Question

subject: operations research

When customers arrive at Cool's Ice Cream Shop, they take a number and wait to be called to purchase ice cream from one of the counter servers. From experience in past summers, the store's staff knows that customers arrive at a rate of 150 per hour on summer days between 3:00 p.m. and 10:00 p.m., and a server can serve 1 customer in 1 minute on average. Cool's wants to make sure that customers wait no longer than 10 minutes for service. Cool's is contemplating keeping three servers behind the ice cream counter during the peak summer hours.

(i)        Will this number be adequate to meet the waiting time policy?

(ii)       What will be the probability that 3 to 4 customers in Shop?

(iii)      In winter season, arrival rate of customer is reduced to half from 3:00 p.m. and 10:00 p.m. What decision should be taken by the owner according to cost cutting point of view?

Question 4b:

Analysis of arrivals at a PSO gas station with a single pump (filler) has shown the time between arrivals with a mean of 10 minutes. Service times were observed with a mean time of 6 minutes.

(i)        What is the probability that a car will have to wait?

(ii)       What is the mean number of customers at the station?

(iii)      What is the mean number of customers waiting to be served?

(iv)      PSO is willing to install a second pump when convinced that an arrival would expect to wait at least twelve minutes for the gas. By how much the flow of arrivals is increased in order to justify a second booth?                                      

please chow the calculations and formula.

Answer 1

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Answer 4a :

solution :-) rate of 150 per howy () Customer worrive at on Summer day between 3:00 pm and 10500PM. ora 2=150 => Server can serve 1 customer in 1 minute on average -> There are three server, 3 Server can serve 3 Customer in minute On average

.: Se ven : pere howus.is 3x60 = 180 oo Rate of service U = 180 per hows: => Cools want to make sure that customers wait no longer than 20 minutes for service. O fiorst we find average waiting time of a customer in the queue : Wq = ucu-a) => Wq = 150 180 (180-150) =) Wq = 0.03 how = 0.03 X 60 3) Wq = 1. 67 minutes 1.67 minutes is lower than 10 minutes .. yes, Number be adequate to meet the waiting time policy. i) probability of n customers in the queuing system is below Pn = ( "(1-1) for 3 to 4 customers in shop, take • Required probability = P + Pop (150 Ja (1-15 => Wq Oo we D=3 and n>4. 150 180 150 18 0,0 9645 + 0,08037 = 0.17682 o's probability that 3 to h customers in shop = 0.17682 000 In winter arrival rate of customers reduced to halt. ó: New arrival late a= 150 = 75 customers per hours. => So for making profit. (cast cutting, is gedeced) and three service counter should be reduced to 1 042 to make stability of profit. 2

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Answer 4b :

Average arrival rate, λ = 1 per 10 minutes = 6 per 60 minutes = 6 per hr.
Average service rate, μ = 1 in 6 minutes = 10 per 60 minutes = 10 per hr.

(i)

The probability of an arriving car waits = 1 - λ/μ = 1 - 6/10 = 0.4 or 40%

(ii)

The average number of customers in the station, L_s = λ / (μ - λ) = 6 / (10 - 6) = 1.50

(iii)

The average number of customers waiting, L_q = λ² / {μ.(μ - λ)} = 6^2/(10*(10-6)) = 0.9 customrs

(iv)

The required average waiting time (W_q) is 12 minutes or 0.20 hours.

So,

W_q = λ / {μ.(μ - λ)} >= 0.2
or, λ / {10*(10 - λ)} >= 0.2
or, λ = 20 - 2λ
or, λ = 6.667

So, the arrival rate has to be 6.667 per hour (which is 1 customer in 60/6.667 i.e. 9 minutes) or more to justify adding another server.

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