almost every field of science and engineering,ranging from efficiency maximization in job-shop scheduling problems [54] to cost minimization in robotic path planning[47]. There are also various optimization problems in our daily life, including minimizing charging the cost of Electrical Vehi-cles (EV) [22], maximizing profits from investments [6], etc. In recent years, a variety of Evolutionary Algorithms (EA) havebeen developed as feasible and effective methods for optimization problems [1,2,15,29,31,50], especially for Non-determin-istic Polynomial (NP) problems [12]. Evolutionary Algorithms have several advantages such as robustness, reliability, globalsearch capability and the fact that there is little or no prior knowledge required. [30]. Many of them, such as EvolutionaryStrategies (ES) [14,16,17], Genetic Algorithms (GA) [7,18,19,32,41], Differential Evolution (DE) [29,34,35,42,55], have beensuccessfully implemented in practical problems.

The main ideas of most EAs are inspired from nature. During the past decades, researchers have proposed novel heuristicalgorithms by drawing inspiration from different natural phenomena. Genetic Algorithm (GA) mimics the biological processthrough producing generations of chromosomes [7,18,19,41]. Simulated Annealing (SA) was inspired by the annealing inmetallurgy [21]. The idea of Particle Swarm Optimization (PSO) comes from the flocking behavior of birds[4,9,20,24,31,45]. Ant Colony Optimization (ACO) simulates the ecological behavior of ants in foraging [8,10], and Artificial

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2 W. Guo et al. / Information Sciences xxx (2013) xxx–xxx

Bee Colony (ABC) was inspired by the behavior of honeybees in collecting nectar [1,33]. With the development of these algo-rithms, several hybrid methods have been proposed as well [44,51,53]. This shows that nature does not only provide us withresources, but that it may also serve as inspiration for developing new methodologies.

Biogeography is the science of the distribution of species and ecosystems in geographic space and through geologic time[27,28,46]. The distribution of species across geographical areas can usually be explained by a combination of environmentalreasons. In the natural world, species tend to explore more suitable environments. Islands that are suitable for species have ahigh habitat suitability index (HSI) while islands that not suitable for species have a low HSI. Throughout the progress ofevolution, species instinctively migrate to the islands with high HSI. Over time, the islands with high HSI come to have morespecies, while those with low HIS come to have fewer. According to this idea, Dan Simon proposed a novel evolutionary algo-rithm named Biogeography-Based Optimization (BBO) [37]. BBO uses mathematical models of biogeography to describe suchattributes as migration, mutation and the distribution of species.

In recent years, BBO has been studied and developed comprehensively. For some well-known benchmarks, BBO performsbetter than other widely used heuristic algorithms like GAs, ACO, PSO, DE, SA [13,25,26,37–39,45]. There are also several suc-cessful applications to practical problems, such as sensor selection problems for aircraft engine health diagnostics [37],groundwater possibility retrieval systems [23], and power flow problems [36]. Previous research has shown the strengthof BBO.

Paper [25] discussed factors that can affect the performance of BBO by empirical studies including migration models andequilibrium species amount. However to date, the migration models have not been analyzed using mathematical studies.Although experience and empirical studies can be relied on sometimes, they are not long-term solutions to algorithm devel-opment. Therefore, further research regarding the migration models of BBO was conducted. According to our analysis, insome cases, the analysis of migration models of paper [25] is false. Therefore, we extend the work of paper [25] by mathe-matically analyzing the migration model of BBO, which is helpful to develop BBO further.

The rest of this paper is organized as follows. Section 2 briefly introduces the mathematical model of BBO. In this section,the transition probability matrix of BBO is investigated and relevant theorems are proved for subsequent discussion. Sec-tion 3 analyzes six different migration models cited from paper [25]. Based on the theorems introduced in Section 2, weuse the migration models to clarify that the transition probability of median number of species is not the only key reasonto affect BBO’s performance. Subsequently, we discuss the effects of migration rates on the performance. The numerical sim-ulation results and relevant analysis between different migration models and several other classical evolutionary algorithmsare investigated in Section 4. Further, two new migration models are proposed based on the analysis in Section 4. We endthis paper with conclusions and present our future work in Section 5.

2. Brief of BBO model and mathematical preliminary

Biogeography-based optimization mimics the species distribution in nature biogeography. In science of biogeography,one of criteria to see whether geographical areas are well suited as residences for biological species is habitat suitability in-dex (HSI). It is affected by various factors including climate, temperature, humidity and topographic features. All the factorsthat influence habitability are called suitability index variables (SIVs). Fertile areas with a high HSI tend to have a large num-ber of species, while the barren ones with low HSI have a small number of species. Habitats with a high HSI have a low immi-gration and high emigration rate since they are nearly saturated with species. In contrast, habitats with low HSI have a highimmigration and low emigration rate. In BBO, a good solution equals to an area with a high HSI in natural biogeography andthe poor solution is analogous an area with a low HSI. All the features in solutions are considered as SIVs. Good solutions tendto share their features with poor solutions, and the poor solutions will highly probably accept the features. This is very sim-ilar to species migrating between fertile areas and barren areas. In [37] that term ‘‘island’’ presents the area which is geo-graphically isolated from other habitats. We follow the expression in this paper. The pseudo codes of BBO are given inTable 1.

The work in [37] states that there are k species in the habitat throughout the evolutionary process with immigrants enter-ing the habitat at an immigration rate kk and emigrants leaving the habitat at an emigration rate lk. The largest possible spe-cies count that the habitat can support is n. Simon considered the probability Pk that the habitat contains exactly k species[37]. The term Pk changes from time t to time t + Dt as follows [25],

Pleasehttp:/

Pkðt þ DtÞ ¼ PkðtÞð1� kkDt � lkDtÞ þ Pk�1ðtÞkk�1Dt þ Pkþ1ðtÞlkþ1Dt ð1Þ

According to [37], (1) holds because in order to have k species at time (t + Dt), one of the following conditions must hold: (i)There were k species at time t, and no immigration or emigration occurred between t and (t + Dt); or, (ii) There were k � 1species at time t, and one species immigrated; or, (iii) There were k + 1 species at time t, and one species emigrated.

By assuming that Dt is small enough, the probability of more than one immigration or emigration can be ignored. Aftertaking the limit of Eq. (1) as Dt ? 0, we have

_Pk ¼�k0P0 þ l1P1; k ¼ 0�ðkk þ lkÞPk þ kk�1Pk�1 þ lkþ1Pkþ1; 1 6 k 6 n� 1�lnPn þ kn�1Pn�1; k ¼ n

8><>: ð2Þ

cite this article in press as: W. Guo et al., An analysis of the migration rates for biogeography-based optimization, Inform. Sci. (2013),/dx.doi.org/10.1016/j.ins.2013.07.018


Table 1Psuedo codes of biogeography-based optimization.

Algorithm of BBO

Set IsIndex1 and IsIndex2 as the index of islandSet K as the maximum of counts of IslandsInitialization of Islands and SpeciesCompute the Fitness of each IslandRank the islands from the best to the worst according to FitnessWhile the terminal condition is not met

For IsIndex1 = 1 to KSelect Island (IsIndex1) with immigration rate kIsindex1

If Island (IsIndex1) is selected to be immigratedFor IsIndex2 = 1 to K

Select Island (IsIndex2) with emigration rate lIsindex2

If Island (IsIndex2) is selected to emigrate to Island (IsIndex1)Species in Island (IsIndex2) migrates to Island (IsIndex1)Break For IsIndex2

End ifEnd For

Next IsIndex2End if

End ForNext IsIndex1Recompute FitnessRank the islands from the best to the worst according to Fitness

End While

W. Guo et al. / Information Sciences xxx (2013) xxx–xxx 3

It is noted that (2) is valid for k = 0, . . . , n, l0 = 0 and kn = 0.By defining P = [P0, . . . , Pn]T for notational simplicity, we obtain

Pleasehttp:/

_P ¼ AP ð3Þ

where A is given as

A ¼

�k0 l1 0 � � � � � � � � � 0

k0 �ðk1 þ l1Þ l2. .

. . .. . .

. ...

. .. . .

. . .. . .

. . .. . .

. ...

. .. . .

. . .. . .

.kn�2 �ðkn�1 þ ln�1Þ ln

0 � � � � � � � � � 0 kn�1 �ln

26666666664

37777777775

ð4Þ

Theorem 1. The steady state probability of the number of each species is given by

Pk ¼ P0

Yk

i¼1

ki�1

li

� �; k 2 f1;2; . . . ;ng ð5Þ

where li – 0 for i = 1, 2, . . . , n.

Proof. If the species count probabilities are in steady state, then from (3), we obtain
_P ¼ AP ¼ 0 ð6Þ
From (2), we obtain

P1 ¼k0

l1P0 ð7Þ

P2 ¼ðk1 þ l1ÞP1 � k0P0

l2¼ðk1 þ l1ÞP0

k0

l1

� �� k0P0

l2¼ P0

Y2

i¼1

ki�1

lið8Þ

P3 ¼ðk2 þ l2ÞP2 � k1P1

l3¼ðk2 þ l2ÞP0

�k0k1

l1l2

�� k0k1

l1P0

l3¼ P0

Y3

i¼1

ki�1

lið9Þ




Assuming (10) holds,

Pleasehttp:/

Pk�1 ¼ P0

Yk�1

i¼1

ki�1

li

� �; k 2 f3;4; . . . ;ng ð10Þ

then it follows that,

Pk ¼ðkk�1 þ lk�1ÞPk�1 � kk�2Pk�2

lk

¼ ðkk�1 þ lk�1ÞP0

lk

Yk�1

i¼1

ki�1

li

� �� kk�2P0

lk

Yk�2

i¼1

ki�1

li

� �

¼ P0

Yk�2

i¼1

ki�1

li

� �ðkk�1 þ lk�1Þkk�2

lklk�1� kk�2lk�1

lklk�1

� �

¼ P0

Yk�2

i¼1

ki�1

li

� �kk�1kk�2

lklk�1

� �¼ P0

Yk

i¼1

ki�1

li

� �; k 2 f3;4; . . . ;ng

ð11Þ

Therefore,

Pk ¼ P0

Yk

i¼1

ki�1

li

� �; k 2 f3;4; . . . ; ng ð12Þ

According to the principle of mathematical induction, Eq. (5) holds. h

Theorem 2. Assuming that

(1) ki�1 and li are bounded, i.e. supi {ki�1} <1 and supi{li} <1,(2) li – 0 for i = 1, . . . , n.

The necessary and sufficient conditions for the existence of stationary probability distribution of Pk, where k 2 {0, 1, . . . , n}, is

Xn

j¼1

Yj

i¼1

ki�1

li

� �< þ1 ð13Þ

where the steady state probabilities are given as follows,

Pk ¼ 1þXn

j¼1

Yj

i¼1

ki�1

li

� �" #�1Yk

i¼1

ki�1

li

� �; n ¼ 1;2; . . . ð14Þ

P0 ¼ 1þXn

j¼1

Yj

i¼1

ki�1

li

� �" #�1

ð15Þ

where li – 0 for i = 1, . . . , n and ki�1 and li are bounded, i.e. supi{ki�1} <1 and supi{li} <1

Proof (Necessity). Suppose the steady state probability distribution exists, from Eq. (5), we obtain

1 ¼Xn

j¼0

Pj ¼ P0 þ P0

Xn

j¼1

Yj

i¼1

ki�1

li

� �¼ P0 1þ

Xn

j¼1

Yj

i¼1

ki�1

li

� �" #ð16Þ

Due to the boundedness of P0;Pn

j¼1

Qji¼1

ki�1li

� �is bounded, i.e.

Pnj¼1

Qji¼1

ki�1li

� �< þ1. From Eq. (16), Eq. (15) holds. According to

Eq. (15) and Theorem 1, Eq. (14) holds.(Sufficiency).AsPn

j¼1Qj

i¼1ki�1li

� �converges, we have the following expressions of probabilities,

P0 ¼ 1þXn

j¼1

Yj

i¼1

ki�1

li

� �" #�1

ð17Þ

Pk ¼ 1þXn

j¼1

Yj

i¼1

ki�1

li

� �" #�1Yk

i¼1

ki�1

li

� �ð18Þ




Because the series {Pk} satisfies the two conditions of being probability distribution,

1 The

Pleasehttp:/

1: Pk P 0; k ¼ 0;1; . . . n ð19Þ

2:Xn

k¼0

Pk ¼ 1 ð20Þ

the series {Pk} is a probability distribution. Furthermore, {Pk} is the unique1 solution of the following equations,

AP ¼ 0Xn

i¼0

Pi ¼ 1

8><>: ð21Þ

The stationary probability distribution exists. h

Remark 1. Theorems 1 and 2 are mathematically same as the Theorem 1 in [25]. Based on the two theorems, we presentTheorem 3 which gives another format of the steady state probability of the number of each species.

