Question

Customers arrive at a hair salon according to a Poisson process with an average of 16 customers per hour. The salon has just one worker due to covied-19 restriction. Therefore, the salon must close whenever the worker leaves. assume that customers who arrive while the salon is closed leave immediately and don’t wait until the worker returns. The salon is closed on weekends.

a. What is the probability that at most (less than) four customers arrive in the hour before closing?

b. If the worker takes a 15-minute coffee break once a day, find the expected number, and variance, of customers lost per week due to the salon being shut while the worker is out.

c. Market research reveals that each customer will make an average purchase from the salon of $25. How much, on average, does it cost the salon worker in sales to take their 15-minute coffee break?

d. The worker returns from a 15-minute coffee break to be told by an attendant in a neighbouring store that they missed exactly one customer. What is the probability that the owner missed the customer by less than 5 minutes?

e. What is the probability that the worker can take a single 15-minute coffee break once a day for a week and not miss a single customer?

f. What is the probability that the worker can take at least two 15-minute coffee breaks in a day before losing a customer?

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Answer 1

Since there is only one barber, the number of service channels is one. Also, since any number of persons can enter the barber shop, the capacity of the system is infinity. Hence this problem comes under the model (M M FCFS / /1 : / . ) (∞ ) (Single server / infinite Queue) 1 Mean service rate = 10 1 per minute 10 µ µ = ∴ = 1 Mean arrival rate , 12 1 per minute 12 λ λ = ∴ = Step 2: Given data: Step 3: To find the following: (a) Expected number of customers in the system i.e., L s Expected number of customers in the queue i.e., L q (b) P (a customer straightly goes to the barber’s chair) i.e., P0 (c) The time can a customer expect to spend in the barber’s shop i.e.,Ws (d) The new arrival rate λr (say) , if Ws > 75 (e) Average time customers spend in the queue i.e., Wq (f)Probability that the waiting time in the system is greater than 30 minutes i.e., P(W>30) (g) Percentage of customers who have to wait prior to getting into the barber’s chair i.e., P(W>0) (h) The probability that more than 3 customers are in the system 1 12 1 1 10 12 5 customers L s λ µ λ = = − − = Step 4: Required computations a) Expected number of customers in the system Expected number of customers in queue 2 1 144 ( ) 11 1 10 10 12 4.17 customers L q λ µµ λ = = −     −   = 1 12 1 1 1 10 1 6 Percentageof timeon arrival need not wait 16.7 P o λ µ =− =− = ∴ = b) P( a customer straightly goes to the barber’s chair) = P(no customer in the system) (i.e., the system is idle) 1 1 ) 1 1 10 12 60 minutes or1hour c Ws µ λ = = − − = 1 ) 75, if 75, where is the new arrival rate. 1 (i.e)if 75 s r r r d W λ µ λ λ µ > > − > − 1 1 (i.e)if 10 75 13 (i.e)if 150 Hence to warrant a second barber, theaveragearrival rate must increase by 13 1 1 per minute 150 12 300 r r λ λ > − > − = ( ) 1 12 ) 50 minutes 11 1 10 10 12 q e W λ µµ λ = = = −     −   ( ) 1 1 30 10 12 0.5 )( ) ( 30) 0.6065 t f PW t e PW e e − − µ λ   −−×     − > = > = = = ) ( ) ( 0) 1 ( 0) 1 ( 0) 1 1 12 5 1 6 10 5 Percentageof customers who have to wait 100 83.33 6 o g P a customer has to wait P W P W P Number of customers P λ µ = > =− = = − = = − = = = ∴ =× = { } 4 56 01 2 3 2 3 4 4 ( ) ( 3) ..... 1 1 (1 ) 1 Since 1 , for 0 5 6 0.4823 n n h PN P P P PPP P P n λ λλ λ µ µµ µ λ λ µ µ λ µ >= ++ + =− + + +     =− − + + +                 =− ≥             = =         =