The C-Center has one single ticket station open in person throughout the day. It

Question

The C-Center has one single ticket station open in person throughout the day. It takes a customer service representative 5 minutes to start and finish a ticket order. Patrons arrive to purchase tickets at a rate of 6 people per hour.

1) Determine the average number of people that will stack up waiting in line.

2) Find the average time a person must wait before they can be helped

3) The C-center has a rule that, on average, a customer service representative can spend a maximum of 5 minutes helping out every customers. There must be 5 minutes of idle time available to relieve tension. Will the ticket office have to schedule an extra person?

Explanation / Answer

= 6 per hour µ = 12 per hour (5 minutes per patron) Average Utilization of the system p = / µ = 6/12 = 0.50 Probability that there are 0 patron in the system P0 = 1-p = 1 - 0.5 = 0.50 Average number of patrons in waiting line = Lq = pL = / µ * / (µ - ) = 6/12 * 6 /12-6 = 0.5 * 1 = 0.5 customers Average waiting time of patrons in line = Wq = pW = / µ * 1 / (µ - ) = 6/12 * 1 /12-6 = 0.0833 hours or 5 minutes Since the waiting time is 5 minutes, another person should be added = 6 per hour µ = 12 per hour (5 minutes per patron) s = 2 Average Utilization of the system p = / Sµ = 6/(2*12) = 0.25 Probability that there are 0 patron in the system (P0 = 1 / [1 + / µ + {( / µ)^s / s!} * (1 / 1-p)] = 1 / [ 1 + 6/12 + (6/12^2 / 2!) * (1 / 1 - 0.25) ] = 1 / [ 1 + 0.50 + (0.5 * 1.33)] = 1 / (1.5 + 0.6667) = 1 / 2.1667 = 0.461538 Average number of patients in waiting line = Lq = P0*(/µ)^s*p / s! (1 -p)^2 = (0.461538 * 0.5^2 * 0.25) / (2! * 0.75 ^2) = 0.028846 / 1.125 = 0.02564 customers Average waiting time of patients in line = Lq / = 0.02564/6 = 0.0043 hours or 0.256 minutes

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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