Theorem 3. The steady-state probability of the number of each species is given by

Pk ¼1

1þPk

j¼1

Qki¼j

1aiþPn

l¼kþ1

Qli¼kþ1ai

; k 2 f1;2; . . . ;ng ð22Þ

where ai ¼ ki�1li

, ki�1 – 0 and li – 0 for all i.

Proof. From Theorem 2, we obtain

Pk ¼ 1þXn

j¼1

Yj

i¼1

ki�1

li

� �" #�1Yk

i¼1

ki�1

li

� �; k ¼ 1;2; . . . ;n ð23Þ

By defining ai ¼ ki�1li

, Eq. (24) holds.

1Pk¼

1þPn

i¼1

Qij¼1ajQk

i¼1ai

¼ 1Qki¼1ai

þQ1

j¼1ajQki¼1ai

þQ2

j¼1ajQki¼1ai

þ � � � þQk�1

j¼1 ajQki¼1ai

þQk

j¼1ajQki¼1ai

þQkþ1

j¼1 ajQki¼1ai

þ � � � þQn

j¼1ajQki¼1ai

¼ 1Qki¼1ai

þ 1Qki¼2ai

þ � � � þ 1Qki¼kai

þ 1þYkþ1

i¼kþ1

ai þ � � � þYn

i¼kþ1

ai

¼ 1þXk

j¼1

Yk

i¼j

1aiþXn

l¼kþ1

Yl

i¼kþ1

ai

ð24Þ

Hence,

Pk ¼1

1þPk

j¼1

Qki¼j

1aiþPn

l¼kþ1

Qli¼kþ1ai

: � ð25Þ

Corollary 1. If we assign a large value to ai when 1 6 i 6 k, and assign a small value to ai when k + 1 6 i 6 n, then value of Pk islarge.

Proof. Based on Theorem 3, Corollary 1 holds. h

proof can be found in Appendix A.




3. Analysis of migration rates

3.1. Equilibrium analysis

Paper [25] investigated 6 migration models given from (26)–(37), where I is the maximum immigration rate, E is the max-imum emigration rate, k is the number of species in an island and n is the largest possible species count that the habitat cansupport.

Model 1 (const immigration and linear emigration model)

Pleasehttp:/

kk ¼I2

ð26Þ

lk ¼kn

E ð27Þ

Model 2 (linear immigration and const emigration model)

kk ¼ I 1� kn

� �ð28Þ

lk ¼E2

ð29Þ

Model 3 (linear migration model)

kk ¼ I 1� kn

� �ð30Þ

lk ¼kn

E ð31Þ

This model is presented in the original BBO paper [37], and the migration rates are linear functions of species amount inthe habitat.

Model 4 (Trapezoidal migration model)

kk ¼I; k 6 i0

2I 1� kn

� ; i0 < k 6 n

(ð32Þ

lk ¼2En k; k 6 i0

E; i0 < k 6 n

(ð33Þ

where i0 is the smallest integer that is larger than or equal to ðnþ1Þ2 , namely, i0 ¼ ceil ðnþ1Þ

2

� �.

Model 5 (quadratic migration model)

kk ¼ Ikn� 1

� �2

ð34Þ

lk ¼ Ekn

� �2

ð35Þ

The migration rates are convex quadratic functions of the number of species.Model 6 (sinusoidal migration model)

kk ¼I2

coskpn

� �þ 1

� �ð36Þ

lk ¼E2� cos

kpn

� �þ 1

� �ð37Þ

The migration rates are sinusoidal functions of the number of species.In [37], the author illustrated migration model in a single habitat as in Fig. 1.We suppose the habitats are ordered as Habitat 1, Habitat 2, . . . , Habitat n and assume there are n species living in Habitat

1, n � 1 species living in Habitat 2, . . . , and the rest deduced by analogy. For the consistency of [25], we set I as 1, E as 1 and nas 50. The six migration models are drawn in Fig. 2.

The values of ai for i = 1, 2, . . . , n are given from (38)–(43) and the curves are shown in Fig. 3.



Fig. 1. Species model of a single habitat.

10 20 30 40 500

0.2

0.4

0.6

0.8

1

Habitat Index

Mig

ratio

n R

ates

of M

odel

1 λμ

Fig. 2.1. The relationship between habitat index and the migration rates of Model 1.

10 20 30 40 500

0.2

0.4

0.6

0.8

1

Habitat Index

Mig

ratio

n R

ates

of M

odel

2 λμ



Pleasehttp:/

Model 1 : ai ¼ki�1

li¼

12i

50

¼ 25i

ð38Þ


li¼

1� i�150

12

¼ 51� i25

ð39Þ


li¼

1� i�150

i50

¼ 51i� 1 ð40Þ



10 20 30 40 500

0.2

0.4

0.6

0.8

1

Habitat IndexM

igra

tion

Rat

es o

f Mod

el 3 λ

μ


10 20 30 40 500

0.2

0.4

0.6

0.8

1

1.2

1.4

Habitat Index

Mig

ratio

n R

ates

of M

odel

4 λμ


10 20 30 40 500

0.2

0.4

0.6

0.8

1

Habitat Index

Mig

ratio

n R

ates

of M

odel

5 λμ



Pleasehttp:/

Model4 : ai ¼ki�1li¼ 1

2i50¼ 25

i 1 6 i 6 26

ki�1li¼ 2ð1�i�1

50 Þ1 ¼ 51�i

25 26 < i 6 50

8<: ð41Þ


li¼

1� i�150

� 2

i50

� 2 ¼ ði� 51Þ2

i2 ð42Þ


li¼

12 cos ði�1Þp

50

� �þ 1

� �12 � cos ip

50

� þ 1

� ¼ cos ði�1Þp50

� �þ 1

� cos ip50

� þ 1

ð43Þ



10 20 30 40 500

0.2

0.4

0.6

0.8

1

Habitat IndexM

igra

tion

Rat

es o

f Mod

el 6 λ

μ


Fig. 3. The value of a in Model 1, 2, 3, 5, 6 (The curve of Model 4 overlaps the curve of Model 1 when i 6 26 and overlaps the curve of Model 2 when i > 26.)


In Fig. 3, by defining k0 as the equilibrium number point of species, where the immigration rate and the emigration rateare equal, it is obvious that when i < k0, the curve of Model 5 is always above the others and when i > k0, the curve of Model 5is always below the others. According to Corollary 1, Pk0 of Model 5 is the largest.

By calculation, the maximum steady state probabilities of Model 5 and Model 6 are 0.1584 and 0.1398 respectively, whichmeans the maximum probability in Model 5 is larger than the probability in Model 6. However, according to the numericaltest in [25], Model 6 performs better than Model 5. This shows that the probability of species amount cannot determine theperformance of migration models. Moreover, the probabilities Pk with species k in Model 5 and Model 6 are drawn in Fig. 4. Inthe figure, we use the same population size and migration models as in [25] to draw the curves of probabilities in Model 5and Model 6 and very similar curves as shown in [25], but the labels are different. Considering that the probabilities are un-ique, we guess paper [25] just accidentally switches the labels of Model 5 and Model 6.

3.2. Analysis of Markov model for BBO

To date, the migration models are designed generally according to the models of biogeography by empirical studies.Although [25] compared different migration models by numerical simulation and presented the importance of the probabil-ity Pk of median species, it did not give determinant factors to design BBO. Hence there lacks the mathematical proofs toreveal how migration models affect the performance of BBO. To complement the work, we investigate the migration mech-anism in this section. [39,40] discussed BBO’s abilities to obtain and retain good solutions in binary space and we use the ideato investigate migration models in a real number search space.

Definition 1. Define ki and li are immigration rate and emigration rate respectively for solution (island) i.

Definition 2. By defining yi,t as the ith individual in a population at generation t and v as the dimension of problem, wherei 2 [1, . . . , m] and m is the amount of candidate solutions, a solution can be expressed in (44).

Pleasehttp:/

yi;t ¼ fyið1Þ;t; yið2Þ;t ; . . . ; yiðvÞ;tg: ð44Þ




Definition 3. By defining Fitnessyi;tas the fitness of solution yi,t and fyiðsÞ;t as the contribution of yi(s),t to yi,t, where s 2 [1, . . . , v],

we obtain

Pleasehttp:/

Fitnessyi;t¼ F fyið1Þ;t ; fyið2Þ;t ; . . . ; fyiðvÞ;t

� �ð45Þ

where F is the function to calculate Fitnessyi;t.

Definition 4. Define fyjðsÞ;t P fyiðsÞ;t to denote that yj(s),t contributes more than or equally with yi(s),t to Fitnessyi;t, where equality

holds when yj(s),t contributes equally with yi(s),t to Fitnessyi;t.

Remark 2. Definition 4 denotes that if fyjðsÞ;t > fyiðsÞ;t , then yj(s),t is considered more suitable than yi(s),t to be the sth element insolution.

Definition 5. Define t� as the time to begin one generation and t+ as the time to end the same generation.

Definition 6. Define a set as follows,

JiðsÞ ¼ fj 2 ½1;n� : fyjðsÞ;t P fyiðsÞ;tg: ð46Þ

Next, we will analyze the expectation of the probability that fyiðsÞ;tþP fyiðsÞ;t� .

On one hand, if yi(s) is not selected for immigration during generation t,

yiðsÞ;tþ ¼ yiðsÞ;t� ðImmigration did not occurÞ ð47Þ

Meanwhile,

fyiðsÞ;tþ¼ fyiðsÞ;t� ð48Þ

According to [37], the probability for non-immigration on Island i is

Prðnon-immigrationÞ ¼ ð1� kiÞ ð49Þ

If there is no immigration on Island i, we obtain

PrðfyiðsÞ;tþ ¼ fyiðsÞ;t� Þ ¼ 1 ð50ÞPrðfyiðsÞ;tþ > fyiðsÞ;t� Þ ¼ 0 ð51Þ

Hence, if there is no immigration on Island i,

Prnon-immigrationðfyiðsÞ;tþP fyiðsÞ;t� Þ ¼ ð1� kiÞ ð52Þ

On the other hand, the probability that immigration occurs on Island i is

PrðimmigrationÞ ¼ ki ð53Þ

0 10 20 30 40 500

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Population Index

Pro

babi

lity

Model 5Model 6

Fig. 4. Comparison of relation of the probability Pk with species k.




In this case, the probability that fyiðsÞ;tþP fyiðsÞ;t� is proportional to the combined emigration rates of all individuals whose

fitness of the sth feature is larger than fyiðsÞ;tþ. By assuming that any species can migrate from one island to the same island, we

obtain,

Pleasehttp:/

Primmigration fyiðsÞ;tþP fyiðsÞ;t�

� �¼P

j2JiðsÞljPjlj

ð54Þ

Combining (49), (52), (53) and (54), we obtain the total probability for fyiðsÞ;tþP fyiðsÞ;t� as follows,

Pr fyiðsÞ;tþP fyiðsÞ;t�

� �¼ Prðnon-immigrationÞPrnon-immigration fyiðsÞ;tþ

P fyiðsÞ;t�

� �þ PrðimmigrationÞPrimmigration fyiðsÞ;tþ

P fyiðsÞ;t�

� �

¼ ð1� kiÞ þ ki

Pj2JiðsÞljP

jljð55Þ

Lemma 1. For an arbitrary solution yi,t, we define E

Pj2Ji ðsÞ

ljPn

j¼1lj

� �as the expectation of the probability that yi,t replaces an old feature

with a new feature which contributes not less than the old one to Fitnessyi;t. Then we obtain,

E

Pj2JiðsÞljPn

j¼1lj

" #¼Pn

i¼1

Pj2JiðsÞlj

nPn

j¼1lj

: ð56Þ

Proof.

