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<article xmlns:xlink="http://www.w3.org/1999/xlink">
  <front>
    <journal-meta />
    <article-meta>
      <title-group>
        <article-title>ANALYSIS OF AN APPROACH TO INCREASE ENERGY EFFICIENCY OF A CLOUD COMPUTING SYSTEM</article-title>
      </title-group>
      <contrib-group>
        <contrib contrib-type="author">
          <string-name>Daraseliya A.V.</string-name>
          <xref ref-type="aff" rid="aff1">1</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Sopin E.S.</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
          <xref ref-type="aff" rid="aff1">1</xref>
        </contrib>
        <aff id="aff0">
          <label>0</label>
          <institution>Federal Research Center Computer Science and Control of the Russian Academy of Sciences</institution>
          ,
          <addr-line>Moscow</addr-line>
          ,
          <country country="RU">Russia</country>
        </aff>
        <aff id="aff1">
          <label>1</label>
          <institution>Peoples' Friendship University of Russia</institution>
          ,
          <addr-line>Moscow</addr-line>
          ,
          <country country="RU">Russia</country>
        </aff>
      </contrib-group>
      <fpage>79</fpage>
      <lpage>87</lpage>
    </article-meta>
  </front>
  <body>
    <sec id="sec-1">
      <title>1. Introduction</title>
      <p>Energy efficiency of cloud system is very important issue for cloud providers. The servers can be put into
standby state in order to improve the energy efficiency of a cloud system in case of light load. On the one hand, the
switching to standby mode allows us to reduce power consumption, and on the other hand, it leads to extra power
usage to turn on/off the server. Therefore, it is important to understand under what conditions it will be
advantageous to put the server in standby state, and under what conditions it is more profitable to leave it in the
operating mode.</p>
      <p>In [1] we have considered a cloud system taking into account switch on and switch off periods of servers, and
it was assumed that the server switches off immediately as it remains empty. In this paper, we consider a model,
in which the server does not switch off immediately after it is empty, but waits for an exponentially distributed
time. For simplicity, we consider only one server with a number of virtual machines working on it.</p>
    </sec>
    <sec id="sec-2">
      <title>2. Mathematical model of a cloud system</title>
      <p>We consider a multi server queuing system with C servers. Customers arrive according to the Poisson law with
rateλ. Service times, switch on and switch off durations are exponentially distributed with the parametersμ, ɑ and
β, respectively. The system state is described by the vector (s, k), where k is the number of customers in the system,
s is the server state. Here s = 0 means that the system is in the standby mode, s=1 reflects switch-on mode and s=2
and s=3 represent operating and switch off modes, respectively. Arrival of a customer in an empty system causes
change of the system state to the switch on mode. After exponentially distributed time with rate ɑ, the syste
enters the operating mode, in which serving of customers is started. When the system remains empty in the
operating mode, it does not switch off immediately, but waits exponentially distributed time with rate γ. If a
customer arrives during that waiting period, then the system starts serving. Otherwise, the state is changed to the
switch off mode. If a customer arrives during the switch off mode, then the system turns to the switch on mode
immediately after the completion of the switch off. Otherwise, the system falls to the stand by mode. Figure 1
shows the transition intensities diagram.
p1,С   p1,С1   p3,С ;</p>
      <p>  
p2,0   p3,0;</p>
      <p> 
p2,1   p2,0;
3 V1
  pi, j  1.</p>
      <p>i0 j0</p>
      <p>
        Taking into account the normalization condition (12) and using matrix methods, the system of equations of
equilibrium (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) – (11) can be solved numerically, but below we represent the analytical solution of (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) – (11).
      </p>
      <p>
        The expression for p3,0 follows directly from formula (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ):
      </p>
      <p>p3,0   p0,0.</p>
      <p>
        The stationary probabilities p3,k follow from formula (
        <xref ref-type="bibr" rid="ref2 ref6">2</xref>
        ) taking into account formula (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ):
      </p>
      <p>  k  k1
p3,k     p3,k1       p3,0     k p0,0, 1  k  С 1.</p>
      <p>
        The equation for stationary probabilityp3,V can be represented by formula (
        <xref ref-type="bibr" rid="ref3 ref7">3</xref>
        ) and expression (14):
      </p>
      <p>    C1    C1    C1
p3,C   p3,C1        p3,0         p0,0       p0,0.</p>
      <p>
        The expression for p1,1 follows from formula (
        <xref ref-type="bibr" rid="ref4 ref8">4</xref>
        ) and the expression for p3,1, is obtained from (14):
p3,1       p3,0   
 
