1. M/M/R/GD/K/K By: Ryan Amigliore Parisay’s comments are in purple
Finite Source Models
M/M/R/GD/K/K
By: Ryan Amigliore
Parisay’s comments are in purple

2. 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 System capacity (K) - Finite Population
System Capacity = Population

3. 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

4. G = Machine in Working Condition
B = Machine in need of Repair

5. Changes With Population
λ = Break Down (Arrival) Rate
μ = Service Rate
Break Down Rate
Changes With Population

6. λj = (K – j)λ (K – j) λ’s Thus…
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

7. μj = jμ For (R < j ≤ K) μj = Rμ For (j ≤ R)
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)

8. 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)

9. Refresher . . . . .
- so -
For (j ≤ R)
For (R < j ≤ K)

10. 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:

11. Gotham Township Police Department has 5 patrol cars
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.

12. 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

13. To solve for L we need to find π0.

14. Now we can solve for π0
And plug this value back in to solve for

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

16. 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

17. Plug in the given equations
&
Plug in the given variables and solve.

18. 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

19. Plug in the variables and solve
π0 + (0.5)π1 = (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 ( ).

20. 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.
Costing
Worker Salary: $55,000/yr ≈ $211/day
Police Officer Salary: $73,000/yr ≈ $280/day
Facility Mortgage: $3500/month ≈ $23.33/day/car

21. Original System with
Costing Figures
WinQSB Input & Output Tables

22. Scenario #1 Parameter: Population Range: 5 - 10
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:

23. Increased Car Population
With Increased Capacity

24. Scenario #2 Parameter: Repair Workers Range: 1 & 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

25. 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?
NUMBER OF WORKERS 1 2
ACCEPTABLE
COMMENTS
UTILIZATION
0.44
0.22
70-83%
LOW
AVG # IN SYSTEM (L)
0.64
0.46
<2
GOOD
AVG # IN QUEUE (Lq)
0.2
0.01
<1
AVG TIME IN QUEUE (Wq)
1.38
0.075
COST OF WORKER(S)
211
422 -
AVERAGE BALK COST
$46.99
$49.49
$0.00
HIGH
TOTAL COST PER DAY
$506.22
$671.63
MIN

26. Scenario #3 Parameter: Break Down Rate Range: 1/30 to 15/30
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

28. Scenario #4 Parameter: Queue Capacity Range: 2 - 5
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

29. Reduce Queue Capacity

30. Scenario #5 Parameter: Worker Experience Range: New to Field & Expert
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

31. 2 Less Experienced & Cheaper Workers

33. While the expert worker was the most expensive
ORIGINAL (2)
INEXPERIENCED (2)
EXPERIENCED
COST 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 DAYS
AVG L
0.47
0.75
0.28
Utilization
22.68%
35.40%
23.60%
Total Cost
$622.13
$510.44
$500.36
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.

34. Questions?