(M/ek/1):fcfs

Introduction:Customers are generated by limited pool of potential customers i.e. finite population. The total customer’s population is M and n represents the number of customers already in the system (waiting line), any arrival must come from M - n number that is not yet in the system.

 (M / Ek / 1 ): (First Come First Served)

We assume that Arrival of one unit means addition of .k. phases in the system and Departure of one unit implies reduction of .k. phases in the system.

1)      λn = λ and k μn = μ

2)      k= number of phases.

3)      System length = E (n) = [(k 1) / 2 k] × (λ /μ)×[λ /(μ − λ)] (λ /μ)

4)      Length of the queue = E (w) = [(k 1) / 2k] × (λ /μ)×[λ /(μ − λ)]

5)      Waiting time in the system = E (v) = [(k = 1) / 2k] × [λ /μ(μ − λ)] (1/μ)

6)      Waiting time in the queue = E (w) = [(k 1) / 2k] × [λ /μ(μ − λ)]

7)      For constant service time equating .k. to ∝, we get: E (n) System length = (1 / 2) (λ/μ)[λ/ (μ − λ)] (λ/μ)

8)      E (w) Length of the queue = (1 / 2) (λ/μ)[λ/ (μ − λ)]

9)   Waiting time in the system = E (v) = (1 / 2) [λ/μ(μ − λ)] (1/μ)

10)   Waiting time in the queue = E(w) = (1 / 2) [λ/μ(μ −λ)].11)   When k = 1 Erlang service time distribution reduces to exponential distribution.

EXAMPLE-1

Repairing a certain type of machine, which breaks down in a given factory, consists of 5 basic steps that must be performed sequentially. The time taken to perform each of the 5 steps is found to have an exponential distribution with a mean of 5 minute and is independent of the other steps. If these machines breakdown in Poisson fashion at an average rate of two per hour and if there is only one repairman, what is the average idle time for each machine that has broken down?

Solution

Data: Number of phases = k = 5, Service time per phase = 5 minutes, λ = 2 units per hour,

Service time per unit = 5 × 5 = 25 minutes, hence μ = 1 / 25 minutes per minute.

Average idle time of the machine = E (v) = [(k 1) / 2k] × (λ/μ)×[1/ (μ − λ) (1/μ)

= 9(5 1) / (2 × 5) × (2 × 5) / 12] × 1 / [(12 / 5 ) . 2] (5 / 12)

= (1 / 2) × (5 / 2) (5 / 12) = (20 / 12 = 5 / 3 hours = 100 minutes.

EXAMPLE-2

A colliery working one shift per day uses a large number of locomotives which breakdown at random intervals, on the average one failing per 8 - hour shift. The fitter carries out a standard maintenance schedule on each faulty locomotive. Each of the five main parts of this schedule takes an average of 1 / 2 an hour but the time varies widely. How much time will the fitter have for other tasks and what is the average time a locomotive is out of service.

Solution

Data: k = 5, λ = 1/8 per hour, Service time per part = 1 / 2 an hour.

Service time per locomotive = 5 / 2 hours. Hence μ = 2 / 5 hours.

Fraction of time the fitter will have for other tasks = Fraction of time for which the fitter is idle =

1 / (λ /μ) = 1 . [(1 / 8) / (2 / 5)] = 1 . (5 / 16) = 11 / 16.

Therefore, time the fitter will have for other tasks in a day = (11 / 16) × 8 = 5.5 hours

Average time a locomotive is out of service = Average time spent by the locomotive in the system = [(k 1) / 2k] × (λ /μ)×[1/ (μ − λ)] (1/μ) = [(5 1) / (2 × 5)] × [(1 / 8) / (2 / 5)] × {1 / [(2 / 5) . (1 / 8)] (5 / 2) = (6 / 10) × (5 / 16) × (40 / 11) (5 / 2) = (15 / 22) (5 / 2) = 70 / 22 = 3.18 hours.