      </p>
      <p>     p0,0
p1,1     p0,0     p3,1   1  p0,0  2 p0,0    2     2
  
       p0,0.</p>
      <p>
        The stationary probabilities p1,k follow from (
        <xref ref-type="bibr" rid="ref5">5</xref>
        ) by substituting expression (14) and the simple algebraic
transformations:
      </p>
      <p> 
p1,k     p1,k1     p3,k , 2  k  C 1
</p>
      <p>    p1,k1    2kk1  p0,0 
    2k2   2k1
      p1,k2     k1   p0,0      k    p0,0 
   k1</p>
      <p>   
   k1
   </p>
      <p>k1  2k1
p1,1  
i1    k1i   i p0,0 
 2k1 k1    i1
p1,1     k    p0,0 </p>
      <p>i1   i1 


 k1  2     2 
  k     p0,0 
k12     2 
  k     p0,0 
    k1 p1,1    2kk1  p0,0 ki11 11k </p>
      <p>
   
 2k1   k     k 
   k       k 
     </p>
      <p> 
 2 k1       k     k </p>
      <p> p0,0.
   k   k
p0,0 </p>
      <p>The expression for p1,V follows from (6) by substituting expression (15) and the simple algebraic
transformations:</p>
      <p> 
p1,C   p1,C1   p3,C 
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
   k1  2     2   2 k1      k     k      C1 (18)</p>
      <p>    k         k   k   p0,0        p0,0.</p>
      <p>
        The expression for calculating stationary probabilities for the operating mode p2,0 follows from formulas (7)
and (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ):
      </p>
      <p>      
p2,0   p3,0   p0,0.</p>
      <p>The equation of stationary probability p2,1 can be represented by the formula (8) and the expression (19):
        
p2,1   p2,0   p0,0.</p>
      <p>The expression for p2,2 is obtained by substituting expressions (16), (19) and (20) into formula (9):
p2,2      p2,1   p2,0   p1,1          p0,0     p0,0 22 p0,0 
           p0,0  22 p0,0 </p>
      <p>     2       2     2 
  2 p0,0         p0,0.
(19)
(20)
(21)
The stationary probabilities p2,k follow from (10) and the simple algebraic transformations:
    p2,k  p1,k   p2,k1  p2,k1, 2  k  C 1;
 p2,k1      p2,k  p2,k1  p1,k , 2  k  C 1;
 p2,k      p2,k1  p2,k2  p1,k1, 3  k  C;
p2,k   p2,k1   p2,k2  p1,k1, 3  k  C;</p>
      <p>   
p2,k        p2,k2   p2,k3  p1,k2    p2,k2  p1,k1 
    2 p2,k2  2   p2,k3    2   p1,k2   p2,k2   p1,k1 </p>
      <p> 
    2 2    p2,k2  2   p2,k3    p1,k1  2 p1,k2  
   22  p2,k2  2   p2,k3    p1,k1  2 p1,k2  

2     2   
 2  </p>
      <p>p2,k3   p2,k4   p1,k3   2   p2,k3    p1,k1  2 p1,k2  
 3  2 3 2   3 p2,k3  2  3  2 </p>
      <p>p2,k4 

3  2   2   3   
 3  
   p1,k1  2  p1,k2  2  2   2 </p>
      <p>
p1,k3  </p>
      <p>
   2     2 
p2,k4   p2,k5   p1,k4    3</p>
      <p>p2,k4 

   p1,k1  2  p1,k2  2  2   2 
3  2   2   3     2     2 
 4</p>
      <p>
p1,k3  
p2,k4 
3  2   2   3 
 4</p>
      <p>p2,k5
   p1,k1  2  p1,k2  2  2   2 
p1,k3 
3  2   2   3 </p>
      <p> 3
  4  3 24 2  3   4 p2,k4   3  2 4 2   3 </p>
      <p>p2,k5
  p1,k1  2  p1,k2 
 2    2 
 2
p1,k3 
 3  2  2   3 
 3</p>
      <p>
p1,k4  </p>
      <p>
p1,k4 .</p>
      <p>

We represent the resulting expression (22) as sum of sequences:
4
 i 4i
III   ki03 il0 i i p1,kl1   ki03 il0  i p1,kl1   kl03 p1,kl1 il0  i 
In turn, we represent the third part (26) of expression (23) in the form of four parts:</p>
      <p>III1  k3 kl2 k2 k3   l  k2
l0   kl1    k1 </p>
      <p>l0      k1
  k1 1    k2</p>
      <p>
  
    
</p>
      <p>    k2 
    1     ;
III2  k3 1 1 k3    l 1</p>
      <p>  
l0  l1   kl1    k1  l0      k1 </p>
      <p>  k2
1  
  </p>
      <p>
  
1  
  </p>
      <p>   k1
1    ;</p>
      <p>
 
1

(23)
(24)
(25)
(26)
(27)
(28)
   1 k1 11 k2    1 k1 11 k2 .</p>
      <p>
Then we substitute the obtained expressions (27) – (30) into expression (26):</p>
      <p>III3  kl03 kl2 klk1l1 kl1kl1  kl03  klk2l1  k3 kl2
 k2 k3    l  k2 k3   ll0  kl1 </p>
      <p>   k1 l0      k1 l0   
</p>
      <p>1
  k1 1  </p>
      <p>
1   k1 </p>
      <p>
  2     k1    k2 1
     
