Finite Source Models M/M/R/GD/K/K

By: Ryan AmiglioreParisay’s comments are in purple

(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 System capacity

(K) - Finite Population

M/M/R/GD/K/K

System Capacity = Population

Arrivals are drawing from a small finite population.

Machine Repair Model

Finite Source Models

Shop CapacityPopulation

K=R=

K= # of Machines

R= # of Repair People

G = Machine in Working ConditionB = Machine in need of Repair

λ = Break Down (Arrival) Rateμ = Service Rate

Break Down RateChanges With Population

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

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

μj = jμ

μj = Rμ For (j ≤ R)

For (R < j ≤ K)

the system is in steady state system…

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

For (R < j ≤ K)

For (j ≤ R)

Refresher

.....

- so -

For (R < j ≤ K)

For (j ≤ R)

There are no Simple formulas forL, Lq, W, & Wq

Now the Bad News

For:

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

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.

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

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

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

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

To solve for L we need to find π0.

Now we can solve for π0

And plug this value back in to solve for

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

Plug in the probabilities and solve for L

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.

Given

L = 0.465

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

&

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.

Given Earlier it was shown that in state 0 and in State 1 there is at least one repair workerIdle. Since in state 0 both workers are idleThe percent of time either could be idle in That state is 100%. In state 1 only one workerIs idle; so the percent of time that eitherWorker could be idle is 50%. In any other State neither worker is idle so all other statesHave a 0% chance of either worker being idle.Thus to find the fraction of time a particularWorker is idle we use:

Idle time per worker = π0 + (0.5)π1

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

Plug in the variables and solve

A particular repair worker is Idle 77.4% of their work dayTherefore the utilization of each worker is (1-0.774).

Costing◦ Worker Salary: $55,000/yr ≈ $211/day◦ Police Officer Salary: $73,000/yr ≈ $280/day◦ Facility Mortgage: $3500/month ≈ $23.33/day/car

Sensitivity Analysis

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

Original System withCosting FiguresWinQSB Input & Output Tables

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

Increased Car Population

Increased Car PopulationWith Increased Capacity

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

NUMBER OF WORKERS

1 2 ACCEPTABLE COMMENTSUTILIZATION 0.44 0.22 70-83% LOWAVG # IN SYSTEM (L) 0.64 0.46 <2 GOODAVG # IN QUEUE (Lq) 0.2 0.01 <1 GOODAVG TIME IN QUEUE (Wq) 1.38 0.075 <2 GOODCOST OF WORKER(S) 211 422 - -AVERAGE BALK COST $46.99 $49.49 $0.00 HIGHTOTAL COST PER DAY $506.22 $671.63 MIN -

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?

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

0 1/30 1/15 1/10 2/15 1/6 1/5 7/30 4/15 3/10 1/3 11/30 2/5 13/30 7/15 1/2 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Utilization vs. Arrival Rate

UtilizationAcceptable (min)Acceptable (max)

Arrival Rate(λ)

Uti

lizati

on

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

Reduce Queue Capacity

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

2 Less Experienced & Cheaper Workers

ORIGINAL (2) INEXPERIENCED (2) EXPERIENCEDCOST OF CUSTOMER BEING SERVED $126.99 $198.25 $65.97

COST OF CUSTOMER WAITNG $3.16 $12.19 $14.39

P(WAIT) 7.21% 16.40% 23.56%AVG W 3.07 DAYS 5.01 DAYS 1.83 DAYSAVG L 0.47 0.75 0.28Utilization 22.68% 35.40% 23.60%Total Cost $622.13 $510.44 $500.36

While the expert worker was the most expensiveBased on salary, their rapid service time allowed The system as a whole to save on total cost as wellAs decrease the average total time in system. Needsmore info on your assumptions.

Questions?