Case Study: Load Balancing...exponential arrival rate distribution with μ=4 hyperexponential...

© Bertoli Marco A.A. 2005/2006 Dimensionamento degli impianti Informatici LoadBal - 1

Case Study:Load Balancing

Thursday, 01 June 2006


IntroductionOptimize the utilization of resources to reduce the user response time In telecom load balancing goal is: search the optimal routing given a performance index valueIn computer networks lb goal is: find the optimalallocation of processes given distributed resources (e.g. minimize response time, maximize throughput)


Load Balancing categoriesStatic

Computed during capacity planning process (batch mode)

DynamicModified at runtime using dedicated hardware or software to collect performance indices (on-line mode)


Network topology – Example A

open single-class modelexponential arrival rate distribution with μ=1FIFO queue strategyservice time distribution const(1), exp(1), hyperexp(1,2)


Model definition - JMODELSee example Load_Balancing_Single_Class.jmodel


Performance indicesResponse Time at each stationSystem Response Time


1 – Random routingStatic routing strategyAll stations have the same probability to be selected


2 – Round Robin routingStatic routing strategySelect stations according to a cyclic algorithm


3 – Shortest Queue Length routingDynamic routing strategySelect station with the smallest queue length


4 – Least Utilization routingDynamic routing strategySelect station with the smallest average utilization


Results – Example A

0

0,5

1

1,5

2

2,5

3

3,5

4

4,5

Syst

em R

espo

nse

Tim

e

const(1) exp(1) hyper(1,2)Service Time Distribution

Random Round Robin Least Utilization Shortest Queue Length


Network topology – Example B

open single-class modelexponential arrival rate distribution with μ=1FIFO queue strategyhyperexponential service time distribution with μ=1 and increasing coefficient of variation c


Results – Example B

0

10

20

30

40

50

60

1 3 5 7 9 11 13 15 17 19

Coefficient of variation c

Syst

em R

espo

nse

Tim

e

Random Round Robin Least Utilization Shortest Q Length


QuestionIf we are forced to use a random routing strategy, how can we improve system response time in the case of hyperexponential service times with high values of c?


Network topology – Example C

exponential arrival rate distribution with μ=4hyperexponential service time distribution:

Two servers with μ = (1 + δ) and c = 4Two servers with μ = (1 - δ) and c = 4One server with μ = 1 and c = 4

from Asymptotic analysis we expect saturation with Random and Round Robin algorithms:

λi = λ / 5 = 4 / 5λi < 1 / DMAX 4 / 5 < 1 / (1 + δ)δ < 0.25

μ = (1 + δ)

μ = (1 - δ)


Results – Example C

0

5

10

15

20

25

30

0% 5% 10% 15% 20% 25% 30%δ

Syst

em R

espo

nse

Tim

e

Random Round Robin Least Utilization Shortest Q Length


Problem definitionFind optimal load balancing in a replicated web server farm

n serversRequests from a Poisson process with mean rate λService times exponentially distributed

Mean response time of server i is:Ri = 1 / (μi- λi) i = 1,2,…,n

μi and λi are the service rate and the arrival rate of device i

Mean response time of the subsystem is:

Compute optimal λ*i that minimize R given:

0 ≤ λi < μi

( ) ∑= −

=n

i ii

inR

121

1,,,λμ

λλ

λλλ K

∑=

=n

ii

1λλ


Chen algorithm (1)

∑

∑ ∑

=

= =

=+

=−=

n

jj

k

j

k

jjkj

nL

nkkL

1

1 1

)1(

,,2,1)(

μ

μμμ K

Order servers by increasing service times:μ1 ≥ μ2 ≥ … ≥ μn

Compute quantities L(k), given by:


Chen algorithm (2)Compute the value of k such that:

L(k + 1) > λ ≥ L(k)

The optimal values are given by:

ni

i

i

pp

pp

iii

,,10

,,2,1

*1

1*

K

K

+==

=−

−=

∑

∑

=

=

k

kk

k

λ

μ

λμμμλ


Chen algorithm – Example (1)

Find optimal load subdivision for replicated web servers:

two new servers (S1,2 = 1/μ1,2 = 8ms) four old servers (S3,4,5,6 = 1/μ3,4,5,6 = 12ms)

System arrival rate is λ = 88 req/s


Chen algorithm – Example (2)We compute L(k):

∑

∑ ∑

=

= =

=+

=−=

n

jj

k

j

k

jjkj

nL

nkkL

1

1 1

)1(

,,2,1)(

μ

μμμ K

L(1) = L(2) = 0L(3) = L(4) = L(5) = L(6) = 45.87L(7) = 585.33

Since λ = 88 req/sL(7) > 88 > L(6)k = 6


Chen algorithm – Example (3)Now we apply:

λ*1=λ*

2= 30.9 req/sλ*

3=λ*4= λ*

5=λ*6= 6.55 req/s

System response time R = 11.34msU1,2 = 24.72%U3,4,5,6= 7.84%

ni

i

i

pp

pp

iii

,,10

,,2,1

*1

1*

K

K

+==

=−

−=

∑

∑

=

=

k

kk

k

λ

μ

λμμμλ


Questions

What happen if arrival rate is λ = 44 req/s ?

What happen if arrival rate is λ = 586 req/s ?


ReferencesBERTOLI, M. jSIM: Un Tool per la Simulazione di Modelli a Reti di Code. Tesi presso il Politecnico di Milano, 2005.BUZEN, J.P., AND CHEN, P.P. Optimal load balancing in memory hierarchies. In Information Processing 74. North-Holland, New York, 1974, pp. 271-275. CHEN, P.P. Optimal file allocation in multi-level storage systems. In AFIPS Conference Proceedings (New York, June 1973), vol. 42. AFIPS Press, Reston, Va., 1973, pp. 277-282. SERAZZI G., "The dynamic behavior of computer systems", in Experimental Computer Performance Evaluation, pp. 127-163, Elsevier, 1981.

Case Study: Load Balancing...exponential arrival rate distribution with μ=4 hyperexponential...

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