Question

All airplane passengers at the Lake City Regional Airport must pass through a security screening area before proceeding to the boarding area. The airport has three screening stations available, and the facility manager must decide how many to have open at any particular time. The service rate for processing passengers at each screening station is 6 passengers per minute. On Monday morning the arrival rate is 7.2 passengers per minute. Assume that processing times at each screening station follow an exponential distribution and that arrivals follow a Poisson distribution.

Note: Use P₀ values from Table 11.4 to answer the questions below.

1. Suppose two of the three screening stations are open on Monday morning. Compute the operating characteristics for the screening facility.
   
   Round your answer to four decimal places.
   
   P₀ = ??
   
   Round your answers to two decimal places.
   
   L_q = ??
   
   L = ??
   
   W_q = ??min
   
   W = ??min
   
   Round your answer to four decimal places.
   
   P_w = ??
   
2. Because of space considerations, the facility manager's goal is to limit the average number of passengers waiting in line to 10 or fewer. Will the two-screening-station system be able to meet the manager’s goal?
   
   Yes
   
3. What is the average time required for a passenger to pass through security screening? Round your answer to two decimal places.
   
   ?? min

Answer 1

Given:

Arrival rate λ = 7.2 Passengers/minute

Service rate µ = 6 Passengers/minute at each screening station

Number of screening station M = 3, but only 2 servers are open. So we will use 2 when calculating the operating characteristics.

(a)

For the value of P0, first we need to find λ/µ.

λ/µ = 7.2/6 = 1.2

For utilization factor 1.2 and number of servers 2, P0 = 0.2500

Lq = 0.68

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Now, L = Lq + (λ/µ) = 0.675 + (7.2/6) = 0.675+ 1.2 = 1.875 = 1.88

L = 1.88

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Wq = Lq/λ = 1.875/7.2 = 0.2604 minutes= 0.26 minutes

Wq = 0.26 minutes

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W = Wq + (1/µ) = 0.2604 + (1/6) = 0.2604 + 0.16667 = 0.4271= 0.43 minutes

W = 0.43 minutes

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Pw

Pw = 0.4500

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(b)

Lq = 0.68, which means number of passengers waiting in line are fewer than 10. Two screening system will be able to meet the manager’s goal.

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(c)

Average time to pass through security screening is W, which includes waiting time in service and waiting time in queue= 0.43 minutes