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1. Distinguish between a channel and a phase.
A channel is the initial service point of a queuing system. A phase refers to the number of stages that the service points provide. It is possible to have single to multiple service channels and single to multiple service phases.
2. In what way might the first-come, first-served rule be unfair to the customer waiting for service in a bank or hospital?
In a bank, FCFS may be perceived to be unfair by customers who have large accounts, but who must wait while the less "important” customers obtain service.
In a hospital, especially in an emergency room, FCFS is probably the exception rather than the rule. FCFS would be unfair when a patient with a minor problem is treated before another experiencing severe pain.
3. Define, in a practical sense, what is meant by an exponential service time.
An exponential service time means that most of the time, the service requirements are of short time duration, but there are occasional long ones. Exponential distribution also means that the probability that a service will be completed in the next instant of time is not dependent on the time at which it entered the system. We can see that a barber, for example, does not fit an exponential distribution in either case. The barber has an average time for cutting hair, and a person who has been sitting in the chair getting a haircut for the past 15 minutes has a higher probability of being completed in the next minute than a person who just walked in and sat down.
4. Would you expect the exponential distribution to be a good approximation of service times for
a. Buying an airline ticket at the airport?
Yes. Although certain customers will require special routings and payment methods, most ticketing is pretty straightforward and entails a short service time.
b. Riding a merry-go-round at a carnival?
No. The Merry-go-round has a fixed cycle time and hence constant services rate.
c. Checking out of a hotel?
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No. Probably a better approximation would be a normal distribution since fast and slow checkouts are likely to be fairly equally balanced.
d. Completing a midterm exam in your OM class?
No. From our experience, students require the entire class period (and then some) to finish the typical mid-term. Thus, the service rate is close to constant.
5. Would you expect the Poisson distribution to be a good approximation of
a. Runners crossing the finish line in the Boston Marathon?
Yes. The arrival pattern typically shows a few runners arriving “early” and the majority arriving in a bunch and the remainder spread out along the tail of the distribution.
b. Arrival times of the student in your OM class?
No. Arrivals are not random since there is a schedule to be met.
c. Arrival times of the bus to you stop at school?
No. Again, arrivals are not random since the bus follows a set schedule.
6. What it the major cost trade-off that must be made in managing waiting line situations?
The classic trade-off is between the cost of waiting for service versus the cost of providing additional service capacity, e.g., the cost of idle WIP versus the cost of adding more workers and machines to process the inventory.
7. Which assumptions are necessary to employ the formulas given for Model 1?
Poisson arrival rates, exponential service rates, which imply a purely random process, but with a known mean (and hence known variance). Also assumed is that the process has reached a point of stochastic equilibrium. In other words, steady state conditions prevail. Infinite calling population and unlimited queue length.
8. Why is simulation often called a technique of last resort?
Simulation is called a technique of last resort because simulation models are time consuming to build (flow charting, coding, etc.) and do not “guarantee” an optimal solution or indeed any solution. Therefore, it makes sense to investigate other problem solving methods such as linear programming or waiting line theory before embarking on simulation.
9. Must you use a computer to get good information from a simulation? Explain.
A computer is a must for any but the most simple simulation problems. Because simulation is a sampling process, it stands to reason that a large number of observations is desirable, and the computer is the only practical way of providing them. Of course, computerization is no guarantee of “good” information. Simulating an invalid model on the computer will only provide a larger volume of questionable data.
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10. What methods are used to increment time in a simulation model? How do they work?
Time incrementing methods include fixed time increments and variable time increments. With fixed time increments, uniform clock times are specified (minutes, hours, days, etc.) and the simulation proceeds by fixed intervals from one time period to the next. At each point in clock time, the system is scanned to determine if any events have occurred and time is advanced; if none have, time is still advanced by one unit.
With variable time increments, the clock time is advanced by the amount required to initiate the next event. It is interesting to note that variable time incrementing generally is more difficult to program unless one is using a special simulation language such as GPSS.
