Page 1081 No 1:

Parisay’s comments are in red.

Solve problem 1 on page 1081 manually then use WinQSB. Answer the following questions: a) What is the probability of having less than 4 people in the system? b) What is the probability of having more than 6 people in the system? c) If on average 8 passengers arrive per minute (instead of 10) how the performance measure will change (just comment). Do not solve.

Problem: Each airline passenger and his or her luggage must be checked to determine whether he or she is carrying weapons onto the plane. Suppose that at Gotham City Airport, an average of 10 passengers per minute arrive (interarrival times are exponential). To check passengers for weapons, the airport must have a checkpoint consisting of a metal detector and baggage X-Ray machine. Whenever a checkpoint is in operation, two employees are required. A checkpoint can check an average of 12 passengers per minute (the time to check a passenger is exponential). Under the assumption that the airport has only one checkpoint, answer the questions (a-g):

a) What is the probability that a passenger will have to wait before being checked for weapons? b) On the average how many passengers are waiting in line to enter the checkpoint? c) On the average, how long will a passenger spend at the checkpoint? d) What is the probability that exactly 10 passengers arrive in one minute? e) What is the probability that exactly 25 passengers will arrive in the next two minutes? f) What is the probability that less than 3 passengers arrive in one minute? g) A passenger has just arrived. What is the probably that it will be between 1 and 3 minutes before the next passenger arrives?

Given:  = 10 passengers/minute,  = 12 passengers/minute

M/M/1 Check for steady state condition:  = / µ = 10/12 = 5/6, which is less than 1 a) PW = probability of an arriving passenger has to wait before being checked for weapons = P (j > S)?

In M/M/1, PW =  /  = 10/12 = 0.83

b) Lq = Average Number of passenger waiting in the line?

2 2 Lq = ρ / (1- ρ) =  /  ( - ) = 100 / 12 (12-10) ≈ 4.17 passengers are waiting in line.

c) WS = Average time a passenger spend at checkpoint?

W = L / ,

L =  / ( - ) = 10/(12-10) = 5 passengers

W = 5/10 = 0.5 minute of average time spent in checkpoint

d) P(n = 10) = probability that exactly 10 passengers arrive in one minute?

P (n) = (e-t * (t)n)/n!

= (e-10 * 1010)/10! = 0.125 probable that exactly 10 pass. Arrive in a min.

e) P (n = 25) = probability that exactly 25 passengers arrive in two minute?

t = 10(2) = 20

= (e-20 * 2025)/25! = 0.045

f) P (n < 3) = probability that less than 3 passengers arrive in one minute?

= P (n = 0) + P (n = 1) + P (n = 2) = [(e-10 * 100)/0!] + [(e-10 * 101)/1!] + [(e- 10 * 102)/2!] = 0.0028

g) P(1≤ X ≤ 3) = probability that it will be between 1 and 3 minutes before the next passenger arrives?

X=3 -10t -10 -30 -5 = X=1 ∫ ( 10e )dt = e - e = 4.539x10

Or you can use Cumulative Function P(1≤ X ≤ 3) = F(X=3) – F(X=1) = …. = 4.539x10-5

Should print probability output of WinQSB and answer questions using output and manually. Answer questions a to c in your textbook.

1 2 3 4 a) P (j<4) = п0 + п1 + п2 + п3 = (1-) +  (1-) +  (1-) +  (1-) = (5/6) (1 – 5/6) = 0.167 + 0.139 + 0.116 + .097 = ,519 b) P (j>6) = 1 – P (j<6) = 1 – (п0 + п1 + п2 + п3 + п4 + п5) = 1 – (0.167 + 0.139 + 0.116 + .097 + 0.080 + 0.067) = 1 – 0.6658 = 0.334 c) Suppose that the passenger arrival rate per minute is decreased from 10 passengers to 8 passengers, the following results to the performance measures indicated below will occur:

1) The number of customers present in the system (L) would decrease. 2) The number of customers waiting in line (Lq) and in service (Ls) would decrease. 3) The time that a customer spends in the system (W) would decrease. 4) The time that a customer spends in the line (Wq) and { this is wrong: in service (Ws)} would decrease. 5) Traffic intensity () would decrease since there will be less customers in the system.

 Effective service rate will decrease.  The probability that all servers are idle will increase  The probability that an arriving customer waits will reduce.