University:
California State University, Northridge
Course:
ECON 500 | Survey of Economics
Academic year:

2024
Views:

107

Pages:

23
Author:

RoatoVotHaima97n

Waiting Lines Waiting Lines Example 1 1. Students arrive at the Administrative Services Office at an average of one every 15 minutes, and their request take on average 10 minutes to be processed. The service counter is staffed by only one clerk, Judy Gumshoes, who works eight hours per day. Assume Poisson arrivals and exponential service times. M/M/1 System λ = 4 customers/hour µ = 6 customers/hour Waiting Lines Waiting Lines Example 2 M/M/1 System λ = 4 customers/hour µ = 6 customers/hour a) What percentage of time is Judy idle? 1-λ/µ = 33.33% b) How much time, on average, does a student spend waiting in line? Wq = λ / µ(µ-λ) = 1/3 hours or 20 minutes Waiting Lines Waiting Lines Example 3 M/M/1 System λ = 4 customers/hour µ = 6 customers/hour c) How long is the (waiting line) on average? Lq = λWq = 4/3 customers d) What is the probability that an arriving student (just before entering the Administrative Services Office) will find at least one other student waiting in line? =1-P0-P1 = 1 – (1-λ/µ) - λ/µ(1-λ/µ) = 4/9, or 0.44 Waiting Lines Waiting Lines Example 4 2. At the California border inspection station, vehicles arrive at the rate of 10 per minute in a Poisson distribution. For simplicity in this problem, assume that there is only one lane and one inspector, who can inspect vehicles at the rate of 12 per minute in an exponentially distributed fashion. The California border inspection station is considering the addition of a second inspector. The vehicles would wait in one lane and then be directed to the first available inspector. Arrival rates would remain the same (10 per minute) and the new inspector would process vehicles at the same rate as the first inspector (12 per minute). Waiting Lines Waiting Lines Example 5 M/M/S System λ = 10 vehicles/minute µ = 12 vehicles/minute a) What would be the average length of the waiting line? 0.175 vehicles b) What would be the average time that a vehicles must wait to get through the system? 0.101 minutes, or 6.06 seconds Waiting Lines Waiting Lines Example 6 If a second lane was added (one lane for each inspector): 2 M/M/1 Systems λ = 5 vehicles/minute µ = 12 vehicles/minute c) What would be the average length of the waiting line? 0.298 vehicles per line  0.298*2 = 0.596 d) What would be the average time that a vehicle must wait to get through the system? 0.143 minutes, or 8.58 seconds Waiting Lines Waiting Lines Example 7 3. You are working at a bank and doing resource requirements planning. You think that there should be six tellers working in the bank. Tellers take 15 minutes per customer with a standard deviation of 5 minutes. On average one customer arrives in every three minutes according to an exponential distribution (recall that the standard deviation is equal to the mean). Waiting Lines Waiting Lines Example 8 S=6 λ = 20 customers/hour µ = 4 customers/hour ρ = λ / Sµ = 20 / (4*6) = 0.83 Ca = 1, Cs = 5/15 = 0.33 ρ 2 ( S +1) Ca2 + Cs2 Lq = × 1− ρ 2 a) On average how many customer would be waiting in line? Lq = 1.68 b) On average how long would a customer spend in the bank? Ws = Wq + (1/µ) = (Lq/λ) + (1/µ) = 0.334 hours = 20.05 minutes Waiting Lines Waiting Lines Example 9 4. Consider a call center that employs 8 agents. Past data collected on the customer inter-arrival times has shown that the mean time between customer arrivals is 1 minute, and has a standard deviation of 1/2 minute. The amount of time in minutes the past 10 callers have spent talking to an agent is as follows: − 4.1, 6.2, 5.5, 3.5, 3.2, 7.3, 8.4, 6.3, 2.6, 4.9. a) What is the coefficient of variation for the inter-arrival times? (standard deviation) / mean = 0.5/1 = 0.5 b) What is the mean time a caller spends talking to an agent? = 5.2 minutes Waiting Lines Waiting Lines Example 10 c) What is the standard deviation of the time a caller spends talking to an agent? = 1.88 minutes d) What is the coefficient of variation for the times a caller spends talking to an agent? (standard deviation) / mean = 5.2/1.88 = 2.77 (standard deviation) / mean = 1.88/5.2 = 0.192 Waiting Lines Waiting Lines Example 11 e) What is the expected number of callers on hold, waiting to talk to an agent? λ= 1 per minute μ = 1/5.2 = 0.192 per minute ρ 2( S +1) Ca2 + Cs2 Lq = × S=8 1− ρ 2 ρ = λ/(Sμ ) = 1/(0.192*8) = 0.65 Upart = 0.46 Vpart = 0.19  Lq = Upart*Vpart = 0.087 Lq = 1.8 Waiting Lines Waiting Lines Example 12 f) What is the expected number of callers either on hold or talking to an agent? Ls = Lq + Sρ = 7 (5.287) g) What is the expected amount of time a caller must wait to talk to an agent? Wq = Lq / λ = 1.8 minutes h) What is the expected amount of time between when a caller first arrives to the system, and when that caller finishes talking to an agent? Ws = Ls / λ = 7 minutes Waiting Lines Waiting Lines Example 13 