HW2_604

jbrnbrg

September 7, 2017

HW2 604:

2, 3, 5, and 12

2:

For an M/M/1 queue with mean inter-arrival time 1.25min and mean service time 1 minute, find:\(W_{q}\), \(W\), \(L_{q}\), \(L\), and \(\rho\) but this time assume that service times are uniformly distributed between \(a = 0.1\) and \(b = 1.9\) with expected value of service time \(E(S)=\frac{a+b}{2}=1\).

JB:

Given that that the model has exponential interarrival times but service times are not, we can use the model format \(M/G/1\) which can be used for situations when the service time distribution is “anything.” Further, we are given that the standard deviation of the continuous uniform distribution takes the form: \(\sigma = \sqrt{(b-a)^2/12}\) with mean \(svc_t=\frac{1}{2}(a+b)\)

\(W_{q}=\frac{\lambda(\sigma^2+1/\mu^2)}{2(1-\lambda/\mu)}\)

\(\rho=\frac{\lambda}{\mu}\)

\(W=W_{q}+E(S)\)

\(L=\frac{\rho}{(1-\rho)}\)

\(L_{q}=\lambda W_{q}\)

See the below function in R that shows how this calculation will be made:

# with mean interarrival time 1.25min and mean service time 1 minute, find:

my_MG1_func <- function(t_inter, t_svc, sd_t, nm){
  lambda <- 1/t_inter                                # inter-arrival times
  mu <- 1/t_svc                                      # svc rate
  Wq <- lambda*(sd_t^2+1/mu^2)/(2*(1-lambda/mu))     # avg time in queue 
  rho <- lambda/mu                                   # utilization of enti
  L <- rho/(1-rho)                                   # steady-state avg of enti
  W <- Wq + t_svc                                    # steady-st avg time in sy
  Lq <- lambda*Wq                                    # steady-st avg enti in q
  p0 <- 0
  
  results <- t(c(nm, lambda, mu, Wq, W, Lq, L, rho, p0)) %>% 
    as.data.frame() %>% 
    rename(name = V1, lambda=V2, mu=V3, Wq = V4, W = V5, Lq = V6, L = V7, rho = V8, p0=V9)
  
  return(results)

}

Using the given \(a=.1\) and \(b=1.9\), I can find the average time for service:

a <- .1
b <- 1.9
inter_arrival_t <- 1.25
svc_avg <- (a+b)/2                         # aka E(S)

arrival_sd <- sqrt((b-a)^2/12)
svc_avg <- (a+b)/2                         # aka E(S)

x<-my_MG1_func(inter_arrival_t, svc_avg, arrival_sd, "2:M/G/1")
Problem 2 results
name lambda mu Wq W Lq L rho
2:M/G/1 0.8 1 2.54 3.54 2.032 4 0.8

Below, we’ll look at the compared (note that problem 1 results are shown on line 1) and see that changes have occurred where expected, namely the times in queue \(W_q\), time in system \(W\) and number of entities in queue \(L_q\) have decreased.

Problem 2 Compare
name lambda mu Wq W Lq L rho
1:M/M/1 0.8 1 4 5 3.2 4 0.8
2:M/G/1 0.8 1 2.54 3.54 2.032 4 0.8

3:

For an M/M/1 queue with mean inter-arrival time 1.25min and service time triangularly distributed between \(a = 0.1\) and \(b = 1.9\), find:\(W_{q}\), \(W\), \(L_{q}\), \(L\), and \(\rho\).

JB:

As this is similar to problem 2, above, I’ll simply reuse the created function with these altered inputs as this is also a \(M/G/1\) system: \(SvcAvg = \frac{a+m+b}{3}\) and \(\sigma = \sqrt{(a^2 + m^2 + b^2-am-ab-bm)/18}\)

a <- .1
b <- 1.9
m <- 1.0
svc_avg <- (a+b+m)/3                         
arrival_sd <- sqrt(( a^2 + m^2 + b^2 - a*m - a*b - b*m)/18)

combo2<-rbind(combo, my_MG1_func(inter_arrival_t, svc_avg, arrival_sd, "3:M/G/1")[1:8])

Again, the times in queue \(W_q\), time in system \(W\) and number of entities in queue \(L_q\) have decreased.

Problem 3 Compare
name lambda mu Wq W Lq L rho
1:M/M/1 0.8 1 4 5 3.2 4 0.8
2:M/G/1 0.8 1 2.54 3.54 2.032 4 0.8
3:M/G/1 0.8 1 2.27 3.27 1.816 4 0.8

3:

Repeat probem 1 as a \(M/M/3\) queue with mean interarrival time 1.25 minutes and mean service time 3 minutes at each of the 3 servers.

