An airline ticket counter forecasts that 220 people per hour will need to check in. It takes an average of 2 minutes to service a customer. Assume that arrivals are poisson distributed and service times exponentially distributed and that customers wait in a single queue for the first available agent.

a) If we want the average time a customer spends in the queue and in service to be 10 minutes or less, how many ticket agents should be on duty?
b) There airline wants to minimize the cost of idle ticket agents and the cost of customers waiting in the queue. The salary of a ticket agent is £12 per hour and the cost of a customer waiting in the queue is £5 per customer per hour. How many ticket agents should be on duty?

Since the waiting time ( 1.428 minutes ) is less than the original waiting time ( 2 minutes ) the number of agents that will achieve a wait time of 10 minutes or less is = 8 agents

B) Determine the number of ticket agents that should be on duty to minimize cost

salary of ticket agent = £12 per hour

cost of customer waiting in queue = £5 per hour per customer

i) When 8 agents are used

waiting time of customers = 0.0238 * 220 = 5.236

waiting cost for customers = 5.236 * 5 = £26.18

employee cost = 8 * 12 = £96

∴ Total cost = 96 + 26.18

= £ 122.18

ii) When 9 agents are used

waiting time for customers = 0.0074 * 220 = 1.628

Wq = 1.6367 / 220 = 0.0074

waiting cost for customers = 1.6367 * 5 = £ 8.1835

assuming Lq = 1.6367

employee cost = 9 * 12 = £ 108

∴ Total cost = 108 + 8.1835 = £ 116.18

From the calculations in ( i ) and ( ii ) the Ideal number of ticket agents that should be on duty to minimize cost should be 9 agents.

Hence we can conclude that A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents and The number of agents required on duty to reduce cost = 9 agents.

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