Presentation on theme: "M/M/R/GD/K/K By: Ryan Amigliore Parisay’s comments are in purple"— Presentation transcript:
1M/M/R/GD/K/K By: Ryan Amigliore Parisay’s comments are in purple Finite Source ModelsM/M/R/GD/K/KBy: Ryan AmiglioreParisay’s comments are in purple
2M/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 PopulationSystem Capacity = Population
3Finite Source ModelsArrivals are drawing from a small finite population. Machine Repair ModelPopulationShop CapacityK=R=K= # of MachinesR= # of Repair People
4G = Machine in Working Condition B = Machine in need of Repair
5Changes With Population λ = Break Down (Arrival) Rateμ = Service RateBreak Down RateChanges With Population
6λj = (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’sThus…λj = (K – j)λ
7μj = 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 = jμFor (R < j ≤ K)μj = RμFor (j ≤ R)
8the 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)
10Now 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’sFor:
11Gotham 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.
121. Determine the average number of police cars in good condition. GivenThe average number of good carscan be shown by the total population minusthe average number of cars in the systemK – L = Average # GoodK is given so we have to find L
14Now we can solve for π0And plug this value back in to solve for
15K – L = 5 – 0.465 = 4.535; Average number of cars in good condition Plug in the probabilities and solve for LL = 0(0.69) + 1(0.310) + 2(0.062) + 3(0.009) + 4(0.001) + 5(0) = 0.465Plug in K and solve for average number of cars in good condition.K – L = 5 – = 4.535; Average number of cars in good condition
162. Find the average down time for a police car that needs repairs. GivenSince a police car is only in the system whenit is either being serviced or waiting to beserviced, we can conclude that the averagedown time will be the average time inthe system (W).W= Average Down Time&L = 0.465
17Plug in the given equations &Plug in the given variables and solve.
183. Find the fraction of the time a particular repair worker is idle. GivenEarlier it was shown that in state 0 and inState 1 there is at least one repair workerIdle. Since in state 0 both workers are idleThe percent of time either could be idle inThat state is 100%. In state 1 only one workerIs idle; so the percent of time that eitherWorker could be idle is 50%. In any otherState 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
19Plug in the variables and solve π0 + (0.5)π1 = (0.5)(0.310) = 0.774A particular repair worker is Idle 77.4% of their work dayTherefore the utilization of each worker is ( ).
20Sensitivity AnalysisIn this Sensitivity Analysis a common unit of cost was applied. This will allow for the analysis and comparison of multiple performance measure.CostingWorker Salary: $55,000/yr ≈ $211/dayPolice Officer Salary: $73,000/yr ≈ $280/dayFacility Mortgage: $3500/month ≈ $23.33/day/car
21Original System withCosting FiguresWinQSB Input & Output Tables
22Scenario #1 Parameter: Population Range: 5 - 10 Total number of police cars is currently at 5 vehicles, the departmentis contemplating adding as many as 5 more vehicles to help keep therising crime rate in check, but they do not want to hire anymore repairpeople. The range to check is 5-10 vehicles.Parameter: PopulationRange:
23Increased Car Population With Increased Capacity
24Scenario #2 Parameter: Repair Workers Range: 1 & 2 In order to cut costs the department is wondering if, without addinganymore vehicles, the maintenance of the vehicles could be performed by1 repair person. Calculate and compare 2 repair people to 1.Parameter: Repair WorkersRange: 1 & 2
25In 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 WORKERS12ACCEPTABLECOMMENTSUTILIZATION0.440.2270-83%LOWAVG # IN SYSTEM (L)0.640.46<2GOODAVG # IN QUEUE (Lq)0.20.01<1AVG TIME IN QUEUE (Wq)1.380.075COST OF WORKER(S)211422-AVERAGE BALK COST$46.99$49.49$0.00HIGHTOTAL COST PER DAY$506.22$671.63MIN
26Scenario #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 inmore often. Analyze and compare the system with arrival rates from1car/30days up to 15cars/30days.Parameter: Break Down RateRange: 1/30 to 15/30
28Scenario #4 Parameter: Queue Capacity Range: 2 - 5 Another possibility to cut costs will be to move the repair shop to asmaller facility, compare the system with a capacity from 5 cars to 2cars. In this type of problems the queue capacity is decided by thenumber of Cars and the number of repairers.Therefore this analysis is not relevant.Parameter: Queue CapacityRange: 2 - 5
30Scenario #5 Parameter: Worker Experience Range: New to Field & Expert Look at two different scenarios where less experienced and cheaperrepair persons are hired, and a scenario where one very experiencedmore expensive employee is hired, which employee has a more beneficialeffect on the cost of the system. Mention how the service rate may changeBecause of low experience of the worker. Provide explanation on dataUsed in next slides.Parameter: Worker ExperienceRange: New to Field & Expert
33While the expert worker was the most expensive ORIGINAL (2)INEXPERIENCED (2)EXPERIENCEDCOST OF CUSTOMER BEING SERVED$126.99$198.25$65.97COST OF CUSTOMER WAITNG$3.16$12.19$14.39P(WAIT)7.21%16.40%23.56%AVG W3.07 DAYS5.01 DAYS1.83 DAYSAVG L0.470.750.28Utilization22.68%35.40%23.60%Total Cost$622.13$510.44$500.36While the expert worker was the most expensiveBased on salary, their rapid service time allowedThe system as a whole to save on total cost as wellAs decrease the average total time in system. Needsmore info on your assumptions.