# Queueing questions- operations (poisson dist, wait time, processing time, idle time, utilization rates)

supply chain multi-part question and need an explanation and answer to help me learn.

3 questions for an assignment on queueing- looking for the calculations shown (as screenshots / written work) with the final answers please.
(Involves analysis with poisson distribution, variable arrival and wait time, processing time, idle time, utilization rates)
Providing example questions w/ answers (for reference of result seeking) and the calculator to use for the calculations on the questions.
Requirements: Multiple parts– show calculation work please and final responses in doc.
The check-in line at Alpha Airlines has two agents who serve all passengers during peak hours. Arrivals to the check-in counter are Poisson with a mean of 50 passengers per hour. Approximately, thirty percent of the arrivals are Priority Class (e.g., First Class, Gold members, etc) while the remainder holds coach tickets. The time that a passenger spends at the ticket check-in counter with an agent is exponentially distributed with a mean of 1.44 minutes. Alpha promises that if a passenger waits in line for more than 3 minutes, a coach passenger will be awarded 1000 frequent flyer miles while a Priority Class passenger will be awarded 1500 miles. Currently, all passengers go through the same queue for check-in.
How many total miles are awarded per hour on average currently?
Management is considering the creation of two separate lines, one for each of the two classes of passengers. In this situation, each check-in line will have one agent who will serve their respective passengers; line switching will not be allowed. Find the average total miles awarded hourly under this configuration.
Hint: To calculate the miles awarded (i.e. cost due to waits exceeding 3 minutes), please refer to Example 2 in the “example problems and solution” posted on Canvas alongside this assignment.
The Acme Vending Company services vending machines for a large university on the west coast. Because students kick the machines at every opportunity out of anger and frustration, Acme management has a constant repair problem. According to MBA students who studied the machines as part of their Operations Management course, vending machines on the campus break down on an average of four per hour, and the breakdowns are distributed in a Poisson manner. Downtime costs \$45/hour per machine, and each maintenance worker earns \$20 per hour. Acme owns one repair truck which can be staffed by one, two, or three workers. One worker can service machines at an average rate of five per hour, distributed exponentially; two workers, working together, can service seven per hour, distributed exponentially; and a team of three workers can service eight per hour. Acme earnings have not been strong in recent times and they cannot afford another repair truck at this time.
What is the optimal maintenance crew for servicing vending machines?
Hint 1: No matter how many workers there are in a crew, they always work as one unit. Hence, the question is about how selecting m, the number of servers; rather it is about the size of the crew – different options have different service speeds and costs.
Hint 2: To calculate the total “waiting cost” (i.e. cost due to machine downtime), please refer to Examples 5, 6 and 7 in the “example problems and solution” posted on Canvas alongside this assignment.
Stanley Cup Café is a small mom-and-pop store in Seattle. Its store layout and process flow are shown below in Figure 1.
During peak time, customers arrive to the shop according to a Poisson process with rate of 45 per hour.
Upon arrival, all customers queue in front of the cashier station. There is only one cashier, who is responsible for answering questions on food or coffee taking orders, serving food, tea, and drip coffee, heating up food orders, and receiving payment. As a result, the service time is highly variable and is assumed to have an exponential distribution. On average, the cashier can process 1.25 customers per minute. After completing transaction at the cashier station, 30% will leave or sit down; 20% will go to the condiments station; and the rest 50% will go to the barista station.
The arrival of all the people going to the barista station can be modeled as Poisson process again. Due to the variety of drinks (espresso, etc.) and the number of drinks ordered by each customer, the service time at the barista station is again modeled as exponential with an average of 2 minutes. There is only one barista. After completing service at the barista station, 20% will leave or sit down and the rest 80% will go to the condiments station.
The arrival of all the people going to the condiments station can be modeled as Poisson and the service time is exponential with an average of 20 seconds. The condiments counter is wide enough to accommodate two customers at a time. After getting the condiments, all customers leave or sit down.
30%

50%

One Barista

20%

Leave or sit down
Leave or
arrive

20%

Figure 1

80%

sit down
Leave or sit down
Assume that all customers follow the process above (i.e. they don’t go to the condiments first and then barista). Please perform the following analysis. Show all your work.
What is the total time spent in the system, from arrival to departure (or sit-down), for an average customer?
The owner is concerned about the service area being too crowded. Assume that customers in all three areas stand there until service is completed. What is the average total number of customers in all three areas?
To reduce congestion in the service area, the owner is considering adding a cashier or a barista during peak time. Which one will achieve a bigger reduction of the total number of people in the service area (as calculated in part b)?
Example 1: Customers arrive to a store with one server at an average rate of 15 per hour (Poisson). Service time is approximately 3 minutes (assume exponential service time).
What is the probability that the server is idle?
What is the average number of customers waiting for service?
What is the average customer’s time in system?
What is the daily mean revenue of this store if an average customer spends \$10 (assume 10 hours a day)?
Solution:
This is M/M/1 with  = 15/hour and  = 3 minutes= 3/60 hours. Therefore, we use the “MMm” spreadsheet and set cells E2-D4 are 15, 20, and 1.
1-utilization = 1 – .75 = .25.
Lq =2.25. (The question asks for number of people, we need to calculate L not W (which is the waiting time. Moreover, it asks for the number waiting for service; it does not include the customer who may be in service. Hence Lq, not Ls (which would include everyone including the one in service)).
Ws = 0.20 hours. (As explained above, this question is asking for W, not L. Moreover, it says “time in system” – end to end. So it is Ws, not Wq.)
The teller can serve 20 people per hour, but on average, only 15 arrive per hour. The teller thus services 15 people per hour on average. Hence,
Daily average revenue=15*10*10=\$1500.

