2.6.1 - Please see below to read the abbreviations in my code chunk:

IAR = mean interarrival time AR = average arrival time AS = average service time

We know that the AR = 1/IAR. Once we AR, we can solve p, L & W.

IAR = 1.25
AS = 1
AR = 1/IAR

p = AR/AS
L = p/(1-p)
W = L/AR


p
## [1] 0.8
L
## [1] 4
W
## [1] 5

In a one server system, the mean or average service rate is equal to 1/E(S). Knowing that we can substitute our “AS” into the \(W_q = L/\lambda - E(S)\)

Wq = (L/AR)-AS
Wq
## [1] 4
Lq = AR * Wq
Lq
## [1] 3.2

Problems

2.6.2

This problem contains an exponentially distributed inter-arrival time, but the server’s service time can be any distribution (M/G/1), in this case it’s constant:

Also from the text to calculate the standard deviation with a constant distribution:

\(\sigma=\sqrt{(b-a)^2/12}\)

For this exercise I will calcluate Lq, using the M/G/1:

\(L_q = \lambda^2\sigma^2+p^2/2(1-p)\)

In my code chunk Lq for M/G/1 will be noted as (Lq1)

All times are in minutes

## [1] 0.5196152
## [1] 0.27
## [1] 0.2368

Once Lq (Lq1) is derived for M/G/1, we can plug values back into Little’s Law to solve the other parameters.

Wq1 = Lq1/AR
Wq1
## [1] 0.296
W = Wq1 + AS
W
## [1] 1.296
L = AR * W
L
## [1] 1.0368

2.6.3

This problem contains an exponentially distributed inter-arrival time, but the server’s service time can be any distribution (M/G/1), in this case triangular:

From the text to calculate the standard deviation with a constant distribution:

\(\sqrt{(a^2 + m^2 + b^2 - am - ab - bm)/18)}\)

Also, for this distribution, we will need to calcualte a new expected value for the mean service time.

\((a + m + b)/3\)

For this exercise I will again calcluate Lq, using the M/G/1:

\(L_q = \lambda^2\sigma^2+p^2/2(1-p)\)

In my code chunk Lq for M/G/1 with a triangular distribution will be noted as (Lq2)

All times are in minutes

## [1] 1
## [1] 0.3674235
## [1] 0.135
## [1] 0.1504

Once Lq is derived for M/G/1, we can plug values back into Little’s Law to solve the other parameters. The average or expected serivce time (AS), is still 1 in the triangular distribution.

All times are in minutes

Wq = Lq/AR
Wq
## [1] 0.188
W = Wq + AS
W
## [1] 1.188
L = AR * W
L
## [1] 0.9504

2.6.5

This time our model is M/M/3

\(p = \lambda/c\mu\)

All times are in minutes

## [1] 3
## [1] 0.8
## [1] 1.003469
## [1] 0.1234441

M/M/3 Little’s Law

Wq = Lq/AR
Wq
## [1] 0.1543051
W = Wq + AS
W
## [1] 3.154305
L = AR * W
L
## [1] 2.523444

2.6.12

Our queue network below is called a Jackson Network. It assumes that we have independent, discrete arrival times that are exponentially distributed (Poisson). We also assume that the service times are exponentially distributed, and independent. We also assume the p is strictly less than 1, meaning we don’t have a overflowing queue anywhere.

“Local” Traffic Intensity or \(P_{station}\), is calculated as follows:

\(x\lambda_{station}/y\mu_{exam}\)

This gives use the utlization at EACH server within a node/station. All times are in minutes.

ARnetwork = 6


ASSignin = 3
ASReg = 5
ASTrauma = 90
ASExam = 16
ASTreat = 15

Signinserver = 2
Regserver = 1
Traumaserver = 1
Examserver = 3
Treatserver = 2

ARSignin = 1
ARReg = .9
ARTrauma = .1
ARExam = .9
ARTreat = .64


SigninLocal = (ARSignin * ARnetwork)/(Signinserver * ASSignin)
SigninLocal
## [1] 1
RegLocal = (ARReg * ARnetwork)/(Regserver * ASReg)
RegLocal
## [1] 1.08
TraumaLocal = (ARTrauma * ARnetwork)/(Traumaserver * ASTrauma)
TraumaLocal
## [1] 0.006666667
ExamLocal = (ARExam * ARnetwork)/(Examserver * ASExam)
ExamLocal
## [1] 0.1125
TreatLocal  = (ARTreat * ARnetwork)/(Treatserver * ASTreat)
TreatLocal
## [1] 0.128

SourceSIMUL8 Tip – What’s The Difference Between Arrival Rates and Inter Arrival Times

SourceQueuing Formulas