11. What are the pros and cons of starting a simulation with the system empty? With the system in equilibrium?
The pros of starting a simulation with the system empty are that this enables evaluation of the transient period in terms of time to reach steady-state and the activities which are peculiar to the transient period.
One con is that it takes a longer period of time to perform the simulation. A second is that the model will be biased by the set of initial values selected, since the time to achieve steadystate and the activities which take place during the transient period will be affected by the initial values. Steps must be taken to remove these initial values if steady-state results are needed.
The advantages of starting the system in equilibrium are that the run time may be greatly reduced, and that the aforementioned bias may be eliminated.
The disadvantage of starting the simulation in equilibrium is, in essence, that it assumes that the analyst has some idea of the range of output he is looking for. This, in a sense, constitutes “beating” the model and may lead to incorrect conclusions from the simulation run.
12. Distinguish between known mathematical distributions and empirical distributions. What information is needed to simulate using a known mathematical distribution?
A “known mathematical distribution” is one that can be generated mathematically and is amenable to the laws of statistical probability. Examples of such distributions are the normal, binomial, Poisson, Gamma, and hypergeometric.
An empirical distribution is one that is obtained from observing the probability of occurrence of phenomena relating to a specific situation. While it may be possible to define the moment generating function for such distributions, their applicability to other situations is likely to be small.
The information required to simulate using a known distribution, of course, depends on the known distribution selected. Generally speaking, however, at a minimum, the analyst must be able to estimate the mean and standard deviation of the population to be sampled since
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they are parameters of distributions. Alternatively, the analyst can use a discrete probability table to represent the empirical distribution.
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1. The exponential distribution is often used to model what in a queuing system?
The time between customer arrivals and/or service times
2. If the average time between customer arrivals is 8 minutes, what is the hourly arrival rate?
7.5 customers per hour
3. How much time on average would a server need to spend on a customer to achieve a service rate of 20 customers per hour?
Three minutes
4. What is the term used for the situation where a potential customer arrives at a service operation and upon seeing a long line decides to leave?
Balking
5. What is the most commonly used priority rule for setting queue discipline, likely because it is seen as most fair?
First-come, first-served (FCFS)
6. Use model 1.
d. At least one other student waiting in line is the same as at least two in the system. This probability is 1-(P
+P1).
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Probability of at least one in line is 1-(.3333 + .2222) = .4444
7. Use model 2. = 60/50 per minute = 60/45 per minute minutes.
8. Use Model 1.
c. Using model 3,
100 per hour = 120 per hour
, from spreadsheet, q L = .1756
9. Use model 2. = 10 per hour = 12 per hour
a. 10) 2(12)(12 10 ) ( 2
2 2 = = q L = 2.083 people
b. + = q s L L = 2.083 + 10/12 = 2.917 people
c. 2083 10 2083 = = = q q
L W hours.
d. 10 2.917 = = s s
L W = .2917 hours
e. It will cause it to increase, at = 12 per hour,
10. Use model 1 = 3 per minute = 4 per minute
a. 3 4 3 = =
s L = 3 customers
b. 3 3 = = s s L W = 1 minute
c. 4 3 = = = .75 or 75%
d. Probability of 3 or more is equal to 1 – probability of 0, 1, 2
Total of P0 + P1 + P2 = (.2500 + .1875 + .1406) = .5781
Therefore, the probability of three or more is 1 - .5781 = .4219
e. If a automatic vendor is installed, use model 2.
By converting to constant service time, the number in line is reduce from 3 to 1.875 people (a reduction of 1.125, and time in system is reduced from 1 minute to .625 minutes (a reduction of .375 minutes or 22.5 seconds).