E

Pj2JiðsÞljPn

j¼1lj

" #¼P

j2J1ðsÞljPnj¼1lj

Prði ¼ 1Þ þP

j2J2ðsÞljPnj¼1lj

Prði ¼ 2Þ þ � � � þP

j2JnðsÞljPnj¼1lj

Prði ¼ nÞ ð57Þ

Since the expectation relates to an arbitrary solution yi,t in one generation, the probabilities are the same. Hence we canobtain

Prði ¼ 1Þ ¼ Prði ¼ 2Þ ¼ � � � ¼ Prði ¼ nÞ ¼ 1n

ð58Þ

Thus,

E

Pj2JiðsÞljPn

j¼1lj

" #¼ 1

n

Pni¼1

Pj2JiðsÞljPn

j¼1lj

� ð59Þ

Lemma 2. Suppose there are no duplicate solutions in the same generation, then in any generation, the probability thatyiðsÞ;tþ ¼ yiðsÞ;t� is ð1� kiÞ þ ki

liPn

i¼1li

, namely,

PrðyiðsÞ;tþ ¼ yiðsÞ;t� Þ ¼ ð1� kiÞ þ kiliPni¼1li

ð60Þ

Proof. We discuss this problem in two cases, no immigration occurs and immigration occurs respectively. Hence,

Pr yiðsÞ;tþ ¼ yiðsÞ;t��

¼ Prnon-immigration yiðsÞ;tþ ¼ yiðsÞ;t��

þ Primmigration yiðsÞ;tþ ¼ yiðsÞ;t��

ð61Þ

On one hand, from (49), the probability that immigration does not occur is (1 � ki). If immigration does not occur,

Prnon-immigrationðyiðsÞ;tþ ¼ yiðsÞ;t� Þ ¼ 1: ð62Þ

On the other hand, from (53), the probability that immigration occurs is ki. If immigration occurs, yiðsÞ;tþ ¼ yiðsÞ;t� only whenyi(s) migrates to yi(s), which means species migrate from one island to the same island. In this case, the probability thatyiðsÞ;tþ ¼ yiðsÞ;t� is

Primmigration yiðsÞ;tþ ¼ yiðsÞ;t��

¼ liPnj¼1lj

: ð63Þ

Thus,




Pleasehttp:/

PrðyiðsÞ;tþ ¼ yiðsÞ;t� Þ ¼ Prðnon-immigrationÞPrnon�immigration yiðsÞ;tþ ¼ yiðsÞ;t��

þ Pr immigrationð ÞPrimmigration yiðsÞ;tþ ¼ yiðsÞ;t��

¼ ð1� kiÞ þ kiliPni¼1li

� ð64Þ

Theorem 4. For yi,t, by defining Eobtain Pr fyiðsÞ;tþP fyiðsÞ;t�

� �h ias the expectation of probability of obtaining a new feature that is not

worse than the old one. Eobtain can be as large as possible if the following conditions are met,

a.

Pn

i¼1

Pj2Ji ðsÞ

lj

nPn

j¼1lj

is as large as possible;

b. ki is as small as possible.

where n is the largest possible species count that the habitat can support.Proof. From (55), we obtain

Eobtain Pr fyiðsÞ;tþP fyiðsÞ;t�

� �h i¼ ð1� kiÞ þ kiE

Pj2JiðsÞljP

jlj

" #ð65Þ

Combine Lemma 1 and (65), we obtain

Eobtain Pr fyiðsÞ;tþ P fyiðsÞ;t�

� �h i¼ ð1� kiÞ þ ki

Pni¼1

Pj2JiðsÞlj

nPn

j¼1lj

¼ 1þ ki

Pni¼1

Pj2JiðsÞlj

nPn

j¼1lj

� 1

!ð66Þ

Pn P Pn P
From (57), Eobtain increases as i¼1 j2Ji ðsÞ
lj

nPn

j¼1lj

increases. Since i¼1 j2Ji ðsÞlj

nPn

j¼1lj� 1 < 0, Eobtain increases as ki decreases.

To sum up, Eobtain Pr fyiðsÞ;tþ P fyiðsÞ;t�

� �h ican be as large as possible if the following conditions are met,

a.

Pn

i¼1

Pj2Ji ðsÞ

lj

nPn

j¼1lj

is as large as possible;

b. ki is as small as possible. h

Remark 3. According to Theorem 4, enlarging

Pn

i¼1

Pj2Ji ðsÞ

lj

nPn

j¼1lj

and diminishing ki can help solutions in BBO obtain a new featurewhich is not worse than the old one.

Next, we will discuss how BBO retains good features. According to Lemma 2, the probability that a feature can be retained is

Pr fyiðsÞ;tþ¼ fyiðsÞ;t�

� �¼ ð1� kiÞ þ ki

liPnj¼1lj

¼ 1þ kiliPnj¼1lj

� 1

!ð67Þ

Hence, if we choose large liPn

j¼1lj

and small ki, it is more likely to retain the feature.

Note that Theorem 4 discusses the case that the algorithm obtains a new feature which is not worse than the old one, but

it does not mean the new feature is better than the old one. Although Eobtain Pr fyiðsÞ;tþP fyiðsÞ;t�

� �h i¼ 1 when ki = 0, it does not

make any sense to enhance BBO. Therefore we define Eobtainbetter Pr fyiðsÞ;tþ> fyiðsÞ;t�

� �h ito denote the expectation of obtaining a

new feature which is better than the old one. It will reveal how to design the migration model to enhance BBO.

Theorem 5. Suppose that l is a continuous and monotonically increasing function, li – 0, 0 6 ki 6 1 for i = 1, 2, . . . , n. EnlargingPn

i¼1ðki

Pj2Ji ðsÞ

ljÞPn

j¼1lj

is helpful to obtain a new feature which is better than the old one.

Proof

Eobtainbetter Pr fyiðsÞ;tþ > fyiðsÞ;t�

� �h i¼ E Pr fyiðsÞ;tþ

P fyiðsÞ;t�

� �� Pr fyiðsÞ;tþ

¼ fyiðsÞ;t�

� �h i¼ E 1� kið Þ þ ki

Pj2JiðsÞljPn

j¼1lj

!� 1þ ki

liPnj¼1lj

� 1

! !" #

¼ E ki

Pj2JiðsÞlj � liPn

j¼1lj

" #

¼ k1

Pj2J1ðsÞlj � l1Pn

j¼1lj

Prði ¼ 1Þ þ � � � þ kn

Pj2JnðsÞlj � lnPn

j¼1lj

Prði ¼ nÞ

ð68Þ




According to (58), we obtain,

2 We

Pleasehttp:/

Eobtainbetter Pr fyiðsÞ;tþ> fyiðsÞ;t�

� �h i¼ k1

Pj2J1ðsÞlj � l1Pn

j¼1lj

Prði ¼ 1Þ þ � � � þ kn

Pj2JnðsÞlj � lnPn

j¼1lj

Prði ¼ nÞ

¼ 1n

Xn

i¼1

kiP

j2JiðsÞlj � li

� �� Pn

j¼1lj

24

35 ð69Þ

Since l is a continuous and monotonically increasing function, we have liPn

j¼1lj� 0 when n is a large value. Thus,P P

Eobtainbetter Pr fyiðsÞ;tþ > fyiðsÞ;t�

� �h i� 1

n

ni¼1ðki j2JiðsÞljÞPn

j¼1lj

: ð70ÞP P P P
From (70), larger value of
n

i¼1ðki j2Ji ðsÞ

ljÞPn

j¼1lj

can help BBO obtain better features. Meanwhile, ki andn

i¼1 j2Ji ðsÞlj

nPn

j¼1lj

are positivelyrelated to Eobtainbetter. h

Remark 4. According to Theorem 4 and Theorem 5, enlarging

Pn

i¼1

Pj2Ji ðsÞ

lj

nPn

j¼1lj

and ki is helpful to BBO to obtain a better feature.

However, according to (66), a large ki compromises the ability to retain good features. The balance of ki will be investigated inour future work.

Next we will present the approaches to design migration models based on above analysis. After each generation, we rankthe solutions according to the fitnesses, and redefine the best solution as y1, the second best solution as y2, the third bestsolution as y3, and the rest deduced by analogy. It is necessary to retain most features in better solutions. According to (67),for y1,

Pr fy1ðsÞ;tþ¼ fy1ðsÞ;t�

� �¼ ð1� k1Þ þ k1

l1Pnj¼1lj

¼ 1þ k1l1Pnj¼1lj

� 1

!: ð71Þ

If we are supposed to retain the features of y1, it is necessary to enlarge Pr fy1ðsÞ;tþ ¼ fy1ðsÞ;t�

� �. According to (71), since

l1Pn

j¼1lj� 1 < 0, small k1 is helpful to retain the features of y1. Here we use indicator R to denote k1 in different models, namely

R ¼ k1: ð72Þ

According to (71), a large l1Pn

k¼1lk

is helpful to retain a good feature. Hence we use indicator S to denote l1Pn

k¼1lk

in differentmodels, namely,

S ¼ l1Pnk¼1lk

: ð73ÞPn P
According to Theorem 5, enlarging ki and i¼1 j2Ji ðsÞ
lj

nPn

j¼1lj

is helpful to obtain large Eobtainbetter. Hence we define2

T ¼Xn

i¼1

ki; ð74Þ

and

Q ¼Pn

i¼1

Pj2JiðsÞlj

nPn

j¼1lj

: ð75Þ

Next, we will discuss how to calculate Ji(s) based on following pseudo-codes of BBO.According to the Table 1, it supposes that a good solution should contain more good features. Namely, BBO uses Fitnessy to

evaluate fyiðsÞ;t . Thus, we consider JiðsÞ ¼ fj 2 ½1;n� : fyjðsÞ;t P fyiðsÞ;tg as

JiðsÞ ¼ fj 2 ½1;n� : fyjðsÞ;t P fyiðsÞ;tg ¼ fj 2 ½1;n� : FitnessyjP Fitnessyi

g: ð76Þ

Since good solutions will be ranked front, the index of good solution is smaller than the bad solution. Hence we obtain,

JiðsÞ ¼ fj 2 ½1;n� : fyjðsÞ;t P fyiðsÞ;tg ¼ fj 2 ½1;n� : FitnessyjP Fitnessyi

g ¼ fj 2 ½1;n� : j 6 ig ð77Þ

According to (76), Q in (75) can be given as follows,

Q ¼Pn

i¼1

Pj2JiðsÞlj

nPn

j¼1lj

¼Pn

i¼1

Pij¼1lj

nPn

j¼1lj

: ð78Þ

with n = 50, we obtain Table 2

usePn

i¼1ki to denote the comprehensive ability to obtain better features.



Table 2Values of R, T, S, Q in Model 1, 2, 3, 4, 5, 6. (The bold values are the best performance for each row)

Model 1 Model 2 Model 3 Model 4 Model 5 Model 6

R 0.5 0.02 0.02 0.04 0.0004 0.001T 25 24.5 24.5 37.04 16.17 24.5S 0.04 0.02 0.04 0.027 0.0594 0.0408Q 0.6733 0.5100 0.6733 0.6162 0.7574 0.7085


Remark 5. According to Theorem 5, large Q can help BBO obtain a better solution. According to (71), large S can help BBOretain good features. Table 2 shows that S and Q of Model 5 are larger than other models, so it is advisable to choose theemigration model of Model 5. As to how immigration rates influence the performance of BBO, we will analyze the simulationresults in Section 4. In the following analysis, Ri indicates the value of R in Model i. So does Ti, Si and Qi respectively.

4. BBO performance analysis

In this section, we investigate the performance of BBO with different migration models. The simulation includes twoparts. First, a representative set of 14 benchmarks in [37] has been used to validate our proposed analysis. Second, weuse all 25 benchmarks in [43] to test algorithms.

4.1. Influence of population size and problem dimension on BBO

In this subsection, we use the benchmarks in Table 3 to evaluate the performance of BBO with different dimension, pop-ulation size. More information about these functions, including their domain, can be found in [3,5,52]. For the sake of con-venience, we let F1, F2, . . . , F14 denote the 14 benchmark functions respectively.

In order to investigate the impact of both population size and problem dimension on BBO, we use the original version ofBBO (BBO with migration Model 3) to conduct simulation. In the simulation, we use a population size of 10, 20 and 50 withdimensions 10, 20 and 50 respectively. The iteration limit is 1000. We run 100 Monte Carlo simulations on each benchmark.The mutation probability is 0. The results are shown in Table 4. ‘Mean’ is an average value of 100 test results, and ‘Best’ is thebest solution in the 100 test results for each benchmark.

On one hand, from Table 4, we know that for problems with dimension 10, BBO with population size of 10 outperformsBBO with population size of 20 and 50 except F1 and F4. For the problems with dimension 10, the performance of BBO dropswith a large population size. This means if the scale of the problem is small, increment in population size cannot benefit theperformance of BBO. The large population size can increase the complexity of computation so that BBO cannot obtain a bet-ter solution. On the other hand, for problems with dimension 50, BBO with population size of 50 has a comprehensive victoryover BBO with population size of 10 and 20 on all benchmarks. This means a large population size is helpful to solve prob-lems with large dimension. Hence we can draw the following conclusion: (i) For problems with small dimension, a smallpopulation size is suitable and a large population size will jeopardize the performance. (ii) For the problems with largedimension, a large population size in BBO is helpful.

