The Friendly Neighbor Grocery Store has a single check-out stand with a full-tim

Question

The Friendly Neighbor Grocery Store has a single check-out stand with a full-time cashier. Customers arrive randomly at the stand at a mean rate of 20 per hour. The service-time distribution is exponential, with a mean of 2.5 minutes. This situation has resulted in occasional long lines and complaints from customers. Therefore, because there is no room for a second checkout stand, the manager is considering the alternative of hiring another per help the cashier by bagging the groceries. This help would reduce the expected time required to process a customer to 1.5 minutes, but the distribution still would be exponential. The manager would like to have the percentage of time that there are more than two customers at the checkout stand down below 25 percent. She also would like to have no more than 5 percent of the customers needing to wait at least 5 minutes before beginning service, or at least 7 minutes before finishing service. Compute L, L_q, W, W_q, P_0, P_1, P_2 for both the current mode of operation and the considered alternative.

Explanation / Answer

The question follows: M/M/1:/FIFO

Where first M denotes the distribution of the number of arrivals per unit time(poisson distribution)

Second M denotes the type of distribution of the service time(exponential distribution)

1 denotes the number of servers

denotes the capacity of the system

FIFO denotes the queuing discipline (First in first out)

For current mode of operation:

(mean of poisson distribution) = 20 per hour

1/µ(mean of exponential distribution) = 2.5/minute

µ=1/2.5*60 = 24 per hour

Average number Ls of customers in the system: L = /(µ-) = 20/(24-20) = 5 customers.

Average waiting time of a customer in the system: W = 1/(µ-) = 1/(24-20)=0.25 hours

Average waiting time of a customer in the queue: Wq = /[µ(µ-)] = 20/[20*4] = 0.25 hours

Average number Lq of customers in the queue: Lq = * Wq = 20*0.25 = 5 customers

Pn(t) is the probability that there are n customers in the system at time t (n>0).

P0 = 1 – (/µ) = 0.16667

Pn = (/µ)n (P0)

P1 = 0.138892

P2 = 0.115743

For alternative mode of operation:

(mean of poisson distribution) = 20 per hour

1/µ(mean of exponential distribution) = 1.5/minute

µ=1/1.5*60 = 40 per hour

Average number Ls of customers in the system: L = /(µ-) = 20/(40-20) = 1 customers.

Average waiting time of a customer in the system: W = 1/(µ-) = 1/(40-20)=0.05 hours

Average waiting time of a customer in the queue: Wq = /[µ(µ-)] = 20/[40*20] = 0.025 hours

Average number Lq of customers in the queue: Lq = * Wq = 20*0.025 = 0.5 customers ~ 1 customer

Pn(t) is the probability that there are n customers in the system at time t (n>0).

P0 = 1 – (/µ) = 0.5

Pn = (/µ)n (P0)

P1 = 0.25

P2 = 0.125

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.