Example 2: Sea-View Bank is a new startup bank that has only one teller. During peak demand periods, bank officials have noticed that an average of six customers come to the bank per hour and that the arrivals appear to follow a Poisson distribution. It takes the teller an average of four minutes to serve each customer (following an exponential distribution).
In response to their competition, Sea-View bank officials have decided to institute a new policy that will pay customers a bonus of \$5 if they have to wait more than 4 minutes in line. Approximately how much will Sea-View Bank have to pay per hour in bonuses to their customers?
Solution:
We did this in class. Here is a more detailed explanation.
This is M/M/1 with  = 6/hour and =15/hoursince average service time  = 4 minutes = 4/60 hours). In the “MMm” spreadsheet we set cells E2-D4 are 6, 15, and 1.
To find the probability that a customer will have to wait more than 4 minutes in queue, we set cell E22 to be “=4/60”. (This may have tripped up some people: we used “hours” for  and , so we must use “hours” here as well. 4 minutes = 4/60 hours.)
So, from cell E25, =0.2195.
Average total payment to customers/hour
= 5 *  * Prob[waiting in queue for more than 4 minutes]
= 5 * * 0.2195 = \$6.59/hour.
Example 3: The law firm of LD and Associates specializes in the practice of waste disposal law. Data regarding the cases received in a year and the times to complete a case was recently compiled. Suppose that the firm works on one case at a time. Cases are received according to a Poisson distribution with a mean of one case every 30 days. The data on the time on the completion times of the last 10 cases are 27, 26, 26, 25, 27, 24, 27, 23, 22 and 23. Assuming that cases are handled one at a time on a first come first served basis, determine the number of clients waiting for their case to be processed and the average time a client has to wait until his/her case is completed (wait in queue plus service).
Solution:
We did this in class. Here is a more detailed explanation.
Here we have an M/G/1 queue with:
1/ = 30 days =>  =1/30 per day,
The average and standard deviation of service time are not given directly. We need to compute them from the given data:
1/ = (27+26+26+25+27+24+27+23+22+23)/10 = 25 days => = 1/25 per day
=3.55, where xi is duration of case i. Note that this is the variance. To find the standard deviation, you take the square root.
Any statistical software should be able to calculate the average, the standard deviation, and the variance for you. You can also use Microsoft Excel. The built-in functions are “=average(range),” “=stdev(range),” and “=var(range)” respectively, where “range” is where you input the raw data.
We will use the “MG1” spreadsheet and set cell E2 to be “=1/30”, cell E4 to be 25, and cell E5 to be “=sqrt(3.55)” (this calculates the square root), respectively. The question is asking for we get Lq and Ws. Note that you do NOT need to set cell E6 () – it will be automatically calculated for you from cell E4.
From cell F10, Lq = Ave. number of cases waiting to be processed = 2.10 cases.
From cell F13, Ws = Ave. time to complete a case = 88 days.

Example 4: There is currently one tollbooth at one of the exits of a state turnpike. On the average, it takes approximately 40 seconds for the toll collector to take the money from a driver and, if necessary, return change; data analysis indicates that these service times follow an exponential distribution. Cars arrive at the tollbooth at an average rate of 70 cars per hour (assume a Poisson distribution).