4.2. Performance comparison between BBO and other EAs

In this subsection, we compare BBO with other classical EAs. In paper [25,37], the authors use the original versions ofother EAs to compare with BBO. To replenish the work, we compare BBO with several improved EAs, including ArtificialBee Colony (ABC), Cuckoo Search algorithm (CSA) [49,50], Differential Evolution (DE), Evolutionary Strategy (ES), GeneticAlgorithm (GA), Particle Swarm Optimization (PSO).

In addition, we also compare BBO with different migration models to verify the analysis in Section 3. To reassemble tomodels in paper [25], we considered Model X and Model Y as the fundamental migration models, and Model X–Y is the mod-el with the immigration rate of Model X and the emigration rate of Model Y. Since in paper [25], Model 3, Model 5 and Model6 perform better than other migration models, we adopt them for reassemble purpose in simulations.

For BBO, all parameters are the same as Table 3 in paper [25]. The differences are just the migration models. For ABC, thecolony size is 50. The number of food sources equals the half of the colony size. The maximum cycle is 100, and limit gen-eration is 100. For ES, we produce 10 offspring in each generation, and standard deviation is 1 for changing solutions. ForPSO, we use only global learning (no local neighborhood). The inertial constant is 0.3, a cognitive constant being 1, and asocial constant for swarm interaction being 1. For DE, we use a weighting factor 0.5 and a crossover constant 0.5. For CuckooSearch Algorithm (CSA), the amount of nests is 50 and the discovery rate of alien eggs is 0.25. For the GA, we use roulettewheel selection, single point crossover with a crossover probability of 1, and a mutation probability being 0.01. Thedimension of each benchmark is 20.

Please cite this article in press as: W. Guo et al., An analysis of the migration rates for biogeography-based optimization, Inform. Sci. (2013),http://dx.doi.org/10.1016/j.ins.2013.07.018


Table 3Benchmark functions.

Function Name Dimension Domain

F1 Ackley’s Function 20 [�30,30]D

F2 Fletcher-Powell 20 [�p,p]D

F3 Generalized Griewank’s function 20 [�600,600]D

F4 Generalized Penalized function 1 20 [�50,50]D

F5 Generalized Penalized function 2 20 [�50,50]D

F6 Quartic function 20 [�1.28,1.28]D

F7 Generalised Rastrigin’s function 20 [�5.12,5.12]D

F8 Generalized Rosenbrock’s function 20 [�2.048,2.048]D

F9 Schwefel’ Problem 1.2 20 [�65.535,65.535]D

F10 Schwefel’ Problem 2.21 20 [�100,100]D



F13 Sphere Model 20 [�5.12,5.12]D

F14 Step Function 20 [�200,200]D

Table 4Mean values and Best values obtained from BBO (Model 3) algorithm with different population sizes for14 benchmarks with dimensions 10, 20 and 50.

Population size Fun. D = 10 D = 20 D = 50

Mean Best Mean Best Mean Best

10 F1 8.10E�01 4.51E�01 1.01E+00 5.28E�01 5.01E+00 4.22E+00F2 1.58E+03 4.46E+02 3.68E+03 6.23E+02 3.77E+05 2.66E+05F3 1.01E+00 1.01E+00 1.02E+00 1.01E+00 3.37E+00 2.79E+00F4 9.01E�03 6.01E�04 4.55E�02 2.09E�03 3.92E+00 2.49E+00F5 7.86E�02 5.11E�02 9.92E�02 5.61E�02 3.45E+01 1.02E+01F6 1.00E�07 1.00E�08 3.54E�07 4.00E�08 1.72E�02 7.28E�03F7 0.00E+00 0.00E+00 4.04E�01 0.00E+00 4.38E+01 3.91E+01F8 5.79E+00 3.59E+00 4.48E+00 1.27E+00 2.35E+02 1.37E+02F9 2.64E+00 1.47E+00 6.03E+00 2.88E+00 8.38E+02 6.16E+02F10 9.32E+01 9.02E+00 2.20E+02 2.22E+01 2.16E+04 1.96E+04F11 1.20E�01 0.00E+00 2.60E�01 0.00E+00 9.00E+00 7.50E+00F12 2.12E+00 1.20E+00 5.86E+00 4.10E+00 6.13E+01 4.90E+01F13 0.00E+00 0.00E+00 8.06E�03 0.00E+00 1.12E+00 9.57E�01F14 1.20E+00 0.00E+00 1.00E+00 0.00E+00 3.44E+02 2.61E+02

20 F1 1.56E+00 7.23E�01 2.64E+00 2.17E+00 3.52E+00 3.22E+00F2 1.35E+04 5.74E+03 2.96E+04 9.73E+03 2.17E+05 1.43E+05F3 1.06E+00 1.03E+00 1.26E+00 1.18E+00 2.33E+00 1.91E+00F4 1.08E�01 1.27E�02 2.47E�01 1.63E�01 1.18E+00 9.68E�01F5 4.59E�01 2.70E�01 1.41E+00 8.68E�01 6.78E+00 4.49E+00F6 2.83E�06 7.90E�07 1.67E�05 4.69E�06 2.63E�03 3.81E�04F7 2.02E�01 0.00E+00 3.82E+00 3.03E+00 2.39E+01 2.01E+01F8 4.88E+01 1.46E+01 5.04E+01 1.32E+01 1.61E+02 1.01E+02F9 2.38E+01 1.38E+01 6.03E+01 3.01E+01 4.03E+02 2.48E+02F10 1.42E+03 8.63E+02 2.62E+03 1.72E+03 1.63E+04 1.37E+04F11 6.20E�01 3.00E�01 1.34E+00 9.00E�01 4.38E+00 3.90E+00F12 1.08E+01 6.00E+00 1.47E+01 1.07E+01 5.43E+01 4.25E+01F13 1.21E�02 0.00E+00 7.26E�02 1.01E�02 4.98E�01 3.63E�01F14 8.80E+00 3.00E+00 2.50E+01 1.40E+01 1.58E+02 1.07E+02

50 F1 5.20E�01 4.51E�01 5.11E+00 4.33E+00 2.13E+00 1.78E+00F2 2.30E+04 5.70E+03 2.53E+05 1.87E+05 1.55E+05 1.02E+05F3 1.02E+00 1.01E+00 3.49E+00 2.49E+00 1.27E+00 1.17E+00F4 7.27E�03 2.31E�03 3.13E+00 2.90E+00 1.87E�01 8.38E�02F5 9.39E�02 5.76E�02 3.37E+02 6.88E+00 1.48E+00 9.46E�01F6 2.12E�07 1.30E�07 1.07E�02 5.01E�03 4.30E�05 1.46E�05F7 0.00E+00 0.00E+00 4.34E+01 3.68E+01 4.43E+00 2.02E+00F8 1.95E+01 1.14E+01 2.30E+02 1.72E+02 1.14E+02 1.00E+02F9 5.64E+00 2.08E+00 9.14E+02 6.06E+02 7.60E+01 5.30E+01F10 5.09E+02 2.26E+02 2.12E+04 1.52E+04 1.07E+04 9.36E+03F11 1.80E�01 1.00E�01 9.16E+00 8.90E+00 1.92E+00 1.60E+00F12 6.10E+00 3.40E+00 6.87E+01 6.34E+01 4.33E+01 3.89E+01F13 0.00E+00 0.00E+00 8.18E�01 6.05E�01 1.11E�01 4.03E�02F14 1.60E+00 0.00E+00 3.91E+02 3.42E+02 3.64E+01 1.60E+01




Table 5Mean results obtained through 20 independent runs on basic functions. (The bold values are the best performance for each column)

Fun. F1 F2 F3 F4 F5 F6 F7

ABC 1.15E+01 5.31E+05 6.19E+00 7.64E�02 1.22E�01 4.42E�01 5.10E+01CSA 1.96E+01 9.30E+05 3.69E+02 1.01E+07 4.25E+07 3.76E+00 2.01E+02DE 5.11E+00 1.68E+05 2.42E+00 1.06E+01 1.64E+03 7.71E�03 1.10E+02ES 1.87E+01 6.18E+05 1.07E+02 4.09E+07 1.26E+08 1.59E+01 2.17E+02GA 1.57E+01 1.60E+05 2.28E+01 3.91E+03 5.67E+05 2.95E�01 1.20E+02PSO 1.56E+01 4.88E+05 7.40E+01 5.11E+06 2.24E+07 2.89E+00 1.55E+02BBO1 4.34E+01 3.27E+05 2.45E+01 1.30E+01 2.78E+03 1.09E�01 1.92E+02BBO2 7.00E+00 5.71E+05 6.76E+01 3.28E+02 5.15E+05 3.04E�01 3.38E+02BBO3 5.54E+00 4.01E+04 3.77E+00 4.33E+00 2.66E+03 1.08E�02 1.98E+01BBO4 7.67E+00 4.24E+04 7.95E+00 1.51E+01 7.62E+04 1.94E�02 2.49E+01BBO5 5.93E+00 4.01E+04 3.84E+00 2.26E+02 7.29E+03 1.54E�02 2.06E+01BBO6 5.18E+00 3.74E+04 3.17E+00 3.31E+00 6.91E+02 6.01E�03 1.71E+01BBO3–6 5.38E+00 3.83E+04 3.22E+00 4.94E+01 3.68E+03 6.78E�03 1.83E+01BBO6–3 5.32E+00 4.05E+04 3.24E+00 3.62E+00 3.76E+03 7.31E�03 1.82E+01BBO3–5 4.98E+00 3.37E+04 2.59E+00 5.66E+00 9.54E+02 4.16E�03 1.76E+01BBO5–3 6.32E+00 4.41E+04 4.58E+00 1.64E+03 1.95E+04 2.24E�02 2.36E+01BBO5–6 6.28E+00 4.48E+04 4.30E+00 3.21E+02 3.72E+04 1.73E�02 2.16E+01BBO6–5 5.04E+00 3.16E+04 2.56E+00 2.64E+00 2.05E+02 4.02E�03 1.61E+01

F8 F9 F10 F11 F12 F13 F14ABC 1.28E+02 9.12E+01 5.28E+01 8.20E�01 6.36E+03 1.89E+00 2.19E+03CSA 6.40E+02 1.59E+01 8.06E+01 3.34E+06 1.08E+04 6.19E+01 1.63E+05DE 8.69E+01 2.55E+03 9.06E+03 4.70E+00 3.26E+01 4.40E�01 1.54E+02ES 2.51E+03 3.76E+03 1.29E+04 7.68E+01 6.37E+01 6.86E+01 1.52E+04GA 2.33E+02 1.12E+03 7.86E+03 2.77E+01 4.66E+01 1.16E+01 2.34E+03PSO 5.11E+02 4.83E+03 8.23E+03 4.12E+01 4.53E+01 2.11E+01 8.18E+03BBO1 5.69E+01 5.00E+02 6.40E+03 3.26E+00 2.04E+02 5.08E�01 1.72E+02BBO2 9.36E+01 6.47E+02 5.83E+03 6.22E+00 3.88E+02 1.58E+00 9.37E+02BBO3 7.04E+01 4.37E+02 3.44E+03 3.85E+00 3.80E+01 6.52E�01 2.61E+02BBO4 1.01E+02 8.09E+02 4.69E+03 6.26E+00 4.26E+01 2.12E+00 7.02E+02BBO5 7.67E+01 4.88E+02 3.71E+03 4.13E+00 4.26E+01 8.18E�01 3.23E+02BBO6 6.75E+01 3.74E+02 3.44E+03 3.02E+00 4.15E+01 5.42E�01 2.06E+02BBO3–6 7.24E+01 3.98E+02 3.52E+03 3.56E+00 3.90E+01 6.72E�01 2.55E+02BBO6–3 7.47E+01 4.01E+02 3.41E+03 3.64E+00 4.11E+01 6.36E�01 2.41E+02BBO3–5 6.74E+01 3.79E+02 3.71E+03 3.19E+00 3.72E+01 5.17E�01 1.95E+02BBO5–3 8.04E+01 5.42E+02 3.53E+03 4.67E+00 4.26E+01 1.03E+00 4.21E+02BBO5–6 7.59E+01 5.12E+02 3.79E+03 4.30E+00 4.25E+01 9.51E�01 3.47E+02BBO6–5 5.04E+00 3.16E+04 2.56E+00 2.64E+00 2.05E+02 4.02E�03 1.61E+01


The Mean Value in Table 5 presents the general performance of different algorithms. In the table, BBO 3–5 performs betterthan others for F1. BBO 6–5 outperforms others on F2, F6, F7, F8, F10, F13 and F14. DE wins for F3 and F12. ABC wins for F4,F5 and F11. CSA wins for F9.

