Finite source models
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Finite Source Models. M/M/R/GD/K/K By: Ryan Amigliore Parisay’s comments are in purple. M/M/ R /GD/ K / K. (M) - Arrival rate follows Exponential Distribution (M) - Service rate follows Exponential Distribution (R) - Finite number of Workers (GD) - Queuing principal is general discipline

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Finite Source Models

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Finite source models

Finite Source Models

M/M/R/GD/K/K

By: Ryan Amigliore

Parisay’s comments are in purple


M m r gd k k

M/M/R/GD/K/K

(M) - Arrival rate follows Exponential Distribution

(M) - Service rate follows Exponential Distribution

(R) - Finite number of Workers

(GD) - Queuing principal is general discipline

(K) - Finite Systemcapacity

(K) - Finite Population

SystemCapacity = Population


Finite source models1

Finite Source Models

Arrivals are drawing from a small finite population.

Machine Repair Model

Population

Shop Capacity

K=

R=

K= # of Machines

R= # of Repair People


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G = Machine in Working Condition

B = Machine in need of Repair


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λ = Break Down (Arrival) Rate

μ = Service Rate

Break Down Rate

Changes With Population


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  • To determine the Rate of Break Down (λ) at a given state (j) we can sum all of the λ’s remaining in the population.

    λj = λ + λ + λ… + λ

(K – j) λ’s

Thus…

λj = (K – j)λ


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  • Determining the Service Rate (μ) is done the same as a M/M/S/GD/∞/∞ system.

μj = jμ

For (R < j ≤ K)

μj = Rμ

For (j ≤ R)


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the system is in steady state system…

It is possible to find the probability of the system being in a certain state (j)

For (j ≤ R)

For (R < j ≤ K)


Refresher

Refresher

.

.

.

.

.

- so -

For (j ≤ R)

For (R < j ≤ K)


Now the bad news

Now the Bad News

There are no Simple formulas for

L, Lq, W, & Wq

The best we can do is express them in terms of πj’s

For:


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Gotham Township Police Department has 5 patrol cars. A patrol car breaks down and requires service once every 30 days. The police department has two repair workers, each of whom takes and average of 3 days to repair a car. Breakdown times and repair times are exponential.

Determine the average number of police cars in good condition.

Find the average down time for a police car that needs repairs.

Find the fraction of the time a particular repair worker is idle.


1 determine the average number of police cars in good condition

1. Determine the average number of police cars in good condition.

Given

The average number of good cars

can be shown by the total population minus

the average number of cars in the system

K – L = Average # Good

K is given so we have to find L


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To solve for L we need to find π0.


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Now we can solve for π0

And plug this value back in to solve for


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Plug in the probabilities and solve for L

L = 0(0.69) + 1(0.310) + 2(0.062) + 3(0.009) + 4(0.001) + 5(0) = 0.465

  • Plug in K and solve for average number of cars in good condition.

  • K – L = 5 – 0.465 = 4.535; Average number of cars in good condition


2 find the average down time for a police car that needs repairs

2. Find the average down time for a police car that needs repairs.

Given

Since a police car is only in the system when

it is either being serviced or waiting to be

serviced, we can conclude that the average

down time will be the average time in

the system (W).

W= Average Down Time

&

L = 0.465


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Plug in the given equations

&

Plug in the given variables and solve.


3 find the fraction of the time a particular repair worker is idle

3. Find the fraction of the time a particular repair worker is idle.

Given

Earlier it was shown that in state 0 and in

State 1 there is at least one repair worker

Idle. Since in state 0 both workers are idle

The percent of time either could be idle in

That state is 100%. In state 1 only one worker

Is idle; so the percent of time that either

Worker could be idle is 50%. In any other

State neither worker is idle so all other states

Have a 0% chance of either worker being idle.

Thus to find the fraction of time a particular

Worker is idle we use:

Idle time per worker = π0 + (0.5)π1


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Plug in the variables and solve

π0 + (0.5)π1 = 0.619 + (0.5)(0.310) = 0.774

A particular repair worker is Idle 77.4% of their work day

Therefore the utilization of each worker is (1-0.774).


Sensitivity analysis

Sensitivity Analysis

  • Costing

    • Worker Salary: $55,000/yr ≈ $211/day

    • Police Officer Salary: $73,000/yr ≈ $280/day

    • Facility Mortgage: $3500/month ≈ $23.33/day/car

In this Sensitivity Analysis a common unit of cost was applied. This will allow for the analysis and comparison of multiple performance measure.


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Original System with

Costing Figures

WinQSB Input & Output Tables


Scenario 1

Scenario #1

  • Total number of police cars is currently at 5 vehicles, the department

  • is contemplating adding as many as 5 more vehicles to help keep the

  • rising crime rate in check, but they do not want to hire anymore repair

  • people. The range to check is 5-10 vehicles.

Parameter: Population

Range: 5 - 10


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Increased Car Population

Increased Car Population

With Increased Capacity


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Scenario #2

  • In order to cut costs the department is wondering if, without adding

  • anymore vehicles, the maintenance of the vehicles could be performed by

  • 1 repair person. Calculate and compare 2 repair people to 1.

Parameter: Repair Workers

Range: 1 & 2


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In this situation, since it was thought that balking (this system does not have balking) may be a factor, the balking cost is equal to 150% of the daily cost. This figure is equal to the overtime pay that will need to be paid to the officer covering the balked car. It is assumed that the balked car will go to a separate maintenance shop to be fixed free of charge. Rewrite the problem. The last column of comment, is that for 1 operator or two?


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Scenario #3

  • As the vehicles get older it is probable that they will need to come in

  • more often. Analyze and compare the system with arrival rates from

  • 1car/30days up to 15cars/30days.

Parameter: Break Down Rate

Range: 1/30 to 15/30


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Scenario #4

  • Another possibility to cut costs will be to move the repair shop to a

  • smaller facility, compare the system with a capacity from 5 cars to 2

  • cars. In this type of problems the queue capacity is decided by the

  • number of Cars and the number of repairers.

  • Therefore this analysis is not relevant.

Parameter: Queue Capacity

Range: 2 - 5


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Reduce Queue Capacity


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Scenario #5

  • Look at two different scenarios where less experienced and cheaper

  • repair persons are hired, and a scenario where one very experienced

  • more expensive employee is hired, which employee has a more beneficial

  • effect on the cost of the system. Mention how the service rate may change

  • Because of low experience of the worker. Provide explanation on data

  • Used in next slides.

Parameter: Worker Experience

Range: New to Field & Expert


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2 Less Experienced & Cheaper Workers


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While the expert worker was the most expensive

Based on salary, their rapid service time allowed

The system as a whole to save on total cost as well

As decrease the average total time in system. Needs

more info on your assumptions.


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Questions?


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