Data 604 - Week 4

Create a M/D/1 model for a service time distribution that is degerate and compare it to a distribution with some variaiton.

Since the distribution will only give a single value, the standard deviation \(\sigma\) is 0. Therefore:

Eq.1 \[W_q = \frac{\lambda(\sigma^2+1/\mu^2)}{2(1-\lambda/\mu)} \rightarrow \frac{\lambda/\mu^2}{2(1-\lambda/\mu)} = \frac{\lambda}{2\mu(\mu-\lambda)}\]

So, using a M/D/1 model:

• \(\rho = \lambda/\mu\)
  • For steady-state, \(\rho < 1\)
• \(L_q = \lambda W_q\)
  • Using Eq.1, \(L_q = \frac{\lambda^2}{2\mu(\mu - \lambda)}\)
• \(W = W_q + 1/\mu\)
  • Using Eq.1, \(W = \frac{\lambda}{2\mu(\mu-\lambda)} + \frac{1}{\mu} = \frac{2\mu-\lambda}{2\mu(\mu - \lambda)}\)
• \(L = \lambda(W_q + 1/\mu)\)
  • Using Eq.1, \(L = \frac{\lambda^2}{2\mu(\mu-\lambda)} + \frac{\lambda}{\mu} = \frac{2\lambda \mu-\lambda^2}{2\mu(\mu-\lambda)}\)

For a M/M/1 distribution with \(\lambda\) and \(\mu\) being the same form the M/D/1 model.

• \(\rho = \lambda/\mu\)
  • For steady-state, \(\rho < 1\)
• \(L = \frac{\lambda}{\mu-\lambda} \rightarrow\) Eq.2
• \(W_q = L/\lambda - \frac{1}{\mu}\)
  • Using Eq.2, \(W_q = \frac{1}{\mu-\lambda} - \frac{1}{\mu} = \frac{\lambda}{\mu(\mu-\lambda)}\rightarrow\) Eq.3
• \(W = W_q + \frac{1}{\mu}\)
  • Using Eq.3, \(W = \frac{1}{\mu-\lambda}\)
• \(L_q = \lambda W_q\)
  • Using Eq.3, \(L_q = \frac{\lambda}{\mu-\lambda} - \frac{\lambda}{\mu} = \frac{\lambda \mu - \lambda(\mu - \lambda)}{\mu(\mu-\lambda)} = \frac{\lambda^2}{\mu(\mu-\lambda)}\)

Comparing the M/D/1 and M/M/1 models:

• \(L_q\) and \(W_q\) of the M/D/1 model is half that of the M/M/1 model.
• \(L\) and \(W\) of the M/D/1 model is smaller by a factor of \(\frac{2\mu - \lambda}{2\mu} \rightarrow 1 - \rho/2\) relative to the M/M/1 model.
  • Since \(0 < \rho <1\), L and W will be smaller by a factor between (not including) 1 and 0.5

This is a very unusual conclusion. With the steady state utilization \(\rho\) remaining constant across both models. The M/D/1 model has half the average queue time and half the average queue length when compared to the M/M/1 model. In addition the average time and length in the system is at least as short as and maybe almost up to half as short as the M/M/1 system.

Using the model created above (with \(\lambda = 1\)/minute and \(\mu = 1/0.9\) per minute), test the theory with a simulaiton in Simio and discuss the process and results.Create a M/D/1 model for a service time distribution that is degerate and compare it to a distribution with some variaiton.

• Global (M/D/1 & M/M/1) method:
  • Warm-up period of 10 hours.
  • Simulation is ran for 12 weeks.
  • The simulation is replicated 10 times for each case.
  • For the arrival time, \(1/\lambda\), the distribution is set Random.Exponential(1).
• Designing M/D/1 in Simio:
  • For the service time, \(1/\mu\), since it is degenerate/constant, the distribution was set to discrete with a cumulative probability of one occurrence for the service time, 0.9 minutes. This was accomplished by Random.discrete(0.9,1).
• Designing M/M/1 in Simio:
  • For the service time, , \(1/\mu\), the distribution was set to exponential distribution for 0.9 minutes. This was accomplished by Random.exponential(0.9).

Running the M/D/1 Simio model:

Using the process described above, the following results were produced:

The following table compares the theoretical vs the simulation.

c_names <- c("MetricEstimated", "Queueing", "Simulaiton", "Halfwidth")
r_names <- c("Utilizaiton (p)", "Number in System (L)", "Number in queue (Lq)",
             "Time in system (W)", "Time in queue (Wq)")
l <- 1
m <- 1/0.9

md1 <- data.frame(matrix(NA, nrow = length(r_names), ncol = length(c_names)))

names(md1) <- c_names
md1$MetricEstimated <- r_names
md1$Queueing <- c(l/m, (2*m*l-l^2)/(2*m*(m-l)), l^2/(2*m*(m-l)),
                 (2*m-l)/(2*m*(m-l)), l/(2*m*(m-l)))
md1$Simulaiton <-  c(0.9007, 4.9713, 4.0408, 4.9671, 4.0373)
md1$Halfwidth <- c(0.2447, 0.1355, 0.1334, 0.1244, 0.1244)
(md1)

##        MetricEstimated Queueing Simulaiton Halfwidth
## 1      Utilizaiton (p)     0.90     0.9007    0.2447
## 2 Number in System (L)     4.95     4.9713    0.1355
## 3 Number in queue (Lq)     4.05     4.0408    0.1334
## 4   Time in system (W)     4.95     4.9671    0.1244
## 5   Time in queue (Wq)     4.05     4.0373    0.1244

The 95% confidence interval catches the theoretical value almost on-center.

Running the M/M/1 Simio model:

Using the process described above, the following results were produced:

The following table compares the theoretical vs the simulation.

mm1 <- data.frame(matrix(NA, nrow = length(r_names), ncol = length(c_names)))

names(mm1) <- c_names
mm1$MetricEstimated <- r_names
mm1$Queueing <- c(l/m, l/(m-l), (l^2)/(m*(m-l)), 1/(m-l), l/(m*(m-l)))
mm1$Simulaiton <-  c(0.8977, 8.9315, 8.012, 8.9410, 8.004)
mm1$Halfwidth <- c(0.3115, 0.5117, 0.5087, 0.4890, 0.4882)
(mm1)

##        MetricEstimated Queueing Simulaiton Halfwidth
## 1      Utilizaiton (p)      0.9     0.8977    0.3115
## 2 Number in System (L)      9.0     8.9315    0.5117
## 3 Number in queue (Lq)      8.1     8.0120    0.5087
## 4   Time in system (W)      9.0     8.9410    0.4890
## 5   Time in queue (Wq)      8.1     8.0040    0.4882

Again, the 95% confidence interval catches the theoretical value almost on-center.

In conclusion, the theory matches almost exactly with the simulation under these circumstances.