5. Wells Fargo operates one ATM machine in a certain Trader Joe’s. There is on average 8 customers that use the ATM every hour, and each customer spends on average 6 minutes at the ATM. Assume customer arrivals follow a Poisson process, and the amount of time each customer spends at the ATM follows as exponential distribution. a) What is the percentage of time the ATM is in use? λ = 8 per hour μ = 10 per hour ρ = 8/10 = 0.8 =80% Waiting Lines Waiting Lines Example 14 b) Suppose that the percentage of time the ATM is in use is 0.8. (This may or may not be the answer you found in (a).) b1) What is the probability there is more than one customer waiting in the line to use the ATM? X = number of customers either using the ATM or waiting to use the ATM P( X=n ) = ρ n(1- ρ) P( X > 2 ) = 1 – P(X=0) – P(X=1) – P(X=2) = 1 – (1- ρ) – ρ(1- ρ) – ρ2(1- ρ) = 1-(1- ρ + ρ - ρ 2 + ρ 2 - ρ 3 ) = ρ 3 = 0.512 Waiting Lines Waiting Lines Example 15 b) Suppose that the percentage of time the ATM is in use is 0.8. (This may or may not be the answer you found in (a).) b2) On average, how many customers are in line waiting to use the ATM? L ρ 2 ( S +1) Ca2 + Cs2 0.82 × = = 3.2 Lq = 1− ρ 2 0.2 Waiting Lines Waiting Lines Example 16 c) Suppose that the number of customers in line waiting to use the ATM is 3. (This may or may not be the answer you found in parts b or c). All information remains as originally stated. What is the average time a customer must wait to use the ATM? State your answer in minutes. Wq = Lq/λ = 3/8  [3(60)]/8 = 22.5 minutes Waiting Lines Waiting Lines Example 17 6. Matador housing office has one customer representative for walk-in students. The arrival rate 10 customers per hour and the average service time is 5 minutes. Both inter-arrival time and service time follow exponential distributions. a) What is the average waiting time in line? λ = 10 customers/hour, μ = 20 customers/hour Wq = λ/(μ(μ – λ)) = 0.05 hour = 3 minutes or Wq = ma ρ2/(1 – ρ)*(1+1)/2 = 0.05 hour b) What is the probability that an arriving student (just before entering the housing office) will find at least two other student waiting in line? 1 – P0 – P1 = 0.25 Waiting Lines Waiting Lines Example 18 7. Monterey post station has 7 tellers from Monday to Saturday. Customers arrive to the station following a Poisson process with rate 36 customers per hour. The service time is exponentially distributed with mean 10 minutes. a) What is the utilization rate of the tellers? λ = 36 customers/hour, μ = 6 customers/hour ρ = λ/Sμ = 36/42 = 6/7 = 85.7% b) What is the average number of customers waiting in line? (Lq =ρ4/(1 – ρ)*(1+1)/2 = 3.78 Waiting Lines Waiting Lines Example 19 On Sunday, instead of tellers, the post station only opens 3 automail machines to provide automatic service. Each machine can weight different size of package, print self-adhesive labels and accept payments. Arrival is Poisson with rate 20 customers per hour. The service time is 4 minutes with probability 0.75 and 20 minutes with probability 0.25. c) What is the mean service time? ms = 4*0.75 + 16 * 0.25 = 8 minutes d) What is the coefficient of variation of service time? Ss = sqrt (0.75 * 32 + 0.25 * 92 ) = 6.93 Cs = 6.93 / 7 = 0.866 Waiting Lines Waiting Lines Example 20 e) What is the utilization rate? λ = 20 customers/hour, μ = 60/8 = 7.5 customers/hour ρ = λ/Sμ = 20/(3*7.5) = 88.9% f) What is the average number of customers waiting in line? ρ 2 ( S +1) Ca2 + Cs2 × = 5.64 Lq = 1− ρ 2 Waiting Lines Waiting Lines Example 21 8. Bank of San Pedro has only one teller. On average, one customer comes every 6 minutes, and it takes the teller an average of 3 minutes to serve a customer. To improve customer satisfaction, the bank is going to implement a unique policy called, “We Pay While You Wait.” Once implemented, the bank will pay each customer $3 per minute while she or he waits in line. (So the clock starts when a customer comes to the end of the line, and stops when he or she begins to talk to the teller.) Bank of San Pedro hired you as a consultant and you are responsible for estimating how much the “We Pay While You Wait” program will cost. Your preliminary study indicates there are, on average, 0.5 customers waiting in line Assume linear cost. If a customer waits for 20 seconds in line, Bank of San Pedro will pay $1. Waiting Lines Waiting Lines Example 22 a) Calculate the capacity of the teller. State the unit. 60/3=20 customers per hour. b) Calculate proportion of the time the teller is busy. 0.5 or 50% c) How long, on average, does a customer wait in line? State the unit. 0.5/10 = 0.05 hours or 3 minutes. Waiting Lines Waiting Lines Example 23 d) Calculate the expected hourly cost of the “We Pay While You Wait” program. Each customer waits, on average, 3 minutes. So he or she receives, on average, 3*3=$9. There are 10 customers arriving per hour. So the overall cost of this program is 9*10=$90

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