All airplane passengers at the Lake City Regional Airport must pass through a se
ID: 448369 • Letter: A
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.
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.
Lq =
L =
Wq = min
W = min
Round your answer to four decimal places.
Pw =
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
What is the average time required for a passenger to pass through security screening? Round your answer to two decimal places.
min
Explanation / Answer
CHANNNELS,k =3
L=LAMBDA=7.2
m=MU=6
HENCE LAMBDA/MU=7.2/6=1.2
Hence P0=0.2941 (from table 11.4 anderson and sweeney)
Lq= ((L/m)^k *L*m*Po)/((k-1)!(km-L)^2)=22.103/233.28=0.0947
L=Lq+L/m=0.047+1.2=1.247
Wq=Lq/L=0.0947/7.2=0.01316
W=Wq+1/m=0.01316+1/6=0.1798
Pw=(1/k!)*(L/m)^k *(km/(km-L)*Po=0.141168
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
What is the average time required for a passenger to pass through security screening? Round your answer to two decimal places. which is W=0.1798 mins=0.18 mins
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