Single Center Single Server Platform: 1184359

question 1. (Single center single server platform)

Given a single server platform with APWP _ls  discipline and two priority levels,calculate and tabulate the T_w   values for increasing values of p (p = 0.4,0.6,0.8).Assume  the channel capacity be C=10⁵ op.ns/sec,the average number of op.ns job be Z=10₅ op.ni.job  and the level 1 arrival rate be =35%.

solution

T_w  = 0.4,0.6,0.8

C= 10₅ op.ns/sec

Z = 105 op.ni.job

rate = 35% λ

P= 0.4+0.6+0.8

L q = P 0 ( μ λ ) c ρ /  c(1 − ρ) 2

where

c−1 X

(cρ) c

(cρ) m

P 0 = 1/

c(1 − ρ) /

m=0 m!

Note that P 0 denotes the probability that there are 0 customers in the system.

ρ =  λ  /c μ

p= 1.8 x 35/100/10⁵

 ans = 0.63×10⁵

question 2 (Single center Multiple server platform)

Given a multiple sever platform with p=0.4,compute and tabulate the  T_w(M,y,C)  M/M/LIFO values  for a increasing number of channels m(m=1,2,3). Assume the global channel capacity of the platform be C=10₅ op.ns/sec and the average number of op.ns per job be Z=10⁵ op.ni.job.

solution

T_w(M,λ,C)

M= 1,2,3

C= 10₅  op.ns/sec

L q =

ρ 2 /1 − ρ

T_w(M,λ,C) = P(Z(C(1-P))

T_w(M,λ,C) =BZ(C(1-P))

T_q(I,λ,C) < T_q(m,λ,C) for m= 1,2,3

The values of increasing channels are T_q(I,λ,C) < T_q(m,λ,C)

question 3.(Multiple center Platform)

Given the Internet Platform P in figure,Consisting of N Clients(T1…..TN) that launched internet service requests to the HTTP Server that, in turn ,when needed,iteratively  launches access requests to the Application Server or to the Database Server.

After a number of Appl or DB access iterations,the HTTP server returns the internet request answer to the Client.

Assuming the routing parameters have the values  below.

uclient=0.067job/sec

uHTTP =200 job/sec

u Appi = 37 job/sec

UDB  = 33 job/sec

number of N clients = 2

service time= exponential

service discipline = abstract

obtain the following:

1. Give the solution to P by use of the markovian method

solution

�����=+1,≥

��1, 1≤��<

���,1, ≤��<1

���,2,1≤��<2,

��,≥�

����1(!)20,0≤��������<,1!!−���0,≤����<1,1−��11−�����1+12!!−���0,1≤����<2,1−��112−���12−�����2+1!!−���0,≥�2,

�={{{��(!)20,0≤��������<,1!!−���0,≤����<1,1−��11−�����1+12!!−���0,1≤����<2,1−��112−���12−�����2+1!!−���0,≥�������2,(3)where1=/1,2=/2,and ��� /.The idle probability � can be obtained as ��0=[−1∑�����=01(!)2+1−1∑��������=1!!−��+2−1∑����=11−��11−�����1+12!!−�+∞∑����=21−��112−���12−����2+1!!

2. Obtain the following performance indices:

2.1 The marginal probabilities of the population.

solution

n�, 1≤����<,, ≥�

����(!)20,0≤��������<,!!−���0,≥�

the probability = ��0=[−1∑����=0(!)2+∞∑��������=!!−�]−1

2.2 The average local response time of the Appl for N=2

solution

n=2

u= 0.067

n = u = 0.0134 x u

2.08 mins per client

2.3 The average global responce time of platform P for N =2

solution

   global responce time  w ��(2−��122−������2/!1−��+∑���=1(−��1+11(−�1)!!−��2−���12−��2+1(−���1)!!−��)+1−1∑����=+1(−�1+11−��2−���12−��2+1(−���1)!!−��)+2−1∑����=1(−�2+12−��−��2+1(−���1)!!−���)2−�12) ⋅��((2−��12−����2(/−��1)!−�−�1+−1∑���=0(−���1+11(+1)!!−��2−���12−�����2+1(+1)!!−��)+1−1∑����=(−�1+11−��2−���12−�����1+1(+1)!!−��)+2−1∑����=1(−�2+12−��−�����2+1(+1)!!−���)2−��12))−

����=((2−��122−������2/!1−��+∑����=1(−��1+11(!)2−��2−���12−��2+1(−���1)!!−��)+1−1∑����=+1(−�1+11−��2−���12−��2+1(−���1)!!−��)+2−1∑����=1(−�2+12−��−��2+1(−���1)!!−���)2−���12)1−�110−�(1−���0)) ((2−��12−����2(/−��1)!−�−1

n=2

u= 0.067 job/sec

global responce time

n/u =p

g responce time = 29.850 mins