The Best Value in Table 6 evaluates the performance of algorithms to get a good solution. For F1 and F12, the best valuesare obtained by BBO 6. For F2, F6, F8, F13, BBO 6–5 performs the best. ABC outperforms other algorithms for F3, F4, F5, F10,F11 and F14. BBO 3–6 wins for F7 and CSA wins for F9.

Based on the numerical simulation results, BBO 6–5 outperforms other EAs for several benchmarks. However, accordingto NFL (No Free Lunch) Theorem [48], all meta-heuristics which search for extreme are exactly the same in performancewhen averaged over all possible objective functions. In other word, in statistics, all meta-heuristics perform identically onsolving computational problems. Hence, the results show that BBO, as a novel optimization approach, has a certain potentialto explore.

4.3. Simulations with different migration models

In order to compare different migration models in detail, we analyze the performance of different migration models ofBBO based on empirical studies. According to Tables 5 and 6, we rank the different versions of BBO from the best to the worst.The ranking tables are shown in Tables 7 and 8. In addition, besides the migration models used in Tables 5 and 6, we addModel 4–5 and Model 1–5. This is because R and T of the two models are larger than others’ according to Table 2. The otherparameters of Model 4–5 and Model 1–5 are the same as in Tables 5 and 6. The simulation results of Model 4–5 and Model 1–5 can be found in Appendix B. We evaluate the migration models based on the data in Tables 7 and 8. We assign 11 points forthe best model, 10 points for the second best model,. . ., and 1 point for the worst model. Then we add the points for allbenchmarks and obtain Table 9. According to Table 9, we draw Fig. 5 to show the scores of different migration models.

According to Fig. 5, we know that,



Table 6Best results obtained through 25 independent runs on basic functions. (The bold values are the best performance for each column)

Fun. F1 F2 F3 F4 F5 F6 F7

ABC 5.91E+00 3.00E+05 1.07E+00 1.58E�03 1.30E�02 1.56E�01 2.99E+01CSA 1.72E+01 4.17E+05 2.45E+02 1.80E+06 9.15E+06 9.90E�01 1.50E+02DE 3.92E+00 6.83E+04 1.64E+00 4.24E+00 1.85E+01 2.03E�03 8.09E+01ES 1.74E+01 3.64E+05 6.44E+01 5.57E+06 4.44E+07 8.73E+00 1.73E+02GA 1.23E+01 2.89E+04 8.37E+00 3.60E+00 3.72E+01 2.42E�02 7.67E+01PSO 1.37E+01 3.29E+05 4.39E+01 4.71E+05 3.16E+06 3.46E�01 1.08E+02BBO1 1.32E+01 1.42E+05 4.22E+01 8.52E+04 5.38E+06 9.36E�01 1.04E+02BBO2 5.79E+00 3.28E+04 5.02E+00 8.72E+00 2.31E+03 1.18E�02 2.63E+01BBO3 4.15E+00 1.32E+04 1.89E+00 1.30E+00 6.40E+00 2.47E�04 9.93E+00BBO4 6.71E+00 2.67E+04 5.86E+00 1.03E+01 7.00E+02 8.37E�03 1.68E+01BBO5 4.00E+00 1.41E+04 1.75E+00 1.81E+00 7.12E+00 4.25E�04 8.92E+00BBO6 3.41E+00 1.32E+04 1.49E+00 7.13E�01 4.03E+00 4.71E�04 9.01E+00BBO3–6 3.54E+00 1.04E+04 1.73E+00 1.14E+00 5.28E+00 3.51E�04 7.91E+00BBO6–3 3.79E+00 1.31E+04 1.46E+00 1.42E+00 3.27E+00 3.63E�04 9.01E+00BBO3–5 3.69E+00 9.32E+03 1.56E+00 1.02E+00 2.16E+00 2.71E�04 8.66E+00BBO5–3 4.36E+00 8.46E+03 1.89E+00 1.96E+00 1.01E+01 8.04E�04 1.16E+01BBO5–6 4.55E+00 1.05E+04 1.78E+00 2.10E+00 7.26E+00 1.20E�03 9.93E+00BBO6–5 3.82E+00 6.18E+03 1.34E+00 5.46E�01 4.27E+00 2.19E�04 8.84E+00

F8 F9 F10 F11 F12 F13 F14ABC 2.64E+01 1.94E�04 3.09E+01 1.89E�01 5.61E+03 1.62E�01 3.80E+01CSA 1.03E+02 1.98E�10 6.25E+01 6.94E+01 9.36E+03 2.78E+01 1.01E+05DE 4.77E+01 2.23E+03 6.93E+03 2.93E+00 2.45E+01 2.47E�01 6.90E+01ES 1.12E+03 2.91E+03 6.63E+03 4.09E+01 5.06E+01 4.55E+01 8.71E+03GA 9.83E+01 2.68E+02 4.41E+03 1.34E+01 2.47E+01 1.47E+00 5.33E+02PSO 1.34E+02 3.75E+03 4.31E+03 2.30E+01 3.15E+01 9.41E+00 5.06E+03BBO1 4.61E+02 2.26E+03 1.21E+04 2.30E+01 4.48E+01 1.50E+01 6.76E+03BBO2 3.66E+01 5.92E+02 3.84E+03 4.40E+00 2.62E+01 1.16E+00 5.20E+02BBO3 2.18E+01 1.60E+02 1.29E+03 1.70E+00 1.79E+01 1.91E�01 6.30E+01BBO4 5.15E+01 5.90E+02 3.19E+03 2.80E+00 3.88E+01 1.29E+00 5.00E+02BBO5 2.63E+01 1.08E+02 1.31E+03 1.70E+00 1.98E+01 2.12E�01 7.90E+01BBO6 1.96E+01 1.31E+02 9.94E+02 1.50E+00 1.76E+01 1.61E�01 7.00E+01BBO3–6 2.16E+01 1.14E+02 1.10E+03 1.60E+00 1.92E+01 2.32E�01 7.80E+01BBO6–3 2.39E+01 1.46E+02 1.11E+03 1.40E+00 2.28E+01 2.22E�01 6.90E+01BBO3–5 1.96E+01 1.17E+02 1.57E+03 1.30E+00 1.83E+01 2.02E�01 5.00E+01BBO5–3 2.23E+01 1.79E+02 1.14E+03 1.90E+00 2.27E+01 3.12E�01 1.37E+02BBO5–6 2.67E+01 2.28E+02 1.11E+03 2.30E+00 2.21E+01 2.42E�01 1.21E+02BBO6–5 1.61E+01 1.19E+02 1.36E+03 1.70E+00 2.00E+01 1.31E�01 5.80E+01

Table 7Comparison of different migrations on each benchmark (Mean value).

F From Best to Worst

F1 M3–5 M6–5 M6 M6–3 M3–6 M3 M5 M5–6 M5–3 M4–5 M1–5F2 M6–5 M3–5 M6 M3–6 M3 M5 M6–3 M5–3 M5–6 M4–5 M1–5F3 M6–5 M3–5 M6 M3–6 M6–3 M3 M5 M5–6 M5–3 M4–5 M1–5F4 M6–5 M6 M6–3 M3 M3–5 M3–6 M5 M5–6 M5–3 M4–5 M1–5F5 M6–5 M6 M3–5 M3 M3–6 M6–3 M5 M5–3 M5–6 M4–5 M1–5F6 M6–5 M3–5 M6 M3–6 M6–3 M3 M5 M5–6 M5–3 M4–5 M1–5F7 M6–5 M6 M3–5 M6–3 M3–6 M3 M5 M5–6 M5–3 M4–5 M1–5F8 M6–5 M3–5 M6 M3 M3–6 M6–3 M5–6 M5 M5–3 M4–5 M1–5F9 M6–5 M6 M3–5 M3–6 M6–3 M3 M5 M5–6 M5–3 M4–5 M1–5F10 M6–5 M6–3 M3 M6 M3–6 M5–3 M3–5 M5 M5–6 M4–5 M1–5F11 M6 M6–5 M3–5 M3–6 M6–3 M3 M5 M5–6 M5–3 M4–5 M1–5F12 M3–5 M3 M3–6 M6–5 M6–3 M6 M5–6 M5 M5–3 M4–5 M1–5F13 M6–5 M3–5 M6 M6–3 M3 M3–6 M5 M5–6 M5–3 M4–5 M1–5F14 M3–5 M6 M6–5 M6–3 M3–6 M3 M5 M5–6 M5–3 M4–5 M1–5


In Fig. 5.1, Model 6–5 has the highest score. This verifies the analysis in Section 3. From the analysis, we know a migrationmodel with large S, large Q, proper R and proper T can help BBO perform better, where S, Q, R and T are defined in Table 3 andS5 and Q5 are the largest, and R6 and T6 are moderate. In Fig. 5.1, Model 3–5 performs the second best on the average values.In the analysis in Section 3, we draw the conclusion that large k can help BBO obtain a good solution. Meanwhile, large k isharmful for BBO to retain a good solution. We deem only the moderate k is suitable. The balance of k is not clear, so we can-not compare different immigration models quantitatively. However, the performance of Model 3–5 shows the emigrationmodel of Model 5 can help BBO perform better.



Table 8Comparison of different migrations on each benchmark (Best value).

F From Best to Worst

F1 M6 M3–6 M3–5 M6–3 M6–5 M5 M3 M5–3 M5–6 M4–5 M1–5F2 M6–5 M5–3 M3–5 M3–6 M5–6 M6–3 M3 M6 M5 M4–5 M1–5F3 M6–5 M6–3 M6 M3–5 M3–6 M5 M5–6 M3 M5–3 M4–5 M1–5F4 M6–5 M6 M3–5 M3–6 M3 M6–3 M5 M5–3 M5–6 M4–5 M1–5F5 M3–5 M6–3 M6 M6–5 M3–6 M3 M5 M5–6 M5–3 M4–5 M1–5F6 M6–5 M3 M3–5 M3–6 M6–3 M5 M6 M5–3 M5–6 M4–5 M1–5F7 M3–6 M3–5 M6–5 M5 M6 M6–3 M3 M5–6 M5–3 M4–5 M1–5F8 M6–5 M6 M3–5 M3–6 M3 M5–3 M6–3 M5 M5–6 M4–5 M1–5F9 M5 M3–6 M3–5 M6–5 M6 M6–3 M3 M5–3 M5–6 M4–5 M1–5F10 M6 M3–6 M5–6 M6–3 M5–3 M3 M5 M6–5 M3–5 M4–5 M1–5F11 M3–5 M6–3 M6 M3–6 M6–5 M5 M3 M5–3 M5–6 M4–5 M1–5F12 M6 M3 M3–5 M3–6 M5 M6–5 M5–6 M5–3 M6–3 M4–5 M1–5F13 M6–5 M6 M3 M3–5 M5 M6–3 M3–6 M5–6 M5–3 M4–5 M1–5F14 M3–5 M6–5 M3 M6–3 M6 M3–6 M5 M5–6 M5–3 M4–5 M1–5

Table 9The score of different models for all benchmarks.

M3 M5 M6 M3–5 M3–6 M5–3 M5–6 M6–3 M6–5 M1–5 M4–5

Best 93 84 120 125 114 62 64 121 99 14 28Mean 99 68 129 131 103 47 55 103 147 14 28

0

20

40

60

80

100

120

140

160

M3 M5 M6 M3-5 M3-6 M5-3 M5-6 M6-3 M6-5 M1-5 M4-5

Model

Scor

e

Fig. 5.1. Comparison of different models of BBO for function 1–14 (Mean value).

0

20

40

60

80

100

120

140

M3 M5 M6 M3-5 M3-6 M5-3 M5-6 M6-3 M6-5 M1-5 M4-5

Model

Sco

re

Fig. 5.2. Comparison of different models of BBO for function 1–14 (Best value).




