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 or no?
   
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

Part b:

The given problem can be model as multiple server queue model.

Question no.	Description	Notation	Formula	Value
	Arrival rate	λ		7.2	Passengers per hour
	Service rate	µ		6	Passengers per hour
	Number of servers	m		2.0
	Individual server utilization	r	λ/µ	7.2/6 = 1.2
	Prob. of zero customers from table for r = 1.2 and m = 2	Po		0.25
	Average number in customers waiting in line	Lq		0.6750	customers
	Average Number of CUSTOMERS waiting plus served	Ls	Lq + r	0.675 + 1.2 = 1.8750	customers
	Average waiting time for truck in line	Wq	Lq/λ	0.6750/7.2 = 0.0938	Hours per customer
	Average waiting time for truck in line and at station	Ws	Wq + (1/µ)	0.0938+(1/6) = 0.2604	Hours per customer
	Probability the both inspector are busy	Pw		0.1662

Part c:

The probability of n customer waiting in line is given as follows:

The probability of 3 to 10 customers in line is tabulated using above equation as follows:

Number of customers in line (n)	Probability (Pn)
3	0.0900
4	0.0450
5	0.0225
6	0.0113
7	0.0056
8	0.0028
9	0.0014
10	0.0007
11	0.0004

Since the probability of waiting customer more than 10 units is 0.0004, which is very less thus, the two-screening-station system be able to meet the manager’s goal

ANS: yes