Arrival Rate \(\lambda\)=\((1/1.25)=0.8\)
Service Rate \(\mu\)=1
As per formula given on page-28 \(\rho\) ==0.8
As per formula given on page-29 => L = \(\rho\) / (1-\(\rho\) ) = 0.8/(1-0.8)=4
As per Little’s law => L = \(\lambda\)W => W = L / \(\lambda\) =>4/0.8=5
Now on page 28 \(\mu\)=1/E(s) =>E(s) =>1/\(\mu\) =1/1=1. W=Wq+E(s)=>Wq=W-E(s)= 5 - 1 = 4
\(L_q=\lambda\;W_q\)=> 0.8 * 4 = 3.2
require(queueing)## Loading required package: queueingMM1<-function(Arrival_Rate,Service_Rate,servers)
{
lambda <- 1/Arrival_Rate
mu <- 1/Service_Rate
p <- lambda / (mu*servers)
L <- p/(1-p)
W <- L/lambda
es <- 1/mu
Wq <- W-es
Lq <- lambda*Wq
df<-data.frame(p,L,W,Wq,Lq)
return (df)
}
MM1(1.25,1,1)## p L W Wq Lq
## 1 0.8 4 5 4 3.2Counting same values with service time are (continuously) uniformly distributrd between a=0.1 and b=1.9 Other formulas same as from Little’s law only the way we find Wq will change. Now we will use Pollaczek-Khinchine formula.
MG1_continuous<-function(Arrival_Rate,Service_Rate,servers)
{
standard_Deviation<-sqrt(((Service_Rate[2]-Service_Rate[1])^2)/12)
lambda <- 1/Arrival_Rate
es <- (Service_Rate[1]+Service_Rate[2])/2
mu <- 1/es
p <- lambda / (mu*servers)
# As per formula on page -30
Wq <- (lambda*((standard_Deviation^2)+(1/(mu^2))))/(2*(1-(lambda/mu)))
W <- Wq+es
L <- lambda*W
Lq <- lambda*Wq
df<-data.frame(p,L,W,Wq,Lq)
return (df)
}
MG1_continuous(1.25,c(0.1,1.9),1)## p L W Wq Lq
## 1 0.8 2.832 3.54 2.54 2.032Steady state average time in queue (Wq) and steady state average number of entities in queue(Lq) is decreased which is expected because decreased wait time at server.
MG1_tri<-function(Arrival_Rate,Service_Rate,servers){
es <- (Service_Rate[1]+Service_Rate[3]+Service_Rate[2])/3
standard_Deviation<-sqrt(((Service_Rate[1]^2)+(Service_Rate[2]^2)+(Service_Rate[3]^2)-(Service_Rate[1]*Service_Rate[3])-(Service_Rate[1]*Service_Rate[2])-(Service_Rate[2]*Service_Rate[3]))/18)
lambda <- 1/Arrival_Rate
mu <- 1/es
p <- lambda / (mu*servers)
# As per Pollaczek-Khinchine formula on page -30
Wq <- (lambda*((standard_Deviation^2)+(1/(mu^2))))/(2*(1-(lambda/mu)))
W <- Wq+es
L <- lambda*W
Lq <- lambda*Wq
df<-data.frame(p,L,W,Wq,Lq)
return (df)
