Question

Customers arrive at the coffee shop in Bilkent FBA atrium, with a mean rate of 80 per hour (assume Poisson). It takes the barista, on average 30 seconds per cup of coffee (assume Exponential). Assume also that each customer buys only one cup. Determine: (a) The average number of customers waiting in line. (b) The average time customers spend in the system. (c) The average number of customers in the system. (d) The probability that a customer will not have to wait. (e) The probability that there is at most 2 customers waiting. (f) The probability that a customer will have to wait at most two minutes.

Answer 1

Service rate, µ         1/service time= 2 cups per minute * 60- minutes=            120       cups per hour

Arrival rate , l = = 80           per hour

(a) The average number of customers waiting in line.

Average number of customers in the waiting line, Lq = 1.33 customer

(b) The average time customers spend in the system.

Wait in the System, W =0.03            hrs

(c) The average number of customers in the system.

Average number of customers in the system (waiting and being served),L =2.00

(d) The probability that a customer will not have to wait.

This means the server is free/idle and no customers are in the system. the customer coming to barista can directly receive the service without waiting in queue

Probability that the server is idle and a customer can be served,= Probability that no customers are in the system (either in the queue or being served)=          0.33

(f) The probability that a customer will have to wait at most two minutes.

Probability that the waiting time of a customer in the system is less than 2 minutes = 0.74

     

Font Alignment Styles Cells Editing U16 MM1 SINGLE SERVER, SINGLE QUEUE,with Poisson Arrivals and Exponential Service Times rrival rate Service rate, 1/time between arrivals80 per hour 1/service time per hour Probability that the server is busy and the customer has to wait (utilizstion of server) Probability that the server is idle and a customer can be served Utilization rate, the proportion of time the system is in use,p-i 7 Probability that the server is idle and a customer can be served,Po 1-p Average number of customarsın the waiting line,Lq-A2 / μ(μ-4 Average number ofcustomers in the systemwaiting and being served),LEiJ 10 12 Wait in the Queue,Wqi Wait in the System,W-1/u-) H4-H3 14 15 16 Probability that no customers are in the system (either in the queue or being served),Po-1- Number of customars,N Probability of n customers in the system, Pnpnpl 18 19 20 service time,st Total time to serve n customer,T time,t Probability that the waitingtime of a customer in the system exceeds t. PXt)-H Probebility that the waiting time of a customer in the system is less thant. 23

r Format Painter | B u Clear Filter Sele Editing Formatting as Table Styles Clipboard Font Alignment Number Styles L14 MM1-SINGLE SERVER, SINGLE QUEUE with Poisson Arrivals and Exponential Service Times Arrival rate ,2. Service rate, μ 1/time between arrival 1/service time 30 per hour per hour 6 Probability that the server is busy and the customer has to wait (utilization of server) 7 Probability that the server is idle . nd a customer can be served Utilization rate, the proportion oftime the system is in use,p 0.67 Probability that the server is idle and a customer can be served,Po-1-p 0.33 Average number of customers in the waiting line,Lq i2/l- Average numbar of customers in the systam(waiting and being served),L-/HH-) 1.33 wait in the queue,wa-a / μ(-2) wait in the System ,W .1 / wa) 0.02 hrS hrs Probability that no customers are in the system (aither in the queua or baing served),Po-1- Number of customers, N Probability of n customers in the system,Pn pAn [1-pl 0.33 service time,St Total time to serve n customer,T time,t Probability that the waiting time of a customer in the Probability that the waiting time of a customer in the ystem exceeds t. aP(Xt)-e system is less thant. -(u-hrt) 0.26 74 25

(e) The probability that there is at most 2 customers waiting.

Probability of n customers in the system, P_n = r^ⁿ * [1 -   r]

The probability that there is at most 2 customers waiting.

=probability (x<=2)

=P₀+P₁+P₂

=0.33 +0.22+0.15

=0.70

J16 0 Arrival rate. Service rate,u Number of customers,N 2"60 Utilization rate; the proportion oftime the system is in use,p /H Probability of n customers in the system,POpAn 1p Probability of n customers in the system,P1 pAn 1p Probability of n customers in the system,P2 pAn [1-pl -03/04 07 0 11-07 -071 (1-07 07A2 (1-07 10 12 13 Probabaility of atmost 2 customers in the system P(KS2) 08+09+010 15 16 17 18 19 20

Clipboard Font Alignment Number L14 N O Arrival rate service rate, μ Number of customers,N 80 120 Utilization rate; the proportion oftime the system is in use,p /H Probability of n customers in the system,PO pAn [1-pl Probability of n customers in the system,P1 pAn [1- pl Probability of n customers in the system,P2 pAn 1-pl 0.67 0.33 0.22 0.15 Probabaility of atmost 2 customers in the system 0.70 14

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