</p>
      <p>  k2 
k1     1
      



  2    k1
    k1 



1    k2 


1    </p>
      <p>k1
  k1 
1   k2 </p>
      <p> 
1    </p>
      <p> 
 
 p0,0 1 2  2    1   k2    k1 11k1  

  2     k1      k2 1 1    k2 </p>
      <p>   1      
   k1      k2 1 1    k2 </p>
      <p>   1    .</p>
      <p>Then we substitute the expressions (31), (24) and (25) into the formula (23):
p2,k  1  k1      2       22 p0,0   1  k2        </p>
      <p>
1    2 
1   </p>
      <p> p0,0 
(29)
(30)
(31)
</p>
      <p>
 p0,0 1  2  2    1    k2      
 
k1 1    k1 </p>
      <p> 
  2        k1      k2 1  11 k2      k1     k2 1  11k2 </p>
    </sec>
    <sec id="sec-3">
      <title>3. Energy consumption indicators</title>
      <p>After receiving the system stationary distribution, we calculate the energy consumption indicators. We will
assume that in the switch on / off mode, the power consumption is constant and equal to the average value. In the
operating mode, the power consumption depends on the server occupancy. By analogy with the formula given in
[3,4,5], we derive the formula for the average server power consumption:</p>
      <p>P  P0 C p0,i  P1 iC0 p1,i  P3 iC0 p3,i  iC0 P2,i p2,i , (33)</p>
      <p>i0
where</p>
      <p>P2,k  P2,min  P2,max  P2,min k. (34)</p>
      <p>V1</p>
      <p>The values of Pi were taken from [2], according to which P0 = 10 W, P1 = 170 W, P3 = 120 W,P2, min = 105 W and
P2, max = 268 W.</p>
      <p>The average number N of customers in the system is equal to the average effective arrival rate λ(1-π) multiplied
by the average sojourn time T. Expressed algebraically the law is</p>
      <p>N   (1 )T,
where blocking probability π is</p>
      <p>  p1,C  p2,C  p3,C.</p>
      <p>The average number N of customers is given by</p>
      <p>The results of numerical analysis for the values C=7, =µ20, ɑ=1, β=2 are presented in figures 2 – 4.
(32)
(35)
(36)
(37)
(38)
The plots of the server’s power consumption (fig. 2) for thoautrthemcoodneslumsehdopwower increases
very fast for small values of the arrival flow intensity λ, also note that with the increase of
which the system doesn’t go into standby mode, the power consumption also increases.</p>
      <p>In Fig. 3, we note that the largest drop for power consumption occurs at small values of γ that co
large values of waiting time before the system goes to the standby mode, and that with the increase of the load
intensity λ, energy consumption also increases.</p>
      <p>In Fig. 4, we note that at small values of γ difference in sojourn time is not big.
sojourn time also increases. It means, the larger the waiting time before the system goes to the standby mode, the
greater sojourn time.</p>
    </sec>
    <sec id="sec-4">
      <title>5. Conclusion</title>
      <p>In the paper, we considered a cloud computing system in which the server switches off after a random time
after it was left empty. The intensity γ monotonically affects the average consumed power, so
investigate its effect on performance measures also.</p>
      <sec id="sec-4-1">
        <title>With inc we nee</title>
        <p>Благодарности
Acknowledgement</p>
        <p>Публикация подготовлена
РФФИ в рамках научных</p>
        <p>при поддержке
проекто-в07-№0305115 и
Программы
№-071-053608.</p>
        <p>Р-1У0Д0Н»
и«5 при
финансовой
поддержке</p>
        <p>The publication was prepared with the support of the “RUDN
according to the research projects No. 15-07-03051 and No. 15-07-03608.</p>
      </sec>
      <sec id="sec-4-2">
        <title>Uni-v1e0r0si”ty anPdrogfruanmded5 by</title>
      </sec>
      <sec id="sec-4-3">
        <title>RFBR</title>
        <p>References</p>
      </sec>
    </sec>
    <sec id="sec-5">
      <title>Note on the authors:</title>
      <p>Sopin Eduard S., professor at the Applied Probability and Informatics departmentof Peoples’
University of Russia; PhD, senior researcher at the Institute of Informatics Problems, Federal Research
Center Computer Science and Control of the Russian Academy of Sciences, sopin_es@rudn.university
Daraseliya Anastasia V., student of Applied Probability and Informatics department, Peoples’
University of Russia, nastyadar6@gmail.com</p>
      <sec id="sec-5-1">
        <title>Friendsh</title>
        <p>Friend
Об авторах:
Сопин Эдуард Сергеевич, кандидат физик-оматематических нау,к доцент кафедры прикладной
информатики и теории вероятно, стРеойссийский университет дружбы народ;освтарший
научный сотрудник Института прикладной информ, аФтиедкиеральный исследовательский
центр «Информатика и упрлаевние» Российской академии ,нsаoуpкin_es@rudn.university
Дараселия Анастасия Валерьевна, студент кафедры прикладной информатики и теории
Российский университет дружбы народnоaвs,tyadar6@gmail.com</p>
      </sec>
    </sec>
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