Table 10Comparison of Model 4–5 with different strategies on each benchmark. (The bold values are the best performance for each row)

F Model 4–5 with storage Model 4–5 with elitism Model 4–5

MeanMin BestMin MeanMin BestMin MeanMin BestMin

F1 1.4871E+01 9.2328E+00 6.1797E+00 4.5158E+00 1.7462E+01 1.4150E+01F2 2.2228E+05 8.2495E+04 4.8608E+04 1.1816E+04 4.7461E+05 2.3274E+05F3 6.0926E+01 2.0405E+01 4.4808E+00 2.2801E+00 1.2060E+02 6.3412E+01F4 7.6494E+06 3.3993E+04 3.6861E+02 2.1312E+00 3.0875E+07 1.5105E+06F5 2.3830E+07 1.4270E+05 5.2707E+03 9.9520E+00 7.8826E+07 1.5901E+07F6 2.3937E+00 2.0226E�01 1.5226E�02 1.4272E�03 8.2603E+00 2.1381E+00F7 1.0465E+02 5.9452E+01 2.3561E+01 1.0945E+01 1.5225E+02 9.0715E+01F8 4.7202E+02 1.5058E+02 8.4383E+01 2.5216E+01 1.1242E+03 3.5270E+02F9 2.9349E+03 1.8730E+03 5.3730E+02 1.8247E+02 3.8628E+03 2.2981E+03F10 1.1645E+04 4.0515E+03 3.5025E+03 1.1867E+03 1.7493E+04 8.1080E+03F11 2.5791E+01 1.5000E+01 4.8500E+00 2.5000E+00 4.0898E+01 2.1600E+01F12 6.3724E+01 3.9400E+01 3.3165E+01 1.4400E+01 6.8355E+01 5.3500E+01F13 1.7482E+01 6.4503E+00 1.0541E+00 3.7291E�01 3.5932E+01 1.6257E+01F14 6.6684E+03 2.0760E+03 3.9360E+02 1.3400E+02 1.3443E+04 5.4270E+03

0 10 20 30 40 500

0.2

0.4

0.6

0.8

1

Habitat IndexM

igra

tion

Rat

es o

f Mod

el 7

λμ


0 10 20 30 40 500

0.2

0.4

0.6

0.8

1

Habitat Index

Mig

ratio

n R

ates

of M

odel

8

λμ



In Fig. 5.1, although Model 4–5 and Model 1–5 adopt the emigration model of Model 5, Model 1–5 and Model 4–5 bothperform worse. The reason is R1 and T4 are too large to retain good solutions. The good features cannot be retained so that thetwo models perform worse. In Fig. 5.1, Model 5, Model 5–3 and Model 5–6 do not perform satisfactorily. According to theanalysis in Section 3, small k is harmful for BBO to obtain good features of the solution. The results also verify the analysis.

In Fig. 5.1, the following observations are obvious. (i) Model 3–5 performs better than Model 3–6, and Model 3–6performs better than Model 3; (ii) Model 5 performs better than Model 5–6, and Model 5–6 performs better than Model



Table 11Comparison of experimental results over 100 Monte Carlo simulations of Model 7, Model 8 and Model 6–5. (The bold values are the best performance for eachrow)

F Model 7 Model 8 Model 6–5

MeanMin BestMin MeanMin BestMin MeanMin BestMin

F1 4.5984E+00 3.2198E+00 4.4110E+00 3.4403E+00 5.0417E+00 3.8177E+00F2 3.4816E+04 1.0724E+04 4.0008E+04 7.8792E+03 3.1580E+04 6.1777E+03F3 2.5204E+00 1.3651E+00 2.1900E+00 1.2135E+00 2.5603E+00 1.3393E+00F4 1.9129E+00 6.0321E�01 1.5272E+00 3.2998E�01 2.6400E+00 5.4600E�01F5 3.1570E+02 2.2867E+00 1.4491E+01 2.0684E+00 2.0526E+02 4.2699E+00F6 2.5001E�03 1.2605E�04 1.2530E�03 4.4810E�05 4.0174E�03 2.1901E�04F7 1.5989E+01 7.5649E+00 1.5047E+01 6.9015E+00 1.6132E+01 8.8359E+00F8 6.4346E+01 1.9260E+01 7.1615E+01 1.6486E+01 5.9409E+01 1.6148E+01F9 2.7759E+02 8.5538E+01 2.6151E+02 7.3560E+01 3.1851E+02 1.1905E+02F10 3.5210E+03 1.0172E+03 4.3570E+03 1.9816E+03 3.2631E+03 1.3561E+03F11 2.9050E+00 1.1000E+00 2.7510E+00 1.3000E+00 3.1640E+00 1.7000E+00F12 4.3427E+01 1.8400E+01 4.3546E+01 1.3700E+01 4.0766E+01 2.0000E+01F13 3.9236E�01 1.3101E�01 3.4489E�01 1.1086E�01 5.0877E�01 1.3102E�01F14 1.6344E+02 5.1000E+01 1.2147E+02 2.7000E+01 2.0661E+02 5.8000E+01

Table 12.1Performance of restart CMA-ES, BBO (original), BBO (Model 7) when FEs = 1e5 with dimension = 30 on function 15–19. (The bold values are the bestperformance for each column)

F15 F16 F17 F18 F19

Restart Mean 5.42E�09 6.22E�09 5.55E�09 1.27E+04 1.08E�07CMA-ES Std 9.80E�10 8.95E�10 1.09E�09 3.59E+04 4.99E�07BBO Mean 4.15E+01 1.31E+03 8.72E+06 5.38E+03 4.16E+03(original) Std 4.56E+01 4.44E+02 4.76E+06 2.23E+03 1.39E+03BBO Mean 2.74E+00 9.70E+01 6.43E+05 3.56E+03 3.05E+03(Model 7) Std 3.75E+00 4.18E+01 5.36E+04 1.79E+03 1.19E+02


F20 F21 F22 F23 F24

Restart Mean 4.78E�01 5.31E�09 2.07E+01 6.89E+00 6.96E+00CMA-ES Std 1.32E+00 1.41E�09 4.28E�01 2.22E+00 2.45E+00BBO Mean 2.02E+02 4.50E+01 2.21E+01 2.05E+01 1.88E+02(original) Std 2.89E+02 4.19E+01 1.19E�01 1.66E+01 5.09E+01BBO Mean 1.50E+02 2.86E+01 1.53E+01 7.38E+01 1.38E+02(Model 7) Std 5.33E+02 1.60E+01 1.86E�01 4.95E+00 1.02E+01


F25 F26 F27 F28 F29

Restart Mean 9.10E+00 5.95E+04 2.89E+00 1.35E+01 2.25E+02CMA-ES Std 3.10E+00 2.74E+05 3.59E�01 3.17E�01 4.10E+01BBO Mean 3.84E+01 1.18E+05 9.31E+00 1.42E+01 4.09E+02(original) Std 5.62E+00 7.25E+04 3.81E+00 3.75E�01 1.01E+02BBO Mean 2.74E+01 7.89E+04 5.61E+00 1.03E+01 2.73E+02(Model 7) Std 1.44E+00 8.25E+04 2.15E+00 1.30E�01 6.08E+02


5–3; (iii) Model 6–5 performs better than Model 6, and Model 6 performs better than Model 6–3. They show that when themodels have the same immigration rates, to enlarging Q and S can enhance BBO. In Fig. 5.2, we can see that the models withmoderate immigration rates, such as immigration model of Model 3 and Model 6, have the similar ability to obtain a goodsolution. However, models which have immigration model of Model 5 or have immigration model of Model 1 and Model 4,cannot help BBO obtain good solutions. The reason is that R5 and T5 are the smallest in the six models. They are not useful forBBO to obtain good features. R1 and T4 are too large so it cannot retain good features. From the above analysis, the resultsvalidate Theorem 4 and Theorem 5.




F30 F31 F32 F33 F34

Restart Mean 5.34E+01 2.92E+02 9.04E+02 9.04E+02 9.04E+02CMA-ES Std 2.84E+01 1.94E+02 7.73E�01 6.07E�01 6.09E�01BBO Mean 2.83E+02 3.35E+02 9.94E+02 9.85E+02 9.96E+02(original) Std 8.14E+01 8.90E+01 1.47E+01 2.79E+01 3.00E+01BBO Mean 2.02E+02 3.24E+02 9.59E+02 9.12E+02 9.64E+02(Model 7) Std 1.24E+01 4.19E+01 1.58E+00 8.66E+00 5.40E+01


F35 F36 F37 F38 F39

Restart Mean 5.00E+02 8.27E+02 5.82E+02 9.13E+02 2.11E+02CMA-ES Std 1.19E�13 1.72E+01 1.55E+02 1.49E+02 9.27E�01BBO Mean 7.38E+02 1.06E+03 8.73E+02 6.95E+02 5.69E+02(original) Std 1.30E+02 3.55E+01 1.47E+02 2.32E+02 1.89E+02BBO Mean 5.11E+02 7.62E+02 6.45E+02 4.94E+02 3.56E+02(Model 7) Std 2.14E+00 2.81E+01 1.35E+02 2.46E+02 2.63E+01


4.4. Simulations with elitism and storage strategy

Since large k is helpful to obtain a good solution but is harmful to retain the good solution, we cannot draw the conclusionon how to design k quantitatively. Therefore, in this subsection we discuss two strategies to retain good solutions. First, weuse elitism to retain good features. Second, we store the best solution during each generation off-line.

We adopt the immigration model of Model 1 and Model 4 based on the value of R1 and T4 in Table 3. Moreover, because of thegood performance of the emigration model of Model 5, we assemble them together, which gives us Model 1–5 and Model 4–5.

By defining the first habitat which has the smallest immigration rate k0 as 0 and the largest emigration rate l0 as 1 andsetting other parameters as same as Tables 4 and 5, we obtain the simulation results in Table 10.

On one hand, according to Table 10, it is obvious that Model 4–5 performs the worst and Model 4–5 with elitism performsthe best. The storage strategy can enhance BBO, but it does not outperform the elitism strategy. The elitism strategy can en-hance BBO overwhelmingly.

On the other hand, although the elitism strategy makes Model 4–5 performs better, the performance with elitism is stillworse than Model 6–5. The reason is that in the evolution, the features of elites may be replaced due to the large k. Therefore,the good features still cannot be retained, let alone sharing with others. For the offline strategy, although we retain the goodfeatures off-line, the good features have few chances to share due to largek.

4.5. Simulations with large Q

In this subsection, we propose two novel migration models to validate the analysis in Section 3. From Theorem 4 and The-

orem 5, a large

Pn

i¼1

Pj2Ji ðsÞ

lj

nPn

j¼1lj

can help BBO perform better, so we design Model 7 and Model 8 as follows. The curves of migra-

tion rates of Model 7 and Model 8 are shown in Fig. 6.1 and Fig. 6.2 repectively.Model 7

Pleasehttp:/

kk ¼I2

coskpn

� �þ 1

� �ð79Þ

lk ¼ Ekn

� �4

ð80Þ

Model 8

kk ¼I2

coskpn

� �þ 1

� �ð81Þ

lk ¼ Ekn

� �16

ð82Þ

In simulations, Model 7 and Model 8 have the same immigration model with Model 6–5 and Q7 = 0.8415 and Q8 = 0.9534and other parameters are designed as same as in Table 5. The results are given in Table 11 and show that Model 7 and Model8 are generally superior to Model 6–5. Model 8 performs the best among the three models, which validates the our analysis.




4.6. Simulations on composite functions

This subsection is the second part of the simulation. Here we test several migration models on 25 real parameter bench-marks. The detail about the benchmarks can be found in paper [43]. The index of the benchmarks in this paper will be theindex in [43] adds 14. For example, F15 in this paper represents Function 1 in [43], F16 being Function 2 in [43],. . ., and F39being Function 25 in [43]. We set the dimension of the benchmarks as 10, 30 and 50 respectively. For BBO, we use the pop-ulation size of 20. The mutation rate is 0. For the consistency of paper [43], we run 25 simulations on each benchmark toobtain representative performance. We record not only the best (1st) and the worst (25th) error values, but also the 7th,13th (Median) and 19th error value. The maximum number of fitness evaluations (FEs) is set to 1E+3, 1E+4 and 1E+5 respec-tively. In paper [11], the author proved that if the difference of error rates is less than 1E�7, the difference can be ignored.Therefore, in this simulation, it is reasonable that when the error value is equal to or less than 1E-8, we terminate the sim-ulation. The test results can be found in Tables A2–A4 in Appendix C. According to the results, BBO with migration Model 3only reaches the given accuracy For F15. In Table A4, BBO cannot reach the accuracy. According to the test results in paper[1,2], restart CMA-ES and the improved ABC generally perform better than BBO. This shows BBO has a huge space to exploreand develop.