}
MG1_tri(1.25,c(0.1,1.9,1),1)## p L W Wq Lq
## 1 0.8 2.616 3.27 2.27 1.816As we can see value of all variables decreased which would be considered as improvement.
summary(QueueingModel(NewInput.MMC(lambda=(1/1.25), mu=(1/3), c=3)))## The inputs of the model M/M/c are:
## lambda: 0.8, mu: 0.333333333333333, c: 3, n: 0, method: Exact
##
## The outputs of the model M/M/c are:
##
## The probability (p0, p1, ..., pn) of the n = 0 clients in the system are:
## 0.05617978
## The traffic intensity is: 2.4
## The server use is: 0.8
## The mean number of clients in the system is: 4.98876404494382
## The mean number of clients in the queue is: 2.58876404494382
## The mean number of clients in the server is: 2.4
## The mean time spend in the system is: 6.23595505617978
## The mean time spend in the queue is: 3.23595505617978
## The mean time spend in the server is: 3
## The mean time spend in the queue when there is queue is: 5
## The throughput is: 0.8summary(QueueingModel(NewInput.MMC(lambda=1/6, mu=1/3, c=2)))## The inputs of the model M/M/c are:
## lambda: 0.166666666666667, mu: 0.333333333333333, c: 2, n: 0, method: Exact
##
## The outputs of the model M/M/c are:
##
## The probability (p0, p1, ..., pn) of the n = 0 clients in the system are:
## 0.6
## The traffic intensity is: 0.5
## The server use is: 0.25
## The mean number of clients in the system is: 0.533333333333333
## The mean number of clients in the queue is: 0.0333333333333333
## The mean number of clients in the server is: 0.5
## The mean time spend in the system is: 3.2
## The mean time spend in the queue is: 0.2
## The mean time spend in the server is: 3
## The mean time spend in the queue when there is queue is: 2
## The throughput is: 0.166666666666667summary(QueueingModel(NewInput.MM1(lambda=(1/6)*0.9, mu=1/5)))## The inputs of the M/M/1 model are:
## lambda: 0.15, mu: 0.2, n: 0
##
## The outputs of the M/M/1 model are:
##
## The probability (p0, p1, ..., pn) of the n = 0 clients in the system are:
## 0.25
## The traffic intensity is: 0.75
## The server use is: 0.75
## The mean number of clients in the system is: 3
## The mean number of clients in the queue is: 2.25
## The mean number of clients in the server is: 0.75
## The mean time spend in the system is: 20
## The mean time spend in the queue is: 15
## The mean time spend in the server is: 5
## The mean time spend in the queue when there is queue is: 20
## The throughput is: 0.15summary(QueueingModel(NewInput.MMC(lambda=(1/6*0.1), mu=1/90, c=2)))## The inputs of the model M/M/c are:
## lambda: 0.0166666666666667, mu: 0.0111111111111111, c: 2, n: 0, method: Exact
##
## The outputs of the model M/M/c are:
##
## The probability (p0, p1, ..., pn) of the n = 0 clients in the system are:
## 0.1428571
## The traffic intensity is: 1.5
## The server use is: 0.75
## The mean number of clients in the system is: 3.42857142857143
## The mean number of clients in the queue is: 1.92857142857143
## The mean number of clients in the server is: 1.5
## The mean time spend in the system is: 205.714285714286
## The mean time spend in the queue is: 115.714285714286
## The mean time spend in the server is: 90
## The mean time spend in the queue when there is queue is: 180
## The throughput is: 0.0166666666666667summary(QueueingModel( NewInput.MMC(lambda=(1/6*0.9), mu=1/16, c=3)))## The inputs of the model M/M/c are:
## lambda: 0.15, mu: 0.0625, c: 3, n: 0, method: Exact
##
## The outputs of the model M/M/c are:
##
## The probability (p0, p1, ..., pn) of the n = 0 clients in the system are:
## 0.05617978
## The traffic intensity is: 2.4
## The server use is: 0.8
## The mean number of clients in the system is: 4.98876404494382
## The mean number of clients in the queue is: 2.58876404494382
## The mean number of clients in the server is: 2.4
## The mean time spend in the system is: 33.2584269662921
## The mean time spend in the queue is: 17.2584269662921
## The mean time spend in the server is: 16
## The mean time spend in the queue when there is queue is: 26.6666666666667
## The throughput is: 0.15summary(QueueingModel(NewInput.MMC(lambda=(1/6)*((0.9*0.6)+0.1), mu=1/15, c=2)))## The inputs of the model M/M/c are:
## lambda: 0.106666666666667, mu: 0.0666666666666667, c: 2, n: 0, method: Exact
##
## The outputs of the model M/M/c are:
##
## The probability (p0, p1, ..., pn) of the n = 0 clients in the system are:
## 0.1111111
## The traffic intensity is: 1.6
## The server use is: 0.8
## The mean number of clients in the system is: 4.44444444444444
## The mean number of clients in the queue is: 2.84444444444444
## The mean number of clients in the server is: 1.6
## The mean time spend in the system is: 41.6666666666667
## The mean time spend in the queue is: 26.6666666666667
## The mean time spend in the server is: 15
## The mean time spend in the queue when there is queue is: 37.5
## The throughput is: 0.106666666666667If we can add one more server then I think we should add at “Exam Room” because it has highest traffic intensity.