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11. Use model 4. N = 4, population of 4 engineers, S = 1, one technical specialist, T = 1, average time to help engineer, U = 7, time between requests for help
a. 7 1 1 + = + = U T T X = .125, look up value of F in Exhibit 10.10 F = .945, therefore, L = N(1-F) = 4(1-.945) = .22 engineers waiting
b. = + = + = .22 4 7) .22(1 ) ( L N U T L W .466 hours or 28 minutes
c. From Exhibit 10.10 at X=.125, and S=1, D = .362. In other words, 36.2% of the time an engineer will have to wait for the specialist.
12. Use model 1. = 20 per hour = 30 per hour
a. 20 30 20 = =
s L = 2 people in the system
b. 20 2 = = s s L W = .10 hours or 6 minutes
c. Probability of 3 or more is equal to 1 – probability of 0, 1, 2
Total of P0 + P1 + P2 = (.3333 + .2222 + .1481) = .7036
Therefore, the probability of three or more is 1 - .7036 = .2964
d. 30 20 = =
= .67 or 67%
e. Use model 3. 30 20 = =
=.6667
From Exhibit 10.9, Lq = .0093
L = .0093 + 20/30 = .676
.676 = =
W = .0338 hours or 2.03 minutes
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13. Use model 4.
N = 4, population of 4 pieces of equipment, S = 1, one repairperson, T = .0833, average time for repair, U = .5, time between requests for repair (hours)
a. .5 .0833 0833 + = + = U T T X = .1428, look up value of F in Exhibit 10.10
F = .928, therefore, L = N(1-F) = 4(1-.928) = .288 machines waiting
b. J=NF(1-X) = 4(.928)(1-.1428) = 3.18 machines operating
c. H = FNX =.928(4).1428 = .53 machines being serviced
d. n =L + H = .288 + .53 = .818 machines in the system
Cost of downtime is .818 times $20 per hour = $16.36 per hour
Cost of one serviceperson = $ 6.00 per hour
Total cost per hour = $22.36
With 2 repairpersons, S =2, 5 0833 .0833 + = + = U T T X = .1428, look up value of F in Exhibit 10.10
F = .995, therefore, L = N(1-F) = 4(1-.995) = .020 machines waiting
H = FNX =.995(4).1428 = .568 machines being serviced
n = L + H = .020 + .568 = .588
Cost of downtime is .588 times $20 per hour = $11.76 per hour
Cost of two serviceperson = $12.00 per hour
Total cost per hour = $23.76
No, added cost of $1.42 per hour would be added for second repairperson
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15. Use model 1.
16. Use model 1. = 6 per hour = 10 per hour
a. 6 10 6 = =
s L = 1.5 people 6 1.5 = =
s s L W = .25 hours or 15 minutes
b. 10 6 = =
= .60 or 60%
c. Probability of more than 2 people is equal to 1 – probability of 0, 1, or 2
Total of P0 + P1 + P2 = (.4000 + .2400 + .1440) = .7840
Therefore, the probability of three or more is 1 - .7840 = .2160
d. Use model 3.
= .60, from Exhibit 10.9, q L = .0593
10-10
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17. Use model 1. = 25 per hour = 30 per hour
a. 30 25 = =
= .833 or 83.3%
b. 25 30 25 = =
s L = 5.00 document in the system
c. 25 5.00 = =
s s L W = .20 hours or 12 minutes
d. Probability of 4 or more is equal to 1 – probability of 0, 1, 2, 3
18. Use model 1.
4 per hour = 6 per hour
a. 6 4 = =
= .667 or 66.7%
b. 2 6 4 = =
s L = 2.00 students in the system
c. 4 00 2 = = s s L W = .50 hours or 30 minutes
d. Probability of 4 or more is equal to 1 – probability of 0, 1, 2, or 3
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Total of P0 + P1 + P2 + P3 = (.3333 + .2222 + .1481 + .0988) = .8024
Therefore, the probability of four or more is 1 - .8024 = .1976 or 19.76%
Use model 1.