In Table 12, we compare restart CMA-ES, BBO (original) and BBO (Model 7) on the 25 benchmarks and analyze the resultsin the following three parts.

First, BBO (original) vs restart CMA-ES. According to the results, BBO (original) only wins on two benchmarks, F18 andF38. For the other benchmarks, BBO (original) is far behind restart CMA-ES, which indicates that BBO (original) has muchroom for improvement.

Second, BBO (Model 7) vs restart CMA-ES. According to the results, on F18, F22, F28, F36, F38, better results are achievedby BBO (Model 7). Restart CMA-ES outperforms BBO on other benchmarks.

Third, BBO (original) vs BBO (Model 7). In Table 12, it is obvious that BBO (Model 7) has a comprehensive victory over BBO(original). It indicates migration model is indeed influential in the performance of BBO and also proves that the analysis inSection 3 is correct.

5. Conclusion

In this paper, we have investigated the factors that could influence the performance of BBO. First, we investigated thetransition probability matrix of BBO. By proving the basic mathematical theorems, we concluded that the transition prob-ability of median number of species cannot entirely explain the performance of migration models. We further investigatedsix basic migration models based on previous work, and we calculated the transition probabilities of median number ofspecies. The calculation results support the above conclusion as well. Hence, we cannot solely rely on the transition prob-abilities to assess migration models. Second, we analyzed the migration rates theoretically to evaluate migration models ofBBO. For emigration rates, increasing S and Q as defined in Table 2 can enhance BBO. For immigration rates, the analysis isqualitative, showing that increasing the immigration rate can help the solution obtain better features. However, largeimmigration rates may be harmful for BBO to retain the features. To solve the problem, we attempted two strategies, elit-ism strategy and off-line storage strategy, to retain good features, but the performance is still not satisfactory. Thus, thebalance of immigration rates should be investigated in the future. In the simulation section, we compare BBO with severalclassical algorithms. Different BBO migration models were also compared. The results demonstrate BBO has great potentialto solve optimization problems. There still exist many aspects of BBO that deserve further exploration, including mutationrates, and crossover method. According to different problems, BBO for multi-objectives and parallel computation is worthyof investigation.

Acknowledgements

This work was sponsored by the National Natural Science Foundation of China under Grant Nos. 70871091, 61075064,61034004 and 61005090, Program for New Century Excellent Talents in University of Ministry of Education of China,Ph.D. Programs Foundation of Ministry of Education of China (20100072110038). In addition, the authors would liketo express their sincere gratitude to the inventor of BBO, Professor Dan Simon, in Cleveland State University for his help-ful comments and suggestions in improving the quality of this paper. The authors also appreciate Professor Shuzhi SamGe, in National University of Singapore, Professor Hongliang Ren, in National University of Singapore, Jasmin Isabella Kaj-opoulos, in Ludwig-Maximilians-Universität München, Matthias Samland, in Universität Heidelberg, Rohit Bharadwaj, inNational University of Singapore, Junjie Hu and his research team, in Technical University of Denmark, for the help inimproving the English language. Last but not least, thanks to Wang Jiali, my beloved girlfriend, for her continuous strongsupport!.




Appendix A

Suppose that {Pk} is the solution of the following equations, where k = 1, 2, . . . , n.

Table ACompar

F

F1F2F3F4F5F6F7F8F9F10F11F12F13F14

Pleasehttp:/

AP ¼ 0Xn

i¼0

Pi ¼ 1

8><>: ðA:1Þ

The solution is unique and the proof is given as follows.First, by defining

A ¼

�k0 l1 0 � � � � � � � � � 0

k0 �ðk1 þ l1Þ l2. .

. . .. . .

. ...

. .. . .

. . .. . .

. . .. . .

. ...

. .. . .

. . .. . .


0 � � � � � � � � � 0 kn�1 �ln

26666666664

377777777752 Rðnþ1Þ�ðnþ1Þ ðA:2Þ

we obtain Lemma A.

Lemma A.

det A ¼ 0 ðA:3Þ

Proof (Mathematical Induction). Here for the sake of convenience, we define An+1 as (A.4).

Anþ1 ¼

�ðk0 þ l0Þ l1 0 � � � � � � � � � 0

k0 �ðk1 þ l1Þ l2. .

. . .. . .

. ...

. .. . .

. . .. . .

. . .. . .

. ...

. .. . .

. . .. . .


0 � � � � � � � � � 0 kn�1 �ðkn þ lnÞ

26666666664

377777777752 Rðnþ1Þ�ðnþ1Þ ðA:4Þ

where l0 = kn = 0. It is obvious that An+1 = A.If we define detA�1 = 0 and detA0 = 1, then it is obvious that

det Ak ¼ �ðkk þ lkÞdet Ak�1 � lkkk�1 det Ak�2 k ¼ 1;2; . . . ;n ðA:5Þ

According to (A.5), we can obtain

det A1 ¼ ð�1Þðk0 þ l0Þ ðA:6Þ

.1ison of experimental results of Best value and Mean value of Model 1–5, 4–5 on F1–14.

Model 1–5 Model 4–5

MeanMin BestMin MeanMin BestMin

1.7538E+01 1.3320E+01 1.5245E+01 1.1289E+014.9469E+05 2.4643E+05 2.3244E+05 6.8857E+041.2133E+02 4.1614E+01 6.1933E+01 1.9534E+013.2844E+07 4.2630E+05 6.9932E+06 4.3640E+047.6118E+07 9.5123E+06 2.3520E+07 1.0494E+068.4634E+00 3.0336E+00 2.5716E+00 2.4583E�011.5031E+02 1.1120E+02 1.0886E+02 7.3157E+011.0577E+03 4.8529E+02 5.2059E+02 1.4928E+024.0237E+03 2.6962E+03 2.9655E+03 1.8337E+031.7379E+04 7.4590E+03 1.1673E+04 5.8639E+034.1865E+01 2.3200E+01 2.6816E+01 1.4800E+016.6894E+01 4.7200E+01 6.3541E+01 4.7000E+013.6805E+01 1.8524E+01 1.9152E+01 6.8333E+001.2978E+04 6.9500E+03 6.9767E+03 2.0920E+03



Table APerform

Fes

1E+0

1E+0

1E+0

Table APerform

FEs

1E+0

1E+0

1E+0


Pleasehttp:/

det A2 ¼ ð�1Þ2ðk0k1 þ l0k1 þ l0l1Þ ¼ ð�1Þ2Y2�1

i¼0

ki þY2�1

i¼0

li þX2�1

k¼1

Yk�1

i¼0

li

Y2�1

j¼k

kj

!" #ðA:7Þ

det A3 ¼ ð�1Þ3ðk0k1k2 þ l0k1k2 þ l0l1k2 þ l0l1l2Þ ¼ ð�1Þ3Y3�1

i¼0

ki þY3�1

i¼0

li þX3�1

k¼1

Yk�1

i¼0

li

Y3�1

j¼k

kj

!" #ðA:8Þ

Here we assume that

det An ¼ ð�1ÞnYn�1

i¼0

ki þYn�1

i¼0

li þXn�1

k¼1

Yk�1

i¼0

li

Yn�1

j¼k

kj

!" #ðA:9Þ

According to (A.5), we obtain

2.1ance of BBO on Problems F15–F22 when FES = 1e3, FES = 1e4, FES = 1e5 with dimension = 10 on function 15–22.

Prob F15 F16 F17 F18 F19 F20 F21 F22

3 Best 6.88E+02 2.30E+03 6.56E+06 2.18E+03 3.65E+03 2.92E+07 1.96E+02 2.18E+017th 1.35E+03 4.05E+03 2.48E+07 4.70E+03 5.35E+03 6.70E+07 3.19E+02 2.21E+01Median 1.68E+03 5.25E+03 3.54E+07 6.20E+03 6.07E+03 9.26E+07 4.11E+02 2.22E+0119th 2.08E+03 7.07E+03 4.64E+07 8.15E+03 6.85E+03 1.46E+08 5.03E+02 2.23E+01Worst 3.23E+03 1.03E+04 8.35E+07 1.38E+04 9.70E+03 2.45E+08 7.51E+02 2.24E+01Mean 1.77E+03 5.67E+03 3.70E+07 6.57E+03 6.19E+03 1.11E+08 4.33E+02 2.21E+01Std 8.08E+02 2.52E+03 2.28E+07 3.39E+03 1.89E+03 6.93E+07 1.78E+02 2.06E�01

4 Best 3.41E�03 2.45E+01 3.35E+05 1.29E+02 8.65E+01 1.52E+01 5.10E�01 2.06E+017th 1.25E�02 6.95E+01 1.47E+06 3.05E+02 1.67E+02 7.27E+01 8.94E�01 2.08E+01Median 1.81E�02 1.48E+02 2.01E+06 4.98E+02 2.54E+02 2.81E+02 1.36E+00 2.09E+0119th 2.89E�02 2.17E+02 2.57E+06 7.15E+02 6.27E+02 4.06E+02 1.59E+00 2.09E+01Worst 8.29E�02 4.85E+02 5.17E+06 1.10E+03 2.33E+03 2.87E+03 2.81E+00 2.11E+01Mean 2.47E�02 1.61E+02 2.11E+06 5.55E+02 4.90E+02 5.60E+02 1.36E+00 2.08E+01Std 2.68E�02 1.56E+02 1.53E+06 3.21E+02 7.73E+02 1.10E+03 7.13E�01 1.41E�01

5 Best 2.45E�09 9.61E�03 4.07E+04 8.07E�01 1.29E+01 1.24E�02 2.92E�02 2.10E+017th 5.31E�09 8.28E�02 1.20E+05 1.62E+01 4.14E+01 1.77E+00 2.63E�01 2.11E+01Median 5.89E�09 1.67E�01 2.66E+05 2.76E+01 1.04E+02 3.26E+00 6.12E�01 2.12E+0119th 6.59E�09 3.81E�01 3.80E+05 7.75E+01 1.65E+02 3.04E+01 8.97E�01 2.13E+01Worst 6.95E�09 1.57E+00 4.56E+05 1.44E+02 5.10E+02 2.07E+03 2.27E+00 2.14E+01Mean 5.66E�09 3.19E�01 2.25E+05 4.69E+01 1.22E+02 1.12E+02 6.98E�01 2.12E+01Std 1.58E�09 5.88E�01 1.32E+05 4.62E+01 1.94E+02 7.10E+02 7.95E�01 1.37E�01


Prob F23 F24 F25 F26 F27 F28 F29 F30 F31

3 Best 2.86E+01 5.88E+01 9.51E+00 1.55E+04 5.92E+01 4.07E+00 4.08E+02 2.93E+02 2.84E+027th 3.78E+01 8.91E+01 1.15E+01 3.68E+04 3.36E+02 4.47E+00 5.95E+02 3.31E+02 3.52E+02Median 4.33E+01 9.56E+01 1.20E+01 4.46E+04 6.36E+02 4.58E+00 6.29E+02 3.54E+02 3.82E+0219th 4.76E+01 1.03E+02 1.25E+01 5.91E+04 1.09E+03 4.65E+00 7.50E+02 3.76E+02 4.04E+02Worst 5.72E+01 1.16E+02 1.36E+01 8.67E+04 2.87E+03 4.75E+00 8.04E+02 4.52E+02 5.04E+02Mean 4.27E+01 9.33E+01 1.19E+01 4.74E+04 8.10E+02 4.54E+00 6.51E+02 3.55E+02 3.82E+02Std 8.80E+00 2.13E+01 1.28E+00 2.38E+04 1.10E+03 2.16E�01 1.25E+02 4.60E+01 7.21E+01

4 Best 8.13E�01 2.44E+01 7.36E+00 1.69E+03 2.82E+00 3.46E+00 2.82E+02 1.51E+02 1.71E+027th 1.94E+00 3.54E+01 8.74E+00 4.77E+03 3.75E+00 3.86E+00 2.95E+02 1.79E+02 2.01E+02Median 3.10E+00 3.84E+01 9.21E+00 5.92E+03 4.29E+00 3.96E+00 3.08E+02 1.92E+02 2.18E+0219th 5.29E+00 4.39E+01 9.83E+00 7.86E+03 4.99E+00 4.10E+00 4.75E+02 2.05E+02 2.33E+02Worst 1.04E+01 5.25E+01 1.09E+01 1.37E+04 6.23E+00 4.21E+00 5.12E+02 2.31E+02 2.53E+02Mean 3.83E+00 3.93E+01 9.25E+00 6.50E+03 4.34E+00 3.95E+00 3.58E+02 1.91E+02 2.16E+02Std 3.58E+00 1.01E+01 1.30E+00 3.91E+03 1.24E+00 3.07E�01 1.01E+02 2.72E+01 2.65E+01

