Operations Research

(M/m/1):fcfs/n/n (Limited Population Model)

Introduction: This model differs from the other model in the sense that the maximum number of customers in the system is limited to N. 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 / M / 1) : FCFS / N /N ( Limited Population or Source Model)

In this model, we assume that 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. The formulae for this model are:

Average number of customers in the system = E (n) =

Average number in the queue = E (L) = M . [(μ λ) / λ]× (1− p0 )

Example:

A mechanic repairs 4 machines. The mean time between service requirements is 5 hours for each machine and forms an exponential distribution. The mean repair time is 1 hour and also follows the same distribution pattern. Machine down time costs Rs. 25/- per hour and the mechanic costs Rs. 55/- per day. Find

(a) Expected number of operating machines,

(b) the expected down time cost per day,

(c) Would it be economical to engage two mechanics, each repairing only two machines?

Solution

Data: Finite population, λ = Arrival rate = (1 / 5) = 0.2, μ = Service rate = μ = (1 / 1) = 1

Probability of the empty system = p0 =

p0 =  1 / 1 (4 × 0.2) (4 × 3 × 0.22) (4 × 3 × 2 × 0.23) (4 × 3 × 2 × 1 × 0.24) = 0.4 i.e. 40 percent

of the time the system is empty and 60 percent of the time the system is busy.

(a)Expected number of breakdown machines I the system = E (n) = M . (μ / λ) (1 . p0)

= 4 . (1 / 0.2) (1 . 0.4) = 4 . 5 × 0.6 = 4 . 3 = 1. i.e.

 Expected number of operating machines in the system = 4 . 1 = 3.

(b)Expected down time cost per day of 8 hours = 8 × (expected number of breakdown machines × Rs. 25 per hour) = 8 × 1 × 25 = Rs. 200 / - day.

(c)When there are two mechanics each serving two machines, M = 2 ,  p0 =  1 / 1 (2 × 0.2) (2 × 1 × 0.22) = 1 / 1.48 = 0.68 i.e. 68

percent of the time the system is idle. It is assumed that each mechanic with his two machines constitutes a separate system with no interplay. Expected number of machines in the system = M . (μ / λ) × (1 / p0) = 2 . (1 / 0.2) × (1 . 0.68) = 0.4.

Therefore expected down time per day = 8 × 0.4 × Number of mechanics or machine in system = 8 × 0.4 × 2 = 6.4 hours per day. Hence total cost involved = Rs. 55 × 2 6.4 × Rs. 25/- = Rs. (110 160) = Rs. 270 per day.

But total cost with one mechanic is Rs. (55 200) = Rs. 255/- per day, which is cheaper compared to the above. Hence use of two mechanics is not advisable.