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M/M/R/GD/K/K By: Ryan Amigliore Parisays comments are in purple

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(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

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Arrivals are drawing from a small finite population. Machine Repair Model Shop CapacityPopulation 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 λs remaining in the population. λ j = λ + λ + λ … + λ (K – j) λ s Thus… λ j = (K – j)λ

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Determining the Service Rate (μ) is done the same as a M/M/S/GD// system. μ j = jμ μ j = Rμ For (j R) For (R < j K)

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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 (R < j K) For (j R)

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so - For (R < j K) For (j R)

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There are no Simple formulas for L, L q, W, & W q For: The best we can do is express them in terms of π j s

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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. 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.

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GivenThe 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 π 0.

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Now we can solve for π 0 And plug this value back in to solve for

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L = 0(0.69) + 1(0.310) + 2(0.062) + 3(0.009) + 4(0.001) + 5(0) = Plug in the probabilities and solve for L Plug in K and solve for average number of cars in good condition. K – L = 5 – = ; Average number of cars in good condition

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Given L = 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 &

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

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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 = π 0 + (0.5)π 1

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π 0 + (0.5)π 1 = (0.5)(0.310) = Plug in the variables and solve A particular repair worker is Idle 77.4% of their work day Therefore the utilization of each worker is ( ).

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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 Original System with Costing Figures WinQSB Input & Output Tables

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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: Parameter: Population Range:

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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 Parameter: Repair Workers Range: 1 & 2

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NUMBER OF WORKERS 12ACCEPTABLECOMMENTS UTILIZATION %LOW AVG # IN SYSTEM (L) <2GOOD AVG # IN QUEUE (Lq) <1GOOD AVG TIME IN QUEUE (Wq) <2GOOD COST OF WORKER(S) AVERAGE BALK COST$46.99$49.49$0.00HIGH TOTAL COST PER DAY$506.22$671.63MIN- 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 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: 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 Parameter: Worker Experience Range: New to Field & Expert

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2 Less Experienced & Cheaper Workers

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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 W3.07 DAYS5.01 DAYS1.83 DAYS AVG L Utilization22.68%35.40%23.60% Total Cost$622.13$510.44$ 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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