5 Best 3.32E�09 1.31E+01 1.73E+00 7.02E+01 3.02E�01 2.47E+00 1.94E+02 1.18E+02 1.09E+027th 3.47E�01 1.91E+01 2.91E+00 1.20E+02 5.66E�01 3.04E+00 2.13E+02 1.40E+02 1.35E+02Median 3.47E�01 2.18E+01 4.22E+00 2.24E+02 8.80E�01 3.27E+00 2.17E+02 1.49E+02 1.58E+0219th 3.47E�01 2.58E+01 6.92E+00 5.47E+02 1.41E+00 3.44E+00 3.69E+02 1.58E+02 1.71E+02Worst 1.39E+00 3.48E+01 8.84E+00 2.16E+03 1.98E+00 3.77E+00 4.76E+02 1.82E+02 1.92E+02Mean 4.02E�01 2.24E+01 4.86E+00 4.88E+02 1.00E+00 3.21E+00 2.80E+02 1.48E+02 1.54E+02Std 4.09E�01 6.72E+00 2.44E+00 7.91E+02 5.10E�01 3.76E�01 1.16E+02 2.05E+01 2.49E+01



Table APerform

FEs

1E+0

1E+0

1E+0

Table APerform

FEs

1E+0

1E+0

1E+0


Pleasehttp:/

det Anþ1 ¼ �ðkn þ lnÞdet An � lnkn�1 det An�1 ðA:10Þ

On right hand side of (A.10),

�ðkn þ lnÞdet An ¼ ð�1Þnþ1Yn

i¼0

ki þ kn

Yn�1

i¼0

li þXn�1

k¼1

Yk�1

i¼0

li

Yn

j¼k

kj

!þ ln

Yn�1

i¼0

ki þYn

i¼0

li þ ln

Xn�1

k¼1

Yk�1

i¼0

li

Yn�1

j¼k

kj

!( )

¼ ð�1Þnþ1Yn

i¼0

ki þXn

k¼1

Yk�1

i¼0

li

Yn

j¼k

kj

!þ ln

Yn�1

i¼0

ki þYn

i¼0

li þ ln

Xn�1

k¼1

Yk�1

i¼0

li

Yn�1

j¼k

kj

!( )ðA:11Þ

�lnkn�1 det An�1 ¼ ð�1Þnþ1 �ln

Yn�1

i¼0

ki � lnkn�1

Yn�2

i¼0

li � ln

Xn�2

k¼1

Yk�1

i¼0

li

Yn�1

j¼k

kj

! !

¼ ð�1Þnþ1 �ln

Yn�1

i¼0

ki � ln

Xn�1

k¼1

Yk�1

i¼0

li

Yn�1

j¼k

kj

! !ðA:12Þ


Prob F32 F33 F34 F35 F36 F37 F38 F39

3 Best 1.04E+03 1.04E+03 1.06E+03 1.19E+03 9.63E+02 1.29E+03 9.68E+02 9.01E+027th 1.16E+03 1.14E+03 1.18E+03 1.38E+03 1.02E+03 1.40E+03 1.10E+03 1.18E+03Median 1.19E+03 1.19E+03 1.20E+03 1.41E+03 1.09E+03 1.42E+03 1.20E+03 1.29E+0319th 1.21E+03 1.21E+03 1.22E+03 1.43E+03 1.11E+03 1.43E+03 1.30E+03 1.39E+03Worst 1.26E+03 1.26E+03 1.25E+03 1.45E+03 1.18E+03 1.46E+03 1.40E+03 1.51E+03Mean 1.18E+03 1.17E+03 1.19E+03 1.39E+03 1.07E+03 1.41E+03 1.19E+03 1.28E+03Std 6.92E+01 7.04E+01 5.55E+01 1.04E+02 6.67E+01 5.22E+01 1.40E+02 1.78E+02




Prob F15 F16 F17 F18 F19 F20 F21 F22







From A.10, A.11 and A.12, we obtain

Table APerform

FEs

1E+0

1E+0

1E+0

Table APerform

FEs

1E+0

1E+0

1E+0

Pleasehttp:/

det Anþ1 ¼ ð�1Þnþ1Yn

i¼0

ki þYn

i¼0

li þXn

k¼1

Yk�1

i¼0

li

Yn

j¼k

kj

! !ðA:13Þ

According to mathematical induction, we obtain

det A ¼ det Anþ1 ¼ ð�1Þnþ1Yn

i¼0

ki þYn

i¼0

li þXn

k¼1

Yk�1

i¼0

li

Yn

j¼k

kj

! !ðA:14Þ

Since in (A.4), l0 = kn = 0, detA = det An+1 = 0. h


Prob F23 F24 F25 F26 F27 F28 F29 F30 F31


4 Best 1.15E+02 2.39E+02 3.99E+01 3.74E+05 3.69E+01 1.43E+01 4.68E+02 3.09E+02 3.68E+027th 1.37E+02 2.63E+02 4.15E+01 4.18E+05 5.22E+01 1.44E+01 5.70E+02 3.35E+02 4.03E+02Median 1.46E+02 2.87E+02 4.26E+01 4.75E+05 7.42E+01 1.45E+01 6.11E+02 3.63E+02 4.45E+0219th 1.63E+02 3.13E+02 4.43E+01 5.71E+05 1.21E+02 1.47E+01 6.72E+02 5.10E+02 5.46E+02Worst 1.74E+02 2.91E+02 4.44E+01 4.74E+05 6.68E+01 1.46E+01 6.28E+02 4.51E+02 5.11E+02Mean 4.01E+01 1.00E+02 1.19E+01 4.75E+04 7.60E+00 4.74E+00 1.63E+02 1.17E+02 1.47E+02Std 4.20E+01 6.92E+01 1.08E+01 1.56E+05 3.32E+01 3.83E+00 1.74E+02 1.20E+02 1.27E+02



Prob F32 F33 F34 F35 F36 F37 F38 F39







Second, by defining

Table APerform

FEs

1E+0

1E+0

1E+0

Table APerform

FEs

1E+0

1E+0

1E+0

Pleasehttp:/

Bnþ1 ¼

�ðk0 þ l0Þ l1 0 � � � � � � � � � 0

k0 � k1 þ l1

� l2

. .. . .

. . .. ..

.

. .. . .

. . .. . .

. . .. . .

. ...

. .. . .

. . .. . .

.kn�2 � kn�1 þ ln�1

� ln

1 1 � � � � � � 1 1 1

26666666664

377777777752 Rðnþ1Þ�ðnþ1Þ ðA:15Þ

we obtain Lemma B.


Prob F15 F16 F17 F18 F19 F20 F21 F22



5 Best 7.63E�01 8.11E+03 2.70E+07 1.55E+04 1.47E+04 1.19E+04 4.87E+00 2.42E+017th 1.47E+02 1.04E+04 3.57E+07 2.57E+04 1.72E+04 2.43E+04 4.72E+01 2.44E+01Median 2.09E+02 1.21E+04 4.25E+07 3.00E+04 1.84E+04 3.21E+04 6.40E+01 2.44E+0119th 2.45E+02 1.39E+04 5.80E+07 3.22E+04 2.06E+04 8.03E+04 8.67E+01 2.45E+01Worst 2.58E+02 1.75E+04 6.87E+07 4.76E+04 2.37E+04 1.68E+05 1.65E+02 2.45E+01Mean 1.75E+02 1.21E+04 4.62E+07 3.00E+04 1.88E+04 5.26E+04 7.03E+01 2.44E+01Std 8.66E+01 3.23E+03 1.39E+07 1.07E+04 2.80E+03 5.24E+04 5.55E+01 1.24E�01


Prob F23 F24 F25 F26 F27 F28 F29 F30 F31






Table A4.3Performance of BBO on Problems F32–F39 when FES = 1e3, FES = 1e4, FES = 1e5 with dimension = 50 on function 32–39.

FEs Prob F32 F33 F34 F35 F36 F37 F38 F39

1E+03 Best 1.35E+03 1.38E+03 1.39E+03 1.48E+03 1.40E+03 1.49E+03 8.94E+02 1.04E+037th 1.41E+03 1.41E+03 1.44E+03 1.52E+03 1.47E+03 1.53E+03 9.24E+02 1.27E+03Median 1.43E+03 1.45E+03 1.46E+03 1.54E+03 1.51E+03 1.55E+03 9.53E+02 1.52E+0319th 1.46E+03 1.47E+03 1.49E+03 1.56E+03 1.57E+03 1.57E+03 1.00E+03 1.64E+03Worst 1.50E+03 1.53E+03 1.52E+03 1.61E+03 1.63E+03 1.62E+03 1.05E+03 1.75E+03Mean 1.43E+03 1.46E+03 1.46E+03 1.54E+03 1.52E+03 1.55E+03 9.61E+02 1.48E+03Std 5.36E+01 5.34E+01 4.19E+01 4.54E+01 7.33E+01 4.07E+01 5.47E+01 2.77E+02




Lemma B.

Pleasehttp:/

det Bnþ1 – 0 ðA:16Þ

Proof. From (A.15), we obtain

det Bnþ1 ¼ 1� det An � ln � det Bn ðA:17Þ

From (A.9), det An ¼ ð�1ÞnQn�1

i¼0 ki since l0 = 0. Hence,

det Bnþ1 ¼ ð�1ÞnYn�1

i¼0

ki � ln � det Bn

¼ ð�1ÞnYn�1

i¼0

ki � ln ð�1Þn�1Yn�2

i¼0

ki � ln�1 � det Bn�1

!

¼ ð�1ÞnYn�1

i¼0

ki þ ln

Yn�2

i¼0

ki þ lnln�1 � det Bn�1

!

¼ . . .

¼ ð�1ÞnYn�1

i¼0

ki þ ln

Yn�2

i¼0

ki þ lnln�1

Yn�3

i¼0

ki þ . . .þYn

j¼3

lj

Y1

i¼0

ki

!

ðA:18Þ

Since li > 0 for i = 3, 4, . . . , n and ki > 0 for i = 1, 2, . . . , n � 1, we obtain detBn+1 – 0. h

Theorem 6. The solution of (A.1) is unique.

Proof. By defining

D ¼ 1 1 � � � 1½ � 2 R1�ðnþ1Þ;

C ¼A

D

� �¼

�k0 l1 � � � � � � � � � � � � 0

k0 � k1 þ l1

� l2

. .. . .

. . .. ..

.

. .. . .

. . .. . .

. . .. . .

. ...

. .. . .

. . .. . .

.kn�2 � kn�1 þ ln�1

� ln

0 � � � � � � � � � � � � kn�1 �ln

1 � � � � � � � � � � � � 1 1

2666666666664

37777777777752 Rðnþ2Þ�ðnþ1Þ;




Pleasehttp:/

T ¼

00...

..

.

01

266666666664

3777777777752 Rðnþ2Þ�1 and P ¼

P0

P1

..

.

..

.

Pn

266666664

3777777752 Rðnþ1Þ�1

we rewrite (A.1) as following format

CP ¼ T ðA:19Þ

According to Lemma A, we obtain the following equation by using elementary transformation,

C ! C 0 ¼

�k0 l1 � � � � � � � � � � � � 0

k0 � k1 þ l1

� l2

. .. . .

. . .. ..

.

. .. . .

. . .. . .

. . .. . .

. ...

. .. . .

. . .. . .

.kn�2 � kn�1 þ ln�1

� ln

1 � � � � � � � � � � � � 1 10 � � � � � � � � � � � � 0 0

2666666666664

3777777777775¼

B

F

� �2 Rðnþ2Þ�ðnþ1Þ ðA:20Þ

where F ¼ 0 0 . . . 0½ � 2 R1�ðnþ1Þ. Then we obtain

BP ¼ T1; ðA:21Þ

where T1 ¼ 0 0 . . . 1 0½ �0 2 Rðnþ1Þ�1. According to Lemma B,

rankðCÞ ¼ rankðCÞ ¼ rankðBÞ ¼ rankðBÞ ¼ nþ 1

where C is the augmented matrix of C and B is the augmented matrix of B. Hence, according to Cramer’s Law, the solution of(A.21) is unique. Since the solutions of (A.21), (A.19) and (A.1) are the same, we can draw the conclusion that the solution of(A.19) and (A.1) is unique. h

Appendix B

Table A.1.

Appendix C

Tables A2.1–A4.3.

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