= 4.17 people
b. 10
s L = 5,
c. 12 10 = =
=
s s L W = .5 hours or 30 minutes
= .833 or 83.3%
d. Probability of 3 or more is equal to 1 – probability of 0, 1, 2
Total of P0 + P1 + P2 + P3 = (.1667 + .1389 + .1157) = .4213
Therefore, the probability of three or more is 1 - .4213 = .5787 or 57.87%
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20. Use model 1. [Students may want to use model 4 because of the limited number of cars in such an attraction, but there is insufficient information in the problem to solve it under model 4. Kudos to students who note this.]
= 2 per hour
With one repair person: = 2
With two repair people:
With three repair people: = 4
Note: 1 = cost of waiting is number in system times downtime cost of $40 per hour. 2 = cost of service is number of repair personnel times wage rate ($20 per hour).
We should use three repair persons.
21. Use model 2.
750 per hour
900 per hour
W = = 2.92/750 = .003889 hours or .2333 minutes or 14 seconds
s
22. Use waiting line approximation because the service times do not follow an exponential distribution. [Note: Student answers may vary slightly due to rounding differences.]
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a. On average how many customers would be waiting in line?
b. On average how long would a customer spend in the bank?
c. If a customer arrived, saw the line and decided not to get in line that customer has
d. A customer who enters the line but decides to leave the line before getting service is said to have
23. Use waiting line approximation because the service times do not follow an exponential distribution.
[Note: Student answers may vary slightly due to rounding differences.]
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a. On average, how long will each line be at each of the cashier windows?
b. On average how long will a customer spend in the bank (assume they enter, go directly to one line and leave as soon as service is complete).
You decide to consolidate all the cashiers so they can handle all types of customers without increasing the service times.
c. What will happen to the amount of time each cashier spends idle? (increase, decrease, stay the same, depends on _____ )
Stay the same
d. What will happen to the average amount of time a customer spends in the bank? (increase, decrease, stay the same, depends on _____ )
24. Use model 1.
Average waiting time = 1.0/6 = 1/6 minute, and average teller idle time between customers = 4.0/6 = 4/6 minute or 40 seconds. Average teller idle time percent = 4.0/16 = .25 or 25%.
Average down time is 14.5/5 = 2.9 hours
1. Calculate the average customer arrival rate and service rate per hour.
The customer arrival rate lambda = 0.0212 patients per hour. This is calculated 62/(8 x 365) = 0.0212. The service rate mu = 0.7427 patients per hour. This calculated 60 min/hour divided by 80.79 minute/patient = 0.7427 patients per hour.
2. Calculate the probability of zero patients in the system (P0), probability of one patient (P1), and the probability of two or more patients simultaneously arriving during the night shift.
P(0) = (1 – lambda/mu) = (1 - .0212/.7427) = 0.9714 the probability of no patients in the system is over 97 percent.
P(1) = (1 – lambda/mu)(lambda/mu)1 = (1 - .0212/.7427)(.0212/.7427) = 0.0278 the probability of exactly 1 patient in the system is 2.78 percent.
The probability that 2 or more patients are in the system is 1 – (P(1) + P(0)) = 1 – (.0278 + 0.9714) = 0.0008.
The probability of two or more patients occurring simultaneously on the night shift is less than 0.1% (less than one chance in 1,000).
3. Using a criterion that if the probability is greater than 1 percent, a backup OR team should be employed, make a recommendation to hospital administration.
A second OR is not needed at this hospital.
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1. Draw a diagram of the process using the format in Exhibit 9.5.
2. Consider a base case where a customer arrives every 40 seconds and the Customer Service Champion can handle 120 customers per hour. There are two Food Champions each capable of handling 100 orders per hour. How long should it take to be served by the restaurant (from the time a customer enters the kiosk queue until her food is delivered)? Use queuing models to estimate this.
= 90, order = 120, prep = 100.