What is the capacity utilization of the toll collector?
What is the average waiting time for a car before it pays the toll?
The turnpike authority has decided to install another toll collection station that is equipped with an electronic scanner that will automatically scan passes on the windshields of cars that have signed up with this service. The scanning time is estimated to be 5 seconds (which is constant for all cars with a pass). Once the new tollbooth is installed, it is estimated that 40 percent of all cars coming through the tollbooth will buy and use the passes. What will be the average waiting time for cars with passes and for cars without passes?
Solution:
This is an M/M/1 system (a single tollbooth) with = 70 /hour, and = 90 /hour. We will use the “MMm” spreadsheet and set cells E2-E4 to be 70, 90, and 1.
Utilization = 77.78%
Cars with passes:
Because the service time is constant, it is an M/G/1 system with =0. So we will use the “MG1” spreadsheet and set cells E2, E4, and E5 to be “=28/60”, “=5/60”, and 0. From cell F12, we get (almost no wait in queue! – but remember, it still spends 5 seconds in process).
Cars without passes:
 = 70 * 60% = 42 /hour, and is still 90 /hour. As in part a)-b), we will use the “MMm” spreadsheet and set cells E2-E4 to be 42, 90, and 1. From cell F10, we get
Example 5: The Airport in (which has only one runway) is beginning to experience congestion. At the present time, an average of 12 planes arrive every hour (the actual number of arriving planes follow a Poisson distribution). With the present equipment, the controllers can land and clear a plane in an average of 4 minutes (actual service times follow exponential distribution). The planes are processes on a first come first served basis; planes that are waiting to land are asked to circle the airport.
What is the expected number of airplanes circling the airport at any point in time waiting for clearance to land?
A new ground approach radar just approved by FAA is being considered; if adopted, it will allow planes to be processed exactly every four minutes. If this radar is used, what is the expected number of airplanes that would be circling the airport at any point in time?
The new ground approach radar will cost \$100 per hour to operate and maintain. If the average cost of keeping a plane airborne is \$70 per hour, would you recommend that the radar be adopted?
Solution:
With the current setup, the landing time follows an exponential distribution, so it is an M/M/1 system. = 12 planes per hour, =15 planes per hour. We use “MMm” spreadsheet and set cells E2-E4 to be 12, 15, and 1. Then,
Lq = 3.2 planes.
With the new radar, the landing time is a constant 4 minutes. So this is a M/G/1 system. = 12 planes per hour, 1/ =4 minutes = (4/60) hours. Standard deviation of service time = 0. We use the “MG1” spreadsheet and set cells E2, E4, and E5 to be 12, “=4/60”, and 0. Then,
Lq = 1.6 planes
(Notice that by simply reducing the variability we are able to cut the queue by half).
To calculate waiting cost per hour, there are two approaches:
For each airplane, the average waiting time in queue (waiting while airborne) is Wq, so the average cost of waiting is \$70*Wq. However, for the whole system, there are on average planes every hour, so the average hourly waiting cost is \$70* Wq* .
From the system perspective, to calculate the average waiting cost, it’s almost as if there are Lq planes waiting to land at any time. Since each plane costs \$70 per hour to keep airborne, the total system hourly waiting cost is \$70*Lq.
The two actually are the same, because the there is a theoretical result – the Little’s Law – that says Lq= *Wq.
With the old system, waiting cost per hour=70 Lq=\$224/hour.
With the new system, waiting cost per hour=70 Lq=\$112/hour.
New radar operating cost per hour = \$100/hour.
Therefore, we would recommend that the radar be adopted (\$212<\$224). Note that the cost difference per hour is small, but over a long stretch of time, it is significant. Example 6: To promote its reputation for fast service, Earl’s While-U-Wait Automotive Tune-up Shop promises to reduce a customer’s bill by \$0.20 for every minute the customer must wait until his or her car’s tune-up is finished. Customers arrive according to a Poisson distribution with a mean arrival rate of 5 customers per hour. The time required by a mechanic to perform a tune-up is exponentially distributed with an average time of 30 minutes. Earl pays his mechanics \$20 per hour. If Earl wants to minimize his total costs (defined as the sum of his labor costs and the profit lost per hour because of reductions in customer’s bills), how many mechanics should Earl hire? Solution: This is an M/M/m system where the number of servers, m, needs to be determined. = 5 customers per hour, = 2 customers per hour. Server cost is \$20 per hour, and waiting cost is \$12 per hour. We use the “MMm” spreadsheet and set cells E2-E3 to be 5 and 2. Then we vary cell E4 (first column in the following table): Total Cost = 20*m + 12*Ls. So, he should hire 4 mechanics. Note that in the previous example, waiting cost is assessed only on the “waiting-in-queue” part when planes are airborne, so we use Lq in the calculation of waiting cost. Here the waiting cost is assessed on the total time, so we use Ls, not Lq. Example 7: Cascade Plastics has a large group of molding machines that breakdown at a (Poisson distributed) mean rate of six per (8-hour) day. Each maintenance technician can service an average of one machine per hour (service times are exponentially distributed). Each hour that a machine is down costs the company approximately \$50 in lost profit. If maintenance technicians are paid \$15 per hour, what size maintenance crew should be hired? Solution: This problem is similar to the previous one. So if you feel comfortable you can skip this one. This is an M/M/m system where the number of technicians, m, needs to be determined. = 0.75 machines per hour, = 1 machine per hour. We use the “MMm” spreadsheet and set cells E2-E3 to be 0.75 and 1. Then we vary cell E4 (first column in the following table): Cost of service = \$15 per technician per hour (this is used to calculate cost of mechanics). Cost of waiting = \$50 per hour per machine down (this is used to calculate waiting cost). Total Cost = 15*m + 50*Ls. So, he should hire 2 technicians.