JB:

See the below function in R that shows how this calculation will be made:

# with mean interarrival time 1.25min and mean service time 3 minute, find:

my_MMC_func <- function(t_inter, t_svc, c, prob = 1, nm){
  lambda <- (1/t_inter)*prob                         # inter-arrival times
  mu <- 1/t_svc                                      # svc rate
  rho <- lambda/(c*mu)                               # utilization per enti
  i <- (0:(c-1))                                     # the index to sum over
  
  # prob system is empty i.e. p(0)
  p_empty <- 1/(((c*rho)^c)/(factorial(c)*(1-rho)) + sum(((c*rho)^i)/factorial(i)))
  
  Lq <- (rho*((c*rho)^c)*p_empty)/(factorial(c)*(1-rho)^2)#steady-st avg enti in q
  Wq <- Lq/lambda                                    # avg time in queue
  L <- Lq + rho                                      # steady-state avg of enti
  W <- Wq + t_svc                                    # steady-st avg time in sy
  Lq <- lambda*Wq                                    # steady-st avg enti in q
  
  
  results <- t(c(nm, round(lambda,3), round(mu,3), round(Wq,3), round(W,3), 
                 round(Lq,3), round(L,3), round(rho,3), round(p_empty,3))) %>% 
    as.data.frame(stringsAsFactors=FALSE) %>% 
    rename(name = V1, lambda=V2, mu=V3, Wq = V4, 
           W = V5, Lq = V6, L = V7, rho = V8, p0 = V9) %>% 
    mutate_each(funs(if(is.numeric(.)) as.numeric(.) else .))
  
  return(results)

}

combo3<-rbind(combo2, my_MMC_func(1.25, 3, 3, 1, "5:M/M/3")[1:8])
Problem 5 Compare
name lambda mu Wq W Lq L rho
1:M/M/1 0.8 1 4 5 3.2 4 0.8
2:M/G/1 0.8 1 2.54 3.54 2.032 4 0.8
3:M/G/1 0.8 1 2.27 3.27 1.816 4 0.8
5:M/M/3 0.8 0.333 3.236 6.236 2.589 3.389 0.8

12:

fig2.1

fig2.1

JB:

Calculate local traffic intensity at each station above: e.g. \(\rho_{exam}\)

Given:

sign_in is \(M/M/2\) with arrival rate \(\lambda_{signin}\).
registration is \(M/M/1\) with arrival rate \(0.9\lambda_{signin}\).
trauma is \(M/M/2\) with arrival rate \(0.1\lambda_{signin}\).
exam is \(M/M/3\) with arrival rate \(0.9\lambda_{signin}\).
treat is \(M/M/2\) with arrival rate \(0.9(0.6)\lambda_{signin}+0.1\lambda_{signin}\).

sign_in_t <- 6 #minutes, mean
signin_svc_t <- 3
reg_svc_t <- 5
trauma_svc_t <- 90
exam_svc_t <- 16
treat_svc_t <- 15

# function(t_inter, t_svc, c, prob = 1, nm)
sign_in <- my_MMC_func(sign_in_t, signin_svc_t, 2, 1, "Sign In:M/M/2")
register <- my_MMC_func(sign_in_t, reg_svc_t, 2, 0.9, "Register:M/M/1")
trauma <- my_MMC_func(sign_in_t, trauma_svc_t, 2, 0.1, "Trauma:M/M/2")
exam <- my_MMC_func(sign_in_t, exam_svc_t, 3, 0.9, "Exam:M/M/3")
treat <- my_MMC_func(sign_in_t, treat_svc_t, 2, (.9*.6+.1), "Treat:M/M/2")

all_stations<-rbind(sign_in, register, trauma, exam, treat)

Below, we review the combined results. First we note that the \(\rho_{name}\) for each station is below 1 - no single station appears to be overrun. Further: we see that the probability of no traffic at a given station \(p_0\) is not too close to zero indicating that there’s movement between the stations.

Problem 12 Compare
name lambda mu Wq W Lq L rho p0
Sign In:M/M/2 0.167 0.333 0.2 3.2 0.033 0.283 0.25 0.6
Register:M/M/1 0.15 0.2 0.818 5.818 0.123 0.498 0.375 0.455
Trauma:M/M/2 0.017 0.011 115.714 205.714 1.929 2.679 0.75 0.143
Exam:M/M/3 0.15 0.062 17.258 33.258 2.589 3.389 0.8 0.056
Treat:M/M/2 0.107 0.067 26.667 41.667 2.844 3.644 0.8 0.111

Upon reviewing the table I believe that adding additional servers to the Trauma room would have the most impact. See the table below for the output of the same scenario with 1 additional “servers” in Trauma - this has reduced \(W_{q}\) and \(W\) quite significantly.

Problem 12 with 3 servers in Trauma
name lambda mu Wq W Lq L rho p0
Sign In:M/M/2 0.167 0.333 0.2 3.2 0.033 0.283 0.25 0.6
Register:M/M/1 0.15 0.2 0.818 5.818 0.123 0.498 0.375 0.455
Trauma:M/M/2 0.017 0.011 14.211 104.211 0.237 0.737 0.5 0.211
Exam:M/M/3 0.15 0.062 17.258 33.258 2.589 3.389 0.8 0.056
Treat:M/M/2 0.107 0.067 26.667 41.667 2.844 3.644 0.8 0.111