For the Customer Service Champion, the average service time is .0083 hours (40 seconds) and the average time in line is .0250 hours (from queuing formulas) for a total of .0333 hrs. (120 seconds). This is calculated using Model 1. For preparing the food for the two Food Champions, the average service time is .01111 hours (40 seconds). Using model 3, and with lambda/mu = .9, Lq from Exhibit 10.9 is 0.2285 customers. The time a customer spends in the system for food to be prepared is .01254 hours or 45.1 seconds. The total time is 120 + 45.1 = 165.1 seconds or 2.75 minutes.
3. On average, how busy are the Service Champion and the two Food Champions?
The Service Champion is busy = 90/120 = 75% of the time and the Food Champions are busy () = 90/200 = 45% of the time.
4. On average, how many cars do you expect to have in the drive-thru line? (Include those waiting to place order and waiting for food.)
Lq due to the order taking = λ2/µ(µ - λ) = 8100/3600 = 2.25 customers
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Lq due to the food preparation = .2285 customers
Average total customers waiting in line = 2.4785 customers
Ls due to the order taking = 3 customers
Ls due to the food preparation = 1.1285 customers
Average number of customers in the system = 4.1285
5. If the restaurant runs a sale and the customer arrival rate increases by 20%, how would this change the total time expected to serve a customer? How would this change the average number of cars in the drive-thru line?
The current customer arrival rate is 90 per hour, this would go up to 90(1.2) = 108 per hour.
Ls due to order taking = (λ/µ – λ) = 108/(120-108) = 9.0 cars
Ws = Ls/λ = 9.0/108 = .08333 hours = 5.0 minutes
Ls due to order preparation = Lq + λ/µ = .445 + 108/100 = 1.525 cars
(note λ/µ = 108/100 = 1.08, Lq = .445 from the spreadsheet)
Ws = Ls/λ = 1.525/108 = .01412 hours = .8472 minutes
Average total time in the system = 5.0 + .8742 = 5.8472 minutes
Average total number of cars in the drive thru = 9.0 + 1.525 = 10.525 cars
6. Currently, relatively few customers (less than ½ percent) order the Crunchwrap Supreme. What would happen if we ran the sale and demand jumped on the Crunchwrap Supreme and 30% of our orders were for this item? Take a quantitative approach to answering this question. Assume that the two processes remain independent.
Just to make this interesting, let’s assume that the customer arrival rate went to 108 customers per hour.
Assume the order taking rate stays the same at 120/hour.
For the order preparation, assume that 70% of the orders still take (60*60)/100 = 36 seconds on average to prepare, but 30% of them now take 72 seconds.
The average time to make an order would be .7(36) + .3(72) = 46.8 seconds
So they could prepare (60*60)/46.8 = 76.92 orders per hour (each Food Champ).
Ls due to order taking = λ/(µ – λ) = 108/(120-108) = 9.0 cars
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Ws = Ls/λ = 9.0/108 = .08333 hours = 5.0 minutes → this is the same as before.
Ls due to order preparation = Lq + λ/µ = 1.3449 + 108/76.92 = 2.749 cars (note λ/µ = 108/76.92 = 1.4, Lq = 1.3449 from exhibit 10.9.)
Ws = Ls/λ = 2.749/108 = .02545 hours = 1.527 minutes
Average total number of cars in the drive thru = 11.749 cars
Average total time in the system = 6.527 minutes Not much of an impact.
7. For the type of analysis done in this case, what are the key assumptions? What would be the impact on our analysis if these assumptions were not true?
A big assumption is that there is no interference between the two processes. This would occur if one process was causing a delay in the other process for some reason. It does not appear that this would be true in this process. We are also assuming the Food Champs are working independent, not as a team. Other major assumptions are the distribution associated with arrival and service rates, and that these rates are valid during the peak noon period.
8. Could this type of analysis be used for other service type businesses? Give examples to support your answer.
Yes, many examples can be given.
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